JEE Kinetic Theory of Gases Questions

The mean free time $$\tau$$ between two successive collisions is obtained from the relation

$$\tau = \frac{\lambda}{\bar{c}}$$

where $$\lambda$$ is the mean free path and $$\bar{c}$$ is the average (root-mean-square) speed of the molecules.

The collision frequency $$f$$, i.e. the number of collisions per second, is the reciprocal of the mean free time:

$$f = \frac{1}{\tau} = \frac{\bar{c}}{\lambda}$$

Substituting the given values $$\lambda = 3 \times 10^{-7}\,\text{m}$$ and $$\bar{c} = 600\,\text{m s}^{-1}$$:

$$f = \frac{600}{3 \times 10^{-7}}\;\text{s}^{-1}$$

$$f = \frac{600}{3}\times 10^{7}\;\text{s}^{-1}$$

$$f = 200 \times 10^{7}\;\text{s}^{-1}$$

$$f = 2 \times 10^{9}\;\text{s}^{-1}$$

Therefore, the collision frequency of oxygen molecules at 300 K and 1 atm is $$2 \times 10^{9}\,\text{s}^{-1}$$, which matches Option C.

For an ideal gas, the average translational kinetic energy of one molecule at absolute temperature $$T$$ is given by the expression

$$\overline{E_k} = \frac{3}{2}\,k\,T$$

Here, $$k$$ is the Boltzmann constant and $$T$$ is the thermodynamic temperature. Observe that $$\overline{E_k}$$ depends only on the temperature and the universal constant $$k$$; it is independent of the nature, molar mass, or molecular mass of the gas.

Both helium and argon are in the same flask at the same temperature $$T = 300 \text{ K}$$. Therefore, their average kinetic energies per molecule are equal:

$$\overline{E_k}(\text{He}) = \frac{3}{2}\,k\,T$$
$$\overline{E_k}(\text{Ar}) = \frac{3}{2}\,k\,T$$

The required ratio is

$$\frac{\overline{E_k}(\text{He})}{\overline{E_k}(\text{Ar})} = \frac{\frac{3}{2}\,k\,T}{\frac{3}{2}\,k\,T} = 1$$

Hence, the ratio of average kinetic energies (per molecule) of helium to argon is $$1 : 1$$.

The correct choice is Option D.

Let the volume of the smaller vessel be $$V$$. The larger vessel has double this volume, so its volume is $$2V$$.

Initial data
Large vessel : $$P_1 = 8\text{ kPa},\; T_1 = 1000\text{ K},\; V_1 = 2V$$
Small vessel : $$P_2 = 7\text{ kPa},\; T_2 = 500\text{ K},\; V_2 = V$$

Using the ideal-gas equation $$PV = nRT$$, the initial number of moles in each vessel is

$$n_{1i} = \frac{P_1 V_1}{R T_1} = \frac{8 \times 2V}{R \times 1000} = \frac{16V}{1000R} = 0.016\frac{V}{R}$$

$$n_{2i} = \frac{P_2 V_2}{R T_2} = \frac{7 \times V}{R \times 500} = \frac{7V}{500R} = 0.014\frac{V}{R}$$

Total moles before connection

$$n_{\text{total}} = n_{1i} + n_{2i} = 0.016\frac{V}{R} + 0.014\frac{V}{R} = 0.030\frac{V}{R}$$

After connecting the vessels, both are maintained at a common temperature $$T_f = 600\text{ K}$$. At steady state the common pressure is $$P_f$$.

Moles in the large vessel at this stage:

$$n_{1f} = \frac{P_f \, V_1}{R T_f} = \frac{P_f \times 2V}{R \times 600} = \frac{P_f V}{300R}$$

Moles in the small vessel:

$$n_{2f} = \frac{P_f \, V_2}{R T_f} = \frac{P_f \times V}{R \times 600} = \frac{P_f V}{600R}$$

Total moles remain unchanged, so

$$n_{1f} + n_{2f} = n_{\text{total}}$$

$$\frac{P_f V}{300R} + \frac{P_f V}{600R} = 0.030\frac{V}{R}$$

Combine the terms on the left:

$$P_f V \left(\frac{1}{300} + \frac{1}{600}\right)\!\bigg/\!R = P_f V \left(\frac{2 + 1}{600}\right)\!\bigg/\!R = \frac{P_f V}{200R}$$

Equate this to the right side and solve for $$P_f$$:

$$\frac{P_f V}{200R} = 0.030\frac{V}{R} \;\;\Longrightarrow\;\; P_f = 0.030 \times 200 = 6\text{ kPa}$$

Hence the common steady-state pressure in both vessels is $$6\text{ kPa}$$.

Option B is correct.

For a fixed amount of ideal gas in a closed rigid vessel, the volume remains constant. According to Gay-Lussac’s law (also called the pressure law), at constant volume the pressure $$P$$ of an ideal gas is directly proportional to its absolute temperature $$T$$.
Mathematically, $$\frac{P}{T} = \text{constant} \quad\Longrightarrow\quad \frac{\Delta P}{P} = \frac{\Delta T}{T}$$

The problem states: the pressure rises by $$0.4\%$$ upon heating the gas by $$1^{\circ}\text{C}$$. Writing these changes in symbols,

$$\frac{\Delta P}{P} = 0.4\% = \frac{0.4}{100} = 0.004$$

Since a change of $$1^{\circ}\text{C}$$ equals a change of $$1\text{ K}$$ on the absolute scale,

$$\Delta T = 1\text{ K}$$

Using the proportionality relation:

$$\frac{\Delta T}{T} = \frac{\Delta P}{P}$$

$$\Rightarrow \quad \frac{1\text{ K}}{T} = 0.004$$

Solving for the initial absolute temperature $$T$$:

$$T = \frac{1}{0.004} = 250 \text{ K}$$

Hence the initial temperature of the gas is $$250\text{ K}$$. Converting to the Celsius scale, $$250\text{ K} - 273 \approx -23^{\circ}\text{C}$$, but none of the given options list this negative Celsius value; instead, option C directly offers $$250\text{ K}$$.

Therefore, the correct choice is Option C (250 K).

We need to find the ratio of rms velocities given the ratio of vapour densities is $$\frac{4}{25}$$.

The root-mean-square velocity of a gas is given by $$v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$$, where $$R$$ is the gas constant, $$T$$ is absolute temperature, and $$M$$ is molar mass.

Vapour density (VD) is defined as the ratio of the mass of a volume of gas to the mass of the same volume of hydrogen under identical conditions, and it is related to the molar mass by $$\text{Vapour Density} = \frac{M}{2}$$. Therefore, the ratio of vapour densities equals the ratio of molar masses: $$\frac{VD_1}{VD_2} = \frac{M_1/2}{M_2/2} = \frac{M_1}{M_2} = \frac{4}{25}$$.

Since at the same temperature $$v_{\text{rms}}\propto\frac{1}{\sqrt{M}}$$, it follows that $$\frac{v_1}{v_2} = \sqrt{\frac{M_2}{M_1}} = \sqrt{\frac{25}{4}} = \frac{5}{2}$$.

The correct answer is Option C: $$\frac{5}{2}$$.

To solve this problem, we need to compare the ratio of specific heats, $$\gamma = \frac{C_p}{C_v}$$, for two types of diatomic gases: one with rigid molecules (no vibrational modes) and another with vibrational modes. The specific heats depend on the degrees of freedom of the gas molecules.

First, recall the formula for $$\gamma$$ in terms of degrees of freedom $$f$$:

$$\gamma = 1 + \frac{2}{f}$$

This is derived from the relations $$C_v = \frac{f}{2} R$$ and $$C_p = C_v + R = \frac{f}{2} R + R$$, so:

$$\gamma = \frac{C_p}{C_v} = \frac{\frac{f}{2} R + R}{\frac{f}{2} R} = \frac{\frac{f}{2} + 1}{\frac{f}{2}} = 1 + \frac{2}{f}$$

Now, determine the degrees of freedom for each case:

Case 1: Rigid diatomic molecules (no vibration)

- Translational degrees of freedom: 3 (for motion along x, y, z axes)

- Rotational degrees of freedom: 2 (for diatomic molecules, rotation about axes perpendicular to the bond axis)

- Total degrees of freedom, $$f_1 = 3 + 2 = 5$$

Thus, $$\gamma_1 = 1 + \frac{2}{f_1} = 1 + \frac{2}{5} = 1.4$$

Case 2: Diatomic molecules with vibrational modes

- Translational degrees of freedom: 3

- Rotational degrees of freedom: 2

- Vibrational modes: 1 vibrational mode, but each vibrational mode contributes 2 degrees of freedom (kinetic and potential energy)

- Total degrees of freedom, $$f_2 = 3 + 2 + 2 = 7$$

Thus, $$\gamma_2 = 1 + \frac{2}{f_2} = 1 + \frac{2}{7} \approx 1.2857$$

Comparing $$\gamma_1$$ and $$\gamma_2$$:

$$\gamma_1 = 1.4$$ and $$\gamma_2 \approx 1.2857$$, so $$\gamma_2 < \gamma_1$$.

Now, evaluate the options:

A. $$\gamma_1 = \gamma_2$$ → False

B. $$2\gamma_2 = \gamma_1$$ → $$2 \times 1.2857 = 2.5714 \neq 1.4$$ → False

C. $$\gamma_2 < \gamma_1$$ → True

D. $$\gamma_2 > \gamma_1$$ → False

Therefore, the correct option is C.

For an ideal gas,

rms speed is

$$v_{rms}=\sqrt{\frac{3kT}{m}}$$

where
k = Boltzmann constant,
m = mass of one molecule.

Squaring,

But mean squared velocity is

$$\overline{v^2}=v_{rms}^2$$

So

$$\overline{v^2}\propto T$$

This means mean squared velocity varies linearly with absolute temperature.

So the graph is a straight line through origin.

The room is rectangular with length $$4\,\text{m}$$, breadth $$4\,\text{m}$$ and height $$3\,\text{m}$$.
Hence its volume is
$$V = 4 \times 4 \times 3 = 48\,\text{m}^3$$

The air inside the room may be treated as an ideal gas. For an ideal gas, the equation of state is $$PV = nRT$$ $$-(1)$$

For a diatomic gas (in the ordinary temperature range) the number of degrees of freedom is $$f = 5$$. The molar specific heat at constant volume is therefore $$C_V = \frac{f}{2}R = \frac{5}{2}R$$.

The total internal energy $$U$$ of an ideal gas is given by $$U = nC_VT$$.
Using $$nRT = PV$$ from $$(1)$$, this becomes $$U = C_V \left( \frac{PV}{R} \right) = \frac{C_V}{R} \, PV$$.

Substituting $$C_V = \frac{5}{2}R$$: $$U = \frac{5}{2}\,PV$$ $$-(2)$$

The room is at one atmosphere pressure. Taking $$P = 1\,\text{atm} = 1.0 \times 10^5\,\text{Pa}$$ and $$V = 48\,\text{m}^3$$:

$$PV = 1.0 \times 10^5 \times 48 = 4.8 \times 10^6\,\text{J}$$

Using $$(2)$$: $$U = \frac{5}{2} \times 4.8 \times 10^6 = 12.0 \times 10^6\,\text{J}$$

Therefore, the internal energy of the air in the room is $$12 \times 10^6\,\text{J}$$.

For 1 mole of an ideal monoatomic gas with temperature increase $$\Delta T = 50°C$$ at constant pressure:

Heat added at constant pressure: $$E_1 = nC_p\Delta T$$.

Change in internal energy: $$E_2 = nC_v\Delta T$$.

For a monoatomic ideal gas: $$C_p = \frac{5}{2}R$$ and $$C_v = \frac{3}{2}R$$.

$$\frac{E_1}{E_2} = \frac{C_p}{C_v} = \frac{5/2}{3/2} = \frac{5}{3}$$.

Given $$\frac{E_1}{E_2} = \frac{x}{9}$$:

$$\frac{x}{9} = \frac{5}{3}$$

$$x = 9 \times \frac{5}{3} = 15$$.

The answer is 15.

We need to find the temperature to which a gas must be raised to double its pressure at constant volume.

For a gas at constant volume, from the ideal gas law:

$$\frac{P_1}{T_1} = \frac{P_2}{T_2}$$

Given: $$T_1 = 27°C = 300 \, K$$, $$P_2 = 2P_1$$.

$$T_2 = T_1 \times \frac{P_2}{P_1} = 300 \times \frac{2P_1}{P_1} = 300 \times 2 = 600 \, K$$

$$T_2 = 600 - 273 = 327°C$$

The answer is 327°C.

For any ideal gas, the molar specific heats at constant volume and at constant pressure are given by the equipartition theorem.

If a molecule has $$f$$ degrees of freedom, then
$$C_V = \frac{f}{2}\,R$$
$$C_P = C_V + R = \frac{f+2}{2}\,R$$
Therefore, the specific-heat ratio is
$$\gamma = \frac{C_P}{C_V} = \frac{f+2}{f}$$ $$-(1)$$

Case A: Mono-atomic gas A

Such a molecule can translate along the three axes only, so
$$f_A = 3$$

Using $$(1)$$,
$$\gamma_A = \frac{3+2}{3} = \frac{5}{3}$$ $$-(2)$$

Case B: Poly-atomic gas B

Given data for one molecule of B:
• Translational degrees = 3
• Rotational degrees = 3
• Vibrational modes = 1

A single vibrational mode contributes two degrees of freedom (one kinetic + one potential). Hence
$$f_{\text{vib}} = 2$$

Total degrees of freedom for B:
$$f_B = 3 + 3 + 2 = 8$$

Using $$(1)$$ with $$f_B = 8$$,
$$\gamma_B = \frac{8+2}{8} = \frac{10}{8} = \frac{5}{4}$$ $$-(3)$$

The question states that
$$\frac{\gamma_A}{\gamma_B} = 1 + \frac{1}{n}$$ $$-(4)$$

Substitute $$(2)$$ and $$(3)$$ into $$(4)$$:
$$\frac{\,\frac{5}{3}\,}{\,\frac{5}{4}\,} = 1 + \frac{1}{n}$$
$$\frac{5}{3} \times \frac{4}{5} = 1 + \frac{1}{n}$$
$$\frac{4}{3} = 1 + \frac{1}{n}$$

Isolate $$\frac{1}{n}$$:
$$\frac{1}{n} = \frac{4}{3} - 1 = \frac{1}{3}$$
$$n = 3$$

Hence, the required value of $$n$$ is $$3$$.

N moles of a polyatomic gas (f = 6) mixed with 2 moles of a monoatomic gas (f = 3) results in a mixture with effective degrees of freedom given by the weighted average: $$f_{mix} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2}$$. Since the mixture behaves as a diatomic gas (fₘᵢₓ = 5), substituting the values yields $$5 = \frac{N \times 6 + 2 \times 3}{N + 2}$$. Multiplying both sides by (N + 2) gives $$5(N + 2) = 6N + 6$$, which simplifies to $$5N + 10 = 6N + 6$$. Solving for N leads to $$N = 4$$.

The value of $$N$$ is 4, which corresponds to Option 3.

Find the molar specific heat at constant volume of a mixture of 2 mol monoatomic and 6 mol diatomic gas.

For an ideal gas, the molar specific heat at constant volume is given by $$C_v = \frac{f}{2}R$$, where $$f$$ is the number of degrees of freedom.

Monoatomic gas has $$f = 3$$ (three translational degrees), giving $$C_{v1} = \frac{3}{2}R$$, while diatomic gas has $$f = 5$$ (three translational plus two rotational), giving $$C_{v2} = \frac{5}{2}R$$.

In the mixture, the effective molar specific heat is the mole-fraction-weighted average $$C_{v,\text{mix}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2}$$, where $$n_1 = 2$$ mol of monoatomic and $$n_2 = 6$$ mol of diatomic gas.

Substituting these values yields $$C_{v,\text{mix}} = \frac{2 \times \frac{3}{2}R + 6 \times \frac{5}{2}R}{2 + 6} = \frac{3R + 15R}{8} = \frac{18R}{8} = \frac{9}{4}R$$.

The correct answer is Option A: $$\frac{9}{4}R$$.

We need to identify which parameter remains the same for molecules of all gases at a given temperature. According to the kinetic theory of gases, the average kinetic energy of a gas molecule depends only on the absolute temperature: $$ \overline{KE} = \frac{3}{2}k_B T $$. This expression is independent of the mass or nature of the gas.

The average speed, given by $$\bar{v} = \sqrt{\frac{8k_BT}{\pi m}}$$, depends on the molecular mass $$m$$, so different gases have different speeds at the same temperature. The average momentum, calculated as $$m\bar{v} = m\sqrt{\frac{8k_BT}{\pi m}} = \sqrt{\frac{8mk_BT}{\pi}}$$, also depends on $$m$$ and therefore varies between gases. Mass itself obviously differs for different gas molecules. In contrast, the average kinetic energy, given by $$\frac{3}{2}k_BT$$, depends only on temperature and is thus the same for all gases at a given temperature, regardless of their mass or chemical identity.

The correct answer is Option (1): kinetic energy.

Collision frequency of H$$_2$$ at 27°C is Z. Find collision frequency at 127°C.

The collision frequency $$Z \propto n^2 \sigma^2 \sqrt{T}$$ where $$n$$ is number density and $$\sigma$$ is collision cross-section. In a closed chamber at constant volume, $$n$$ is constant and thus $$Z \propto \sqrt{T}$$.

For $$T_1 = 27 + 273 = 300$$ K and $$T_2 = 127 + 273 = 400$$ K,

$$\frac{Z_2}{Z_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{400}{300}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}},$$ so $$Z_2 = \frac{2}{\sqrt{3}} Z.$$

The correct answer is Option B: $$\frac{2}{\sqrt{3}}Z$$.

For an ideal gas at fixed temperature, the pressure is proportional to the number of moles contained in a vessel of fixed volume: $$P = \dfrac{nRT}{V}$$.

Initially only gas $$X$$ is present.

$$P_1 = 2 \text{ atm}, \quad n_X = \dfrac{m_X}{M_X}, \quad m_X = 10\ \text{g}$$

After adding gas $$Y$$ (at the same temperature and in the same vessel) the total pressure becomes

$$P_2 = 6 \text{ atm}, \quad n_Y = \dfrac{m_Y}{M_Y}, \quad m_Y = 80\ \text{g}$$

Because $$T$$, $$R$$ and $$V$$ remain unchanged, the pressure ratio equals the mole ratio:

$$\dfrac{P_2}{P_1} = \dfrac{n_X + n_Y}{n_X} \Longrightarrow \dfrac{6}{2} = 3 = 1 + \dfrac{n_Y}{n_X}$$

Hence $$\dfrac{n_Y}{n_X} = 2$$ $$-(1)$$.

Substitute the expressions for the moles:

$$\dfrac{m_Y/M_Y}{m_X/M_X} = 2$$

$$\Longrightarrow \dfrac{80/M_Y}{10/M_X} = 2$$

$$\Longrightarrow \dfrac{80}{M_Y} = \dfrac{20}{M_X}$$

$$\Longrightarrow 80M_X = 20M_Y$$

$$\Longrightarrow M_Y = 4M_X$$ $$-(2)$$.

The root‐mean‐square (rms) speed of an ideal gas is given by

$$u_{\text{rms}} = \sqrt{\dfrac{3RT}{M}}$$.

For two gases at the same temperature, the ratio of their rms speeds is inversely proportional to the square root of their molar masses:

$$\dfrac{u_{\text{rms}}(X)}{u_{\text{rms}}(Y)} = \sqrt{\dfrac{M_Y}{M_X}}$$.

Using result $$(2)$$, $$M_Y = 4M_X$$, so

$$\dfrac{u_{\text{rms}}(X)}{u_{\text{rms}}(Y)} = \sqrt{4} = 2$$.

Therefore the required ratio is $$2 : 1$$.

Option D which is: $$2 : 1$$

8 mol Ar (monoatomic): $$U_1=8\cdot\frac{3}{2}RT=12RT$$.

6 mol O₂ (diatomic, no vibration): $$U_2=6\cdot\frac{5}{2}RT=15RT$$.

Total: $$U=12RT+15RT=27RT$$.

The answer is Option (3): $$27RT$$.

For an adiabatic process with $$n = 1$$ mole, $$\gamma = \frac{3}{2}$$, initial temperature $$T$$, volume doubles.

For an adiabatic process: $$TV^{\gamma-1} = \text{constant}$$.

$$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$$

$$T \cdot V^{1/2} = T_2 \cdot (2V)^{1/2}$$

$$T_2 = \frac{T}{\sqrt{2}} = \frac{T\sqrt{2}}{2}$$

Work done in an adiabatic process:

$$W = \frac{nR(T_1 - T_2)}{\gamma - 1} = \frac{1 \cdot R\left(T - \frac{T}{\sqrt{2}}\right)}{\frac{1}{2}} = 2R\left(T - \frac{T}{\sqrt{2}}\right)$$

$$= 2RT\left(1 - \frac{1}{\sqrt{2}}\right) = 2RT\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right) = RT\left(\frac{2(\sqrt{2}-1)}{\sqrt{2}}\right)$$

$$= RT \cdot \frac{2\sqrt{2}-2}{\sqrt{2}} = RT(2 - \sqrt{2})$$

The correct answer is Option 3: $$RT[2 - \sqrt{2}]$$.

The r.m.s. velocity of a gas molecule is given by:

$$v_{rms} = \sqrt{\frac{3RT}{M}}$$

where $$R$$ is the gas constant, $$T$$ is the absolute temperature, and $$M$$ is the molar mass.

We need the r.m.s. velocity of hydrogen ($$M_{H_2} = 2$$ g/mol) to equal that of oxygen ($$M_{O_2} = 32$$ g/mol) at $$47°C = 320$$ K. Setting the two velocities equal:

$$\sqrt{\frac{3RT_{H_2}}{M_{H_2}}} = \sqrt{\frac{3RT_{O_2}}{M_{O_2}}}$$

$$\frac{T_{H_2}}{M_{H_2}} = \frac{T_{O_2}}{M_{O_2}}$$

$$T_{H_2} = T_{O_2} \times \frac{M_{H_2}}{M_{O_2}} = 320 \times \frac{2}{32}$$

$$T_{H_2} = 320 \times \frac{1}{16} = 20 \text{ K}$$

The correct answer is $$20 \text{ K}$$.

We need to find the ratio of r.m.s. speeds of helium to oxygen at the same temperature.

Recall the r.m.s. speed formula.

$$ v_{rms} = \sqrt{\frac{3RT}{M}} $$

where $$R$$ is the gas constant, $$T$$ is absolute temperature, and $$M$$ is the molar mass.

Calculate the ratio.

At the same temperature, $$R$$ and $$T$$ are the same for both gases:

$$ \frac{v_{He}}{v_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{He}}} $$

Molar masses: $$M_{He} = 4$$ g/mol, $$M_{O_2} = 32$$ g/mol

$$ \frac{v_{He}}{v_{O_2}} = \sqrt{\frac{32}{4}} = \sqrt{8} = 2\sqrt{2} $$

The correct answer is Option (2): $$2\sqrt{2}$$.

Statement I:

Mean free path is

$$λ=\frac{1}{\sqrt{\ 2}n\pi\ d^2}$$

So

$$\lambda\propto\frac{1}{d^2}$$

It is inversely proportional to square of molecular diameter.

So Statement I is true.

Statement II:

Average kinetic energy per molecule is

$$\frac{3}{2}kT$$

So

$$\text{Average KE}\propto T$$

directly proportional to absolute temperature.

So Statement II is also true.

Mean free path is

where

• n = number density
• σ\sigmaσ = collision cross section

For molecules of diameter d,

$$\sigma=\pi d^2$$

So

$$λ=\frac{1}{\sqrt{\ 2n\pi\ d^2}}$$

For gases at same temperature,

$$v_{rms}=\sqrt{\frac{3RT}{M}}$$

So

$$v_{rms}\propto\frac{1}{\sqrt{M}}$$

Therefore

Molar masses:

$$M_{H_2}=2$$

$$M_{O_2}=32$$

Given

$$v_{H_2}=2\ \text{km s}^{-1}$$

So

2vO2=322\frac{2}{v_{O_2}}

=

\sqrt{\frac{32}{2}}vO2​​2​=232​​

$$=\sqrt{16}=4$$

Thus

$$v_{O_2}=\frac{2}{4}$$

$$=0.5\ \text{km s}^{-1}$$

For a mixture of gases, the effective $$\gamma$$ is found using:

$$\frac{n_1}{\gamma_1 - 1} + \frac{n_2}{\gamma_2 - 1} = \frac{n_1 + n_2}{\gamma_{mix} - 1}$$

$$\frac{3}{5/3 - 1} + \frac{2}{7/5 - 1} = \frac{5}{\gamma - 1}$$

$$\frac{3}{2/3} + \frac{2}{2/5} = \frac{5}{\gamma - 1}$$

$$\frac{9}{2} + 5 = \frac{5}{\gamma - 1}$$

$$\frac{19}{2} = \frac{5}{\gamma - 1}$$

$$\gamma - 1 = \frac{10}{19}$$

$$\gamma = 1 + \frac{10}{19} = \frac{29}{19} \approx 1.526$$

The answer is Option (3): $$\boxed{1.52}$$.

The average kinetic energy of a monatomic molecule is:

$$KE = \frac{3}{2}k_BT$$

Given $$KE = 0.414$$ eV $$= 0.414 \times 1.6 \times 10^{-19} = 6.624 \times 10^{-20}$$ J.

$$T = \frac{2 \times KE}{3k_B} = \frac{2 \times 6.624 \times 10^{-20}}{3 \times 1.38 \times 10^{-23}} = \frac{13.248 \times 10^{-20}}{4.14 \times 10^{-23}} = \frac{13.248}{0.0414} \approx 3200 \text{ K}$$

The answer is $$3200$$ K, which corresponds to Option (2).

To find the temperature of a gas given its number density and pressure, we start with the ideal gas law in the form $$P = nkT$$, where $$n$$ is the number density (molecules per unit volume), $$k$$ is Boltzmann's constant, and $$T$$ is the temperature.

Converting the given pressure of 1.38 atm to SI units gives $$P = 1.38$$ atm $$= 1.38 \times 1.01325 \times 10^5$$ Pa $$= 1.398 \times 10^5$$ Pa.

Solving for temperature yields $$T = \frac{P}{nk} = \frac{1.398 \times 10^5}{2.0 \times 10^{25} \times 1.38 \times 10^{-23}}$$ and $$= \frac{1.398 \times 10^5}{2.76 \times 10^2} = \frac{1.398 \times 10^5}{276} \approx 506.5$$ K $$\approx 500$$ K.

The correct answer is Option (1): 500 K.

We need to find the total kinetic energy of 1 mole of oxygen gas at 27°C.

Given the universal gas constant $$R = 8.31 \text{ J mol}^{-1} \text{K}^{-1}$$, temperature $$T = 27°C = 27 + 273 = 300$$ K, and amount $$n = 1$$ mole, we proceed as follows.

Oxygen ($$O_2$$) is a diatomic molecule that, at moderate temperatures (around 300 K), has 5 degrees of freedom: 3 translational degrees of freedom (motion along the x, y, and z axes) and 2 rotational degrees of freedom (rotation about two axes perpendicular to the bond axis). Vibrational modes are not significantly excited at room temperature, so $$f = 5$$.

By the equipartition theorem, each degree of freedom contributes $$\frac{1}{2}k_BT$$ of kinetic energy per molecule, or $$\frac{1}{2}RT$$ per mole. Thus for $$n$$ moles with $$f$$ degrees of freedom, the total kinetic energy is

$$KE = \frac{f}{2} \times n \times R \times T$$

Substituting the values,

$$KE = \frac{5}{2} \times 1 \times 8.31 \times 300$$

$$= \frac{5}{2} \times 2493$$

$$= 5 \times 1246.5$$

$$= 6232.5 \text{ J}$$

The correct answer is Option 3: 6232.5 J.

Methane CH4 is a non-linear polyatomic molecule (tetrahedral structure).

For any molecule:

Translational degrees of freedom are always

$$f_t=3$$

because it can move along x,y,z directions.

For rotational degrees of freedom:

• Linear molecules have

$$f_r=2$$

• Non-linear molecules have

$$f_r=3$$

Since $$\text{CH}_4$$ is non-linear,

$$f_r=3$$

We need to find the ratio of pressures $$P_A/P_B$$ where vessel A contains 1 g of hydrogen and vessel B contains 1 g of oxygen, with both vessels being the same size and at the same temperature.

$$PV = nRT$$

Since both vessels have the same volume ($$V$$) and temperature ($$T$$), and $$R$$ is a constant:

$$P \propto n$$

Therefore:

$$\frac{P_A}{P_B} = \frac{n_A}{n_B}$$

The number of moles is given by $$n = \frac{\text{mass}}{\text{molar mass}}$$.

For hydrogen ($$H_2$$), molar mass = 2 g/mol:

$$n_A = \frac{1}{2} = 0.5 \text{ mol}$$

For oxygen ($$O_2$$), molar mass = 32 g/mol:

$$n_B = \frac{1}{32} \text{ mol}$$

$$\frac{P_A}{P_B} = \frac{n_A}{n_B} = \frac{0.5}{1/32} = 0.5 \times 32 = 16$$

The correct answer is Option (1): 16.

Find the ratio $$\frac{\gamma_1}{\gamma_2}$$ where $$\gamma_1$$ corresponds to a monoatomic gas and $$\gamma_2$$ refers to a diatomic gas treated as a rigid rotator.

Since a monoatomic gas has three degrees of freedom (translational only), its molar heat capacity at constant volume is $$C_V = \frac{3}{2}R$$. Therefore, $$C_P = C_V + R = \frac{5}{2}R$$, giving

$$\gamma_1 = \frac{C_P}{C_V} = \frac{5/2}{3/2} = \frac{5}{3}\,.$$

Now for a diatomic gas modeled as a rigid rotator, there are five degrees of freedom (three translational and two rotational, with vibrational modes frozen out). Thus, $$C_V = \frac{5}{2}R$$ and $$C_P = C_V + R = \frac{7}{2}R$$, which leads to

$$\gamma_2 = \frac{C_P}{C_V} = \frac{7/2}{5/2} = \frac{7}{5}\,.$$

Substituting these values into the desired ratio yields

$$\frac{\gamma_1}{\gamma_2} = \frac{5/3}{7/5} = \frac{5}{3} \times \frac{5}{7} = \frac{25}{21}\,.$$

Therefore, the answer is Option C: $$\frac{25}{21}\,.$$

The collision frequency of a gas molecule (number of collisions per unit time for a single molecule) is given by

$$f = \sqrt{2}\;\pi d^2 n \bar{v}$$

where $$d$$ is the molecular diameter, $$n$$ is the number density, and $$\bar{v}$$ is the mean speed.

Since the temperature remains constant, both $$\bar{v}$$ and $$d$$ remain the same (because mean speed depends only on temperature and molecular mass). So $$f \propto n$$.

The number density increases from $$n_1 = 3 \times 10^{19}\;\text{cm}^{-3}$$ to $$n_2 = 12 \times 10^{19}\;\text{cm}^{-3}$$. The ratio of collision frequency before and after is then

$$\frac{f_1}{f_2} = \frac{n_1}{n_2} = \frac{3 \times 10^{19}}{12 \times 10^{19}} = \frac{1}{4} = 0.25$$

Hence, the correct answer is 0.25.

Given that $$PT^2 = \text{constant}$$ for an ideal gas, we need to find the volume expansion coefficient.

We start by noting that

From the ideal gas law: $$PV = nRT \implies P = \frac{nRT}{V}$$

Substituting into $$PT^2 = C$$:

$$\frac{nRT}{V} \cdot T^2 = C \implies \frac{nRT^3}{V} = C$$

$$V = \frac{nRT^3}{C}$$

Next,

The volume expansion coefficient is defined as:

$$\beta = \frac{1}{V}\frac{dV}{dT}$$

$$\frac{dV}{dT} = \frac{3nRT^2}{C}$$

$$\beta = \frac{1}{V} \cdot \frac{3nRT^2}{C} = \frac{C}{nRT^3} \cdot \frac{3nRT^2}{C} = \frac{3}{T}$$

The volume expansion coefficient is $$\frac{3}{T}$$.

The correct answer is Option 4: $$\frac{3}{T}$$.

For an ideal gas with $$n$$ degrees of freedom, the molar heat capacities are

$$C_V=\frac{n}{2}R,\qquad C_P=C_V+R=\left(\frac{n}{2}+1\right)R$$

so that the ratio of specific heats (adiabatic index) is

$$\gamma=\frac{C_P}{C_V}=\frac{\frac{n}{2}+1}{\frac{n}{2}}=\frac{n+2}{n}=1+\frac{2}{n}$$

Hence $$\gamma$$ decreases when $$n$$ increases because the fraction $$2/n$$ gets smaller.

Internal energy per mole
Each quadratic degree of freedom contributes $$\tfrac12 kT$$ per molecule. For one mole this gives

$$U_n=\frac{n}{2}RT$$

Thus $$U_n$$ is directly proportional to $$n$$: more degrees of freedom ⇒ larger internal energy.

Speed of sound
For an ideal gas the speed of sound is

$$v_n=\sqrt{\frac{\gamma RT}{M}}$$

where $$M$$ is the molar mass (taken the same while comparing different values of $$n$$). Because $$R,T,M$$ are fixed,

$$v_n\propto\sqrt{\gamma}=\sqrt{1+\frac{2}{n}}$$

The function $$\sqrt{1+\dfrac{2}{n}}$$ decreases as $$n$$ increases, so the speed of sound is larger for smaller $$n$$.

Case 1: $$n=5$$ (typical diatomic gas)
$$\gamma_5=\dfrac{7}{5}=1.4,\qquad v_5\propto\sqrt{1.4},\qquad U_5=\dfrac{5}{2}RT$$

Case 2: $$n=7$$ (typical nonlinear polyatomic gas)
$$\gamma_7=\dfrac{9}{7}\approx1.286,\qquad v_7\propto\sqrt{1.286},\qquad U_7=\dfrac{7}{2}RT$$

Comparing the two cases:

$$v_5 \gt v_7\quad\text{and}\quad U_5 \lt U_7$$

Only Option C states $$v_5 \gt v_7$$ and $$U_5 \lt U_7$$, so it is the correct choice.

Final answer: Option C which is: $$v_5 \gt v_7$$ and $$U_5 \lt U_7$$

Three low density gases A, B, C have their pressure plotted against temperature (in degree C) at constant volume. All lines converge at point K when extrapolated.

We start by noting that

For an ideal gas at constant volume $$V$$, with $$n$$ moles:

$$ P = \frac{nR}{V} T_{\text{abs}} $$

where $$T_{\text{abs}}$$ is the absolute temperature in Kelvin.

Next,

Since $$T_{\text{abs}} = T(°C) + 273$$:

$$ P = \frac{nR}{V}(T + 273) $$

This is a linear equation in $$T$$ (Celsius). Different gases have different slopes ($$nR/V$$), but all lines have the same x-intercept (where $$P = 0$$).

From this,

Setting $$P = 0$$:

$$ 0 = \frac{nR}{V}(T + 273) $$

$$ T + 273 = 0 \implies T = -273°C $$

This is absolute zero -- the temperature at which an ideal gas would theoretically have zero pressure. All gas lines, regardless of the type or amount of gas, converge to this point when extrapolated.

The temperature corresponding to point K is $$-273°$$C.

The correct answer is Option 1: $$-273°$$C.

We need to find the pressure of air in the tyre when the temperature increases from $$27°$$C to $$36°$$C.

Initial pressure: $$P_1 = 270$$ kPa

Initial temperature: $$T_1 = 27°C = 300$$ K

Final temperature: $$T_2 = 36°C = 309$$ K

Using Gay-Lussac's Law: (pressure is directly proportional to temperature at constant volume):

$$\frac{P_1}{T_1} = \frac{P_2}{T_2}$$

Solving for $$P_2$$:

$$P_2 = P_1 \times \frac{T_2}{T_1} = 270 \times \frac{309}{300}$$

$$P_2 = 270 \times 1.03 = 278.1$$ kPa

The approximate pressure of air in the tyre is $$278$$ kPa.

The correct answer is Option 3: $$278$$ kPa.

We need to find the total internal energy of a gas mixture consisting of 2 moles of oxygen ($$O_2$$) and 4 moles of neon ($$Ne$$) at temperature $$T$$, neglecting vibrational modes.

We recall that the internal energy of an ideal gas is given by the equipartition theorem:

$$U = n \cdot \frac{f}{2} \cdot RT$$

where $$n$$ is the number of moles, $$f$$ is the number of degrees of freedom per molecule, $$R$$ is the universal gas constant, and $$T$$ is the absolute temperature.

We begin by determining the degrees of freedom for oxygen ($$O_2$$, a diatomic gas). It has the following degrees of freedom when vibrational modes are neglected:
- 3 translational degrees of freedom (motion along the x, y, and z axes)
- 2 rotational degrees of freedom (rotation about two axes perpendicular to the bond axis; rotation about the bond axis itself contributes negligibly because the moment of inertia about that axis is extremely small)
Total for $$O_2$$: $$f_{O_2} = 3 + 2 = 5$$

Next, we determine the degrees of freedom for neon ($$Ne$$, a monoatomic gas). A monoatomic gas consists of single atoms and therefore can only translate; it has no rotational or vibrational modes:
- 3 translational degrees of freedom only
Total for $$Ne$$: $$f_{Ne} = 3$$

Substituting into the equipartition formula gives the internal energy of oxygen as:

$$U_{O_2} = n_{O_2} \cdot \frac{f_{O_2}}{2} \cdot RT = 2 \cdot \frac{5}{2} \cdot RT = 5RT$$

Similarly, for neon we obtain:

$$U_{Ne} = n_{Ne} \cdot \frac{f_{Ne}}{2} \cdot RT = 4 \cdot \frac{3}{2} \cdot RT = 6RT$$

Therefore, since internal energy is an extensive property, the total internal energy of the mixture is the sum of the individual contributions:

$$U_{total} = U_{O_2} + U_{Ne} = 5RT + 6RT = 11RT$$

The correct answer is Option 1: $$11RT$$.

We analyze each statement: Statement I: The temperature changes from $$-73°C = 200 \text{ K}$$ to $$527°C = 800 \text{ K}$$. The root mean square speed is given by $$v_{rms} = \sqrt{\frac{3RT}{M}}$$, so $$v_{rms} \propto \sqrt{T}$$. $$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{800}{200}} = \sqrt{4} = 2$$. The rms speed is doubled. Statement I is TRUE.

Statement II: For an ideal gas, $$PV = nRT$$ and the translational kinetic energy is $$KE = \frac{3}{2}nRT$$. Therefore $$PV = \frac{2}{3}KE$$, which means $$PV \neq KE$$. The product of pressure and volume equals $$\frac{2}{3}$$ of the translational kinetic energy, not equal to it. Statement II is FALSE.

The correct answer is Option B: Statement I is true but Statement II is false.

The root mean square speed of gas molecules is:

$$v_{rms} = \sqrt{\frac{3k_BT}{m}}$$

Given that $$T = 27°C = 300$$ K, $$m = 4.6 \times 10^{-26}$$ kg, $$k_B = 1.4 \times 10^{-23}$$ J/K

$$v_{rms} = \sqrt{\frac{3 \times 1.4 \times 10^{-23} \times 300}{4.6 \times 10^{-26}}}$$

$$= \sqrt{\frac{1.26 \times 10^{-20}}{4.6 \times 10^{-26}}}$$

$$= \sqrt{2.739 \times 10^5}$$

$$\approx 523 \text{ m/s}$$

The root mean square speed of gas molecules is given by:

$$v_{rms} = \sqrt{\frac{3RT}{M}}$$

where $$R$$ is the gas constant, $$T$$ is the temperature, and $$M$$ is the molar mass of the gas.

Since all three vessels are at the same temperature:

$$v_{rms} \propto \frac{1}{\sqrt{M}}$$

Molar masses:

Neon (monoatomic): $$M = 20$$ g/mol

Chlorine (diatomic, Cl$$_2$$): $$M = 71$$ g/mol

Uranium hexafluoride (polyatomic, UF$$_6$$): $$M = 238 + 6 \times 19 = 352$$ g/mol

Since $$v_{rms} \propto \frac{1}{\sqrt{M}}$$, the gas with the smallest molar mass has the highest $$v_{rms}$$:

$$M_{\text{Ne}} < M_{\text{Cl}_2} < M_{\text{UF}_6}$$

Therefore:

$$v_{rms}(\text{mono}) > v_{rms}(\text{dia}) > v_{rms}(\text{poly})$$

A flask contains H$$_2$$ and Ar in the ratio 2:1 by mass at 30 degrees C. We need the ratio of average kinetic energy per molecule.

We start by noting that

The average translational kinetic energy per molecule of any ideal gas is given by:

$$ KE = \frac{3}{2}k_B T $$

where $$k_B$$ is Boltzmann's constant and $$T$$ is the absolute temperature. This is a fundamental result of the kinetic theory of gases.

Next,

The formula $$KE = \frac{3}{2}k_B T$$ depends only on the temperature and is completely independent of:

- The type of gas (whether monatomic Ar or diatomic H$$_2$$)

- The molecular mass of the gas

- The mass ratio in the mixture

This is because the kinetic energy distribution depends only on temperature (Maxwell-Boltzmann distribution).

From this,

Since both gases are at the same temperature (30 degrees C = 303 K):

$$ \frac{K_{Ar}}{K_{H_2}} = \frac{\frac{3}{2}k_B T}{\frac{3}{2}k_B T} = 1 $$

The ratio is 1.

The correct answer is Option 2: 1.

We need to determine $$\alpha$$ given that the rms speed of O$$_2$$ at 300 K is $$\sqrt{\frac{\alpha + 5}{\alpha}}$$ times the average speed.

Recalling that the rms speed is given by $$v_{rms} = \sqrt{\frac{3RT}{M}}$$ and the average speed by $$v_{avg} = \sqrt{\frac{8RT}{\pi M}}$$, we can form their ratio.

Thus, $$\frac{v_{rms}}{v_{avg}} = \sqrt{\frac{3RT/M}{8RT/(\pi M)}} = \sqrt{\frac{3\pi}{8}}$$.

Upon substituting $$\pi = \frac{22}{7}$$, the expression becomes $$\frac{3\pi}{8} = \frac{3 \times 22}{7 \times 8} = \frac{66}{56} = \frac{33}{28}$$.

Since the problem states that this ratio equals $$\sqrt{\frac{\alpha + 5}{\alpha}}$$, we set $$\frac{\alpha + 5}{\alpha} = \frac{33}{28}$$ and solve. Multiplying both sides by 28 yields $$28(\alpha + 5) = 33\alpha$$, which simplifies to $$28\alpha + 140 = 33\alpha$$ and hence $$5\alpha = 140$$, giving $$\alpha = 28$$.

Accordingly, the correct choice corresponds to option 2 with a value of $$28$$.

The r.m.s. speed of a gas molecule is given by:

$$v_{rms} = \sqrt{\frac{3RT}{M}}$$

At the same temperature, the ratio of r.m.s. speeds of two gases is:

$$\frac{v_{Ar}}{v_{Cl_2}} = \sqrt{\frac{M_{Cl_2}}{M_{Ar}}}$$

Given: $$v_{Cl_2} = 490$$ m/s, $$M_{Cl_2} = 70.9$$ u, $$M_{Ar} = 39.9$$ u.

$$\frac{v_{Ar}}{490} = \sqrt{\frac{70.9}{39.9}}$$

Compute the ratio:

$$\frac{70.9}{39.9} = 1.7770$$

$$\sqrt{1.7770} = 1.3330$$

Therefore:

$$v_{Ar} = 490 \times 1.3330 = 653.2 \text{ m/s}$$

This is closest to $$651.7$$ m/s. Let me verify with more precision:

$$\frac{70.9}{39.9} = 1.77694...$$

$$\sqrt{1.77694} = 1.33302...$$

$$490 \times 1.33302 = 653.18$$

The closest option is $$651.7$$ m/s (Option B).

The answer is Option B: $$651.7$$ m/s$$^{-1}$$.

Matching degrees of freedom with gas types:

(A) 3 translational degrees of freedom only: This describes monoatomic gases (I), which have no rotational or vibrational modes.

(B) 3 translational + 2 rotational: This describes rigid diatomic gases (III), like N₂, O₂ at normal temperatures.

(C) 3 translational + 2 rotational + 1 vibrational: This describes non-rigid diatomic gases (IV), where vibrational mode is activated.

(D) 3 translational + 3 rotational + more than one vibrational: This describes polyatomic gases (II), like CO₂, H₂O.

Matching: A-I, B-III, C-IV, D-II

The correct answer is Option 1.

According to the kinetic theory of gases, the average kinetic energy of a molecule is given by

$$\langle KE \rangle = \frac{3}{2} k_B T$$

where $$k_B$$ is the Boltzmann constant and $$T$$ is the absolute temperature.

From this expression, the average kinetic energy is directly proportional to the absolute temperature $$T$$ and does not depend on the volume, pressure, or the nature of the gas. This means that at the same temperature, molecules of all gases (regardless of their molecular mass) have the same average kinetic energy.

Hence, the correct answer is that the average kinetic energy of a molecule is proportional to absolute temperature.

The ratio $$\gamma = \frac{C_P}{C_V}$$ for an ideal gas.

For an ideal gas, $$C_P = C_V + R$$, where $$R$$ is the universal gas constant.

$$C_V = \frac{f}{2}R$$ where $$f$$ is the degrees of freedom.

$$\gamma = \frac{C_P}{C_V} = 1 + \frac{2}{f}$$

For an ideal gas, $$f$$ is a constant (depends only on the type of gas - monatomic, diatomic, etc.), and does not depend on temperature.

Therefore $$\gamma$$ is independent of temperature: $$\gamma \propto T^0$$.

(Note: In reality, at very high temperatures, vibrational modes get activated, changing $$f$$, but in the standard kinetic theory treatment, $$\gamma$$ is constant.)

The correct answer is Option (2): $$\boxed{\gamma \propto T^0}$$.

We need to determine the dependence of root mean square (RMS) velocity of gas molecules on temperature.

Write the formula for RMS velocity.

$$v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$$

where $$R$$ is the universal gas constant, $$T$$ is the absolute temperature, and $$M$$ is the molar mass.

Identify the proportionality.

Since $$R$$ and $$M$$ are constants for a given gas:

$$v_{\text{rms}} \propto \sqrt{T}$$

The RMS velocity is proportional to the square root of the absolute temperature.

The correct answer is Proportional to square root of temperature $$\sqrt{T}$$.

The average kinetic energy of gas molecules is given by

$$KE = \frac{3}{2}k_BT$$

where $$T$$ is the absolute temperature in Kelvin.

Since $$KE \propto T$$, for the kinetic energy to become double we need $$T' = 2T$$.

We have $$T = 27°C = 300$$ K, so

$$T' = 2 \times 300 = 600 \text{ K} = 327°C$$

Hence, the correct answer is Option 2.

When the temperature of an ideal gas increases from $$200$$ K to $$800$$ K, the r.m.s. speed follows the relation $$v_{rms} = \sqrt{\dfrac{3kT}{m}}$$, implying that $$v_{rms}\propto\sqrt{T}$$.

Accordingly, the ratio of r.m.s. speeds at $$800$$ K and $$200$$ K becomes

$$\dfrac{v_{rms}(800)}{v_{rms}(200)} = \sqrt{\dfrac{800}{200}} = \sqrt{4} = 2\,. $$

Since the r.m.s. speed at $$200$$ K is given as $$v_0$$, the speed at $$800$$ K is

$$v_{rms}(800) = 2 \times v_0 = 2v_0\,. $$

Therefore, the r.m.s. speed at $$800$$ K is \boxed{2v_0}.

We need to identify which statements about the kinetic theory of gases are correct.

Statement A: "The motion of gas molecules freezes at 0°C."

This is false. The motion of gas molecules would theoretically cease at absolute zero (0 K = $$-273.15°$$C), not at 0°C. At 0°C (273.15 K), gas molecules still possess significant kinetic energy.

Statement B: "The mean free path decreases if the density of molecules is increased."

The mean free path is given by $$\lambda = \frac{1}{\sqrt{2}\pi d^2 n}$$, where $$n$$ is the number density and $$d$$ is the molecular diameter. As density ($$n$$) increases, mean free path decreases. This is true.

Statement C: "The mean free path increases if temperature is increased keeping pressure constant."

At constant pressure, $$n = \frac{P}{k_BT}$$. As temperature increases, $$n$$ decreases. Since $$\lambda = \frac{1}{\sqrt{2}\pi d^2 n}$$, a decrease in $$n$$ leads to an increase in $$\lambda$$. This is true.

Statement D: "Average kinetic energy per molecule per degree of freedom is $$\frac{3}{2}k_BT$$."

This is false. The average kinetic energy per molecule per degree of freedom is $$\frac{1}{2}k_BT$$. The total average kinetic energy for a monoatomic gas (3 degrees of freedom) is $$\frac{3}{2}k_BT$$, but that is per molecule, not per degree of freedom.

The correct statements are B and C.

The correct answer is Option B.

Sound travels in a mixture of 2 moles of helium (He) and $$n$$ moles of hydrogen (H$$_2$$). We are given that the rms speed of gas molecules in the mixture is $$\sqrt{2}$$ times the speed of sound in the mixture.

Relate rms speed and speed of sound to find $$\gamma_{mix}$$.

The rms speed and speed of sound in a gas are:

Given $$v_{rms} = \sqrt{2} \times v_{sound}$$, we square both sides:

Cancelling $$\frac{RT}{M_{mix}}$$ from both sides:

Determine the specific heats of the individual gases.

Helium is monatomic with degrees of freedom $$f_{He} = 3$$:

Hydrogen is diatomic with degrees of freedom $$f_{H_2} = 5$$:

Calculate $$C_{v,mix}$$ and $$C_{p,mix}$$ for the mixture.

For a mixture of ideal gases, the molar specific heats are weighted by the number of moles:

Apply the condition $$\gamma_{mix} = \frac{3}{2}$$.

Cross-multiplying:

Verification: With $$n = 2$$: $$\gamma_{mix} = \frac{10 + 14}{6 + 10} = \frac{24}{16} = \frac{3}{2}$$ $$\checkmark$$

The correct answer is Option B: $$2$$.

We need to find the effect on the root mean square (rms) velocity of oxygen molecules when the temperature is doubled and the oxygen molecule dissociates into atomic oxygen.

Write the formula for rms velocity.

$$v_{rms} = \sqrt{\frac{3RT}{M}}$$

where $$M$$ is the molar mass and $$T$$ is the temperature.

Calculate the initial rms velocity for O$$_2$$.

For molecular oxygen: $$M_{O_2} = 32$$ g/mol, temperature = $$T$$

$$v_1 = \sqrt{\frac{3RT}{32}}$$

Calculate the new rms velocity for atomic oxygen at doubled temperature.

For atomic oxygen: $$M_O = 16$$ g/mol, temperature = $$2T$$

$$v_2 = \sqrt{\frac{3R(2T)}{16}} = \sqrt{\frac{6RT}{16}}$$

Find the ratio of velocities.

$$\frac{v_2}{v_1} = \sqrt{\frac{6RT/16}{3RT/32}} = \sqrt{\frac{6RT \times 32}{16 \times 3RT}} = \sqrt{\frac{192}{48}} = \sqrt{4} = 2$$

The velocity of atomic oxygen is doubled compared to the original velocity of molecular oxygen.

The correct answer is Option B.

We have 0.056 kg = 56 g of Nitrogen ($$N_2$$). The molar mass of $$N_2$$ is 28 g/mol.

Calculate the number of moles:: $$n = \frac{56}{28} = 2 \text{ moles}$$

We need to double the speed of the molecules. Since the rms speed is given by $$v_{rms} = \sqrt{\frac{3RT}{M}}$$, speed is proportional to $$\sqrt{T}$$. To double the speed:: $$\frac{v_2}{v_1} = 2 = \sqrt{\frac{T_2}{T_1}}$$

$$\frac{T_2}{T_1} = 4$$

$$T_2 = 4 \times 400 = 1600 \text{ K}$$

The change in temperature is:: $$\Delta T = T_2 - T_1 = 1600 - 400 = 1200 \text{ K}$$

$$N_2$$ is a diatomic gas, so $$C_v = \frac{5}{2}R$$. The heat required at constant volume is:: $$Q = n C_v \Delta T = n \times \frac{5}{2}R \times \Delta T$$

$$Q = 2 \times \frac{5}{2} \times 2 \times 1200$$

$$Q = 2 \times 5 \times 1200 = 12000 \text{ cal} = 12 \text{ kcal}$$

The answer is 12 kcal.

For a monoatomic gas, the molar specific heat at constant volume is $$C_{v1} = \dfrac{3}{2}R$$ (3 degrees of freedom).

For a diatomic gas with no vibrational mode, $$C_{v2} = \dfrac{5}{2}R$$ (5 degrees of freedom: 3 translational + 2 rotational).

For a mixture of $$n_1 = 1$$ mole of monoatomic gas and $$n_2 = 3$$ moles of diatomic gas, the molar specific heat at constant volume is:

$$C_{v,\text{mix}} = \dfrac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \dfrac{1 \cdot \dfrac{3}{2}R + 3 \cdot \dfrac{5}{2}R}{1 + 3}$$

$$= \dfrac{\dfrac{3R}{2} + \dfrac{15R}{2}}{4} = \dfrac{\dfrac{18R}{2}}{4} = \dfrac{9R}{4}$$

We are given that $$C_{v,\text{mix}} = \dfrac{\alpha^2}{4}R$$. Comparing: $$\dfrac{\alpha^2}{4} = \dfrac{9}{4}$$, so $$\alpha^2 = 9$$, giving $$\alpha = 3$$.

The answer is $$3$$.

We need to find the total internal energy of two moles of a monoatomic ideal gas. The number of moles is $$n = 2$$, the temperature is $$T = 300$$ K, and the gas constant is $$R = 8.31$$ J mol$$^{-1}$$ K$$^{-1}$$. For a monoatomic ideal gas, the degrees of freedom $$f = 3$$, so the internal energy is given by:

$$U = \frac{f}{2} nRT = \frac{3}{2} nRT$$

Substituting the values:

$$U = \frac{3}{2} \times 2 \times 8.31 \times 300$$

$$U = 3 \times 8.31 \times 300$$

$$U = 3 \times 2493$$

$$U = 7479 \text{ J}$$

Hence, the total internal energy is 7479 J.

Temperature increase: $$\Delta T = 1°$$ C $$= 1$$ K

Pressure increase: $$0.4\%$$

For a gas at constant volume:

$$\frac{P}{T} = \text{constant}$$

If the temperature increases from $$T$$ to $$T + 1$$:

$$\frac{P_1}{T} = \frac{P_2}{T + 1}$$

$$P_2 = P_1 \times \frac{T + 1}{T} = P_1\left(1 + \frac{1}{T}\right)$$

The percentage increase in pressure:

$$\frac{P_2 - P_1}{P_1} \times 100 = \frac{1}{T} \times 100 = 0.4\%$$

$$\frac{100}{T} = 0.4$$

$$T = \frac{100}{0.4} = 250 \text{ K}$$

Hence, the initial temperature of the gas is 250 K.

We need to find the ratio of average kinetic energy per molecule for argon and oxygen at the same temperature.

Argon (Ar) is a monoatomic gas with degrees of freedom $$f_1 = 3$$

Oxygen (O$$_2$$) is a diatomic gas with degrees of freedom $$f_2 = 5$$

The average kinetic energy per molecule of a gas is given by:

$$E = \frac{f}{2}k_BT$$

where $$f$$ is the number of degrees of freedom, $$k_B$$ is Boltzmann's constant, and $$T$$ is the temperature.

Since both gases are at the same temperature ($$27°$$C = 300 K):

$$\frac{E_{Ar}}{E_{O_2}} = \frac{\frac{f_1}{2}k_BT}{\frac{f_2}{2}k_BT} = \frac{f_1}{f_2} = \frac{3}{5}$$

Note: The mass ratio (3:2) is irrelevant here as the average kinetic energy per molecule depends only on degrees of freedom and temperature, not on the total mass of the gas.

Hence, the correct answer is Option A.

For an ideal gas, the molar specific heats are related to the degree of freedom $$f$$ as follows:

$$C_v = \frac{f}{2}R$$

$$C_p = C_v + R = \frac{f}{2}R + R = \left(\frac{f}{2} + 1\right)R$$

The ratio of specific heats is:

$$\gamma = \frac{C_p}{C_v} = \frac{\frac{f}{2}R + R}{\frac{f}{2}R} = \frac{\frac{f}{2} + 1}{\frac{f}{2}} = 1 + \frac{2}{f}$$

Hence, the correct answer is Option B.

We need to find the relation between the root mean square speed $$v_{rms}$$ and the most probable speed $$v_p$$ for oxygen gas at $$300$$ K.

Recall the expressions for $$v_{rms}$$ and $$v_p$$: $$v_{rms} = \sqrt{\frac{3RT}{M}}$$ and $$v_p = \sqrt{\frac{2RT}{M}}$$, where $$R$$ is the universal gas constant, $$T$$ is the temperature, and $$M$$ is the molar mass.

Taking the ratio gives $$\frac{v_{rms}}{v_p} = \frac{\sqrt{\frac{3RT}{M}}}{\sqrt{\frac{2RT}{M}}} = \sqrt{\frac{3}{2}}$$, so that $$v_{rms} = \sqrt{\frac{3}{2}} \, v_p$$.

This relation holds for any ideal gas at any temperature, as the ratio depends only on the numerical constants 3 and 2 in the respective formulas.

The correct answer is Option C: $$v_{rms} = \sqrt{\frac{3}{2}}v_p$$.

We need to find the ratio of specific heat at constant volume ($$C_v$$) to the specific heat at constant pressure ($$C_p$$) for a gas with $$n$$ degrees of freedom. According to the equipartition theorem, each degree of freedom contributes $$\dfrac{1}{2}kT$$ of energy per molecule; therefore, for one mole of an ideal gas with $$n$$ degrees of freedom, the internal energy is $$U = \dfrac{n}{2}RT$$, where $$R$$ is the universal gas constant and $$T$$ is the temperature.

Since $$C_v = \dfrac{dU}{dT}$$, it follows that $$C_v = \dfrac{n}{2}R$$. Substituting into Mayer's relation $$C_p - C_v = R$$ gives $$C_p = C_v + R = \dfrac{n}{2}R + R = \left(\dfrac{n}{2} + 1\right)R = \dfrac{n + 2}{2}R$$.

From this, $$\dfrac{C_v}{C_p} = \dfrac{\dfrac{n}{2}R}{\dfrac{n+2}{2}R} = \dfrac{n}{n+2}$$.

Verification: For a monoatomic gas ($$n = 3$$): $$\dfrac{C_v}{C_p} = \dfrac{3}{5}$$, which matches the known value $$\dfrac{1}{\gamma} = \dfrac{3}{5}$$.

The correct answer is Option A: $$\dfrac{n}{n+2}$$.

We have 14 g of nitrogen gas ($$N_2$$, molecular mass = 28 g/mol), so the number of moles is $$n = \frac{14}{28} = 0.5$$ mol. The initial temperature is $$T_1 = 27^\circ C = 300$$ K.

The r.m.s. speed of gas molecules is given by $$v_{rms} = \sqrt{\frac{3RT}{M}}$$. Since $$v_{rms} \propto \sqrt{T}$$, to double the r.m.s. speed, we need the temperature to become 4 times the original. So the final temperature is $$T_2 = 4 \times 300 = 1200$$ K.

The change in temperature is $$\Delta T = 1200 - 300 = 900$$ K.

Nitrogen is a diatomic gas, so the molar heat capacity at constant volume is $$C_v = \frac{5}{2}R$$. The heat transferred (at constant volume, since the vessel is rigid) is:

$$Q = n C_v \Delta T = 0.5 \times \frac{5}{2} \times 8.32 \times 900$$

$$Q = 0.5 \times 2.5 \times 8.32 \times 900 = 1.25 \times 8.32 \times 900 = 10.4 \times 900 = 9360 \text{ J}$$

Hence, the correct answer is Option C.

We need to determine which statements about kinetic theory of gases are correct.

Statement (1): The average kinetic energy of a gas molecule decreases when the temperature is reduced.

The average kinetic energy of a gas molecule is $$KE = \frac{3}{2}k_BT$$.

Since KE is directly proportional to temperature, reducing the temperature reduces the kinetic energy. This is correct.

Statement (2): The average kinetic energy of a gas molecule increases with increase in pressure at constant temperature.

Since $$KE = \frac{3}{2}k_BT$$, the average kinetic energy depends only on temperature. At constant temperature, even if pressure increases, the average kinetic energy remains the same. This is incorrect.

Statement (3): The average kinetic energy of a gas molecule decreases with increase in volume.

Again, average kinetic energy depends only on temperature, not on volume. This is incorrect.

Statement (4): Pressure of a gas increases with increase in temperature at constant volume.

From Gay-Lussac's law: $$\frac{P}{T} = \text{constant}$$ at constant volume. So pressure increases with temperature. This is correct.

Statement (5): The volume of gas decreases with increase in temperature.

From Charles's law at constant pressure: $$V \propto T$$. Volume increases (not decreases) with temperature. This is incorrect.

The correct statements are (1) and (4) only.

Hence, the correct answer is Option A.

We need to evaluate two statements about kinetic theory of gases.

Statement I: The average momentum of a molecule in a sample of an ideal gas depends on temperature.

The molecules in an ideal gas move randomly in all directions. While the average speed depends on temperature, the average velocity (and hence average momentum) is zero because the molecular motion is random and symmetric in all directions. Since the average momentum is always zero regardless of temperature, it does not depend on temperature.

Statement I is FALSE.

Statement II: If the temperature is doubled and O$$_2$$ molecules dissociate into O atoms, the rms speed becomes 2v.

The rms speed is given by:

$$v_{rms} = \sqrt{\frac{3RT}{M}}$$

Initially, for O$$_2$$ (molar mass $$M = 32$$ g/mol) at temperature $$T$$:

$$v = \sqrt{\frac{3RT}{32}}$$

After dissociation, we have O atoms (molar mass $$M' = 16$$ g/mol) at temperature $$2T$$:

$$v' = \sqrt{\frac{3R(2T)}{16}} = \sqrt{\frac{6RT}{16}} = \sqrt{\frac{3RT}{8}}$$

Taking the ratio:

$$\frac{v'}{v} = \sqrt{\frac{3RT/8}{3RT/32}} = \sqrt{\frac{32}{8}} = \sqrt{4} = 2$$

So $$v' = 2v$$.

Statement II is TRUE.

Hence, the correct answer is Option D: Statement I is false but Statement II is true.

Given: $$V = 2000 \text{ cm}^3 = 2 \times 10^{-3} \text{ m}^3$$, $$T = 300$$ K, $$P = 100$$ kPa $$= 10^5$$ Pa, total mass $$= 0.76$$ g, $$R = 8.3$$ J K$$^{-1}$$ mol$$^{-1}$$.

Let the number of moles of hydrogen be $$n_1$$ and oxygen be $$n_2$$.

Using the ideal gas law for the mixture:

$$PV = (n_1 + n_2)RT$$

$$10^5 \times 2 \times 10^{-3} = (n_1 + n_2) \times 8.3 \times 300$$

$$200 = (n_1 + n_2) \times 2490$$

$$n_1 + n_2 = \frac{200}{2490} = \frac{20}{249} \approx 0.08 \text{ mol}$$

From the total mass:

$$2n_1 + 32n_2 = 0.76$$ g

From the ideal gas equation: $$n_1 + n_2 = 0.08$$, so $$n_2 = 0.08 - n_1$$.

Substituting:

$$2n_1 + 32(0.08 - n_1) = 0.76$$

$$2n_1 + 2.56 - 32n_1 = 0.76$$

$$-30n_1 = 0.76 - 2.56 = -1.80$$

$$n_1 = \frac{1.80}{30} = 0.06 \text{ mol}$$

$$n_2 = 0.08 - 0.06 = 0.02 \text{ mol}$$

Therefore:

$$\frac{n_1}{n_2} = \frac{0.06}{0.02} = \frac{3}{1}$$

The correct answer is Option B.

A vessel contains 16 g of hydrogen and 128 g of oxygen at STP.

Moles of H$$_2$$ = $$\frac{16}{2} = 8$$ mol

Moles of O$$_2$$ = $$\frac{128}{32} = 4$$ mol

Total moles = $$8 + 4 = 12$$ mol

At STP (Standard Temperature and Pressure: 273.15 K, 1 atm), 1 mole of ideal gas occupies 22,400 cm$$^3$$.

$$V = 12 \times 22400 = 268800 \text{ cm}^3 = 2.688 \times 10^5 \text{ cm}^3$$

This is approximately $$27 \times 10^4$$ cm$$^3$$.

Hence, the correct answer is Option C: $$27 \times 10^4$$.

Concept:
For an ideal gas:

$$PV=nRT\ \Rightarrow\ \frac{V}{T}=\frac{nR}{P}\ $$

Analysis:
Slope of V-T graph:

slope=VT=nRP$$\text{slope}=\frac{V}{T}=\frac{nR}{P}$$

So, slope $$\propto\frac{1}{P}$$

From graph:
Line $$P_2$$ is steeper ⇒ larger slope ⇒ smaller pressure

$$\frac{nR}{P_2}>\frac{nR}{P_1}$$

$$\Rightarrow P_2$$

Final Answer:

$$P_2\ <\ P_1$$

Same gas in two vessels of same volume and same temperature, with number of molecules in ratio 1:4. We check statements A, B, C, D.

Statement A: RMS velocity is the same in both vessels.

The RMS velocity of gas molecules is:

$$v_{\text{rms}} = \sqrt{\frac{3k_BT}{m}}$$

Since the gas is the same (same molecular mass $$m$$) and the temperature $$T$$ is the same, the RMS velocity is identical in both vessels regardless of the number of molecules.

Statement A is correct.

Statement B: Ratio of pressures is 1:4.

From the ideal gas law: $$PV = Nk_BT$$

Since $$V$$ and $$T$$ are the same:

$$P \propto N$$

$$\frac{P_1}{P_2} = \frac{N_1}{N_2} = \frac{1}{4}$$

Statement B is correct.

Statement C: Ratio of pressure is 1:1.

This contradicts Statement B. Since pressure is proportional to the number of molecules, the ratio is 1:4, not 1:1.

Statement C is incorrect.

Statement D: RMS velocity ratio is 1:4.

As shown in Statement A, RMS velocity is the same for both (ratio 1:1, not 1:4).

Statement D is incorrect.

Statements A and B are correct.

The correct answer is Option C: A and B only.

We have smoke particles of mass $$m = 5 \times 10^{-17}$$ kg undergoing Brownian motion in air at NTP. At NTP the temperature is $$T = 273$$ K. The Boltzmann constant is $$k = 1.38 \times 10^{-23}$$ J K$$^{-1}$$.

The root mean square speed from kinetic theory is given by $$v_{\text{rms}} = \sqrt{\frac{3kT}{m}}$$.

Substituting the values: $$v_{\text{rms}} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 273}{5 \times 10^{-17}}}$$

We first compute the numerator: $$3 \times 1.38 \times 273 = 3 \times 376.74 = 1130.22$$. So the numerator is $$1130.22 \times 10^{-23}$$.

Now $$\frac{1130.22 \times 10^{-23}}{5 \times 10^{-17}} = \frac{1130.22}{5} \times 10^{-23+17} = 226.04 \times 10^{-6} = 2.26 \times 10^{-4}$$.

Therefore $$v_{\text{rms}} = \sqrt{2.26 \times 10^{-4}} = \sqrt{2.26} \times 10^{-2} \approx 1.503 \times 10^{-2}$$ m/s $$\approx 15$$ mm s$$^{-1}$$.

Hence, the correct answer is Option 3.

Let's understand degrees of freedom step by step. Degrees of freedom refer to the independent ways a molecule can store energy, including translational, rotational, and vibrational motions. We'll evaluate each statement one by one.

Starting with statement A: "A molecule with $$n$$ degrees of freedom has $$n^{2}$$ different ways of storing energy." This is incorrect. The number of degrees of freedom (n) directly corresponds to the number of independent quadratic terms in the energy expression (like $$\frac{1}{2}mv_x^2$$ for translational motion). There is no squaring involved; the energy is distributed equally among the n degrees of freedom according to the equipartition theorem. Thus, statement A is false.

Now statement B: "Each degree of freedom is associated with $$\frac{1}{2}RT$$ average energy per mole." This is correct. The equipartition theorem states that each quadratic degree of freedom contributes $$\frac{1}{2}kT$$ per molecule. Since $$R = N_A k$$ (where $$N_A$$ is Avogadro's number), per mole, this becomes $$\frac{1}{2}RT$$. This applies to translational and rotational degrees of freedom at standard temperatures. Therefore, statement B is true.

Statement C: "A monoatomic gas molecule has 1 rotational degree of freedom whereas diatomic molecule has 2 rotational degrees of freedom." This is partially incorrect. A monoatomic molecule (like helium or argon) is a point mass with no significant rotational inertia. It has 0 rotational degrees of freedom. A diatomic molecule (like nitrogen or oxygen) has 2 rotational degrees of freedom, as it can rotate about two axes perpendicular to the molecular axis (rotation along the molecular axis is negligible). Since the monoatomic part is wrong (it should be 0, not 1), statement C is false.

Statement D: "$$CH_4$$ has a total of 6 degrees of freedom." For methane ($$CH_4$$), which is a non-linear polyatomic molecule (tetrahedral structure), we calculate degrees of freedom as follows: Total atoms = 5 (1 carbon + 4 hydrogen). Total degrees of freedom without constraints = 3N = 3 × 5 = 15. However, at standard temperatures, vibrational modes are not excited, so we only consider translational and rotational degrees of freedom. Translational degrees of freedom = 3 (for any molecule in 3D space). Rotational degrees of freedom for a non-linear molecule = 3. Thus, total degrees of freedom = 3 (translational) + 3 (rotational) = 6. Therefore, statement D is true.

Summarizing: A is false, B is true, C is false, D is true. The true statements are B and D.

Now, looking at the options:

• A: B and C only → Incorrect (C is false)
• B: B and D only → Correct (matches our true statements)
• C: A and B only → Incorrect (A is false)
• D: C and D only → Incorrect (C is false)

Hence, the correct answer is Option B.

For an ideal gas, the equation of state is stated as $$pV = nRT,$$ where $$p$$ is the pressure, $$V$$ is the volume, $$n$$ is the number of moles, $$R$$ is the universal gas constant and $$T$$ is the absolute temperature.

We have one mole of hydrogen and two moles of carbon dioxide. So the total number of moles present in the cylinder is

$$n = 1 + 2 = 3 \text{ mol}.$$

The gas constant is given as $$R = 8.3 \text{ J mol}^{-1}\text{ K}^{-1}.$$

The absolute temperature of the mixture is given to be $$T = 400 \text{ K}.$$

The volume of the cylindrical container is

$$V = 4.0 \times 10^{-3} \text{ m}^3.$$

Now we substitute these values into the ideal-gas equation. Rearranging the formula for pressure, we get

$$p = \frac{nRT}{V}.$$

Substituting $$n = 3 \text{ mol},\; R = 8.3 \text{ J mol}^{-1}\text{ K}^{-1},\; T = 400 \text{ K},$$ and $$V = 4.0 \times 10^{-3} \text{ m}^3,$$ we obtain

$$p = \frac{3 \times 8.3 \times 400}{4.0 \times 10^{-3}}.$$

First, multiply the numerator step by step:

$$8.3 \times 400 = 3320,$$

so

$$3 \times 3320 = 9960.$$

Hence the numerator equals $$9960.$$

Next we divide by the volume:

$$p = \frac{9960}{4.0 \times 10^{-3}}.$$

Recognise that dividing by $$4.0 \times 10^{-3}$$ is the same as multiplying by $$\frac{1}{4.0 \times 10^{-3}} = \frac{1}{0.004} = 250.$$ Therefore,

$$p = 9960 \times 250.$$

Carrying out this multiplication:

$$9960 \times 250 = 2\,490\,000.$$

Expressing this in scientific notation,

$$2\,490\,000 \text{ Pa} = 2.49 \times 10^{6} \text{ Pa}.$$

Writing the same value with one decimal place different,

$$2.49 \times 10^{6} \text{ Pa} = 24.9 \times 10^{5} \text{ Pa}.$$

This matches option C of the given choices.

Hence, the correct answer is Option 3.

The root-mean-square (RMS) velocity of a gas molecule is $$v_{\text{rms}} = \sqrt{\frac{3RT}{M_0}}$$, where $$M_0$$ is the molar mass.

The average velocity of a gas molecule is $$v_{\text{avg}} = \sqrt{\frac{8RT}{\pi M_0}}$$.

The ratio of RMS velocity to average velocity is: $$\frac{v_{\text{rms}}}{v_{\text{avg}}} = \frac{\sqrt{\frac{3RT}{M_0}}}{\sqrt{\frac{8RT}{\pi M_0}}} = \sqrt{\frac{3RT}{M_0} \cdot \frac{\pi M_0}{8RT}} = \sqrt{\frac{3\pi}{8}}$$.

This ratio is independent of temperature, molar mass, and the type of gas.

When two ideal gases are mixed with no loss of energy, the total internal energy is conserved. The internal energy of an ideal gas with $$n$$ molecules, each having $$F$$ degrees of freedom, at temperature $$T$$ is $$U = n \cdot \frac{F}{2} k_B T$$, where $$k_B$$ is the Boltzmann constant.

Before mixing: $$U_{\text{total}} = n_1 \cdot \frac{F_1}{2} k_B T_1 + n_2 \cdot \frac{F_2}{2} k_B T_2$$. After mixing at temperature $$T$$: $$U_{\text{total}} = n_1 \cdot \frac{F_1}{2} k_B T + n_2 \cdot \frac{F_2}{2} k_B T$$.

Equating these and cancelling $$\frac{k_B}{2}$$ from both sides: $$n_1 F_1 T_1 + n_2 F_2 T_2 = (n_1 F_1 + n_2 F_2) T$$.

Solving for the mixture temperature gives $$T = \frac{n_1 F_1 T_1 + n_2 F_2 T_2}{n_1 F_1 + n_2 F_2}$$.

The correct answer is option 2.

For an ideal gas with $$f$$ degrees of freedom, the molar specific heat at constant volume is $$C_v = \frac{f}{2}R$$ and at constant pressure is $$C_p = C_v + R = \frac{f}{2}R + R = \frac{f+2}{2}R$$.

The ratio of specific heats is $$\gamma = \frac{C_p}{C_v} = \frac{(f+2)/2}{f/2} = \frac{f+2}{f} = 1 + \frac{2}{f}$$.

Solving for $$f$$: $$\gamma - 1 = \frac{2}{f}$$, so $$f = \frac{2}{\gamma - 1}$$.

For any ideal gas, the average translational kinetic energy per molecule is given first:

$$\text{Average K.E. per molecule} = \dfrac{3}{2}\,k_B\,T,$$

where $$k_B$$ is the Boltzmann constant and $$T$$ is the absolute temperature in kelvin.

An electron that starts from rest and is accelerated through a potential difference $$V$$ acquires kinetic energy according to the work-energy principle. The work done on a charge $$q$$ in moving through a potential difference $$V$$ is

$$\text{K.E. gained} = q\,V.$$

For an electron, the magnitude of the charge is $$e = 1.6 \times 10^{-19}\ \text{C}$$. Given that the accelerating potential difference is $$V = 0.1\ \text{V}$$, we have

$$\text{K.E. of the electron} = eV = (1.6 \times 10^{-19}\ \text{C})(0.1\ \text{V}) = 1.6 \times 10^{-20}\ \text{J}.$$

According to the statement of the question, this electron kinetic energy must equal the average translational kinetic energy of one $$\mathrm{N_2}$$ molecule. Hence we equate the two expressions:

$$\dfrac{3}{2}\,k_B\,T = 1.6 \times 10^{-20}\ \text{J}.$$

Now we substitute the given value $$k_B = 1.38 \times 10^{-23}\ \text{J K}^{-1}$$ and solve algebraically for $$T$$:

$$T = \dfrac{2}{3}\,\dfrac{1.6 \times 10^{-20}}{1.38 \times 10^{-23}}\ \text{K}.$$

Simplifying step by step, we first divide the numerical factors:

$$\dfrac{1.6}{1.38} \approx 1.1594,$$

and we handle the powers of ten separately:

$$10^{-20}\, /\,10^{-23} = 10^{3}.$$

So the fraction becomes

$$\dfrac{1.6 \times 10^{-20}}{1.38 \times 10^{-23}} \approx 1.1594 \times 10^{3} = 1.1594 \times 10^{3}.$$

Multiplying by the factor $$\dfrac{2}{3}$$ gives

$$T = \dfrac{2}{3} \times 1.1594 \times 10^{3}\ \text{K}.$$

Carrying out the multiplication,

$$\dfrac{2}{3} \times 1.1594 \approx 0.7730,$$

and finally,

$$T \approx 0.7730 \times 10^{3}\ \text{K} = 773\ \text{K}.$$

To express this temperature in degrees Celsius, we use the relation

$$T\;({}^{\circ}\text{C}) = T\;(\text{K}) - 273,$$

so

$$T\;({}^{\circ}\text{C}) = 773 - 273 = 500^{\circ}\text{C}.$$

The question asks for the nearest integer value, which is already an integer. Hence, the correct answer is Option 500.

We are asked to compare the average distance travelled between two successive collisions, that is, the mean free path of the two kinds of molecules present in the mixture.

For a dilute gas the mean free path is given by the hard-sphere formula

$$\lambda \;=\;\dfrac{1}{\sqrt{2}\,\pi d^{2} n}\,,$$

where $$d$$ is the molecular diameter and $$n$$ is the number density of that species. We are told that both gases have the same number density

$$n_A \;=\; n_B \;=\; 2\times10^{25}\ \text{m}^{-3}\,.$$

Hence any dependence on $$n$$ will cancel out when we form the ratio of the two mean free paths.

Let

$$d_A = 10\ \text{\AA}=10\times10^{-10}\ \text{m}, \qquad d_B = 5\ \text{\AA}=5\times10^{-10}\ \text{m}.$$

Using the formula separately for the two kinds of molecules we have

$$\lambda_A = \dfrac{1}{\sqrt{2}\,\pi d_A^{2} n_A}, \qquad \lambda_B = \dfrac{1}{\sqrt{2}\,\pi d_B^{2} n_B}.$$

Taking the ratio of $$\lambda_A$$ to $$\lambda_B$$ we get

$$ \dfrac{\lambda_A}{\lambda_B} \;=\; \dfrac{\dfrac{1}{\sqrt{2}\,\pi d_A^{2} n_A}} {\dfrac{1}{\sqrt{2}\,\pi d_B^{2} n_B}} \;=\; \dfrac{d_B^{2}}{d_A^{2}}, $$

because the common factors $$\sqrt{2}\,\pi$$ and the equal number densities cancel out.

Now substituting the diameters,

$$ \dfrac{\lambda_A}{\lambda_B} \;=\; \dfrac{(5\times10^{-10})^{2}}{(10\times10^{-10})^{2}} \;=\; \dfrac{25\times10^{-20}}{100\times10^{-20}} \;=\; \dfrac{25}{100} \;=\; 0.25. $$

We can rewrite $$0.25$$ as

$$0.25 \;=\; 25 \times 10^{-2}.$$

Thus the required ratio, expressed in the form “_____ × 10−2”, has the blank filled by 25.

So, the answer is $$25$$.

A monoatomic gas (molar mass $$M = 4.0\,\text{g/mol}$$, i.e., Helium) is kept in an insulated container moving with velocity $$v = 30\,\text{m/s}$$. When the container is suddenly stopped, the kinetic energy of the bulk motion is entirely converted into internal (thermal) energy of the gas.

For $$n$$ moles of gas, the kinetic energy of bulk motion is $$\frac{1}{2}(nM)v^2$$, where $$M$$ is the molar mass. For a monoatomic ideal gas, the change in internal energy is $$\Delta U = n \cdot \frac{3}{2}R\Delta T$$.

Setting these equal: $$\frac{1}{2}nMv^2 = \frac{3}{2}nR\Delta T$$.

Solving for $$\Delta T$$: $$\Delta T = \frac{Mv^2}{3R}$$.

Substituting $$M = 4$$ and $$v = 30\,\text{m/s}$$: $$\Delta T = \frac{4 \times (30)^2}{3R} = \frac{4 \times 900}{3R} = \frac{3600}{3R}$$.

Comparing with the given expression $$\Delta T = \frac{x}{3R}$$, we get $$x = 3600$$.

The root-mean-square (rms) speed of gas molecules is given by $$v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$$, where $$R$$ is the universal gas constant, $$T$$ is the absolute temperature, and $$M$$ is the molar mass of the gas. Notably, the rms speed depends only on the temperature and the molar mass of the gas, and is independent of pressure.

At $$T_1 = 27°\text{C} = 300$$ K, the rms speed is $$v_1 = 200$$ m s$$^{-1}$$. At $$T_2 = 127°\text{C} = 400$$ K, the rms speed is $$v_2$$. Since the gas is the same (same molar mass $$M$$), we have $$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$$.

Substituting the values: $$\frac{v_2}{200} = \sqrt{\frac{400}{300}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$$.

Therefore, $$v_2 = 200 \times \frac{2}{\sqrt{3}} = \frac{400}{\sqrt{3}}$$ m s$$^{-1}$$.

Comparing with the given expression $$v_2 = \frac{x}{\sqrt{3}}$$ m s$$^{-1}$$, we get $$x = 400$$.

Therefore, the value of $$x$$ is $$400$$.

We analyse each statement separately.

Statement I says that in a diatomic molecule, the rotational energy at a given temperature obeys Maxwell's distribution. This is true because, according to the equipartition theorem and statistical mechanics, each quadratic degree of freedom follows the Maxwell-Boltzmann energy distribution. The rotational kinetic energy of a diatomic molecule, being a sum of quadratic terms in angular velocities, obeys this distribution.

Statement II says that the rotational energy at a given temperature equals the translational kinetic energy for each molecule. A diatomic molecule at moderate temperatures has 2 rotational degrees of freedom (rotation about two axes perpendicular to the bond axis) and 3 translational degrees of freedom. By the equipartition theorem, the average energy per degree of freedom is $$\frac{1}{2}k_BT$$. Therefore, the average translational kinetic energy is $$\frac{3}{2}k_BT$$ while the average rotational energy is $$\frac{2}{2}k_BT = k_BT$$. Since $$\frac{3}{2}k_BT \neq k_BT$$, Statement II is false.

Hence, Statement I is true but Statement II is false.

According to the kinetic theory of gases, gas molecules are in constant random motion and frequently collide with the walls of the container.

When a gas molecule hits the wall, it bounces back. During this collision, the molecule's momentum changes direction. By Newton's second law, the rate of change of momentum equals the force exerted. Since millions of molecules strike the walls every second, the cumulative effect of these momentum changes produces a steady, measurable force on the walls.

The pressure exerted by the gas is this total force per unit area of the container wall. The molecules do not stick to the walls, nor are they attracted by the walls, nor do they lose energy before reaching the walls (collisions are assumed elastic in the kinetic theory model).

Therefore, the gas exerts pressure because its molecules suffer a change in momentum when they impinge on the walls of the container.

The correct answer is that the molecules suffer change in momentum when impinge on the walls of container.

We begin by recalling the ideal-gas equation

$$PV = nRT,$$

where $$P$$ is the pressure, $$V$$ is the volume, $$T$$ is the absolute temperature, $$n$$ is the number of moles and $$R$$ is the universal gas constant.

The problem further tells us that during the expansion the gas obeys the empirical relation

$$P\,T^{3} = \text{constant}.$$

For convenience, let us denote this constant by $$k$$, so that

$$P\,T^{3}=k \quad\Longrightarrow\quad P=\dfrac{k}{T^{3}}.$$

Now we insert this expression for $$P$$ into the ideal-gas equation. Substituting $$P=\dfrac{k}{T^{3}}$$ in $$PV=nRT$$, we have

$$\dfrac{k}{T^{3}}\;V = nRT.$$

Multiplying both sides by $$T^{3}$$ to clear the denominator,

$$kV = nR\,T^{4}.$$

Next we isolate $$V$$: dividing both sides by $$k$$,

$$V = \dfrac{nR}{k}\,T^{4}.$$

This shows explicitly that the volume varies as the fourth power of the temperature along the given path. To find the coefficient of volume expansion we recall its definition.

The (volumetric) coefficient of expansion $$\beta$$ is defined as

$$\beta = \dfrac{1}{V}\,\dfrac{dV}{dT}.$$

Using the expression $$V=\dfrac{nR}{k}\,T^{4}$$, we differentiate with respect to $$T$$:

$$\dfrac{dV}{dT} = \dfrac{nR}{k}\,\dfrac{d}{dT}\bigl(T^{4}\bigr) = \dfrac{nR}{k}\,(4T^{3}) = 4\,\dfrac{nR}{k}\,T^{3}.$$

But from the earlier result $$V=\dfrac{nR}{k}\,T^{4},$$ we observe that

$$\dfrac{nR}{k}\,T^{3}=\dfrac{V}{T}.$$

Substituting this in the derivative, we obtain

$$\dfrac{dV}{dT}=4\,\dfrac{V}{T}.$$

Now we plug this result into the definition of $$\beta$$:

$$\beta = \dfrac{1}{V}\,\dfrac{dV}{dT} = \dfrac{1}{V}\,\Bigl(4\,\dfrac{V}{T}\Bigr)=\dfrac{4}{T}.$$

Thus the coefficient of volume expansion for the gas undergoing the process $$PT^{3}=\text{constant}$$ is

$$\boxed{\dfrac{4}{T}}.$$

Hence, the correct answer is Option D.

The mean free path of a gas molecule is given by $$\lambda = \frac{kT}{\sqrt{2}\,\pi d^2 P}$$, where $$k$$ is Boltzmann's constant, $$T$$ is temperature, $$d$$ is the molecular diameter, and $$P$$ is pressure.

Substituting the given values: $$T = 27°\text{C} = 300$$ K, $$d = 0.3$$ nm $$= 0.3 \times 10^{-9}$$ m, $$P = 1.01 \times 10^5$$ Pa, and $$k = 1.38 \times 10^{-23}$$ J/K.

The numerator is $$kT = 1.38 \times 10^{-23} \times 300 = 4.14 \times 10^{-21}$$ J.

The denominator is $$\sqrt{2}\,\pi d^2 P = 1.414 \times 3.1416 \times (0.3 \times 10^{-9})^2 \times 1.01 \times 10^5$$. Computing $$d^2 = 9 \times 10^{-20}$$ m$$^2$$, so the denominator becomes $$1.414 \times 3.1416 \times 9 \times 10^{-20} \times 1.01 \times 10^5 = 1.414 \times 3.1416 \times 9.09 \times 10^{-15} = 4.037 \times 10^{-14}$$.

Therefore, $$\lambda = \frac{4.14 \times 10^{-21}}{4.037 \times 10^{-14}} = 1.026 \times 10^{-7}$$ m $$\approx 102$$ nm.

A rigid diatomic gas molecule has 5 degrees of freedom: 3 translational and 2 rotational. By the equipartition theorem, the molar heat capacity at constant volume is $$C_V = \frac{f}{2}R = \frac{5}{2}R$$, where $$R$$ is the universal gas constant.

Since no work is done by the gas (constant volume process), the first law of thermodynamics gives $$Q = \Delta U = nC_V\Delta T$$.

With $$n = 4$$ moles and $$\Delta T = 50 - 0 = 50$$ K:

$$Q = 4 \times \frac{5}{2}R \times 50 = 4 \times 2.5 \times 50 \times R = 500R$$

The heat required is $$500R$$.

We recall the expression for the root-mean-square (R.M.S.) speed of the molecules of an ideal gas. For one mole we write

$$v_{\text{rms}}=\sqrt{\frac{3RT}{M}}$$

where $$R$$ is the universal gas constant, $$T$$ is the absolute temperature, and $$M$$ is the molar mass of the gas.

From this formula we clearly see that at a fixed temperature $$T$$ the only variable factor is the molar mass $$M$$ that appears in the denominator under the square root. Hence,

$$v_{\text{rms}}\propto\frac{1}{\sqrt{M}}.$$

In simple words, the lighter the gas (smaller $$M$$), the larger its R.M.S. speed, and vice versa.

Now we list the molar masses of the three given gases:

$$M_{\text{H}_2}=2\;\text{g mol}^{-1},\qquad M_{\text{O}_2}=32\;\text{g mol}^{-1},\qquad M_{\text{CO}_2}=44\;\text{g mol}^{-1}.$$

The order of their molar masses is therefore

$$2 \lt 32 \lt 44.$$

Because the R.M.S. speed is inversely proportional to the square root of the molar mass, the order of the speeds will be exactly the reverse:

$$v_H \gt v_O \gt v_C.$$

This matches Option C.

Hence, the correct answer is Option C.

According to the equipartition theorem, each degree of freedom of a gas molecule contributes $$\frac{1}{2}k_BT$$ to the average energy.

A monoatomic gas (like He, Ne, Ar) has only translational degrees of freedom. Since there are 3 translational degrees of freedom (motion along $$x$$, $$y$$, and $$z$$ axes), the average kinetic energy per molecule is:

$$\langle E \rangle = 3 \times \frac{1}{2}k_BT = \frac{3}{2}k_BT$$

This is the well-known result for the average thermal energy of a monoatomic ideal gas molecule at temperature $$T$$.

We begin with the principle of buoyancy for a gas balloon. The upward (lifting) force equals the weight of the air displaced, while the downward forces are the weight of the gas inside the balloon plus the weight of the load that the balloon can carry. If the material of the balloon itself is light enough to be neglected, the net load that can be supported is proportional to the difference between the density of the outside air and the density of the gas inside the balloon, multiplied by the volume of the balloon and the acceleration due to gravity.

Mathematically, for any set of atmospheric conditions, we may write

$$\text{Load}=Vg(\rho_{\text{air}}-\rho_{\text{gas}}).$$

Because the problem states “assuming the volume constant,” the volume $$V$$ (and of course $$g$$) do not change when the balloon rises. Hence the load varies only with the factor $$(\rho_{\text{air}}-\rho_{\text{gas}})$$.

Both the outside air and the gas inside the balloon obey the ideal-gas equation $$\rho=\frac{PM}{RT},$$ where $$P$$ is the pressure, $$T$$ is the absolute temperature, $$M$$ is the molar mass of the gas concerned and $$R$$ is the universal gas constant.

Inside the balloon the gas (say hydrogen or helium) is at the same pressure and temperature as the surrounding air, so the factor $$\dfrac{P}{T}$$ is common to both $$\rho_{\text{air}}$$ and $$\rho_{\text{gas}}$$. Therefore the difference of the two densities can be expressed as

$$\rho_{\text{air}}-\rho_{\text{gas}}=\frac{P}{RT}\bigl(M_{\text{air}}-M_{\text{gas}}\bigr).$$

Notice that the bracketed term $$(M_{\text{air}}-M_{\text{gas}})$$ is a constant, because the molar masses of air ($$\approx 29\ \text{g mol}^{-1}$$) and of the balloon gas ($$\approx 2\ \text{g mol}^{-1}$$ for hydrogen or $$4\ \text{g mol}^{-1}$$ for helium) do not change with height. Consequently, the entire density difference, and hence the load, varies directly as the ratio $$\dfrac{P}{T}$$.

So, for two different levels (ground and high-altitude) we have the proportionality

$$\frac{\text{Load at height}}{\text{Load at ground}}=\frac{\dfrac{P_2}{T_2}}{\dfrac{P_1}{T_1}}.$$

All the data are now substituted step by step. The ground (initial) conditions are stated to be “normal pressure” and $$27^{\circ}\text{C}$$. Normal pressure is $$P_1=76\ \text{cm of Hg},$$ and the absolute temperature is $$T_1=27^{\circ}\text{C}=27+273=300\ \text{K}.$$

At the new height the given pressure and temperature are

$$P_2=45\ \text{cm of Hg},\qquad T_2=-7^{\circ}\text{C}=-7+273=266\ \text{K}.$$

The balloon supports a load of $$185\ \text{kg}$$ under the ground conditions, so $$\text{Load}_1=185\ \text{kg}$$. Now we compute the new load:

$$\text{Load}_2=\text{Load}_1\;\frac{P_2/T_2}{P_1/T_1}.$$

First evaluate each ratio separately:

$$\frac{P_2}{T_2}=\frac{45}{266}=0.169172,$$

$$\frac{P_1}{T_1}=\frac{76}{300}=0.253333.$$

Now form the quotient of these two ratios:

$$\frac{P_2/T_2}{P_1/T_1}=\frac{0.169172}{0.253333}=0.667751.$$

Multiplying this factor by the initial load gives

$$\text{Load}_2=185\times0.667751=123.543\ \text{kg}.$$

Rounding to two decimal places, the new load is $$123.54\ \text{kg}$$.

Hence, the correct answer is Option B.

For a polyatomic gas, the degrees of freedom include 3 translational, 3 rotational, and the vibrational modes. Each vibrational mode contributes 2 degrees of freedom (one kinetic and one potential), so 2 vibrational modes contribute 4 degrees of freedom. The total degrees of freedom are $$f = 3 + 3 + 4 = 10$$.

The molar specific heat at constant volume is $$C_v = \frac{f}{2}R = \frac{10}{2}R = 5R$$, and at constant pressure $$C_p = C_v + R = 6R$$.

Therefore $$\beta = \frac{C_p}{C_v} = \frac{6R}{5R} = \frac{6}{5} = 1.2$$.

We need to find the total pressure of a mixture of gases in a volume $$V$$ at temperature $$T$$. First, we calculate the number of moles of each gas.

For oxygen ($$O_2$$, molar mass 32 g/mol): $$n_{O_2} = \frac{16}{32} = 0.5$$ mol.

For nitrogen ($$N_2$$, molar mass 28 g/mol): $$n_{N_2} = \frac{28}{28} = 1$$ mol.

For carbon dioxide ($$CO_2$$, molar mass 44 g/mol): $$n_{CO_2} = \frac{44}{44} = 1$$ mol.

The total number of moles is $$n = 0.5 + 1 + 1 = 2.5 = \frac{5}{2}$$ mol.

By Dalton's law of partial pressures (or equivalently, the ideal gas law for the mixture), the total pressure is $$P = \frac{nRT}{V} = \frac{5}{2} \cdot \frac{RT}{V}$$.

We are dealing with a gaseous mixture, so first of all we recall the ideal-gas equation

$$PV = nRT$$

where $$P$$ is pressure, $$V$$ is volume, $$n$$ is the total number of moles present, $$R$$ is the universal gas constant and $$T$$ is absolute temperature.

All the data must be in SI units. We therefore convert each quantity one by one.

Pressure:
$$P = 400\ \text{kPa} = 400 \times 10^{3}\ \text{Pa} = 4 \times 10^{5}\ \text{Pa}$$

Volume:
$$V = 500\ \text{cm}^3 = 500 \times 10^{-6}\ \text{m}^3 = 5 \times 10^{-4}\ \text{m}^3$$

Temperature is already in kelvin, so
$$T = 300\ \text{K}$$

Universal gas constant:
$$R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$$

Substituting these values into $$PV = nRT$$ we get

$$n = \dfrac{PV}{RT} = \dfrac{\left(4 \times 10^{5}\ \text{Pa}\right)\left(5 \times 10^{-4}\ \text{m}^3\right)}{8.314\ \text{J mol}^{-1}\text{K}^{-1}\; \times\; 300\ \text{K}}$$

First multiply the numerator:

$$4 \times 10^{5}\; \text{Pa} \times 5 \times 10^{-4}\; \text{m}^3 = 200\ \text{J}$$

Next multiply the denominator:

$$8.314 \times 300 = 2494.2$$

Hence

$$n = \dfrac{200}{2494.2} \approx 0.08016\ \text{mol}$$

So the mixture contains in total

$$n_{\text{total}} = 0.08016\ \text{mol}$$

Let us denote the individual mole numbers as

$$n_{\text{O}_2} = x,\qquad n_{\text{H}_2} = y$$

Then we have the relation

$$x + y = 0.08016 \qquad\qquad (1)$$

The given total mass of the mixture is $$0.76\ \text{g}$$. Writing the mass in terms of molar masses, we obtain

$$32x + 2y = 0.76 \qquad\qquad (2)$$

To simplify equation (2) we divide every term by 2:

$$16x + y = 0.38 \qquad\qquad (3)$$

Now we solve the simultaneous equations (1) and (3).

From (1) we express $$y$$:

$$y = 0.08016 - x$$

Substituting this value of $$y$$ into (3):

$$16x + \left(0.08016 - x\right) = 0.38$$

Simplifying:

$$16x - x + 0.08016 = 0.38$$

$$15x + 0.08016 = 0.38$$

$$15x = 0.38 - 0.08016 = 0.29984$$

$$x = \dfrac{0.29984}{15} = 0.019989\ \text{mol} \approx 0.020\ \text{mol}$$

Using $$y = 0.08016 - x$$ gives

$$y = 0.08016 - 0.019989 = 0.060171\ \text{mol} \approx 0.060\ \text{mol}$$

Now we determine the individual masses.

Mass of oxygen:

$$m_{\text{O}_2} = 32x = 32 \times 0.019989 \approx 0.63965\ \text{g} \approx 0.64\ \text{g}$$

Mass of hydrogen:

$$m_{\text{H}_2} = 2y = 2 \times 0.060171 \approx 0.12034\ \text{g} \approx 0.12\ \text{g}$$

Hence the required ratio of the masses is

$$m_{\text{O}_2} : m_{\text{H}_2} = 0.64 : 0.12$$

Dividing both numbers by 0.04 gives

$$16 : 3$$

Hence, the correct answer is Option A.

A polyatomic ideal gas has degrees of freedom from three types of motion: translational, rotational, and vibrational. Every gas molecule in three-dimensional space has 3 translational degrees of freedom (motion along $$x$$, $$y$$, and $$z$$ axes). A nonlinear polyatomic molecule also has 3 rotational degrees of freedom (rotation about each principal axis).

Each vibrational mode contributes 2 degrees of freedom: one for kinetic energy and one for potential energy. Since the gas has 24 vibrational modes, the vibrational contribution is $$2 \times 24 = 48$$ degrees of freedom.

The total number of degrees of freedom is therefore $$f = 3 \text{ (translational)} + 3 \text{ (rotational)} + 48 \text{ (vibrational)} = 54$$.

For an ideal gas, the molar specific heat at constant volume is $$C_V = \frac{f}{2}R$$ and at constant pressure is $$C_P = C_V + R = \frac{f}{2}R + R = \frac{f+2}{2}R$$. The ratio of specific heats is $$\gamma = \frac{C_P}{C_V} = \frac{f+2}{f} = 1 + \frac{2}{f}$$.

Substituting $$f = 54$$: $$\gamma = 1 + \frac{2}{54} = 1 + \frac{1}{27} = 1.037$$, which is closest to 1.03.

The correct answer is option 1: 1.03.

The root mean square speed of gas molecules is given by $$v_{rms} = \sqrt{\frac{3RT}{M}}$$, where $$M$$ is the molar mass of the gas and $$T$$ is the absolute temperature.

Since all gases A, B, and C are at the same temperature (normal temperature and pressure), the rms speed is inversely proportional to the square root of the molar mass: $$v_{rms} \propto \frac{1}{\sqrt{M}}$$.

Given that $$m_A < m_B < m_C$$, we have $$v_A > v_B > v_C$$, which means $$\frac{1}{v_A} < \frac{1}{v_B} < \frac{1}{v_C}$$.

The difference between the molar heat capacities of any fluid is given by the exact thermodynamic identity

$$C_P-C_V \;=\; -\,T\; \frac{\left(\dfrac{\partial P}{\partial T}\right)_{V}^{2}} {\left(\dfrac{\partial P}{\partial V}\right)_{T}}.$$

For an ideal gas the equation of state is $$P\,V = R\,T$$ and after inserting this equation in the above identity one gets the familiar Mayer relation $$C_P-C_V=R,$$ which is independent of temperature. A departure of the value of $$C_P-C_V$$ from $$R$$ therefore signals non-ideal behaviour, so we must work with a more realistic equation of state. The simplest and most frequently used one is the van der Waals equation

$$P=\frac{R\,T}{V-b}-\frac{a}{V^{2}},$$

in which the parameters $$a$$ and $$b$$ account, respectively, for intermolecular attraction and the finite size of the molecules.

We now calculate the two derivatives that enter the heat-capacity identity.

First, keeping volume constant,

$$\left(\frac{\partial P}{\partial T}\right)_{V} =\frac{R}{V-b}.$$

Secondly, keeping temperature constant,

$$\left(\frac{\partial P}{\partial V}\right)_{T} = -\frac{R\,T}{(V-b)^{2}}+\frac{2a}{V^{3}}.$$

Substituting these derivatives into the general expression, we obtain

$$C_P-C_V = -\,T\; \frac{\displaystyle\left(\dfrac{R}{V-b}\right)^{2}} {\displaystyle -\frac{R\,T}{(V-b)^{2}}+\frac{2a}{V^{3}}}.$$

The two negative signs in the numerator and denominator cancel, and a factor $$\dfrac{R\,T}{(V-b)^{2}}$$ can be taken out of the denominator to give

$$C_P-C_V = \frac{T\,R^{2}/(V-b)^{2}} {\,R\,T/(V-b)^{2}\, \Bigl[\,1-\dfrac{2a(V-b)^{2}}{R\,T\,V^{3}}\Bigr]} = \frac{R}{1-\dfrac{2a(V-b)^{2}}{R\,T\,V^{3}}}.$$

Write this result in the more compact form

$$C_P-C_V =\;R\; \Biggl[\, 1-\frac{2a\,(V-b)^{2}}{R\,T\,V^{3}} \Biggr]^{-1}. \quad -(1)$$

Observe the structure of the denominator in (1). The fraction

$$\frac{2a\,(V-b)^{2}}{R\,T\,V^{3}}$$

is strictly positive (all the symbols denote positive quantities) and it is divided by $$T$$. Hence

• When the temperature $$T$$ is very high, the term $$\dfrac{2a\,(V-b)^{2}}{R\,T\,V^{3}}$$ becomes small, the bracket approaches $$1$$ and (1) reduces to $$C_P-C_V\approx R$$ - the ideal-gas limit.
• When the temperature $$T$$ is lowered, the same term becomes larger, the entire bracket becomes smaller than 1, and therefore its reciprocal, and with it $$C_P-C_V,$$ becomes larger than $$R$$.

In other words, the difference $$C_P-C_V$$ is inversely related to the temperature for a real gas described by the van der Waals equation: a greater value of $$C_P-C_V$$ corresponds to a lower temperature.

Let us now compare the two given states.

State $$P$$: $$C_P-C_V = R$$ (the ideal value)
State $$Q$$: $$C_P-C_V = 1.10\,R$$ (10 % larger than the ideal value)

Because $$1.10\,R \;>\; R,$$ the above reasoning tells us that state $$Q$$ must be at a lower temperature than state $$P$$. Symbolically,

$$T_Q \;< T_P.$$

So, the correct inequality is

$$T_P \;>\; T_Q.$$

Hence, the correct answer is Option D.

For one mole of an ideal gas the equation of state is stated first: $$PV=\mathcal{R}T$$.

We are also given the differential relation between pressure and volume: $$\frac{dP}{dV}=-aP$$, where $$a$$ is a positive constant.

Now we solve this first-order differential equation. We rewrite it as $$\frac{1}{P}\,dP=-a\,dV$$. Integrating both sides, we obtain $$\int\frac{1}{P}\,dP=-a\int dV$$ which gives $$\ln P=-aV+C$$, where $$C$$ is the constant of integration.

To determine $$C$$ we use the boundary condition $$P=P_0$$ when $$V=0$$. Substituting these values, $$\ln P_0=C$$. Hence $$C=\ln P_0$$, and the pressure as a function of volume is $$P(V)=P_0e^{-aV}$$.

Substituting this expression for pressure in the ideal-gas equation, we have $$P(V)\,V=\mathcal{R}T$$. Therefore $$\mathcal{R}T = P_0e^{-aV}\,V,$$ so the temperature is $$T(V)=\frac{P_0}{\mathcal{R}}\,V\,e^{-aV}.$$

To find the maximum attainable temperature we must maximize the function $$f(V)=V\,e^{-aV}$$ for $$V \ge 0$$.

We differentiate: $$\frac{df}{dV} = e^{-aV} - aV e^{-aV} = e^{-aV}\bigl(1-aV\bigr).$$ Setting $$\frac{df}{dV}=0$$ gives $$1-aV=0 \;\;\Longrightarrow\;\; V=\frac{1}{a}.$$ Because the exponential factor is always positive, this critical point corresponds to a maximum.

Substituting $$V=\frac{1}{a}$$ back into $$T(V)$$, we get $$T_{\max} = \frac{P_0}{\mathcal{R}}\left(\frac{1}{a}\right) e^{-a\left(\frac{1}{a}\right)} = \frac{P_0}{\mathcal{R}}\left(\frac{1}{a}\right) e^{-1} = \frac{P_0}{a e \mathcal{R}}.$$

Hence, the correct answer is Option D.

We recall the ideal-gas expression for root-mean-square (R.M.S.) speed: $$v_{\text{rms}}=\sqrt{\dfrac{3RT}{M}},$$ where $$R$$ is the universal gas constant, $$T$$ is the absolute temperature and $$M$$ is the molar mass of the gas in kilogram per mole.

Because both gases are at the same temperature $$T=0^{\circ}\text{C}=273\ \text{K},$$ the factor $$3RT$$ will be common. Hence, for two gases 1 and 2 we may write the ratio

$$\dfrac{v_{\text{rms,1}}}{v_{\text{rms,2}}} =\sqrt{\dfrac{M_2}{M_1}}.$$

Here, gas 1 is hydrogen $$(\mathrm{H_2})$$ and gas 2 is oxygen $$(\mathrm{O_2}).$$ Therefore we have

$$v_{\text{rms}(\text{H}_2)}=v_{\text{rms}(\text{O}_2)}\sqrt{\dfrac{M_{\mathrm{O_2}}}{M_{\mathrm{H_2}}}}.$$

Now we substitute the given value $$v_{\text{rms}(\text{O}_2)}=160\ \text{m s}^{-1}$$ and the molar masses $$M_{\mathrm{O_2}}=32\ \text{g mol}^{-1}$$, $$M_{\mathrm{H_2}}=2\ \text{g mol}^{-1}$$:

$$v_{\text{rms}(\text{H}_2)} = 160 \times \sqrt{\dfrac{32}{2}}.$$

Inside the square root we simplify first: $$\dfrac{32}{2}=16.$$ Taking the square root gives $$\sqrt{16}=4.$$

So we obtain

$$v_{\text{rms}(\text{H}_2)} = 160 \times 4 = 640\ \text{m s}^{-1}.$$

Hence, the correct answer is Option C.

For an ideal gas, the internal energy is related to pressure and volume by $$U = \frac{f}{2}nRT = \frac{f}{2}PV$$, where $$f$$ is the number of degrees of freedom.

We are given $$U = 3PV + 4$$. The constant 4 does not affect the relationship between changes in $$U$$ and $$PV$$, so the effective relationship is $$U = 3PV + \text{constant}$$, meaning $$\frac{f}{2} = 3$$, giving $$f = 6$$.

For a monoatomic gas, $$f = 3$$. For a diatomic gas (at moderate temperatures), $$f = 5$$. For a polyatomic gas, $$f = 6$$ (3 translational + 3 rotational degrees of freedom).

Since $$f = 6$$, the gas is polyatomic only.

We are given a sample of ideal gas whose volume is one litre. Remember that one litre must first be turned into SI units:

$$1 \text{ litre}=1 \times 10^{-3}\ \text{m}^3$$

The pressure is quoted as $$2$$ atmospheres. To write this in pascals, we recall that

$$1\ \text{atm}=1.013\times 10^{5}\ \text{Pa}.$$

So,

$$P=2\ \text{atm}=2 \times 1.013\times 10^{5}\ \text{Pa}=2.026\times 10^{5}\ \text{Pa}.$$

For an ideal gas, the kinetic theory relation connecting pressure, volume and the average translational kinetic energy per molecule is first stated:

$$PV=\frac{2}{3} N \, \langle E_{\text{kin}}\rangle,$$

where

$$N$$ is the required number of molecules and

$$\langle E_{\text{kin}}\rangle$$ is the mean kinetic energy per molecule.

The problem itself tells us that each molecule has an average kinetic energy

$$\langle E_{\text{kin}}\rangle = 2 \times 10^{-9}\ \text{J}.$$

Now we substitute the known values of $$P$$ and $$V$$ into $$PV$$:

$$PV = \left(2.026\times 10^{5}\ \text{Pa}\right)\left(1\times 10^{-3}\ \text{m}^3\right).$$

Multiplying we obtain

$$PV = 2.026\times 10^{2}\ \text{J}=202.6\ \text{J}.$$

Next we insert this result and the given kinetic energy into the kinetic‐theory equation:

$$202.6\ \text{J} = \frac{2}{3} N \left(2\times 10^{-9}\ \text{J}\right).$$

To isolate $$N$$, we first multiply both sides by $$3$$:

$$3 \times 202.6\ \text{J} = 2 N \left(2\times 10^{-9}\ \text{J}\right).$$

This simplifies to

$$607.8\ \text{J} = 4 \times 10^{-9}\ \text{J}\times N.$$

Finally, we divide both sides by $$4 \times 10^{-9}\ \text{J}$$ to solve for $$N$$:

$$N=\frac{607.8}{4 \times 10^{-9}}.$$

Carrying out the division,

$$N = 1.5195 \times 10^{11}.$$

Rounding appropriately,

$$N \approx 1.5 \times 10^{11}.$$

Hence, the correct answer is Option C.

For an ideal gas, we first recall the general formula for the molar internal energy:

$$U = n \left(\frac{f}{2}\right)RT,$$

where $$n$$ is the number of moles, $$f$$ is the number of degrees of freedom of one molecule, $$R$$ is the universal gas constant, and $$T$$ is the absolute temperature. We shall apply this separately to oxygen and argon, then add the two contributions.

Now, consider oxygen. The question explicitly tells us to treat the oxygen bond as rigid, so vibrational motion is not excited. A rigid diatomic molecule possesses:

3 translational degrees of freedom + 2 rotational degrees of freedom $$\;=\;5$$ total degrees of freedom.

Hence, for oxygen $$f_{\text{O}_2} = 5.$$ Substituting this into the energy expression gives

$$U_{\text{O}_2} = n_{\text{O}_2}\left(\frac{f_{\text{O}_2}}{2}\right)RT = 3\left(\frac{5}{2}\right)RT = \frac{15}{2}RT.$$

Next, consider argon. Argon is a mono-atomic gas, and such a particle can move only translationally. Therefore,

$$f_{\text{Ar}} = 3.$$

Using the same formula, we get

$$U_{\text{Ar}} = n_{\text{Ar}}\left(\frac{f_{\text{Ar}}}{2}\right)RT = 5\left(\frac{3}{2}\right)RT = \frac{15}{2}RT.$$

We now add the internal energies of the two components to obtain the total internal energy of the mixture:

$$U_{\text{total}} = U_{\text{O}_2} + U_{\text{Ar}} = \frac{15}{2}RT + \frac{15}{2}RT = \left(\frac{15}{2} + \frac{15}{2}\right)RT = 15\,RT.$$

The question asks for the numerical value in units of $$RT$$, so we divide by $$RT$$:

$$\frac{U_{\text{total}}}{RT} = 15.$$

Hence, the correct answer is Option A.

For an ideal gas we always start from the ideal‐gas equation

$$PV=nRT.$$

Because the amount of gas does not change, the quantity $$nR$$ is a constant. To find how the volume changes when pressure or temperature change only slightly, we differentiate this equation. The total differential of $$PV=nRT$$ is

$$P\,dV+V\,dP=nR\,dT.$$

Now we divide every term by $$PV$$ so that each differential appears as a fractional (relative) change:

$$\frac{P\,dV}{PV}+\frac{V\,dP}{PV}=\frac{nR\,dT}{PV}.$$

Simplifying each fraction we obtain the useful relation

$$\frac{dV}{V}+\frac{dP}{P}=\frac{dT}{T}.$$

Rearranging this gives

$$\frac{dV}{V}=\frac{dT}{T}-\frac{dP}{P}.$$

This formula tells us how a small change in temperature or pressure affects the volume.

We are told that two different experiments produce equal magnitudes of volume change:

1. A small additional pressure $$\Delta P$$ is applied while the temperature is kept constant (isothermal process).
2. A small decrease in temperature $$\Delta T$$ is made while the pressure is kept constant (isobaric process).

We now evaluate the fractional change in volume for each case.

Isothermal case (constant temperature)

Because the temperature does not change, $$dT=0$$. Substituting $$dT=0$$ into the general differential formula, we have

$$\frac{dV}{V}=0-\frac{dP}{P}=-\frac{dP}{P}.$$

Taking magnitudes (because the statement uses the words “change in magnitude”) gives

$$\left|\frac{\Delta V}{V}\right|=\frac{|\Delta P|}{P}.$$

Isobaric case (constant pressure)

Now the pressure is fixed, so $$dP=0$$. Putting $$dP=0$$ into the same differential relation gives

$$\frac{dV}{V}=\frac{dT}{T}-0=\frac{dT}{T}.$$

Again, taking magnitudes,

$$\left|\frac{\Delta V}{V}\right|=\frac{|\Delta T|}{T}.$$

According to the question these two magnitudes of volume change are equal, so we set them equal:

$$\frac{|\Delta P|}{P}=\frac{|\Delta T|}{T}.$$

The statement $$|\Delta T|=C|\Delta P|$$ is given, so we substitute $$|\Delta T|$$ from the last equation:

$$C|\Delta P| = |\Delta T| = \frac{T}{P}\,|\Delta P|.$$

Because $$|\Delta P|$$ is common on both sides, it cancels, leaving

$$C=\frac{T}{P}.$$

The initial temperature and pressure are provided as $$T=300\ \text{K}$$ and $$P=2\ \text{atm}$$. Substituting these values we get

$$C=\frac{300\ \text{K}}{2\ \text{atm}}=150\ \text{K/atm}.$$

So, the answer is $$150$$.

We have a rigid, perfectly insulated (adiabatic) vessel, so there is no work done (because the volume is constant) and no heat exchanged with the surroundings. Under these conditions the total internal energy of the gas in the vessel remains conserved during the mixing process.

For an ideal gas the molar internal energy depends only on temperature. Specifically, for a mono-atomic ideal gas the internal energy per mole is given by the well-known formula

$$U_{\text{molar}}=\frac{3}{2}RT,$$

where $$R$$ is the universal gas constant and $$T$$ is the absolute temperature.

If a sample contains $$n$$ moles, its total internal energy is therefore

$$U=\frac{3}{2}nRT.$$

Initially the vessel holds

$$n_1 = 0.1\ \text{mol}, \qquad T_1 = 200\ \text{K}.$$

So its internal energy is

$$U_1=\frac{3}{2} n_1 R T_1 = \frac{3}{2} R (0.1)(200).$$

An additional sample is injected containing

$$n_2 = 0.05\ \text{mol}, \qquad T_2 = 400\ \text{K},$$

whose internal energy before mixing is

$$U_2=\frac{3}{2} n_2 R T_2 = \frac{3}{2} R (0.05)(400).$$

After the two portions mix and thermal equilibrium is reached, the total number of moles inside the vessel becomes

$$n_f = n_1 + n_2 = 0.1 + 0.05 = 0.15\ \text{mol},$$

and let the common final temperature be $$T_f$$. Because energy is conserved, we set the sum of the initial internal energies equal to the final internal energy:

$$U_1 + U_2 = U_f.$$

Substituting the expressions for each internal energy, we have

$$\frac{3}{2} R n_1 T_1 + \frac{3}{2} R n_2 T_2 = \frac{3}{2} R n_f T_f.$$

Every term contains the common factor $$\dfrac{3}{2}R$$, so it cancels out, yielding the simple relation

$$n_1 T_1 + n_2 T_2 = n_f T_f.$$

Now we substitute the numerical values step by step:

$$\bigl(0.1\bigr)(200) + \bigl(0.05\bigr)(400) = \bigl(0.15\bigr) T_f.$$

First, calculate each product:

$$0.1 \times 200 = 20,$$

$$0.05 \times 400 = 20.$$

Adding them gives

$$20 + 20 = 40.$$

Hence,

$$40 = 0.15\, T_f.$$

Solving for $$T_f$$, we divide both sides by $$0.15$$:

$$T_f = \frac{40}{0.15}.$$

Carrying out the division,

$$\frac{40}{0.15} = 266.\overline{6}\ \text{K},$$

which is very close to $$267\ \text{K}$$ when rounded to the nearest whole number.

So, the answer is $$267\ \text{K}$$.

We begin with the ideal-gas equation, which is written as $$PV = nRT,$$ where $$P$$ is the pressure, $$V$$ is the volume, $$n$$ is the number of moles, $$R$$ is the universal gas constant and $$T$$ is the absolute temperature.

Initially the gas is diatomic, so let the initial number of moles be $$n$$. The initial state is described by

$$P_1V_1 = nR(250\,\text{K}).$$

Now 25 % of the diatomic molecules dissociate according to the reaction

$$\text{A}_2 \;\longrightarrow\; 2\text{A}.$$

Out of the original $$n$$ moles, the fraction that dissociates is $$0.25n$$. Each mole of dissociating diatomic molecules produces two moles of monatomic gas, so the number of moles of atoms produced is

$$2 \times 0.25n \;=\; 0.50n.$$

The number of moles of diatomic molecules that remain undissociated is

$$n - 0.25n \;=\; 0.75n.$$

Thus the total number of moles after dissociation becomes

$$n_2 = 0.75n + 0.50n = 1.25n.$$

This dissociated mixture is now kept in a volume that is twice the original, namely $$2V_1,$$ and its temperature is raised to $$2000\,\text{K}$$. Applying the same ideal-gas equation to the final state gives

$$P_2(2V_1) = n_2R(2000\,\text{K}).$$

Substituting $$n_2 = 1.25n$$ we write

$$P_2(2V_1) = 1.25nR \times 2000.$$

Now we isolate $$P_2$$:

$$P_2 = \frac{1.25nR \times 2000}{2V_1}.$$

For the ratio $$\dfrac{P_2}{P_1}$$ we divide this expression by the earlier expression for $$P_1$$. From the initial state we have

$$P_1 = \frac{nR \times 250}{V_1}.$$

Taking the ratio,

$$\frac{P_2}{P_1} = \frac{\dfrac{1.25nR \times 2000}{2V_1}}{\dfrac{nR \times 250}{V_1}}.$$

The factors $$n$$, $$R$$ and $$V_1$$ cancel out completely, leaving

$$\frac{P_2}{P_1} = \frac{1.25 \times 2000}{2 \times 250}.$$

Calculating the numerator first, $$1.25 \times 2000 = 2500$$. Dividing by 2 gives $$1250$$. Finally,

$$\frac{P_2}{P_1} = \frac{1250}{250} = 5.$$

Hence, the correct answer is Option 5.

First, we convert the given Celsius temperature of nitrogen gas into the Kelvin scale, because all gas-kinetic formulae are written for absolute temperature. By definition,

$$T\;(\text{in K}) \;=\; T\;({}^\circ\text{C}) + 273.$$

Here the nitrogen temperature is $$300\,{}^\circ\text{C}$$, so

$$T_{\text{N}_2}=300+273=573\;\text{K}.$$

For any ideal gas the root-mean-square (rms) speed is given by the well-known kinetic-theory relation

$$v_{\text{rms}}=\sqrt{\dfrac{3RT}{M}},$$

where $$R$$ is the universal gas constant, $$T$$ is the absolute temperature, and $$M$$ is the molar mass (expressed in the same mass unit for every gas in the comparison).

We are told that the rms speed of a hydrogen molecule $$(\text{H}_2)$$ must equal the rms speed of a nitrogen molecule $$(\text{N}_2)$$. Therefore we set the two expressions equal:

$$\sqrt{\dfrac{3R\,T_{\text{N}_2}}{M_{\text{N}_2}}}\;=\;\sqrt{\dfrac{3R\,T_{\text{H}_2}}{M_{\text{H}_2}}}.$$

First we square both sides to remove the square roots, obtaining

$$\dfrac{3R\,T_{\text{N}_2}}{M_{\text{N}_2}}\;=\;\dfrac{3R\,T_{\text{H}_2}}{M_{\text{H}_2}}.$$

The common factor $$3R$$ cancels out, leaving the simple proportionality

$$\dfrac{T_{\text{N}_2}}{M_{\text{N}_2}}\;=\;\dfrac{T_{\text{H}_2}}{M_{\text{H}_2}}.$$

We now solve for the unknown temperature $$T_{\text{H}_2}$$ of hydrogen:

$$T_{\text{H}_2} = T_{\text{N}_2}\,\dfrac{M_{\text{H}_2}}{M_{\text{N}_2}}.$$

Substituting the numeric values, we use the molar mass of nitrogen as $$M_{\text{N}_2}=28\;\text{g mol}^{-1}$$ (given) and the customary molar mass of hydrogen as $$M_{\text{H}_2}=2\;\text{g mol}^{-1}$$. Hence,

$$T_{\text{H}_2} = 573\;\text{K}\,\times\,\dfrac{2}{28}.$$

Now we simplify the fraction:

$$\dfrac{2}{28} = \dfrac{1}{14}.$$

So,

$$T_{\text{H}_2} = 573\;\text{K}\,\times\,\dfrac{1}{14}.$$

Carrying out the division,

$$T_{\text{H}_2} = \dfrac{573}{14}\;\text{K} = 40.93\;\text{K}.$$

Rounding to the nearest whole number in Kelvin, we have $$T_{\text{H}_2}\approx 41\;\text{K}.$$

Hence, the correct answer is Option 41.

For an ideal gas, the number of molecules is $$N$$ and the container is closed, so its volume $$V$$ is fixed. Therefore the number density is

$$n \;=\;\dfrac{N}{V},$$

and this value $$n$$ does not change when we raise the temperature, because neither $$N$$ nor $$V$$ changes.

Now recall the kinetic-theory formula for the mean free path. We state it first:

$$\lambda \;=\;\dfrac{1}{\sqrt{2}\,\pi d^{2}\,n},$$

where $$d$$ is the effective molecular diameter and $$n$$ is the number density. Since $$d$$ is a property of the molecule and $$n$$ has just been shown to stay constant, we see immediately

$$\lambda \;$$ remains constant when the temperature rises.

So, statement (A) “the mean free path decreases” is false, while statement (C) “the mean free path remains unchanged” is true.

Next we look at the mean collision time, often denoted $$\tau$$. The definition is

$$\tau \;=\;\dfrac{\text{mean free path}}{\text{average speed}},$$ $$\text{so}$$ $$\tau \;=\;\dfrac{\lambda}{v_{\text{avg}}}.$$

For an ideal gas, the average (more precisely the root-mean-square) speed is given by the kinetic-theory result

$$v_{\text{rms}} \;=\;\sqrt{\dfrac{3kT}{m}},$$

where $$k$$ is Boltzmann’s constant, $$T$$ is the absolute temperature, and $$m$$ is the molecular mass. Thus as the temperature increases,

$$v_{\text{avg}} \propto \sqrt{T}$$ $$\Longrightarrow$$ $$v_{\text{avg}}\ \text{increases}.$$

Because $$\lambda$$ is constant while $$v_{\text{avg}}$$ increases, the quotient $$\tau=\lambda/v_{\text{avg}}$$ decreases:

$$\tau \;=\;\dfrac{\lambda}{v_{\text{avg}}}\;\;\Longrightarrow\;\;\tau \propto \dfrac{1}{\sqrt{T}}.$$

Therefore statement (B) “the mean collision time decreases” is true, and statement (D) “the mean collision time remains unchanged” is false.

Collecting the results:

  • (B) is true. Mean collision time decreases.
  • (C) is true. Mean free path remains unchanged.
  • (A) is false. Mean free path does not decrease.
  • (D) is false. Mean collision time does not remain the same.

Hence, the correct answer is Option A.

For any gas molecule, every quadratic degree of freedom contributes an average energy of $$\dfrac{1}{2}kT$$, where $$k$$ is Boltzmann’s constant. This statement is called the equipartition of energy theorem.

For one mole of the gas, we multiply by Avogadro’s number $$N_A$$ and use the relation $$kN_A = R$$, the universal gas constant. Hence the molar internal energy is

$$U \;=\; \dfrac{f}{2}\,RT,$$

where $$f$$ is the total number of quadratic degrees of freedom available to each molecule.

Every molecule certainly possesses three translational degrees of freedom, and, because the question does not restrict the gas to be mon-atomic, we must also count the two rotational degrees of freedom that a linear/diatomic molecule has about the axes perpendicular to the bond. Thus

$$f \;=\; 3\;(\text{translation}) \;+\; 2\;(\text{rotation}) \;=\; 5.$$

Substituting this value in the expression for $$U$$ gives

$$U \;=\; \dfrac{5}{2}\,RT.$$

The molar heat capacity at constant volume is defined as the temperature derivative of the internal energy, so

$$C_v \;=\; \left(\dfrac{\partial U}{\partial T}\right)_V \;=\; \dfrac{5}{2}\,R.$$

For an ideal gas we always have the Mayer relation

$$C_p \;=\; C_v + R.$$

Substituting the value of $$C_v$$ just obtained, we find

$$C_p \;=\; \dfrac{5}{2}\,R + R \;=\; \dfrac{5}{2}\,R + \dfrac{2}{2}\,R \;=\; \dfrac{7}{2}\,R.$$

Now we compute the ratio

$$\gamma \;=\; \dfrac{C_p}{C_v} \;=\; \dfrac{\dfrac{7}{2}\,R}{\dfrac{5}{2}\,R} \;=\; \dfrac{7}{5}.$$

We have thus obtained

$$U = \dfrac{5}{2}\,RT \quad\text{and}\quad \gamma = \dfrac{7}{5}.$$

These correspond exactly to Option C.

Hence, the correct answer is Option C.

For any ideal gas, the law of equipartition of energy tells us that each quadratic degree of freedom contributes an average energy of $$\frac{1}{2}kT$$ per molecule, where $$k$$ is Boltzmann’s constant and $$T$$ is the absolute temperature. If a molecule has a total of $$f$$ quadratic degrees of freedom, the average internal energy per molecule is $$\frac{f}{2}kT$$ and, per mole, the internal energy is $$\frac{f}{2}RT$$, because $$R = N_Ak$$.

The molar specific heat at constant volume is the derivative of the internal energy with respect to temperature, so the formula we use is

$$C_V = \left(\frac{\partial U}{\partial T}\right)_V = \frac{f}{2}R.$$

We now count the degrees of freedom for the two gases.

Gas $$A$$ is a rigid diatomic molecule. A rigid diatomic has

• $$3$$ translational degrees of freedom, and
• $$2$$ rotational degrees of freedom (about the two axes perpendicular to the bond).

There are no vibrational degrees of freedom because the molecule is stated to be rigid. Hence

$$f_A = 3 + 2 = 5.$$

Using the formula, the molar specific heat of gas $$A$$ is

$$ (C_V)_A = \frac{f_A}{2}R = \frac{5}{2}R. $$

Gas $$B$$ is also diatomic, but it possesses one additional vibrational mode that is fully active. A single vibrational mode contributes two quadratic degrees of freedom: one kinetic and one potential. Therefore, the count for gas $$B$$ is

• $$3$$ translational,
• $$2$$ rotational,
• $$2$$ vibrational.

This gives

$$f_B = 3 + 2 + 2 = 7.$$

So the molar specific heat of gas $$B$$ is

$$ (C_V)_B = \frac{f_B}{2}R = \frac{7}{2}R. $$

We are asked for the ratio $$(C_V)_A : (C_V)_B$$. Substituting the expressions we have just obtained,

$$ (C_V)_A : (C_V)_B = \frac{5}{2}R : \frac{7}{2}R. $$

The common factors $$\frac{1}{2}R$$ cancel out, leaving

$$ (C_V)_A : (C_V)_B = 5 : 7. $$

Hence, the correct answer is Option D.

For an ideal gas, the relation between the molar specific heats at constant pressure and volume is written first:

$$C_p = C_v + R$$

Here $$R$$ is the universal gas constant. To convert this into the required ratio, we recall the expression of $$C_v$$ in terms of the number of degrees of freedom $$f$$ of a single molecule:

$$C_v=\dfrac{f}{2}\,R$$

Substituting this value of $$C_v$$ in the earlier relation gives

$$C_p = \dfrac{f}{2}R + R = \left(\dfrac{f}{2}+1\right)R = \dfrac{f+2}{2}\,R$$

Now we build the desired ratio $$\dfrac{C_p}{C_v}$$, usually denoted by $$\gamma$$:

$$\gamma \;=\; \dfrac{C_p}{C_v} \;=\; \dfrac{\dfrac{f+2}{2}R}{\dfrac{f}{2}R} \;=\; \dfrac{f+2}{f} \;=\; 1 + \dfrac{2}{f}$$

Therefore, once we know the degrees of freedom $$f$$ of each type of molecule, we can obtain its $$\gamma$$ value directly.

We list the customary degrees of freedom for various molecular models:

• A monoatomic gas possesses only three translational degrees of freedom, so $$f=3$$.
• A diatomic rigid molecule (rotations allowed about two perpendicular axes) has five degrees of freedom, so $$f=5$$.
• A diatomic non-rigid molecule includes two additional vibrational degrees of freedom, making $$f=7$$.
• A triatomic rigid molecule (linear or planar but counted rigid) enjoys six degrees of freedom, so $$f=6$$.

Now we place each value of $$f$$ into $$\gamma = 1+\dfrac{2}{f}$$ and simplify step by step.

Monoatomic gas ($$f=3$$)

$$\gamma_{\text{mono}} = 1 + \dfrac{2}{3} = \dfrac{3}{3} + \dfrac{2}{3} = \dfrac{5}{3}$$

Diatomic rigid gas ($$f=5$$)

$$\gamma_{\text{dia, rigid}} = 1 + \dfrac{2}{5} = \dfrac{5}{5} + \dfrac{2}{5} = \dfrac{7}{5}$$

Diatomic non-rigid gas ($$f=7$$)

$$\gamma_{\text{dia, non-rigid}} = 1 + \dfrac{2}{7} = \dfrac{7}{7} + \dfrac{2}{7} = \dfrac{9}{7}$$

Triatomic rigid gas ($$f=6$$)

$$\gamma_{\text{tria, rigid}} = 1 + \dfrac{2}{6} = 1 + \dfrac{1}{3} = \dfrac{4}{3}$$

We now match each molecule category with its calculated $$\gamma$$ ratio:

$$\begin{aligned} \text{Monoatomic} &\longrightarrow \dfrac{5}{3}\;(\text{IV})\\[4pt] \text{Diatomic rigid} &\longrightarrow \dfrac{7}{5}\;(\text{I})\\[4pt] \text{Diatomic non-rigid} &\longrightarrow \dfrac{9}{7}\;(\text{II})\\[4pt] \text{Triatomic rigid} &\longrightarrow \dfrac{4}{3}\;(\text{III}) \end{aligned}$$

Reading these links against the options supplied, we obtain the correspondence

(A) - (IV), (B) - (I), (C) - (II), (D) - (III), which is exactly the content of Option C in the list.

Hence, the correct answer is Option 3.

Let us denote the ratio $$\dfrac{C_p}{C_v}$$ of a gas by the symbol $$\gamma$$ (gamma).

For the first ideal gas we are given $$\gamma_1 = \dfrac{5}{3}$$ and the amount present is $$n_1 = 2$$ moles.

For the second ideal gas we are given $$\gamma_2 = \dfrac{4}{3}$$ and the amount present is $$n_2 = 3$$ moles.

For any ideal gas we have the universal relation $$C_p - C_v = R,$$ where $$R$$ is the universal gas constant per mole.

We begin with the first gas.

Using $$C_p - C_v = R$$ and $$\gamma_1 = \dfrac{C_{p1}}{C_{v1}},$$ we can write $$\gamma_1 = \dfrac{C_{p1}}{C_{v1}} = \dfrac{C_{v1} + R}{C_{v1}} = 1 + \dfrac{R}{C_{v1}}.$$ Solving for $$C_{v1}$$ we obtain $$\dfrac{R}{C_{v1}} = \gamma_1 - 1,$$ so $$C_{v1} = \dfrac{R}{\gamma_1 - 1}.$$

Substituting $$\gamma_1 = \dfrac{5}{3}$$ gives $$C_{v1} = \dfrac{R}{\dfrac{5}{3} - 1} = \dfrac{R}{\dfrac{2}{3}} = \dfrac{3R}{2}.$$

Now, $$C_{p1} = C_{v1} + R = \dfrac{3R}{2} + R = \dfrac{5R}{2}.$$

Next we turn to the second gas.

Exactly in the same way we have $$C_{v2} = \dfrac{R}{\gamma_2 - 1} = \dfrac{R}{\dfrac{4}{3} - 1} = \dfrac{R}{\dfrac{1}{3}} = 3R,$$ and $$C_{p2} = C_{v2} + R = 3R + R = 4R.$$

We now compute the total heat capacities of the mixture.

The total heat capacity at constant volume is $$C_{v,\text{mix}} = n_1 C_{v1} + n_2 C_{v2}.$$ Substituting the known numbers, $$C_{v,\text{mix}} = 2 \left(\dfrac{3R}{2}\right) + 3 (3R) = 3R + 9R = 12R.$$

The total heat capacity at constant pressure is $$C_{p,\text{mix}} = n_1 C_{p1} + n_2 C_{p2}.$$ Substituting the values, $$C_{p,\text{mix}} = 2 \left(\dfrac{5R}{2}\right) + 3 (4R) = 5R + 12R = 17R.$$

The total number of moles in the mixture is $$n_{\text{total}} = n_1 + n_2 = 2 + 3 = 5.$$

Hence the molar (per mole) heat capacities of the mixture are $$\overline{C_v} = \dfrac{C_{v,\text{mix}}}{n_{\text{total}}} = \dfrac{12R}{5} = 2.4\,R,$$ and $$\overline{C_p} = \dfrac{C_{p,\text{mix}}}{n_{\text{total}}} = \dfrac{17R}{5} = 3.4\,R.$$

The required ratio for the mixture is then $$\gamma_{\text{mix}} = \dfrac{\overline{C_p}}{\overline{C_v}} = \dfrac{3.4\,R}{2.4\,R} = \dfrac{34}{24} = \dfrac{17}{12} \approx 1.42.$$

Hence, the correct answer is Option D.

We are given a gaseous mixture that contains $$n$$ moles of helium (a mon-atomic gas) and $$2n$$ moles of oxygen (a di-atomic gas that is to be treated as a rigid molecule). We have to evaluate the ratio $$\dfrac{C_P}{C_V}$$ for the whole mixture, treating it as an ideal gas.

First, we recall the basic kinetic-theory relation connecting the molar heat capacity at constant volume with the number of degrees of freedom $$f$$ of a molecule: for any ideal gas,

$$C_V = \dfrac{f}{2}\,R,$$

where $$R$$ is the universal gas constant. Correspondingly,

$$C_P = C_V + R.$$

For helium, which is mon-atomic, there are only the three translational degrees of freedom, so $$f_{\text{He}} = 3$$. Hence

$$C_{V,\text{He}} = \dfrac{3}{2}R, \qquad C_{P,\text{He}} = \dfrac{3}{2}R + R = \dfrac{5}{2}R.$$

For oxygen, which is di-atomic and described as a rigid rotor, the molecule possesses three translational and two rotational degrees of freedom, giving $$f_{\text{O}_2}=5$$. Therefore,

$$C_{V,\text{O}_2} = \dfrac{5}{2}R, \qquad C_{P,\text{O}_2} = \dfrac{5}{2}R + R = \dfrac{7}{2}R.$$

Let us now compute the molar heat capacities of the entire mixture. The total number of moles present is

$$n_{\text{total}} = n + 2n = 3n.$$

The mixture’s molar heat capacity at constant volume is obtained by weighting each component’s heat capacity by the number of moles of that component and then dividing by the total moles:

$$\begin{aligned} C_{V,\text{mix}} &= \dfrac{\,n \,C_{V,\text{He}} \;+\; 2n \,C_{V,\text{O}_2}\,}{\,n_{\text{total}}\,} \\ &= \dfrac{\,n\left(\dfrac{3}{2}R\right) + 2n\left(\dfrac{5}{2}R\right)}{3n}. \end{aligned}$$

Simplifying the numerator inside the fraction:

$$n\left(\dfrac{3}{2}R\right) = \dfrac{3}{2}\,nR,$$

$$2n\left(\dfrac{5}{2}R\right) = 2n \cdot \dfrac{5}{2}R = 5nR.$$

Adding these two contributions gives

$$\dfrac{3}{2}\,nR + 5nR = \left(\dfrac{3}{2} + 5\right)nR = \left(\dfrac{3 + 10}{2}\right)nR = \dfrac{13}{2}\,nR.$$

Dividing by the denominator $$3n$$ produces

$$C_{V,\text{mix}} = \dfrac{\dfrac{13}{2}\,nR}{3n} = \dfrac{13}{6}\,R.$$

Next, the mixture’s molar heat capacity at constant pressure is simply one $$R$$ higher:

$$C_{P,\text{mix}} = C_{V,\text{mix}} + R = \dfrac{13}{6}R + R = \left(\dfrac{13}{6} + \dfrac{6}{6}\right)R = \dfrac{19}{6}\,R.$$

Finally, we take the ratio

$$\dfrac{C_P}{C_V}\Bigg|_{\text{mix}} = \dfrac{\dfrac{19}{6}R}{\dfrac{13}{6}R} = \dfrac{19}{13}.$$

Hence, the correct answer is Option A.

We begin by recalling the two calorimetric relations for an ideal gas:

At constant pressure: $$Q_P = n\,C_P\,\Delta T$$

At constant volume: $$Q_V = n\,C_V\,\Delta T$$

Here $$n$$ is the number of moles of the gas, $$C_P$$ and $$C_V$$ are its molar heat capacities at constant pressure and constant volume respectively, and $$\Delta T$$ is the change in temperature expressed in either °C or K (the difference is the same).

For the first experiment we are told that the temperature is raised by $$50^\circ{\rm C}$$ at constant pressure and that the gas absorbs $$160\ {\rm cal}$$ of heat. Substituting these data into the pressure-process formula we get

$$160 = n\,C_P\,(+50)$$

Dividing both sides by $$50$$ we obtain

$$n\,C_P = \frac{160}{50} = 3.2\ {\rm cal\;K^{-1}}.$$

In the second experiment the same mass of gas is cooled by $$100^\circ{\rm C}$$ at constant volume and releases $$240\ {\rm cal}$$ of heat. Heat released is taken as negative, so we write

$$Q_V = -240\ {\rm cal},\quad \Delta T = -100^\circ{\rm C}.$$

Now substituting in the volume-process formula,

$$-240 = n\,C_V\,(-100).$$

Both minus signs cancel, giving

$$n\,C_V = \frac{240}{100} = 2.4\ {\rm cal\;K^{-1}}.$$

An ideal gas obeys the universal relation

$$C_P - C_V = R,$$

so for our sample

$$n\,C_P - n\,C_V = n\,R.$$

We substitute the values already obtained:

$$3.2 - 2.4 = n\,R \;\Longrightarrow\; n\,R = 0.8\ {\rm cal\;K^{-1}}.$$

Next we express the molar heat capacity at constant volume in terms of the number of degrees of freedom $$f$$. For an ideal gas,

$$C_V = \dfrac{f}{2}\,R \quad\Longrightarrow\quad n\,C_V = \dfrac{f}{2}\,n\,R.$$

We know both $$n\,C_V$$ and $$n\,R$$, so we substitute:

$$2.4 = \dfrac{f}{2}\,(0.8).$$

Solving for $$f$$,

$$\dfrac{f}{2} = \dfrac{2.4}{0.8} = 3 \;\Longrightarrow\; f = 6.$$

The gas molecules therefore possess six degrees of freedom, which corresponds to Option B.

Hence, the correct answer is Option B.

We start with the well-known relation for the root-mean-square (r.m.s.) speed of an ideal gas,

$$v_{\text{rms}}=\sqrt{\dfrac{3RT}{M}},$$

where $$R$$ is the universal gas constant, $$T$$ is the absolute temperature, and $$M$$ is the molar mass of the gas.

The r.m.s. speed is to be doubled. Therefore, if the initial speed is $$v_{\text{rms,1}}$$ at temperature $$T_1$$ and the final speed is $$v_{\text{rms,2}}=2v_{\text{rms,1}}$$ at temperature $$T_2$$, we write

$$\dfrac{v_{\text{rms,2}}}{v_{\text{rms,1}}}=2=\sqrt{\dfrac{T_2}{T_1}}.$$

Squaring both sides,

$$4=\dfrac{T_2}{T_1}\quad\Longrightarrow\quad T_2=4T_1.$$

The gas is initially at $$27^{\circ}\text{C}=27+273=300\ \text{K}$$, so

$$T_2=4\times 300\ \text{K}=1200\ \text{K}.$$

Next, we calculate the number of moles of nitrogen. The mass is 15 g and the molar mass of $$\mathrm{N_2}$$ is 28 g mol−1, hence

$$n=\dfrac{m}{M}=\dfrac{15\ \text{g}}{28\ \text{g mol}^{-1}}=0.5357\ \text{mol}.$$

The vessel is rigid, so the heating takes place at constant volume. For a diatomic gas in the temperature range considered, the molar heat capacity at constant volume is

$$C_V=\dfrac{5}{2}R.$$

Stating the formula for heat supplied at constant volume,

$$Q=nC_V\Delta T,$$

where $$\Delta T=T_2-T_1=1200\ \text{K}-300\ \text{K}=900\ \text{K}.$$

Substituting all numerical values,

$$\begin{aligned} Q&=n\left(\dfrac{5}{2}R\right)\Delta T\\[4pt] &=0.5357\ \text{mol}\times\left(\dfrac{5}{2}\times 8.3\ \text{J mol}^{-1}\text{K}^{-1}\right)\times 900\ \text{K}. \end{aligned}$$

First evaluate the heat capacity factor:

$$\dfrac{5}{2}\times 8.3=20.75\ \text{J mol}^{-1}\text{K}^{-1}.$$

Then multiply by the moles:

$$0.5357\times 20.75=11.12\ \text{J K}^{-1}.$$

Finally multiply by $$\Delta T$$:

$$Q=11.12\ \text{J K}^{-1}\times 900\ \text{K}=10,008\ \text{J}\approx 1.0\times 10^{4}\ \text{J}.$$

Expressing this in kilojoules,

$$Q\approx 10\ \text{kJ}.$$

Hence, the correct answer is Option B.

The problem speaks about an ideal gas, so we recall the relationship between the internal energy $$U$$ of an ideal, mono-atomic gas and its state variables. For such a gas, the internal energy depends only on temperature and is given by the well-known formula

$$U \;=\; \frac{3}{2}\,nRT.$$

Here $$nRT$$ can be replaced by $$PV$$ through the ideal-gas equation $$PV=nRT$$. Substituting $$nRT=PV$$ into the expression for $$U$$, we arrive at

$$U \;=\; \frac{3}{2}\,PV.$$

Now we insert the numerical values supplied in the question. The pressure is

$$P \;=\; 3 \times 10^{6}\ \text{Pa},$$

and the volume is

$$V \;=\; 2\ \text{m}^{3}.$$

First we form the product $$PV$$:

$$PV \;=\; \bigl(3 \times 10^{6}\ \text{Pa}\bigr)\,\bigl(2\ \text{m}^{3}\bigr) \;=\; 6 \times 10^{6}\ \text{Pa·m}^{3}.$$

The unit $$\text{Pa·m}^{3}$$ is equivalent to the joule, because $$1\ \text{Pa}=1\ \text{N\,m}^{-2}$$ and $$1\ \text{N\,m}=1\ \text{J}$$. Hence

$$PV \;=\; 6 \times 10^{6}\ \text{J}.$$

Next we multiply by $$\dfrac{3}{2}$$ as demanded by the formula for internal energy:

$$U \;=\; \frac{3}{2}\,\bigl(6 \times 10^{6}\ \text{J}\bigr) \;=\; 3 \times 10^{6}\ \text{J} \times 1.5 \;=\; 9 \times 10^{6}\ \text{J}.$$

Thus the energy of the gas is $$9 \times 10^{6}\ \text{J}.$$

Hence, the correct answer is Option B.

We have a mono-atomic ideal gas whose mass is given as $$m = 2\ \text{kg}$$. The pressure of the gas is given as $$P = 4 \times 10^{4}\ \text{N m}^{-2}$$ and its density is $$\rho = 8\ \text{kg m}^{-3}$$.

Density is related to mass and volume by the formula $$\rho = \dfrac{m}{V}.$$

Re-arranging, the volume of the gas is obtained as $$ V = \dfrac{m}{\rho}. $$

Substituting the given values, $$ V = \dfrac{2\ \text{kg}}{8\ \text{kg m}^{-3}} = 0.25\ \text{m}^{3}. $$

For an ideal gas, the product of pressure and volume equals $$nRT$$, i.e. $$ PV = nRT. $$ Although the actual value of $$nRT$$ is not required, we still use the equality $$PV = nRT$$ to replace $$nRT$$ in the expression for internal energy.

The total thermal (internal) energy of a mono-atomic ideal gas is given by the formula $$ U = \dfrac{3}{2}\,nRT. $$ Using $$PV = nRT,$$ we can rewrite this as $$ U = \dfrac{3}{2}\,PV. $$

Now we first evaluate $$PV$$: $$ PV = \left(4 \times 10^{4}\ \text{N m}^{-2}\right)\left(0.25\ \text{m}^{3}\right) = 1.0 \times 10^{4}\ \text{J}. $$ (The unit N m−2 × m3 simplifies to joules.)

Substituting this value of $$PV$$ into the internal-energy formula, $$ U = \dfrac{3}{2}\,PV = \dfrac{3}{2}\left(1.0 \times 10^{4}\ \text{J}\right) = 1.5 \times 10^{4}\ \text{J}. $$

The numerical value $$1.5 \times 10^{4}\ \text{J}$$ lies in the order of $$10^{4}\ \text{J}.$$

Hence, the correct answer is Option C.

For an ideal gas, the molar internal energy is linked to the number of degrees of freedom $$f$$ by the well-known expression

$$U_{\text{molar}}=\dfrac{f}{2}RT$$

because each quadratic degree of freedom contributes an average energy of $$\tfrac12kT$$ per molecule or $$\tfrac12RT$$ per mole.

Now we examine the two components of the mixture one by one.

Argon (Ar) is mono-atomic. A mono-atomic molecule can translate along the three Cartesian axes but cannot rotate in a way that stores energy (its moment of inertia about any axis through the centre is negligible). Hence, for argon

$$f_{\text{Ar}} = 3 \quad\Longrightarrow\quad U_{\text{molar, Ar}}=\dfrac{3}{2}RT$$

Oxygen (O2) is diatomic. At the given temperature range we are told to consider only translational and rotational motion; vibrational modes are “frozen out”. A diatomic molecule has three translational and two rotational degrees of freedom (rotation about the internuclear axis contributes negligibly). Thus, for oxygen

$$f_{\text{O}_2}=3+2=5 \quad\Longrightarrow\quad U_{\text{molar, O}_2}=\dfrac{5}{2}RT$$

The mixture contains 5 moles of argon and 3 moles of oxygen, so the total internal energy is the sum of the individual contributions:

$$\begin{aligned} U_{\text{total}} &= \left(5\;\text{mol}\right)\left(\dfrac{3}{2}RT\right) + \left(3\;\text{mol}\right)\left(\dfrac{5}{2}RT\right)\\[4pt] &= \dfrac{15}{2}RT + \dfrac{15}{2}RT\\[4pt] &= \dfrac{30}{2}RT\\[4pt] &= 15RT \end{aligned}$$

Hence, the correct answer is Option A.

For an ideal gas, the mean time between two successive intermolecular collisions (often written as $$\tau$$) is inversely proportional to the collision frequency. The collision frequency, in turn, is proportional to the number density of molecules and to their average speed.

Mathematically, we write the proportionality as

$$\tau \propto \dfrac{1}{n \, \bar v}$$

Here, $$n$$ is the number of molecules per unit volume and $$\bar v$$ is the average (root-mean-square) molecular speed.

For an ideal gas we have two important relations:

1. The ideal-gas equation gives the number density

$$n = \dfrac{P}{k_{\text B}T} \; ,$$

where $$P$$ is pressure, $$T$$ is absolute temperature and $$k_{\text B}$$ is Boltzmann’s constant.

2. The root-mean-square speed varies with temperature as

$$\bar v \propto \sqrt{T}\; .$$

Putting these together in the proportionality for $$\tau$$ we get

$$\tau \propto \dfrac{1}{\left(\dfrac{P}{T}\right)\sqrt{T}} = \dfrac{T}{P\sqrt{T}} = \dfrac{\sqrt{T}}{P}\; .$$

So, the mean collision time is directly proportional to $$\sqrt{T}$$ and inversely proportional to the pressure $$P$$. We can therefore write the ratio of the mean times in two different states as

$$\dfrac{\tau_2}{\tau_1} = \dfrac{\sqrt{T_2}/P_2}{\sqrt{T_1}/P_1} = \dfrac{\sqrt{T_2}\,P_1}{\sqrt{T_1}\,P_2}\; .$$

Now we substitute the numerical data. The initial state has

$$P_1 = 2 \text{ atm}, \qquad T_1 = 300 \text{ K}, \qquad \tau_1 = 6 \times 10^{-8} \text{ s}.$$

The final state has

$$P_2 = 2P_1 = 4 \text{ atm}, \qquad T_2 = 500 \text{ K}.$$

First we compute the square roots of the temperatures:

$$\sqrt{T_1} = \sqrt{300}\; \text{K}^{1/2} \approx 17.3205\; ,$$

$$\sqrt{T_2} = \sqrt{500}\; \text{K}^{1/2} \approx 22.3607\; .$$

Next we form the ratio:

$$\dfrac{\tau_2}{\tau_1} = \dfrac{22.3607 \times 2}{17.3205 \times 4} = \dfrac{44.7214}{69.2820} \approx 0.645\; .$$

Multiplying this ratio by the initial mean time gives the new mean time:

$$\tau_2 = 0.645 \times \left(6 \times 10^{-8}\right)\, \text{s} = 3.87 \times 10^{-8}\, \text{s} \approx 4 \times 10^{-8}\, \text{s}.$$

Hence, the correct answer is Option D.

We begin by recalling the expression for the root-mean-square (rms) speed of the molecules of an ideal gas. For a gas of molar mass $$M$$ kept at absolute temperature $$T$$, the formula is

$$u_{\text{rms}}=\sqrt{\dfrac{3RT}{M}},$$

where $$R$$ is the universal gas constant. Clearly, for a given gas (that is, for fixed $$M$$), the rms speed depends only on the absolute temperature and is independent of the pressure. Hence, if the same gas is taken from one temperature to another, the ratio of the two rms speeds is simply the square root of the ratio of the two absolute temperatures. Symbolically,

$$\dfrac{u_2}{u_1}=\sqrt{\dfrac{T_2}{T_1}}.$$

We are supplied with the following data for the first state:

Initial pressure $$P_1 = 1 \text{ atm},$$

Initial temperature $$T_1 = 127^\circ\text{C}.$$

Because our formula demands absolute temperature, we convert Celsius to Kelvin by using $$T(\text{K}) = T(\!^\circ\text{C}) + 273$$. Thus,

$$T_1 = 127 + 273 = 400 \text{ K}.$$

The rms speed in this state is given as

$$u_1 = 200 \text{ m s}^{-1}.$$

For the second state we are told

Second pressure $$P_2 = 2 \text{ atm},$$

Second temperature $$T_2 = 227^\circ\text{C}.$$

Again converting to Kelvin, we get

$$T_2 = 227 + 273 = 500 \text{ K}.$$

Now we apply the proportionality relation between rms speeds and absolute temperatures. Writing it explicitly,

$$u_2 = u_1 \sqrt{\dfrac{T_2}{T_1}}.$$

Substituting the numerical values, we have

$$u_2 = 200 \times \sqrt{\dfrac{500}{400}}.$$

Simplify the fraction under the square root:

$$\dfrac{500}{400} = \dfrac{5}{4}.$$

Therefore,

$$u_2 = 200 \times \sqrt{\dfrac{5}{4}}.$$

Write the square root of a fraction as the fraction of the square roots:

$$\sqrt{\dfrac{5}{4}} = \dfrac{\sqrt{5}}{\sqrt{4}} = \dfrac{\sqrt{5}}{2}.$$

So,

$$u_2 = 200 \times \dfrac{\sqrt{5}}{2}.$$

Cancel the factor of 2 between 200 and the denominator 2:

$$u_2 = 100 \sqrt{5}\ \text{m s}^{-1}.$$

The numerical value $$100 \sqrt{5}$$ matches option D.

Hence, the correct answer is Option D.

For an ideal gas the molar heat capacities at constant pressure and volume are related by the very first identity we always quote

where $$R$$ is the universal gas constant. In addition, the total molar internal energy is written in the form

with $$f$$ denoting the total number of quadratic degrees of freedom that are actually “active” at the temperature of the experiment. Because

we obtain immediately the useful pair of working formulae

Finally we define the ratio of heat capacities

Now we recall the textbook values for a diatomic molecule:

• 3 translational degrees of freedom.
• 2 rotational degrees of freedom (because the molecule is linear).
• Each normal vibrational mode contributes two quadratic terms (one kinetic, one potential) and therefore counts as two in the energy‐counting scheme.

Hence

• Without any vibrational excitation (a “rigid” diatomic) we have $$f=3+2=5.$$ This gives $$C_v=\dfrac{5}{2}R\approx20.8\ \text{J mol}^{-1}\text{K}^{-1},\qquad C_p=\dfrac{7}{2}R\approx29.1\ \text{J mol}^{-1}\text{K}^{-1},\qquad\gamma=\dfrac{7}{5}=1.40.$$
• If the single vibrational mode is fully excited, two more terms are added, so $$f=5+2=7.$$ Then $$C_v=\dfrac{7}{2}R\approx29.1\ \text{J mol}^{-1}\text{K}^{-1},\qquad C_p=\dfrac{9}{2}R\approx37.4\ \text{J mol}^{-1}\text{K}^{-1},\qquad\gamma=\dfrac{9}{7}=1.29.$$

With the theory organised, let us examine the two experimental gases.

Gas A. The given numbers are

We use the difference to obtain the effective $$R$$ that belongs to the same set of rounded data:

With this value the effective number of degrees of freedom is

This figure is distinctly larger than 5 and lies quite close to the ideal value 7 that one gets when a single vibrational mode is active. We confirm the same tendency by the ratio

which is very near the theoretical 1.29 for a diatomic possessing one vibrational mode, and markedly below 1.40 for a rigid diatomic. Hence Gas A does possess its vibrational mode.

Gas B. The quoted values are

First compute the rounded difference:

Now derive the corresponding $$f$$:

This result is very close to 5 and far from 7, so the only degrees of freedom effectively operating are the three translational and the two rotational ones; the vibrational mode is evidently still “frozen out.” The heat-capacity ratio reinforces the same conclusion:

essentially the theoretical 1.40 expected for a rigid diatomic with no vibrational contribution.

Putting the two discussions together we find that Gas A has its single vibrational mode active, whereas Gas B behaves as a rigid diatomic molecule with no vibrational contribution.

Hence, the correct answer is Option B.

We first recall the general result for an ideal gas: if a molecule has $$f$$ degrees of freedom, then its molar specific heat at constant volume is given by the formula $$C_v = \dfrac{f}{2}\,R,$$ because the internal energy of one mole is $$U = \dfrac{f}{2}\,RT$$ and differentiating with respect to temperature at constant volume gives $$\left(\dfrac{\partial U}{\partial T}\right)_V = \dfrac{f}{2}\,R.$$

For a monatomic gas such as helium, there are only the three translational degrees of freedom, so $$f = 3.$$ Therefore, for each mole of helium, we have

$$C_{v,\text{He}} = \dfrac{3}{2}\,R.$$

A hydrogen molecule ($$\text{H}_2$$) is di-atomic and is stated to be rigid in the question, which means it possesses three translational and two rotational degrees of freedom, giving $$f = 5.$$ Hence, for each mole of hydrogen,

$$C_{v,\text{H}_2} = \dfrac{5}{2}\,R.$$

Now we form the mixture. The numbers of moles are:

Helium: $$n_{\text{He}} = 2,$$   Hydrogen: $$n_{\text{H}_2} = 3.$$

The total heat capacity of the mixture at constant volume is the sum of the individual contributions:

$$C_V^{\text{(total)}} \;=\; n_{\text{He}}\,C_{v,\text{He}} \;+\; n_{\text{H}_2}\,C_{v,\text{H}_2}.$$

Substituting the values just obtained, we get

$$\begin{aligned} C_V^{\text{(total)}} & = 2\left(\dfrac{3}{2}\,R\right) + 3\left(\dfrac{5}{2}\,R\right) \\ & = 2 \times \dfrac{3}{2}\,R + 3 \times \dfrac{5}{2}\,R \\ & = 3R + \dfrac{15}{2}\,R \\ & = 3R + 7.5R \\ & = 10.5R. \end{aligned}$$

The mixture contains a total of

$$n_{\text{total}} = n_{\text{He}} + n_{\text{H}_2} = 2 + 3 = 5 \text{ moles}.$$

The required quantity is the molar specific heat of the mixture, defined as

$$C_v^{\text{(mix)}} = \dfrac{C_V^{\text{(total)}}}{n_{\text{total}}}.$$

Substituting the numbers:

$$\begin{aligned} C_v^{\text{(mix)}} & = \dfrac{10.5R}{5} \\ & = 2.1R. \end{aligned}$$

We are given $$R = 8.3 \text{ J mol}^{-1}\text{K}^{-1},$$ so

$$\begin{aligned} C_v^{\text{(mix)}} & = 2.1 \times 8.3 \\ & = 17.43 \text{ J mol}^{-1}\text{K}^{-1}. \end{aligned}$$

This numerical value matches closest with the option $$17.4 \text{ J/mol K}.$$

Hence, the correct answer is Option A.

We begin with the statement of the equipartition theorem: “For every independent quadratic degree of freedom present in a molecule, the average energy possessed per molecule is $$\frac{1}{2}k_B T.$$”

An HCl molecule is linear, so it indeed possesses three kinds of motion:

• Translational motion along the three mutually perpendicular axes (x, y and z).
• Rotational motion about two axes that are perpendicular to the internuclear axis (a linear molecule cannot rotate about its own bond axis in the classical sense, so only two rotational degrees of freedom are counted).
• Vibrational motion along the bond, which contributes two quadratic terms (one kinetic and one potential) and hence amounts to two degrees of freedom.

Altogether, the molecule has $$3 + 2 + 2 = 7$$ quadratic degrees of freedom. However, the root-mean-square speed $$\bar v$$ is defined solely from the translational kinetic energy, because it involves only the centre-of-mass velocity of the molecule. Therefore, to relate temperature to $$\bar v$$ we need consider only the translational part of the energy.

For translation there are exactly three degrees of freedom, and by the equipartition theorem the average translational kinetic energy per molecule is therefore

$$E_{\text{trans}} \;=\; \frac{3}{2}\,k_B\,T.$$

By definition, the average translational kinetic energy may also be written in terms of the molecular mass $$m$$ and the root-mean-square speed $$\bar v$$ as

$$E_{\text{trans}} \;=\; \frac{1}{2}\,m\,\bar v^{\,2}.$$

Since both expressions represent the same physical quantity, we equate them:

$$\frac{1}{2}\,m\,\bar v^{\,2} \;=\; \frac{3}{2}\,k_B\,T.$$

Now, we cancel the common factor of $$\frac{1}{2}$$ from both sides to obtain

$$m\,\bar v^{\,2} \;=\; 3\,k_B\,T.$$

Solving for $$T$$ gives

$$T \;=\; \frac{m\,\bar v^{\,2}}{3\,k_B}.$$

Thus the temperature corresponding to the given root-mean-square speed is

$$\displaystyle \frac{m\bar{v}^{2}}{3k_B}.$$

Hence, the correct answer is Option C.

We want the root mean square (r.m.s.) speed of a hydrogen molecule to be exactly equal to the escape speed from the surface of the earth. The two speeds are:

$$v_{\text{rms}}=\sqrt{\dfrac{3k_B T}{m}} \qquad\text{and}\qquad v_{\text{esc}}=\sqrt{2gR}$$

Here $$k_B$$ is the Boltzmann constant, $$T$$ the absolute temperature, $$m$$ the mass of one hydrogen molecule, $$g$$ the acceleration due to gravity and $$R$$ the radius of the earth.

We set $$v_{\text{rms}}=v_{\text{esc}}.$$ Doing so and then squaring both sides gives

$$\dfrac{3k_B T}{m}=2gR.$$

Now we isolate $$T$$:

$$T=\dfrac{2gR\,m}{3k_B}.$$

Next we evaluate each quantity. The escape-speed part $$2gR$$ is calculated first. Substituting $$g=10\ \text{m s}^{-2}$$ and $$R=6.4\times10^{6}\ \text{m}$$ we obtain

$$2gR = 2\times10\times6.4\times10^{6} = 1.28\times10^{8}\ \text{m}^2\text{s}^{-2}.$$

For the molecular mass $$m$$ we remember that one hydrogen atom has mass $$\dfrac{1}{N_A}\ \text{kg},$$ because $$N_A$$ particles make up $$1\ \text{kg}$$ (the statement of the Avogadro number in the data). A hydrogen molecule (H$$_2$$) consists of two atoms, so

$$m = \dfrac{2}{N_A} = \dfrac{2}{6.02\times10^{26}} = 3.32\times10^{-27}\ \text{kg}.$$

The denominator in the temperature formula is $$3k_B$$:

$$3k_B = 3\times1.38\times10^{-23} = 4.14\times10^{-23}\ \text{J K}^{-1}.$$

We now substitute every value into $$T=\dfrac{2gR\,m}{3k_B}:$$

$$T =\dfrac{(1.28\times10^{8})\;(3.32\times10^{-27})} {4.14\times10^{-23}} =\dfrac{4.25\times10^{-19}}{4.14\times10^{-23}}.$$

Dividing the numerical coefficients and combining powers of ten step by step, we first handle the coefficients:

$$\dfrac{4.25}{4.14}\approx1.03.$$

Then we combine the exponents:

$$10^{-19}\div10^{-23}=10^{\,4}.$$

So

$$T\approx1.03\times10^{4}\ \text{K}.$$

The temperature that makes the r.m.s. velocity of hydrogen molecules equal to the earth’s escape velocity is therefore about $$10^{4}$$ kelvin, which matches the fourth option.

Hence, the correct answer is Option D.

For any ideal gas, the root-mean-square (rms) speed of its molecules is given by the formula

$$v_{rms}=\sqrt{\frac{3RT}{M}},$$

where $$R$$ is the universal gas constant, $$T$$ is the absolute temperature in kelvin, and $$M$$ is the molar mass of the gas in kilograms per mole. This expression follows directly from equating the kinetic-theory expression for pressure with the ideal-gas law.

We are asked to compare the rms speeds of helium and argon at the same temperature. Because $$R$$ and $$T$$ are identical for both gases, the ratio of their rms speeds will depend only on the inverse square root of their molar masses. Mathematically, we can write

$$\frac{v_{rms}(\text{helium})}{v_{rms}(\text{argon})} =\sqrt{\frac{3RT/M_{\text{He}}}{3RT/M_{\text{Ar}}}} =\sqrt{\frac{M_{\text{Ar}}}{M_{\text{He}}}}.$$

Now we substitute the numerical molar masses. The atomic (and hence molar) mass of helium is given as $$4\ \text{u}=4\ \text{g mol}^{-1}$$, while that of argon is $$40\ \text{u}=40\ \text{g mol}^{-1}$$. Converting grams per mole to kilograms per mole is not necessary for a ratio because the factor of $$10^{-3}$$ cancels out, but we can still show it explicitly:

$$M_{\text{He}} = 4\ \text{g mol}^{-1}=4\times10^{-3}\ \text{kg mol}^{-1},$$

$$M_{\text{Ar}} = 40\ \text{g mol}^{-1}=40\times10^{-3}\ \text{kg mol}^{-1}.$$

Substituting these values into the ratio expression, we obtain

$$\frac{v_{rms}(\text{He})}{v_{rms}(\text{Ar})} =\sqrt{\frac{40\times10^{-3}}{4\times10^{-3}}} =\sqrt{\frac{40}{4}} =\sqrt{10} \approx 3.162.$$

This numerical value is closest to $$3.16$$ in the given options.

Hence, the correct answer is Option D.

We are told that the number density (molecules per unit volume) at a distance $$r$$ from the origin is $$n(r)=n_0 e^{-\alpha r^4}$$. To obtain the total number of molecules $$N$$ present in the whole space, we must integrate this density over the entire volume.

Because the density depends only on the radial distance, the gas is spherically symmetric. In spherical polar coordinates the volume element is $$dV = 4\pi r^2\,dr$$. Hence, using the definition $$N=\displaystyle\int_{\text{all space}} n(r)\,dV$$, we write

$$ N = \int_0^{\infty} n(r)\;4\pi r^2\,dr = 4\pi n_0\int_0^{\infty} r^2 e^{-\alpha r^4}\,dr. $$

Now we evaluate the integral $$I=\displaystyle\int_0^{\infty} r^2 e^{-\alpha r^4}\,dr.$$ To remove the exponential’s quartic argument, we perform the substitution $$x=\alpha r^4.$$ Then $$r = \left(\dfrac{x}{\alpha}\right)^{1/4}, \qquad dr = \dfrac{1}{4}\,\alpha^{-1/4}\,x^{-3/4}\,dx.$$

We also need $$r^2$$ in terms of $$x$$:

$$ r^2 = \left(\dfrac{x}{\alpha}\right)^{1/2} = x^{1/2}\,\alpha^{-1/2}. $$

Substituting $$r^2$$ and $$dr$$ into $$I$$ gives

$$ I = \int_0^{\infty} \Bigl(x^{1/2}\alpha^{-1/2}\Bigr)\, e^{-x}\, \Bigl(\frac{1}{4}\alpha^{-1/4}x^{-3/4}\Bigr)\,dx. $$

Collecting powers of $$x$$ and $$\alpha$$:

$$ x^{1/2}x^{-3/4}=x^{-1/4}, \qquad \alpha^{-1/2}\alpha^{-1/4}= \alpha^{-3/4}. $$

Thus

$$ I = \frac{1}{4}\,\alpha^{-3/4}\int_0^{\infty} x^{-1/4}e^{-x}\,dx. $$

The remaining integral is a standard Gamma-function. Recall the formula $$\displaystyle\Gamma(s)=\int_0^{\infty}x^{\,s-1}e^{-x}\,dx.$$ Here $$x^{-1/4}=x^{\,(-1/4)}=x^{\,\,(3/4-1)}$$, so $$s=\frac{3}{4}$$. Therefore

$$ \int_0^{\infty}x^{-1/4}e^{-x}\,dx = \Gamma\!\left(\frac{3}{4}\right), $$

a pure numerical constant, independent of $$\alpha$$ and $$n_0$$. Consequently,

$$ I = \frac{1}{4}\,\Gamma\!\left(\frac{3}{4}\right)\,\alpha^{-3/4}. $$

Returning to the expression for $$N$$, we substitute this result:

$$ N = 4\pi n_0\,I = 4\pi n_0\left[\frac{1}{4}\Gamma\!\left(\frac{3}{4}\right)\alpha^{-3/4}\right] = \pi\,\Gamma\!\left(\frac{3}{4}\right)\,n_0\,\alpha^{-3/4}. $$

The factors $$\pi$$ and $$\Gamma\!\left(\frac{3}{4}\right)$$ are fixed numerical constants, so we conclude that

$$ N \propto n_0\,\alpha^{-3/4}. $$

Hence, the correct answer is Option C.

We have a vertical cylinder of uniform cross-sectional area $$A$$. A light, frictionless piston of mass $$m$$ separates the cylinder into an upper part of length $$l_1$$ and a lower part of length $$l_2$$, with $$l_1 > l_2$$. Each part contains $$n$$ moles of an ideal gas at the same absolute temperature $$T$$. Because the piston is at rest, the forces acting on it must balance perfectly.

For each gas sample we write the ideal-gas equation. The ideal-gas law states

$$PV = nRT,$$

where $$P$$ is pressure, $$V$$ is volume, $$n$$ is the number of moles, $$R$$ is the universal gas constant and $$T$$ is the absolute temperature.

First for the gas above the piston, whose pressure is $$P_1$$ and volume is $$V_1 = A\,l_1$$, we obtain

$$P_1 V_1 = nRT \;\;\Longrightarrow\;\; P_1 (A\,l_1) = nRT \;\;\Longrightarrow\;\; P_1 = \frac{nRT}{A\,l_1}.$$

Next for the gas below the piston, whose pressure is $$P_2$$ and volume is $$V_2 = A\,l_2$$, we write

$$P_2 V_2 = nRT \;\;\Longrightarrow\;\; P_2 (A\,l_2) = nRT \;\;\Longrightarrow\;\; P_2 = \frac{nRT}{A\,l_2}.$$

Now we look at the forces acting on the piston. The upward force on the piston is the pressure exerted by the lower gas times area $$A$$, namely $$P_2A$$. The downward forces are the pressure exerted by the upper gas, $$P_1A$$, and the weight of the piston, $$mg$$. For mechanical equilibrium (piston stationary) we therefore set

$$\text{Upward force} = \text{Downward force}$$

$$P_2A = P_1A + mg.$$

Solving for the weight term, we write

$$mg = A\,(P_2 - P_1).$$

We now substitute the expressions for $$P_1$$ and $$P_2$$ obtained from the ideal-gas law:

$$mg = A\!\left(\frac{nRT}{A\,l_2} - \frac{nRT}{A\,l_1}\right).$$

The area $$A$$ cancels out:

$$mg = nRT\!\left(\frac{1}{l_2} - \frac{1}{l_1}\right).$$

We combine the fractions inside the parentheses:

$$\frac{1}{l_2} - \frac{1}{l_1} = \frac{l_1 - l_2}{l_1\,l_2}.$$

Substituting this result gives

$$mg = nRT\!\left(\frac{l_1 - l_2}{l_1\,l_2}\right).$$

Finally we isolate the mass $$m$$ by dividing both sides by $$g$$:

$$m = \frac{nRT}{g}\!\left(\frac{l_1 - l_2}{l_1\,l_2}\right).$$

This expression matches option B.

Hence, the correct answer is Option B.

We have been told that every second $$10^{22}$$ gas molecules strike a perfectly rigid surface of area $$1\ \text{m}^2$$ at right angles. Each molecule has a mass $$m = 10^{-26}\ \text{kg}$$ and a speed $$v = 10^{4}\ \text{m s}^{-1}$$. Because the collisions are stated to be perfectly elastic and normal (i.e. perpendicular), each molecule reverses its velocity on impact.

For one molecule the initial momentum component perpendicular to the wall is $$+mv$$, while the final momentum (after rebounding straight back) is $$-mv$$. The change in momentum for a single collision is therefore

$$\Delta p = p_{\text{final}} - p_{\text{initial}} = (-mv) - (+mv) = -2mv.$$

Only the magnitude matters when calculating force, so the momentum transferred to the wall per collision is

$$|\Delta p| = 2mv.$$

Now, $$N = 10^{22}$$ such collisions occur every second. The total momentum imparted to the wall each second is therefore

$$\text{Total momentum per second} = N \times 2mv = 10^{22} \times 2 \times 10^{-26}\ \text{kg} \times 10^{4}\ \text{m s}^{-1}.$$

Multiplying the powers of ten step by step:

$$10^{22} \times 10^{-26} = 10^{(-26+22)} = 10^{-4},$$

and then

$$10^{-4} \times 10^{4} = 10^{(-4+4)} = 10^{0} = 1.$$

So the numerical product simplifies to

$$N \times 2mv = 2 \times 1 = 2\ \text{kg m s}^{-2}.$$

But $$1\ \text{kg m s}^{-2} = 1\ \text{N}$$, hence the force exerted on the wall is

$$F = 2\ \text{N}.$$

By definition, pressure is force per unit area. Stating the formula first, we write

$$P = \frac{F}{A}.$$

Since $$A = 1\ \text{m}^2$$, we obtain

$$P = \frac{2\ \text{N}}{1\ \text{m}^2} = 2\ \text{Pa}.$$

Hence, the correct answer is Option A.

First, let us translate the data into symbols. The capacity of the cylinder is given as 67.2 L, and the gas inside is helium at S.T.P. (Standard Temperature and Pressure).

At S.T.P. the temperature is $$T_0 = 273\ \text{K}$$ and the pressure is $$P_0 = 1\ \text{atm}$$. For an ideal gas, the well-known molar volume relation at S.T.P. is

$$V_m = 22.4\ \text{L mol}^{-1}.$$

We have a total volume $$V = 67.2\ \text{L}.$$ Using $$n = \dfrac{V}{V_m}$$ we find the number of moles present:

$$n = \dfrac{67.2\ \text{L}}{22.4\ \text{L mol}^{-1}} = 3.0\ \text{mol}.$$

The cylinder is rigid, so the volume of gas cannot change; the heating therefore occurs at constant volume. For a process at constant volume, the heat supplied is related to the molar heat capacity at constant volume $$C_V$$ by

$$q = n\,C_V\,\Delta T.$$

Helium is a mono-atomic ideal gas. For such gases the theoretical value of the molar heat capacity is

$$C_V = \dfrac{3}{2}R,$$

where $$R = 8.31\ \text{J mol}^{-1}\,\text{K}^{-1}$$ (given). The required temperature rise is $$\Delta T = 20^{\circ}\text{C} = 20\ \text{K}.$$

Substituting every quantity into the formula, we have

$$q = n\left(\dfrac{3}{2}R\right)\Delta T = 3\ \text{mol} \times \dfrac{3}{2}\,(8.31\ \text{J mol}^{-1}\,\text{K}^{-1}) \times 20\ \text{K}.$$

Proceeding step by step:

$$\dfrac{3}{2} \times 8.31 = 1.5 \times 8.31 = 12.465\ \text{J mol}^{-1}\,\text{K}^{-1},$$

$$12.465 \times 20 = 249.3\ \text{J mol}^{-1},$$

$$3 \times 249.3 = 747.9\ \text{J}.$$

Rounding to the appropriate number of significant digits,

$$q \approx 7.48 \times 10^{2}\ \text{J} \; \approx \; 748\ \text{J}.$$

Hence, the correct answer is Option A.

We start by recalling the molar heat-capacities of the two gases in the temperature region where only translational and rotational degrees of freedom are active.

For a monatomic gas (such as helium) the degrees of freedom are three, so

$$C_{v,\,\text{He}}=\frac{3}{2}R,\qquad C_{p,\,\text{He}}=C_{v,\,\text{He}}+R=\frac{5}{2}R.$$

For a diatomic gas (such as hydrogen, H2) the active degrees of freedom are five, giving

$$C_{v,\,\text{H}_2}=\frac{5}{2}R,\qquad C_{p,\,\text{H}_2}=C_{v,\,\text{H}_2}+R=\frac{7}{2}R.$$

Two moles of helium are mixed with $$n$$ moles of hydrogen, so the total molar heat capacities of the mixture are obtained by adding the contributions of each component:

$$C_p^{\text{(tot)}}=2\!\left(\frac{5}{2}R\right)+n\!\left(\frac{7}{2}R\right)=5R+\frac{7n}{2}R,$$ $$C_v^{\text{(tot)}}=2\!\left(\frac{3}{2}R\right)+n\!\left(\frac{5}{2}R\right)=3R+\frac{5n}{2}R.$$

The ratio $$\dfrac{C_p}{C_v}$$ for the whole mixture is the ratio of these two totals because the common factor $$R$$ cancels out:

$$\frac{C_p}{C_v}=\frac{5+\dfrac{7}{2}n}{3+\dfrac{5}{2}n}=\frac{5+3.5n}{3+2.5n}.$$

According to the problem statement this ratio equals $$\dfrac{3}{2}$$, so we set up the equation

$$\frac{5+3.5n}{3+2.5n}=\frac{3}{2}.$$

Now we solve algebraically, showing every step:

First cross-multiply:

$$2\left(5+3.5n\right)=3\left(3+2.5n\right).$$

Expanding each side gives

$$10+7n=9+7.5n.$$

Subtract $$9$$ from both sides:

$$1+7n=7.5n.$$

Now subtract $$7n$$ from both sides:

$$1=0.5n.$$

Finally divide by $$0.5$$:

$$n=2.$$

Therefore the mixture must contain two moles of hydrogen.

Hence, the correct answer is Option C.

We are asked to find the “thermal velocity” of a helium atom at room temperature. In kinetic theory this quantity is usually taken as the root-mean-square speed, given by the well-known formula

$$v_{\text{rms}} = \sqrt{\dfrac{3k_B T}{m}}.$$

Here $$k_B$$ is Boltzmann’s constant, $$T$$ is the absolute temperature and $$m$$ is the mass of one atom (or molecule) of the gas.

We now substitute the data provided in the question. We have

$$k_B = 1.4 \times 10^{-23}\;{\rm J\,K^{-1}}, \qquad T = 300\;{\rm K}, \qquad m_{He} = 7 \times 10^{-27}\;{\rm kg}.$$

First we calculate the numerator $$3 k_B T$$ step by step.

$$3 k_B T = 3 \times (1.4 \times 10^{-23}) \times 300.$$

Multiply the numbers in the brackets:

$$1.4 \times 300 = 420.$$

So

$$3 k_B T = 3 \times 420 \times 10^{-23}.$$

Now, $$3 \times 420 = 1260,$$ therefore

$$3 k_B T = 1260 \times 10^{-23}\;{\rm J}.$$

We rewrite the power of ten so that there is only one non-zero digit before the decimal point:

$$1260 \times 10^{-23} = 1.26 \times 10^{3} \times 10^{-23} = 1.26 \times 10^{-20}\;{\rm J}.$$

Next we form the ratio with the mass of a helium atom:

$$\dfrac{3 k_B T}{m_{He}} = \dfrac{1.26 \times 10^{-20}}{7 \times 10^{-27}}.$$

We divide the coefficients and subtract the exponents of ten:

$$\dfrac{1.26}{7} \approx 0.18, \qquad 10^{-20} \,/\, 10^{-27} = 10^{7}.$$

So

$$\dfrac{3 k_B T}{m_{He}} \approx 0.18 \times 10^{7} = 1.8 \times 10^{6}\;{\rm (m/s)^2}.$$

Finally, we take the square root to obtain the speed:

$$v_{\text{rms}} = \sqrt{1.8 \times 10^{6}}.$$

We separate the numerical and the power-of-ten parts:

$$\sqrt{1.8} \approx 1.34, \qquad \sqrt{10^{6}} = 10^{3}.$$

Therefore

$$v_{\text{rms}} \approx 1.34 \times 10^{3}\;{\rm m\,s^{-1}}.$$

This value is closest to $$1.3 \times 10^{3}\;{\rm m\,s^{-1}}.$$

Hence, the correct answer is Option D.

We begin with the well-known relation

$$\text{Pressure }(P)=\frac{\text{Force }(F)}{\text{Area }(A)}$$

So, to obtain the pressure, we first have to calculate the force exerted on the wall by the stream of molecules. Force is the rate of change of momentum, therefore

$$F=\frac{\Delta p_{\text{total}}}{\Delta t}$$

Since the data are already “per second”, we may take $$\Delta t = 1\ \text{s}$$, and then

$$F = \text{(change of momentum per molecule)}\times(\text{number of molecules per second}).$$

Let us work out the change of momentum for one molecule. The given velocity of each molecule is $$v = 10^{3}\ \text{m s}^{-1}$$, and the molecule strikes the wall making an angle $$\theta = 45^\circ$$ with the normal. Hence the component of velocity normal to the wall is

$$v_n = v\cos\theta = 10^{3}\cos45^\circ\ \text{m s}^{-1}.$$

Because $$\cos45^\circ = \dfrac{1}{\sqrt2}$$, we have

$$v_n = 10^{3}\times\dfrac{1}{\sqrt2}\ \text{m s}^{-1}.$$

The molecule rebounds elastically, so its speed is unchanged but its normal component reverses direction. Thus

$$p_{\text{initial}} = m v_n,\qquad p_{\text{final}} = -m v_n.$$

Therefore the change in normal momentum for one molecule is

$$\Delta p = p_{\text{final}} - p_{\text{initial}} = (-m v_n) - (m v_n) = -2 m v_n.$$

We shall use the magnitude only, so

$$|\Delta p| = 2 m v_n = 2 m v\cos\theta.$$

Substituting the numerical values one by one:

Mass of one hydrogen molecule

$$m = 3.32\times10^{-27}\ \text{kg},$$

velocity

$$v = 10^{3}\ \text{m s}^{-1},$$

and $$\cos\theta = \dfrac{1}{\sqrt2}\approx0.7071.$$

Hence

$$|\Delta p| = 2 \times (3.32\times10^{-27}) \times (10^{3}) \times 0.7071$$ $$= 2 \times 3.32 \times 0.7071 \times 10^{-24}$$ $$= 4.695\times10^{-24}\ \text{kg m s}^{-1}.$$

Now, the number of molecules striking the wall each second is

$$N = 10^{23}\ \text{s}^{-1}.$$

Therefore the total change of momentum per second (which is the force) becomes

$$F = N\,|\Delta p| = (10^{23})\,(4.695\times10^{-24})$$ $$= 4.695\times10^{-1}\ \text{N}$$ $$= 0.4695\ \text{N}.$$

Next we convert the area of the wall to SI units. The given area is

$$A = 2\ \text{cm}^2.$$

Because $$1\ \text{cm} = 10^{-2}\ \text{m},$$ we have

$$A = 2 \times (10^{-2}\ \text{m})^{2} = 2 \times 10^{-4}\ \text{m}^2.$$

Finally, substituting into the pressure formula, we get

$$P = \frac{F}{A} = \frac{0.4695\ \text{N}}{2\times10^{-4}\ \text{m}^2}$$ $$= \frac{0.4695}{0.0002}\ \text{N m}^{-2}$$ $$= 2347.5\ \text{N m}^{-2}.$$

Written in scientific notation,

$$P \approx 2.35 \times 10^{3}\ \text{N m}^{-2}.$$

Hence, the correct answer is Option B.

Let $$f$$ be the degrees of freedom of the gas molecules. Then,

$$\frac{C_P}{C_V}=1+\frac{2}{f}$$

$$\therefore\ \frac{C_P}{C_V}=1+\frac{2}{5}=\frac{7}{5}$$

For an ideal gas, the relation between its mass-specific heats is first written explicitly:

$$C_{p}-C_{v}=\frac{R}{M}$$

Here $$R$$ is the universal gas constant and $$M$$ is the molar mass of the gas expressed in $$\text{kg mol}^{-1}$$. Because $$R$$ is the same for every gas, the difference $$C_{p}-C_{v}$$ varies only with the reciprocal of the molar mass.

We are told that for hydrogen gas the difference is $$a$$, so

$$a=C_{p}-C_{v}\;(\text{for hydrogen})=\frac{R}{M_{\,\mathrm{H_2}}}.$$

For nitrogen gas the difference is $$b$$, hence

$$b=C_{p}-C_{v}\;(\text{for nitrogen})=\frac{R}{M_{\,\mathrm{N_2}}}.$$

To relate $$a$$ and $$b$$ we divide the first expression by the second:

$$\frac{a}{b}=\frac{\dfrac{R}{M_{\,\mathrm{H_2}}}}{\dfrac{R}{M_{\,\mathrm{N_2}}}} =\frac{M_{\,\mathrm{N_2}}}{M_{\,\mathrm{H_2}}}.$$

Now we substitute the molar masses in kilogram per mole:

$$M_{\,\mathrm{H_2}} = 2\;\text{g mol}^{-1}=2\times 10^{-3}\;\text{kg mol}^{-1},$$

$$M_{\,\mathrm{N_2}} = 28\;\text{g mol}^{-1}=28\times 10^{-3}\;\text{kg mol}^{-1}.$$

Putting these values into the ratio gives

$$\frac{a}{b}=\frac{28\times 10^{-3}}{2\times 10^{-3}}=\frac{28}{2}=14.$$

Therefore we finally obtain

$$a = 14\,b.$$

Hence, the correct answer is Option D.

We start with $$N$$ moles of an ideal diatomic gas at a fixed temperature $$T$$. For an ideal gas the total kinetic energy (translational + rotational) per mole equals $$\dfrac{f}{2}RT$$, where $$f$$ is the number of degrees of freedom.

For a diatomic molecule at ordinary temperatures the three translational and two rotational modes are active, so $$f=5$$. Hence the total kinetic energy of the initial sample is

$$ U_i \;=\; N\left(\dfrac{5}{2}RT\right)\;=\;\dfrac{5}{2}NRT . $$

Now heat is supplied while the temperature is kept constant, and $$n$$ moles (in the sense of diatomic molecules) dissociate completely into mono-atomic molecules. A single diatomic molecule $$XY$$ dissociates as

$$ XY \;\rightarrow\; X + Y , $$

so one mole of the diatomic gas produces two moles of mono-atomic gas. Therefore, after the dissociation:

• Remaining diatomic moles $$=\;N-n$$
• New mono-atomic moles $$=\;2n$$
• Total final moles $$=\;(N-n)+2n \;=\;N+n$$

The mono-atomic gas possesses only the three translational degrees of freedom, so for it $$f=3$$ and the kinetic energy per mole is $$\dfrac{3}{2}RT$$.

Hence the total kinetic energy after the change is

$$ \begin{aligned} U_f &= (N-n)\left(\dfrac{5}{2}RT\right) + (2n)\left(\dfrac{3}{2}RT\right) \\ &= \dfrac{5}{2}NRT - \dfrac{5}{2}nRT + 3nRT \\ &= \dfrac{5}{2}NRT + \left(3 - \dfrac{5}{2}\right)nRT \\ &= \dfrac{5}{2}NRT + \dfrac{1}{2}nRT . \end{aligned} $$

The change in total kinetic energy is therefore

$$ \Delta U = U_f - U_i = \left[\dfrac{5}{2}NRT + \dfrac{1}{2}nRT\right] - \dfrac{5}{2}NRT = \dfrac{1}{2}nRT . $$

Hence, the correct answer is Option D.

We start with the ideal­-gas equation written for a collection of molecules:

$$PV = NkT$$

Here $$P$$ is the pressure, $$V$$ is the volume, $$N$$ is the number of molecules, $$k = 1.38 \times 10^{-23}\,\text{J K}^{-1}$$ is Boltzmann’s constant and $$T$$ is the absolute temperature in kelvin.

The room is open to the atmosphere, so the pressure inside always equals the atmospheric pressure and remains constant at

$$P = 1 \times 10^{5}\,\text{Pa}.$$

The rigid walls keep the volume fixed at

$$V = 30\ \text{m}^{3}.$$

Because both $$P$$ and $$V$$ are constant, the product $$PV$$ is constant, and we can write for two different states:

$$PV = N_i k T_i \quad\text{and}\quad PV = N_f k T_f.$$

Dividing the two equations gives

$$\frac{N_f}{N_i} = \frac{T_i}{T_f}.$$ So

$$N_f = N_i\,\frac{T_i}{T_f}.$$

We want the difference $$N_f - N_i$$, so we first express both $$N_i$$ and $$N_f$$ individually using $$PV = NkT$$.

For the initial state (17 °C):

Temperature in kelvin: $$T_i = 17^{\circ}\text{C} + 273 = 290\ \text{K}.$$

Hence

$$N_i = \frac{PV}{kT_i}.$$

For the final state (27 °C):

Temperature in kelvin: $$T_f = 27^{\circ}\text{C} + 273 = 300\ \text{K}.$$

Thus

$$N_f = \frac{PV}{kT_f}.$$

Now we write the required difference:

$$N_f - N_i = \frac{PV}{k}\left(\frac{1}{T_f} - \frac{1}{T_i}\right).$$

We substitute all the numerical values step by step.

First calculate the factor $$\dfrac{PV}{k}$$:

$$\frac{PV}{k} = \frac{(1 \times 10^{5}\,\text{Pa})(30\,\text{m}^{3})}{1.38 \times 10^{-23}\,\text{J K}^{-1}} = \frac{3.0 \times 10^{6}\,\text{J}}{1.38 \times 10^{-23}\,\text{J K}^{-1}} = 2.1739 \times 10^{29}\,\text{K}^{-1}.$$

Next evaluate the bracket containing the temperatures:

$$\frac{1}{T_f} - \frac{1}{T_i} = \frac{1}{300\ \text{K}} - \frac{1}{290\ \text{K}} = \frac{290 - 300}{300 \times 290} = \frac{-10}{87000} = -1.1494 \times 10^{-4}\,\text{K}^{-1}.$$

Multiplying the two results gives

$$N_f - N_i = \left(2.1739 \times 10^{29}\,\text{K}^{-1}\right) \left(-1.1494 \times 10^{-4}\,\text{K}^{-1}\right) = -2.50 \times 10^{25}.$$

The negative sign tells us that some molecules have left the room so that the pressure could stay constant while the temperature increased.

Hence, the correct answer is Option A.

$$PV = nRT$$

$$PV = \left(\frac{m}{M}\right)RT$$

$$P = \left(\frac{m}{V}\right) \frac{RT}{M}$$

$$P = \rho \frac{RT}{M}$$

The plot of pressure ($$P$$) versus density ($$\rho$$) at a constant temperature results in a straight line starting from the origin, indicating that pressure is directly proportional to density. The correct relationship is represented by the straight-line graph in Option D.

To solve this problem, we need to determine how the average force exerted by a single molecule on the walls of a closed container depends on the temperature $$T$$ for an ideal gas. Specifically, we are told that this force varies as $$T^q$$, and we must find the exponent $$q$$.

First, recall that the pressure $$P$$ exerted by an ideal gas is given by the kinetic theory equation:

$$ P = \frac{1}{3} \rho \overline{v^2} $$

where $$\rho$$ is the density (mass per unit volume) and $$\overline{v^2}$$ is the mean square speed of the molecules. Since $$\rho = \frac{N m}{V}$$, where $$N$$ is the number of molecules, $$m$$ is the mass of one molecule, and $$V$$ is the volume, we can write:

$$ P = \frac{1}{3} \frac{N m}{V} \overline{v^2} $$

From the kinetic theory of gases, the average kinetic energy per molecule is related to the temperature $$T$$ by:

$$ \frac{1}{2} m \overline{v^2} = \frac{3}{2} k T $$

where $$k$$ is Boltzmann's constant. Solving for $$\overline{v^2}$$:

$$ \overline{v^2} = \frac{3 k T}{m} $$

Substitute this into the pressure equation:

$$ P = \frac{1}{3} \frac{N m}{V} \cdot \frac{3 k T}{m} $$

Simplify by canceling $$m$$ and the factor of 3:

$$ P = \frac{N k T}{V} $$

This is the ideal gas law in terms of the number of molecules.

Now, pressure is force per unit area, so the total force on a wall of area $$A$$ is $$F_{\text{total}} = P A$$. Substituting the expression for $$P$$:

$$ F_{\text{total}} = \frac{N k T}{V} A $$

This total force is the result of all molecules colliding with the wall. However, we need the average force exerted by a single molecule. Since the molecules are identical and the gas is isotropic, the total average force on the wall is the sum of the average forces from each molecule. Therefore:

$$ F_{\text{total}} = N \overline{F} $$

where $$\overline{F}$$ is the average force per molecule. Equating the two expressions for $$F_{\text{total}}$$:

$$ N \overline{F} = \frac{N k T}{V} A $$

Divide both sides by $$N$$:

$$ \overline{F} = \frac{k T A}{V} $$

For a closed container, both the area $$A$$ of the wall and the volume $$V$$ are constant. Therefore, the average force per molecule $$\overline{F}$$ is proportional to $$T$$:

$$ \overline{F} \propto T $$

This implies that $$\overline{F}$$ depends on $$T$$ as $$T^q$$ with $$q = 1$$.

Hence, the correct answer is Option C.

According to the equipartition theorem of energy, each quadratic degree of freedom contributes an average energy of $$\dfrac12 k_B T$$ to a system in thermal equilibrium, where $$k_B$$ is the Boltzmann constant and $$T$$ is the absolute temperature.

An atom bound inside a crystalline solid behaves like a three-dimensional harmonic oscillator. Such an oscillator possesses

• three kinetic energy terms (one for motion along each axis), and
• three potential energy terms (one for the restoring force along each axis).

Thus the number of quadratic terms per atom is

Using equipartition, the average energy of one atom is therefore

For one mole we have Avogadro’s number $$N_A$$ of atoms, so the molar internal energy becomes

But the universal gas constant is defined as $$R = N_A k_B,$$ so we can write

The molar heat capacity at constant volume is the temperature derivative of this energy:

At high temperatures the difference between $$C_P$$ and $$C_V$$ for a solid is negligible, so we may take the molar specific heat simply as

The numerical value of the gas constant is

Hence

To convert this molar specific heat to a mass specific heat (J kg$$^{-1}$$ K$$^{-1}$$) we divide by the molar mass (atomic weight expressed in kg per mole). Aluminium has an atomic weight of 27 g mol$$^{-1}$$, which is

Therefore the specific heat per kilogram is

Rounding to a convenient integer gives approximately $$925\ \text{J kg}^{-1}\ \text{K}^{-1}.$$

Hence, the correct answer is Option D.