JEE Thermodynamics Questions

Melting takes place isothermally at $$T = 273\;\text{K}$$ and constant atmospheric pressure $$P = 1\;\text{atm}$$ (≈ $$1.013\times10^{5}\;\text{Pa}$$).

At constant pressure the heat absorbed equals the molar enthalpy of fusion:
$$q_{p}= \Delta H_{fus} \;\;(\text{positive})$$

The first-law relation connecting enthalpy and internal energy is
$$\Delta H = \Delta U + P\,\Delta V \quad -(1)$$

Rearranging, the change in internal energy is
$$\Delta U = \Delta H - P\,\Delta V \quad -(2)$$

When ice converts to water its volume decreases because water is denser than ice:
density of ice ≈ $$0.917\;\text{g cm}^{-3}$$, density of water ≈ $$1.00\;\text{g cm}^{-3}$$.
Hence $$\Delta V \lt 0$$ (final volume < initial volume).

Work done by the system on the atmosphere is
$$w = P\,\Delta V$$.
Since $$\Delta V \lt 0$$, $$w$$ is negative: the system does negative work, that is, the atmosphere does positive work on the system.

Substituting $$\Delta V \lt 0$$ in $$(2)$$:
$$\Delta U = \Delta H - P\,(\text{negative}) = \Delta H + (\text{positive term})$$.
Therefore $$\Delta U \gt 0$$; the internal energy increases, it certainly does not remain unchanged or decrease.

Summarising:
• Internal energy increases ⇒ Options A and C are incorrect.
• Work done on the system is positive ⇒ Option B is incorrect, Option D is correct.

Hence the correct choice is Option D: Positive work is done on the ice-water system by the atmosphere.

For a perfect black body the radiative power transferred per unit area between two surfaces at absolute temperatures $$T_1$$ and $$T_2$$ (with $$T_1 \gt T_2$$) is given by Stefan-Boltzmann law:

$$W = \sigma \left(T_1^{4} - T_2^{4}\right)$$

where $$\sigma$$ is the Stefan-Boltzmann constant.

Case 1: Only plates P and Q are present (Fig. 1).

The power per unit area flowing from P $$\rightarrow$$ Q is therefore

$$W_0 = \sigma\left(T_P^{4} - T_Q^{4}\right) \quad -(1)$$

Case 2: Two additional identical black plates are inserted between P and Q (Fig. 2). Let their steady-state temperatures be $$T_1$$ (next to P) and $$T_2$$ (next to Q), with $$T_P \gt T_1 \gt T_2 \gt T_Q$$. Heat exchange occurs only between adjacent plates, and in steady state the same power per unit area $$W_S$$ must flow across every gap.

Across the three gaps we have

$$W_S = \sigma\left(T_P^{4} - T_1^{4}\right)$$ $$W_S = \sigma\left(T_1^{4} - T_2^{4}\right)$$ $$W_S = \sigma\left(T_2^{4} - T_Q^{4}\right)$$

Divide each equation by $$\sigma$$ to write them purely in terms of temperature differences:

$$T_P^{4} - T_1^{4} = T_1^{4} - T_2^{4} = T_2^{4} - T_Q^{4} = \Delta \quad -(2)$$

Since the same quantity $$\Delta$$ appears three times, add the three equalities:

$$\left(T_P^{4} - T_1^{4}\right) + \left(T_1^{4} - T_2^{4}\right) + \left(T_2^{4} - T_Q^{4}\right) = 3\Delta$$

The middle terms cancel, leaving

$$T_P^{4} - T_Q^{4} = 3\Delta \quad -(3)$$

From $$(3)$$, $$\Delta = \dfrac{T_P^{4} - T_Q^{4}}{3}$$.

Substitute this $$\Delta$$ back into any expression for $$W_S$$ (say the first in $$(2)$$):

$$W_S = \sigma\Delta = \sigma \dfrac{T_P^{4} - T_Q^{4}}{3} = \dfrac{1}{3}\,\sigma\left(T_P^{4} - T_Q^{4}\right) \quad -(4)$$

Compare $$(1)$$ and $$(4)$$:

$$W_0 = \sigma\left(T_P^{4} - T_Q^{4}\right), \qquad W_S = \dfrac{1}{3}\,\sigma\left(T_P^{4} - T_Q^{4}\right)$$

Hence the required ratio is

$$\frac{W_0}{W_S} = \frac{\sigma\left(T_P^{4} - T_Q^{4}\right)}{\,\dfrac{1}{3}\,\sigma\left(T_P^{4} - T_Q^{4}\right)} = 3$$

Therefore, $$\dfrac{W_0}{W_S} = 3$$.

First convert the given dimensions to SI units:
length $$l = 9\text{ cm} = 0.09\text{ m}$$, width $$d = 4\text{ cm} = 0.04\text{ m}$$.

The initial area of the rectangular sheet is
$$A_0 = l \times d = 0.09 \times 0.04 = 0.0036\ \text{m}^2$$ $$-(1)$$

Heat supplied to the sheet is $$Q = 8.1 \times 10^{2}\ \text{J}$$.
Using the relation $$Q = m\,C_v\,\Delta T$$, the rise in temperature is
$$\Delta T = \frac{Q}{m\,C_v} = \frac{8.1 \times 10^{2}}{0.1 \times 900} = 9\ \text{K}$$ $$-(2)$$

For an isotropic solid, the coefficient of superficial (area) expansion is
$$\beta = 2\alpha$$ $$-(3)$$
where $$\alpha = 3.1 \times 10^{-5}\ \text{K}^{-1}$$ is the linear expansion coefficient.

From $$(3)$$,
$$\beta = 2 \times 3.1 \times 10^{-5} = 6.2 \times 10^{-5}\ \text{K}^{-1}$$ $$-(4)$$

The fractional change in area due to heating through $$\Delta T$$ is $$\beta\,\Delta T$$, so the absolute change in area is
$$\Delta A = A_0 \,\beta \,\Delta T$$ $$-(5)$$

Substituting from $$(1)$$, $$(2)$$ and $$(4)$$ into $$(5)$$:
$$\Delta A = 0.0036 \times 6.2 \times 10^{-5} \times 9$$

Compute step by step:
$$6.2 \times 10^{-5} \times 9 = 5.58 \times 10^{-4}$$;
$$0.0036 \times 5.58 \times 10^{-4} = 2.01 \times 10^{-6}\ \text{m}^2$$.

Rounding to two significant figures,
$$\Delta A \approx 2.0 \times 10^{-6}\ \text{m}^2$$.

Hence, the change in area of the rectangular sheet is $$2.0 \times 10^{-6}\ \text{m}^2$$.
Option A is correct.

Let the number of moles, the pressure and the volume in the left compartment be $$n_1,\;P_L,\;V_L$$ and in the right compartment be $$n_2,\;P_R,\;V_R$$.

Given
$$n_1 = \frac32,\qquad n_2 = 1$$
length of each compartment at equilibrium: $$\frac{L}{2}$$
cross-sectional area of the piston: $$A$$

Therefore
$$V_L = A \left(\frac{L}{2}\right)=V_R = A\left(\frac{L}{2}\right)=V$$ where $$V = \frac{AL}{2}\,.$$

The piston is thermally conducting, hence the two gases finally attain the same temperature $$T$$. For an ideal gas, $$PV=nRT$$; therefore

$$P_LV = n_1 RT,\qquad P_RV = n_2 RT$$

Dividing the first relation by the second,

$$\frac{P_L}{P_R} = \frac{n_1}{n_2}= \frac{3/2}{1}= \frac32 \quad\Longrightarrow\quad P_L = \frac32\,P_R\;.$$  -(1)

Next, write the force-balance for the movable piston. Choose the rightward direction as positive.

• Force on piston by left gas (to the right): $$P_LA$$
• Force on piston by right gas (to the left): $$P_RA$$
• Force by the spring (to the left): $$F_s$$

At mechanical equilibrium, net force is zero:

$$P_LA - P_RA - F_s = 0 \quad\Longrightarrow\quad (P_L-P_R)A = F_s\;.$$  -(2)

The spring is fixed to the left wall; its natural length is $$\dfrac{2L}{5}$$ and its actual length equals the left-hand length $$\dfrac{L}{2}$$. Hence the extension is

$$\Delta x = \frac{L}{2} - \frac{2L}{5} = \frac{L}{10}\;.$$

Thus the spring force is

$$F_s = k\,\Delta x = k\left(\frac{L}{10}\right)=\frac{kL}{10}\;.$$  -(3)

Insert $$(1)$$ and $$(3)$$ into $$(2)$$:

$$(P_L-P_R)A = \frac{kL}{10}$$ $$\left(\frac32\,P_R - P_R\right)A = \frac{kL}{10}$$ $$\left(\frac12\,P_R\right)A = \frac{kL}{10}$$ $$P_R = \frac{kL}{10A}\times 2 = \frac{kL}{5A}\;.$$

Therefore the pressure in the right compartment is

$$P = P_R = \frac{kL}{A}\left(\frac15\right)\,.$$

Comparing with the required form $$P = \dfrac{kL}{A}\,\alpha$$ gives

$$\alpha = \frac15 = 0.2\;.$$

Hence, the desired value is 0.2.

Initial state:

For one complete cycle,

So net heat exchanged equals net work done:

$$Q=W$$

Now calculate work in each process.

1. Isothermal expansion from $$V_0\ ​to\ 4V_0$$

For isothermal process,

$$W_1=nRT\ln\frac{4V_0}{V_0}$$

Since initially

$$nRT=P_0V_0$$

So

$$W_1=P_0V_0\ln4$$

1. Isobaric compression

After isothermal expansion pressure becomes

$$P=\frac{P_0V_0}{4V_0}=\frac{P_0}{4}$$

Compression occurs at constant pressure $$\frac{P_0}{4}\ from\ 4V_0​\ to\ V_0​$$.

Work done:

1. Isochoric heating

$$W_3=0$$

Net work:

$$W=P_0V_0\ln4-\frac{3}{4}P_0V_0$$

When water falls freely, its gravitational potential energy is converted into internal energy (heat) on reaching the pool.
Ignoring any heat loss to the surroundings, the gain in internal energy raises the water’s temperature.

For a mass $$m$$ of water falling through height $$h$$, the potential energy lost is
$$\text{Potential energy} = m g h$$

This energy appears as heat: $$m c \Delta T$$, where $$c$$ is the specific heat capacity and $$\Delta T$$ is the rise in temperature.

Equating the two energies:
$$m g h = m c \Delta T$$

Cancel the common factor $$m$$ (mass of water):
$$g h = c \Delta T$$

Hence the temperature rise is
$$\Delta T = \frac{g h}{c}$$

Insert the given values: $$g = 10 \text{ m s}^{-2}$$, $$h = 200 \text{ m}$$, $$c = 4200 \text{ J kg}^{-1} \text{ K}^{-1}$$.

$$\Delta T = \frac{10 \times 200}{4200}$$

$$\Delta T = \frac{2000}{4200} = 0.476 \text{ K} \approx 0.48 \text{ K}$$

Therefore, the rise in temperature of the water is about $$0.48 \text{ K}$$.

Correct option: Option D (0.48 K)

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.

Concept: Extensive vs intensive thermodynamic variables.

Definitions:

Extensive variables depend on the size/amount of the system (they scale with the quantity of matter). Examples: internal energy, volume, mass, enthalpy, entropy.

Intensive variables are independent of the size of the system. Examples: pressure, temperature, density.

Evaluating each statement:

(A) Internal energy, volume (V), and mass (M) are extensive variables.

All three depend on the amount of matter. Statement A is correct.

(B) Pressure (P), temperature (T), and density ($$\rho$$) are intensive variables.

None of these depend on the system size. Statement B is correct.

(C) Volume (V), temperature (T), and density ($$\rho$$) are intensive variables.

Volume is an extensive variable (it doubles when the system size doubles). Statement C is incorrect.

(D) Mass (M), temperature (T), and internal energy are extensive variables.

Temperature is an intensive variable. Statement D is incorrect.

Correct statements: (A) and (B) only, which corresponds to Option 4.

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 an adiabatic process of an ideal gas, the heat exchanged with the surroundings is zero: $$Q = 0$$.

First law of thermodynamics: $$\Delta U = Q + W$$, where $$W$$ is the work done on the gas.
Because $$Q = 0$$, we have

$$\Delta U = W \quad -(1)$$

For a monoatomic ideal gas, the internal energy is $$U = \dfrac{3}{2}nRT$$. Hence

$$\Delta U = \dfrac{3}{2}nR\,(T_2 - T_1) \quad -(2)$$

Using $$(1)$$ and $$(2)$$, the work done on the gas is

$$W = \dfrac{3}{2}nR\,(T_2 - T_1) \quad -(3)$$

Thus, $$W$$ is directly proportional to $$(T_2 - T_1)$$.

For an adiabatic change of an ideal gas, the pressure-volume relation is $$PV^{\gamma} = \text{constant}$$, where $$\gamma = \dfrac{C_P}{C_V}$$. For a monoatomic gas, $$\gamma = \dfrac{5}{3}$$. Consequently

$$TV^{\gamma - 1} = TV^{2/3} = \text{constant}$$,
which is not simply $$TV = \text{constant}$$, and $$PV$$ is also not constant.

Now examine each statement:

Statement (A): “Internal energy is constant.”
From $$(2)$$, $$\Delta U \neq 0$$ when temperature changes, so (A) is false.

Statement (B): “Work done in the process is equal to the change in internal energy.”
Equation $$(1)$$ shows $$W = \Delta U$$ (with the ‘work on the gas’ sign convention). Hence (B) is true.

Statement (C): “The product of temperature and volume is a constant.”
We have $$TV^{2/3} = \text{constant}$$, not $$TV$$. Therefore (C) is false.

Statement (D): “The product of pressure and volume is a constant.”
For adiabatic change, $$PV^{5/3} = \text{constant}$$, so $$PV$$ is not constant. (D) is false.

Statement (E): “The work done to change the temperature from $$T_1$$ to $$T_2$$ is proportional to $$(T_2 - T_1)$$.”
Equation $$(3)$$ confirms this proportionality. Hence (E) is true.

Therefore, only statements (B) and (E) are correct.

Option C (B, E only) is the correct choice.

The process is sudden and the container is thermally insulated, so no heat enters or leaves the gas. Such a process is adiabatic.

For an adiabatic change in an ideal gas, the equation is
$$P V^{\gamma} = \text{constant}$$

Let the initial state be $$(P_1 , V_1)$$ and the final state be $$(P_2 , V_2)$$. Writing the adiabatic relation for the two states gives
$$P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$$

Re-arrange to obtain the pressure ratio:
$$\frac{P_2}{P_1} = \left(\frac{V_1}{V_2}\right)^{\gamma}$$ $$-(1)$$

The gas is compressed to $$\left(\frac{1}{8}\right)^{\text{th}}$$ of its original volume, so
$$V_2 = \frac{V_1}{8}$$ ⇒ $$\frac{V_1}{V_2} = 8$$

Given $$\gamma = \frac{5}{3}$$ for a monoatomic gas, substitute these values in $$(1)$$:
$$\frac{P_2}{P_1} = 8^{\,\frac{5}{3}}$$

Write 8 as $$2^3$$ and simplify:
$$8^{\,\frac{5}{3}} = \left(2^3\right)^{\frac{5}{3}} = 2^{\,3 \times \frac{5}{3}} = 2^5 = 32$$

Hence the ratio of the final pressure to the initial pressure is $$32$$.

Therefore, Option C is correct.

Since the gas is heated isothermally,

$$ΔU=0$$

(for ideal gas internal energy depends only on temperature)

So heat supplied by filament equals work done by gas:

$$Q=W$$

Total work done by gas has two parts:

1. Work done against atmosphere/piston weight
2. Work done in stretching spring

At equilibrium initially,

$$P_0A=Mg$$

where A is piston area.

Gas expands from height $$L_0$$ to $$L_1$$​

Change in volume:

Work against piston weight:

$$W_1=Mg(L_1-L_0)$$

Spring force law:

$$F=kx^3$$

Initially spring is natural length, so initial extension

x=0

Final extension:

$$x=L_1-L_0$$

Work stored in spring:

$$W_2​=\int kx^3dx\ \ from\ 0\ to\ L_{1\ }-L_0$$

$$\frac{k}{4}(L_1-L_0)^4$$

Also because process is reversible isothermal,

$$W=nRT\ln\frac{V_1}{V_0}$$

with

so

$$W_3=nRT\ln\frac{L_1}{L_0}$$

This is the expansion work of ideal gas.

Hence total energy supplied by filament

For an adiabatic process involving an ideal gas, the relation between temperature and volume is
$$T\,V^{\gamma-1}= \text{constant}$$

Given data:
Initial volume $$V_1 = 800\;\text{cm}^3$$
Final volume $$V_2 = 200\;\text{cm}^3$$
Initial temperature $$T_1 = 27^\circ\text{C} = 27 + 273 = 300\;\text{K}$$
Specific-heat ratio $$\gamma = 1.5$$

Applying the adiabatic relation for the final temperature $$T_2$$:
$$T_1\,V_1^{\gamma-1}=T_2\,V_2^{\gamma-1}$$

Solve for $$T_2$$:
$$T_2 = T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}$$

Compute the volume ratio and exponent:
$$\frac{V_1}{V_2} = \frac{800}{200} = 4$$
$$\gamma - 1 = 1.5 - 1 = 0.5$$

Therefore,
$$T_2 = 300\,\text{K}\;\times\;4^{0.5} = 300\,\text{K}\;\times\;2 = 600\,\text{K}$$

The change in temperature is
$$\Delta T = T_2 - T_1 = 600\,\text{K} - 300\,\text{K} = 300\,\text{K}$$

Hence, the temperature rises by $$300\;\text{K}$$.
Option D is correct.

For any thermodynamic process the first law of thermodynamics states
$$dQ = dU + dW$$ where

$$dQ$$ = heat supplied to the system, $$dU$$ = change in internal energy and $$dW$$ = work done by the system.

Adiabatic process: By definition no heat is exchanged with the surroundings, hence
$$dQ = 0$$ $$-(1)$$

Putting $$dQ = 0$$ in the first-law equation $$-(1)$$ gives
$$0 = dU + dW \;\;\Longrightarrow\;\; dU = -\,dW$$ $$-(2)$$

Equation $$-(2)$$ shows that the work done by the gas is numerically equal to the decrease (negative) in its internal energy, or equivalently, the increase in internal energy is numerically equal to the negative of the work done.

Now recall the definition of molar heat capacity for any path:
$$C = \frac{1}{n}\,\frac{dQ}{dT}$$ where $$n$$ is the number of moles.

For an adiabatic process $$dQ = 0$$, therefore
$$C = \frac{1}{n}\,\frac{0}{dT} = 0$$ $$-(3)$$

Hence the molar heat capacity along an adiabatic path is zero.

Let us test each option:

Option A The molar heat capacity is infinite - contradicts $$-(3)$$ (false).

Option B Work done by the gas equals the increase in internal energy - according to $$-(2)$$ the signs are opposite (false).

Option C The molar heat capacity is zero - matches $$-(3)$$ (true).

Option D The internal energy of the gas decreases as the temperature increases - no such rule exists; internal energy is directly related to temperature for an ideal gas (false).

Therefore the correct statement is Option C: the molar heat capacity in an adiabatic process is zero.

For an ideal gas molecule the ratio of specific heats is given by the kinetic-theory formula
$$\gamma = \frac{C_p}{C_v} = 1 + \frac{2}{f}$$
where $$f$$ is the total number of degrees of freedom of one molecule.

Case A : Triatomic rigid gas
A rigid triatomic molecule that is non-linear (e.g. $$H_2O$$) has
 • $$3$$ translational degrees + $$3$$ rotational degrees  $$\Rightarrow f = 6$$.
Therefore $$\gamma = 1 + \frac{2}{6} = 1 + \frac13 = \frac43$$.
So A matches List-II item III.

Case B : Diatomic non-rigid gas
A non-rigid diatomic molecule possesses vibration along the bond in addition to translation and rotation.
 • Translational : $$3$$, Rotational : $$2$$, Vibrational : $$2$$ (one kinetic + one potential)
Hence $$f = 3 + 2 + 2 = 7$$.
Thus $$\gamma = 1 + \frac{2}{7} = \frac{9}{7}$$.
So B matches List-II item IV.

Case C : Monoatomic gas
A monoatomic molecule can only translate.
 • $$3$$ translational degrees, no rotation or vibration  $$\Rightarrow f = 3$$.
Hence $$\gamma = 1 + \frac{2}{3} = \frac{5}{3}$$.
So C matches List-II item I.

Case D : Diatomic rigid gas
A rigid diatomic molecule has no vibrational contribution at ordinary temperatures.
 • Translational : $$3$$, Rotational : $$2$$  $$\Rightarrow f = 5$$.
Therefore $$\gamma = 1 + \frac{2}{5} = \frac{7}{5}$$.
So D matches List-II item II.

Collecting the correspondences:
A - III, B - IV, C - I, D - II

This is exactly Option A.

Final answer: Option A.

Using Newton's law of cooling in the approximate form:

$$ \frac{\Delta T}{t} = k\left(\bar{T} - T_s\right) $$

where $$\bar{T}$$ is the average temperature and $$T_s$$ is the surrounding temperature.

First case: Cooling from 90° to 80° in time $$t$$:

$$\bar{T} = 85°$$, $$\Delta T = 10°$$

$$ \frac{10}{t} = k(85 - 20) = 65k \quad \cdots (1) $$

Second case: Cooling from 80° to 60° in time $$t'$$:

$$\bar{T} = 70°$$, $$\Delta T = 20°$$

$$ \frac{20}{t'} = k(70 - 20) = 50k \quad \cdots (2) $$

Dividing (2) by (1):

$$ \frac{20/t'}{10/t} = \frac{50k}{65k} $$

$$ \frac{2t}{t'} = \frac{10}{13} $$

$$ t' = \frac{2t \times 13}{10} = \frac{26t}{10} = \frac{13t}{5} $$

The correct answer is Option 4: $$\frac{13}{5}t$$.

The first law of thermodynamics gives the relation $$\Delta Q = \Delta U + \Delta W$$ where:
  • $$\Delta Q$$ is the heat supplied to the system,
  • $$\Delta U$$ is the change in internal energy,
  • $$\Delta W$$ is the work done by the system.

For each thermodynamic process we check which of these three quantities is zero or non-zero.

Case A: Isothermal process

Definition: Temperature remains constant.
For an ideal gas, internal energy depends only on temperature, so when temperature is constant $$\Delta U = 0$$.
Heat supplied equals work done, but both are generally non-zero.
Hence the correct description is $$\Delta U = 0$$, i.e. List-II item (IV).

Case B: Adiabatic process

Definition: No heat is exchanged with the surroundings.
Therefore $$\Delta Q = 0$$.
Neither $$\Delta U$$ nor $$\Delta W$$ is automatically zero; they are related by $$\Delta U = -\Delta W$$.
Thus the matching statement is $$\Delta Q = 0$$, i.e. List-II item (II).

Case C: Isobaric process

Definition: Pressure remains constant.
At constant pressure, the gas expands or compresses, so work is done: $$\Delta W \ne 0$$.
Temperature usually changes, so internal energy changes: $$\Delta U \ne 0$$.
Only $$\Delta U \ne 0$$ is listed among the four given statements, therefore we match with List-II item (III).

Case D: Isochoric process

Definition: Volume remains constant.
Work done for a volume change is $$\Delta W = P\,\Delta V$$. With $$\Delta V = 0$$ we get $$\Delta W = 0$$.
Heat added goes entirely into changing internal energy: $$\Delta Q = \Delta U$$, so neither of them is necessarily zero.
Thus the matching statement is $$\Delta W = 0$$, i.e. List-II item (I).

Collecting the matches:
(A) Isothermal → (IV) $$\Delta U = 0$$
(B) Adiabatic → (II) $$\Delta Q = 0$$
(C) Isobaric → (III) $$\Delta U \ne 0$$
(D) Isochoric → (I) $$\Delta W = 0$$

This corresponds to Option C.

Final Answer: Option C   [ (A)-(IV), (B)-(II), (C)-(III), (D)-(I) ]

The first law of thermodynamics for a quasi-static process is written as $$\Delta Q = \Delta U + \Delta W$$, where $$\Delta W$$ is the work done by the gas.

For expansion/compression work at constant external pressure, $$\Delta W = P\,\Delta V$$, so the first law becomes $$\Delta Q = \Delta U + P\,\Delta V$$ $$-(1)$$.

Now analyse each process in List-I.

Case A: Isobaric (constant pressure)

Pressure stays constant, therefore $$P = \text{constant}$$. Substituting in $$(1)$$ gives
$$\Delta Q = \Delta U + P\,\Delta V$$.
This is exactly the expression in List-II (IV). Hence A → IV.

Case B: Isochoric (constant volume)

Here $$\Delta V = 0$$, so no work is done: $$\Delta W = P\,\Delta V = 0$$.
Using the first law: $$\Delta Q = \Delta U + 0 \; \Rightarrow \; \Delta Q = \Delta U$$.
This matches List-II (II). Hence B → II.

Case C: Adiabatic (no heat exchange)

By definition, $$\Delta Q = 0$$ for an adiabatic process.
Thus the relation in List-II (III) fits. Hence C → III.

Case D: Isothermal (constant temperature)

For an ideal gas, internal energy depends only on temperature. Since temperature is constant,
$$\Delta U = 0$$.
First law then gives $$\Delta Q = \Delta W$$.
This is relation (I) in List-II. Hence D → I.

Collecting the matches: (A-IV), (B-II), (C-III), (D-I).

The option with this sequence is Option C.

Final Answer: Option C

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).

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.

For each physical quantity, first recall its definition, then write the factors on which it depends, and finally write the corresponding SI unit.

Case A - Heat capacity of a body
Definition: Heat required to raise the temperature of the entire body by $$1\;{\rm K}$$.
Heat $$Q$$ needed is proportional only to temperature change $$\Delta T$$: $$Q = C\,\Delta T$$ where $$C$$ is heat capacity.
Hence unit of $$C$$ is Joule per Kelvin, $$\rm J\,K^{-1}$$. Match: (A) $$\rightarrow$$ (II).

Case B - Specific heat capacity
Definition: Heat required to raise temperature of unit mass of a substance by $$1\;{\rm K}$$.
Formula: $$Q = m\,c\,\Delta T$$. Here $$c$$ relates heat to both mass and temperature difference, so its unit is Joule per kilogram per Kelvin, $$\rm J\,kg^{-1}\,K^{-1}$$. Match: (B) $$\rightarrow$$ (III).

Case C - Latent heat
Definition: Heat required to change the phase of unit mass of a substance without temperature change.
Formula: $$Q = m\,L$$, where $$L$$ is latent heat.
Only mass appears in the relation, so the unit is Joule per kilogram, $$\rm J\,kg^{-1}$$. Match: (C) $$\rightarrow$$ (I).

Case D - Thermal conductivity
Definition (Fourier’s law for one-dimensional steady conduction):$$\displaystyle \frac{dQ}{dt} = k\,A\,\frac{\Delta T}{\Delta x}$$.
Rearrange for $$k$$: $$k = \frac{1}{A}\,\frac{dQ}{dt}\,\frac{\Delta x}{\Delta T}$$.
Units:
$$\frac{dQ}{dt}$$ has unit $$\rm J\,s^{-1}$$ (power), $$A$$ is $$\rm m^{2}$$, $$\Delta x$$ is $$\rm m$$, $$\Delta T$$ is $$\rm K$$.
Thus $$k$$ has unit $$\dfrac{\rm J\,s^{-1}}{\rm m^{2}}\times \rm m \times \rm K^{-1}=J\,m^{-1}\,K^{-1}\,s^{-1}$$. Match: (D) $$\rightarrow$$ (IV).

Combining all matches:
(A)-(II), (B)-(III), (C)-(I), (D)-(IV).

The option with this sequence is Option D.

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 steady one-dimensional conduction through a rod, the heat current is
$$Q=\frac{K\,A}{L}\,\left(T_{\text{hot}}-T_{\text{cold}}\right)$$

When two rods are joined in series, the same heat current $$Q$$ flows through each of them in the steady state. Let $$T$$ be the interface temperature (at the junction of rods A and B).

Given ratios
$$\frac{L_A}{L_B}=\frac12 \;\Rightarrow\; L_A=\frac{L_B}{2}$$
$$\frac{r_A}{r_B}=2 \;\Rightarrow\; r_A=2r_B$$
$$\frac{K_A}{K_B}=\frac12 \;\Rightarrow\; K_A=\frac{K_B}{2}$$

Cross-sectional areas: $$A=\pi r^{2}$$, so
$$A_A=\pi r_A^{2}=\pi\,(2r_B)^{2}=4\pi r_B^{2}=4A_B$$
thus $$\frac{A_A}{A_B}=4$$.

Heat current through rod A:
$$Q=\frac{K_AA_A}{L_A}\,(400-T)$$

Heat current through rod B:
$$Q=\frac{K_BA_B}{L_B}\,(T-200)$$

Equate the two heat currents:

$$\frac{K_AA_A}{L_A}\,(400-T)=\frac{K_BA_B}{L_B}\,(T-200)$$

Substitute the ratio values:
$$K_A=\frac{K_B}{2},\;A_A=4A_B,\;L_A=\frac{L_B}{2}$$

$$\frac{\left(\tfrac{K_B}{2}\right)\,(4A_B)}{\tfrac{L_B}{2}}\,(400-T)=\frac{K_BA_B}{L_B}\,(T-200)$$

Simplify the fraction on the left:
$$\frac{(\tfrac{K_B}{2})\,(4A_B)}{\tfrac{L_B}{2}}=\frac{2K_BA_B}{\tfrac{L_B}{2}}=\frac{2K_BA_B\cdot2}{L_B}=\frac{4K_BA_B}{L_B}$$

Hence
$$\frac{4K_BA_B}{L_B}\,(400-T)=\frac{K_BA_B}{L_B}\,(T-200)$$

Cancel the common factor $$\frac{K_BA_B}{L_B}$$ from both sides:

$$4(400-T)=T-200$$

Solve for $$T$$:
$$1600-4T=T-200$$
$$1600+200=5T$$
$$1800=5T$$
$$T=360\,\text{K}$$

The temperature at the interface of the two rods is $$\mathbf{360\;K}$$.

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.

The power radiated from a hot surface is given by the Stefan-Boltzmann law
$$P = \varepsilon \,\sigma \,A \,T^{4}$$
where $$\varepsilon$$ is the emissivity, $$\sigma$$ is the Stefan-Boltzmann constant, $$A$$ is the surface area and $$T$$ is the absolute temperature.

Given data:
Length of wire, $$L = 10 \,\text{cm} = 0.10 \,\text{m}$$
Diameter of wire, $$d = 0.5 \,\text{mm} = 0.0005 \,\text{m}$$
Temperature, $$T = 1727^{\circ}\text{C} = 1727 + 273 = 2000 \,\text{K}$$
Power radiated, $$P = 94.2 \,\text{W}$$
Stefan-Boltzmann constant, $$\sigma = 6.0 \times 10^{-8} \,\text{W m}^{-2}\text{K}^{-4}$$

For a cylindrical wire, surface area (neglecting the ends)
$$A = \pi \,d\,L$$
Substituting the values,
$$A = 3.14 \times 0.0005 \,\text{m} \times 0.10 \,\text{m}$$
$$A = 3.14 \times 5.0 \times 10^{-5} \,\text{m}^{2}$$
$$A = 1.57 \times 10^{-4} \,\text{m}^{2}$$

Fourth power of temperature:
$$T^{4} = (2000\,\text{K})^{4} = 2^{4} \times 10^{12} = 16 \times 10^{12} = 1.6 \times 10^{13}$$

Put these numbers into the Stefan-Boltzmann law and solve for $$\varepsilon$$:
$$\varepsilon = \frac{P}{\sigma\,A\,T^{4}}$$
$$\varepsilon = \frac{94.2}{\left(6.0 \times 10^{-8}\right)\!\left(1.57 \times 10^{-4}\right)\!\left(1.6 \times 10^{13}\right)}$$

First multiply the constants in the denominator:
$$6.0 \times 1.57 = 9.42$$
$$9.42 \times 1.6 = 15.072$$
Power of ten: $$10^{-8}\times 10^{-4}\times 10^{13}=10^{1}$$
Hence,
$$\sigma A T^{4} = 15.072 \times 10^{1} = 150.72$$

Therefore,
$$\varepsilon = \frac{94.2}{150.72} = 0.625$$

Expressing $$\varepsilon$$ as a fraction:
$$0.625 = \frac{5}{8}$$

The problem states $$\varepsilon = \frac{x}{8}$$, so
$$x = 5$$

Final answer: $$x = 5$$

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 $$3$$.

Treat it as two conductors $$k_1,k_2$$​ in parallel, connected in series with conductor $$k_3$$

Let area of conductors 1 and 2 be A.

Given for conductor 3:

$$A_3=2A$$

All lengths are same l.

Thermal resistance:

$$R=\frac{l}{kA}$$

For conductor 1:

$$R_1=\frac{l}{60A}$$

For conductor 2:

$$R_2=\frac{l}{120A}$$

These are in parallel:

$$\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{60A}{l}+\frac{120A}{l}=l60A​+l120A​$$

$$=\frac{180A}{l}$$

So

$$R_p=\frac{l}{180A}$$

Equivalent conductivity of left section:

$$k_{eq}=180$$

Now conductor 3:

$$R_3=\frac{l}{135(2A)}=\frac{l}{270A}$$

Suppose junction temperature is θ\thetaθ.

Heat flow through left section:

$$Q_1=\frac{(100-\theta)}{R_p}$$

$$=(100-\theta)\frac{180A}{l}$$

Heat flow through right section:

$$Q_2=\frac{\theta}{R_3}=\theta\frac{270A}{l}$$

Steady state:

$$Q_1=Q_2$$

$$180(100-\theta)=270\theta$$

$$18000=450\theta$$

$$\theta=40^{\circ}C$$

Using area of an ellipse on a $$V-P$$ diagram:

$$\vert{}W\vert{} = \pi a b = 3.14 \times 10^5 \times 10^{-4} = 31.4\text{ J}$$

$$W = 314 \times 10^{-1}\text{ J}$$

We need to find the ratio of specific heats at constant volume for a monoatomic gas and a diatomic (rigid) gas.

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

For a monoatomic gas, $$f = 3$$ (3 translational), so $$C_{v_1} = \frac{3}{2}R$$. For a diatomic gas (rigid, i.e., no vibrational modes), $$f = 5$$ (3 translational + 2 rotational), so $$C_{v_2} = \frac{5}{2}R$$.

Therefore, $$\frac{C_{v_1}}{C_{v_2}} = \frac{\frac{3}{2}R}{\frac{5}{2}R} = \frac{3}{5}$$.

The correct answer is Option (2): $$\frac{3}{5}$$.

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.

For a real gas, the relation between the molar specific heats is

$$C_P - C_V \;=\; T\Bigl(\frac{\partial P}{\partial T}\Bigr)_V \Bigl(\frac{\partial V}{\partial T}\Bigr)_P$$

This identity is obtained from the first & second laws together with exact-differential criteria, and it is valid for one mole of any substance.

The gas in the question obeys the equation of state

$$P\,V^2 \;=\; R\,T \;\;-(1)$$

Step 1: Find $$\Bigl(\dfrac{\partial P}{\partial T}\Bigr)_V$$.

From $$(1)$$: $$P \;=\; \dfrac{R\,T}{V^{2}}$$. Keeping $$V$$ constant and differentiating with respect to $$T$$:

$$\Bigl(\frac{\partial P}{\partial T}\Bigr)_V = \frac{R}{V^{2}} \;\;-(2)$$

Step 2: Find $$\Bigl(\dfrac{\partial V}{\partial T}\Bigr)_P$$.

Rewrite $$(1)$$ to express $$V$$ in terms of $$T$$ at fixed $$P$$:

$$V^{2} = \frac{R\,T}{P} \quad\Longrightarrow\quad 2V\,dV = \frac{R}{P}\,dT$$ so

$$\Bigl(\frac{\partial V}{\partial T}\Bigr)_P = \frac{R}{2\,P\,V} \;\;-(3)$$

Step 3: Compute $$C_P - C_V$$.

Substitute $$(2)$$ and $$(3)$$ into the heat-capacity relation:

$$\begin{aligned} C_P - C_V &= T\left(\frac{R}{V^{2}}\right)\left(\frac{R}{2\,P\,V}\right)\\[4pt] &= \frac{T\,R^{2}}{2\,P\,V^{3}} \;\;-(4) \end{aligned}$$

Step 4: Eliminate $$P$$ using the equation of state.

From $$(1)$$: $$P = \dfrac{R\,T}{V^{2}}$$. Insert this in $$(4)$$:

$$\begin{aligned} C_P - C_V &= \frac{T\,R^{2}}{2\,(\tfrac{R\,T}{V^{2}})\,V^{3}}\\[6pt] &= \frac{T\,R^{2}}{2\,R\,T\,V}\\[4pt] &= \frac{R}{2\,V} \end{aligned}$$

Therefore

$$C_P = C_V + \frac{R}{2\,V}$$

Among the given options this matches Option C.

Answer: Option C

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: $$\frac{2}{\sqrt{3}}Z$$.

The mass of the sample is $$m = 1 \text{ kg}$$ (given).

For a temperature-dependent specific heat capacity $$C = kT$$, the small amount of heat $$dQ$$ needed to raise the temperature by $$dT$$ is
$$dQ = m\,C\,dT = m\,(kT)\,dT = mkT\,dT.$$

Total heat supplied when the temperature rises from $$T_1$$ to $$T_2$$ is obtained by integration:
$$Q = \int_{T_1}^{T_2} mkT\,dT = mk \int_{T_1}^{T_2} T\,dT.$$

Evaluating the integral:
$$\int_{T_1}^{T_2} T\,dT = \left[\frac{T^{2}}{2}\right]_{T_1}^{T_2} = \frac{T_2^{2} - T_1^{2}}{2}.$$

Hence
$$Q = mk\,\frac{T_2^{2} - T_1^{2}}{2}.$$

The temperature limits must be in kelvin. Using the relation $$T(\text{K}) = T(^{\circ}\text{C}) + 273$$:
$$T_1 = -73^{\circ}\text{C} + 273 = 200 \text{ K},$$
$$T_2 = 27^{\circ}\text{C} + 273 = 300 \text{ K}.$$

Substituting $$m = 1 \text{ kg}$$, $$T_1 = 200 \text{ K}$$, $$T_2 = 300 \text{ K}$$:
$$Q = 1 \times k \times \frac{300^{2} - 200^{2}}{2}.$$

Calculate the bracketed term:
$$300^{2} - 200^{2} = 90000 - 40000 = 50000.$$

Therefore,
$$Q = k \times \frac{50000}{2} = 25000\,k.$$

The heat required is given in the problem as $$nk$$, so comparing,
$$n = 25000.$$

Final answer: 25000

The diagram (not shown here) is a rectangle in the $$P$$-$$T$$ plane with the numerical labels

$$J:(P_0,T_0),\;K:(P_0,3T_0),\;L:(2P_0,3T_0),\;M:(2P_0,T_0).$$

One mole of a monatomic ideal gas moves along the four sides JK, KL, LM, MJ in that order.

For one mole the ideal-gas equation is $$PV = RT$$ and the internal energy is $$U = \tfrac32 RT.$$ Work done by the gas in any step is $$W = \int P\,dV,$$ and the first law gives $$Q = \Delta U + W.$$ We treat every leg separately.

Temperatures: $$T_0 \rightarrow 3T_0 \;(\Delta T = 2T_0).$$
Change in internal energy $$\Delta U_{JK} = \tfrac32 R\,(2T_0) = 3RT_0.$$

Volume rises from $$V_J = \frac{RT_0}{P_0}$$ to $$V_K = \frac{3RT_0}{P_0}.$$ Work done $$W_{JK} = P_0\,(V_K-V_J) = P_0 \left(\frac{2RT_0}{P_0}\right)=2RT_0.$$

Pressure $$P_0 \rightarrow 2P_0,$$ so volume $$V_K = \frac{3RT_0}{P_0}$$ compresses to $$V_L = \frac{3RT_0}{2P_0} = \tfrac12 V_K.$$ Work done $$W_{KL}=RT\ln\!\left(\frac{V_L}{V_K}\right)=3RT_0\ln\!\left(\tfrac12\right)=-3RT_0\ln2.$$

Because the temperature is constant, $$\Delta U_{KL}=0,$$ hence $$Q_{KL}=W_{KL}=-3RT_0\ln2.$$

Temperature $$3T_0 \rightarrow T_0 \;(\Delta T = -2T_0).$$
Change in internal energy $$\Delta U_{LM}=\tfrac32 R(-2T_0)=-3RT_0.$$

Volumes: $$V_L=\frac{3RT_0}{2P_0},\;V_M=\frac{RT_0}{2P_0}$$ so $$\Delta V = -\frac{RT_0}{P_0}.$$ Work done $$W_{LM}=2P_0\left(-\frac{RT_0}{P_0}\right)=-2RT_0.$$

Pressure $$2P_0 \rightarrow P_0,$$ volume doubles: $$V_M=\frac{RT_0}{2P_0},\;V_J=\frac{RT_0}{P_0}=2V_M.$$ Work done $$W_{MJ}=RT_0\ln\!\left(\frac{V_J}{V_M}\right)=RT_0\ln2.$$

Again $$\Delta U_{MJ}=0$$ (temperature constant), so $$Q_{MJ}=W_{MJ}=RT_0\ln2.$$

$$W_{\text{cycle}} = 2RT_0 + (-3RT_0\ln2) + (-2RT_0) + RT_0\ln2$$ $$\phantom{W_{\text{cycle}}}=0 - 2RT_0\ln2 = -2RT_0\ln2.$$

Now we can match each quantity with List-II:

(P) Work in the entire cycle: $$-2RT_0\ln2$$ → (4)
(Q) $$\Delta U_{JK}=3RT_0$$ → (3)
(R) Heat in KL: $$Q_{KL}=-3RT_0\ln2$$ → (5)
(S) $$\Delta U_{MJ}=0$$ → (2)

Therefore the correct set of matches is P → 4; Q → 3; R → 5; S → 2, which is Option B.

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: $$2 : 1$$

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 $$20 \text{ K}$$.

For an adiabatic process,

$$PV^{\gamma}=\text{constant}$$

Also ideal gas law gives

PV=nRT

or

$$V=\frac{nRT}{P}$$

Substitute into adiabatic equation:

$$P\left(\frac{nRT}{P}\right)^{\gamma}=\text{constant}$$

$$P\cdot\frac{(nR)^{\gamma}T^{\gamma}}{P^{\gamma}}=\text{constant}$$

$$P^{1-\gamma}T^{\gamma}=\text{constant}$$

Given pressure is proportional to cube of absolute temperature,

$$P\propto T^3$$

So

$$P=kT^3$$

Comparing with

$$P^{1-\gamma}T^{\gamma}=\text{constant}$$

write

$$P\propto T^{\frac{\gamma}{\gamma-1}}$$

Thus,

$$=3\frac{\gamma}{\gamma-1}$$

Solving,

$$γ=3γ−3$$

$$2γ=3$$

$$γ=\frac{3}{2}$$

But for an adiabatic process so,

$$\gamma=\frac{C_P}{C_V}$$

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 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.

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,

$$f_r=3$$

In the graph, the lines have positive slope in the (log⁡V,log⁡P) plane, so they represent

$$\log P=n\log V$$

or

$$P=V^n$$

which can be written as

This is a polytropic process with index

m=−n

For a polytropic process, molar heat capacity is

Substituting m=−n,

$$C=C_V+\frac{R}{1+n}$$

For process A, slope is given by

$$\tan⁡θ=γ$$

so

$$n=γ$$

Hence

$$C_A=C_V+\frac{R}{1+\gamma}$$

For process BBB, angle is $$45^{\circ}$$, so

n=1

and

$$C_B=C_V+\frac{R}{2}$$

Also,

$$C_P=C_V+R$$

Now comparing:

$$C_P>C_B$$

because

$$R>\frac{R}{2}$$

Also since γ>1,

$$\frac{R}{2}>\frac{R}{1+\gamma}$$

so

$$C_B>C_A$$

and clearly

$$C_A>C_V$$

Therefore,

$$CP>CB>CA>CV$$

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.

We need to evaluate two statements about thermodynamics.

Statement I: "If heat is added to a system, its temperature must increase."

This is FALSE. Consider the following counterexamples:

- During a phase change (e.g., melting of ice at 0°C), heat is added but the temperature remains constant.

- In an isothermal expansion of an ideal gas, heat is added to do work, but the temperature stays constant.

Statement II: "If positive work is done by a system in a thermodynamic process, its volume must increase."

This is FALSE. Consider these counterexamples:

- In a cyclic process, the system can perform net positive work while the volume returns to its original value.

- A system may do non-expansion (non-PdV) work, such as electrical work, without any change in volume.

Since both statements are false, the correct answer is Option B: Both Statement I and Statement II are false.

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: $$\frac{25}{21}\,.$$

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 $$U_5 \lt U_7$$

A body cools from 60°C to 40°C in 6 minutes. Temperature of surroundings is 10°C. Find the temperature after the next 6 minutes.

According to Newton's Law of Cooling: $$T(t) - T_s = (T_0 - T_s)e^{-kt}$$ where $$T_s = 10°$$C is the surrounding temperature.

At $$t=6$$ min, $$T=40°$$C, $$T_0=60°$$C, so $$40-10=(60-10)e^{-6k}$$, giving $$30=50e^{-6k}$$ and hence $$e^{-6k}=\frac{3}{5}$$.

At $$t=12$$ min from the start, the temperature satisfies $$T-10=50e^{-12k}=50(e^{-6k})^2=50\left(\frac{3}{5}\right)^2=18$$, so $$T=28°\text{C}$$.

Verification using the interval method: Alternatively, treating the second interval separately with initial temperature 40°C: $$T-10=(40-10)e^{-6k}=30\cdot\frac{3}{5}=18$$, hence $$T=28°\text{C}$$ ✓

Answer: 28°C

For an ideal gas undergoing an adiabatic process, the relation between temperature and volume is
$$T\,V^{\gamma-1}= \text{constant} \quad -(1)$$

Case 1: Adiabatic expansion
Initial state $$\left(T_A,V_0\right)$$ → final state $$\left(T_f,5V_0\right)$$.
Using $$(1):$$
$$T_A\,V_0^{\gamma-1}=T_f\,(5V_0)^{\gamma-1}$$
$$\Longrightarrow\; T_A=T_f\,5^{\gamma-1} \quad -(2)$$

Case 2: Isothermal expansion
Initial state $$\left(T_B,V_0\right)$$ → final state $$\left(T_f,5V_0\right)$$.
For an isothermal process, temperature is constant, so
$$T_B=T_f \quad -(3)$$

Dividing $$(2)$$ by $$(3):$$
$$\frac{T_A}{T_B}= \frac{T_f\,5^{\gamma-1}}{T_f}=5^{\gamma-1}$$

Hence the required ratio is $$T_A/T_B = 5^{\gamma-1}$$.

Option A which is: $$5^{\gamma-1}$$

Two plates A and B are in series (thermal contact). In steady state, the heat flow rate through both is equal.

Let $$T$$ be the temperature at the contact surface. Both plates have the same area $$A$$ and thickness $$L$$.

Heat flow: $$\frac{K_A A (100 - T)}{L} = \frac{K_B A (T - 0)}{L}$$

$$K_A(100 - T) = K_B \times T$$

$$84(100 - T) = 126T$$

$$8400 - 84T = 126T$$

$$8400 = 210T$$

$$T = \frac{8400}{210} = 40°\text{C}$$

The temperature at the contact surface is $$\mathbf{40}$$°C.

A hole of diameter 5 cm in a metal sheet is heated from 27 degrees C to 177 degrees C, and we wish to find the change in diameter. When a metal is heated, all linear dimensions expand uniformly (including holes). The change in a linear dimension is given by $$ \Delta L = L \alpha \Delta T $$ where $$\alpha$$ is the coefficient of linear expansion.

Here, $$D = 5$$ cm, $$\alpha = 1.6 \times 10^{-5}$$ /degrees C, and $$\Delta T = 177 - 27 = 150$$ degrees C. Substituting these values yields $$ \Delta D = 5 \times 1.6 \times 10^{-5} \times 150 = 5 \times 24 \times 10^{-4} = 12 \times 10^{-3}\;\text{cm}$$. Therefore $$d = 12$$. The answer is 12.

We need to find $$d$$ where the change in diameter is $$d \times 10^{-3}$$ cm when temperature changes from 27°C to 177°C.

Given data.

Initial diameter $$D_0 = 5$$ cm, $$\alpha = 1.6 \times 10^{-5}$$ /°C, $$\Delta T = 177 - 27 = 150$$°C.

Apply linear expansion.

When a metal sheet with a hole is heated, the hole expands as if it were filled with the same metal.

$$\Delta D = D_0 \times \alpha \times \Delta T = 5 \times 1.6 \times 10^{-5} \times 150$$

$$= 5 \times 2.4 \times 10^{-3} = 12 \times 10^{-3}$$ cm

Conclusion.

$$d = 12$$.

Therefore, the answer is 12.

Power = 2000 W, efficiency = 70%, mass = 2 kg, $$\Delta T = 50°C$$, specific heat = 4200 J/(kg·K).

Heat required: $$Q = mc\Delta T = 2 \times 4200 \times 50 = 420000$$ J

Effective power: $$P_{eff} = 0.7 \times 2000 = 1400$$ W

Time: $$t = \frac{Q}{P_{eff}} = \frac{420000}{1400} = 300$$ s

The answer is $$\boxed{300}$$.

For an ideal gas kept at constant pressure, the work done is related to the temperature rise as
$$W = P\Delta V = nR\Delta T$$

The mixture contains
- 2 moles of monatomic gas with $$\gamma_1 = \dfrac{5}{3}$$,
- 1 mole of diatomic gas with $$\gamma_2 = \dfrac{7}{5}$$.
Total moles
$$n = 2 + 1 = 3$$

Given work done
$$W = 66 \text{ J}$$, hence from the relation above
$$\Delta T = \frac{W}{nR} = \frac{66}{3R} = \frac{22}{R}$$ $$-(1)$$

For each component we first evaluate its molar specific heat at constant volume $$C_v$$.
The relation between $$\gamma$$, $$C_p$$ and $$C_v$$ is
$$\gamma = \frac{C_p}{C_v}, \qquad C_p - C_v = R$$ $$-(2)$$

Monatomic gas: $$\gamma_1 = \dfrac{5}{3}$$.
Using $$(2)$$,
$$C_{v1} = \frac{R}{\gamma_1 - 1} = \frac{R}{\frac{5}{3} - 1} = \frac{R}{\frac{2}{3}} = \frac{3}{2}R$$

Diatomic gas: $$\gamma_2 = \dfrac{7}{5}$$.
Similarly,
$$C_{v2} = \frac{R}{\gamma_2 - 1} = \frac{R}{\frac{7}{5} - 1} = \frac{R}{\frac{2}{5}} = \frac{5}{2}R$$

The change in internal energy of the mixture is the sum over all moles:
$$\Delta U = n_1 C_{v1}\Delta T + n_2 C_{v2}\Delta T$$
$$\Delta U = 2\left(\frac{3}{2}R\right)\Delta T \;+\, 1\left(\frac{5}{2}R\right)\Delta T$$
$$\Delta U = \left(3R + \frac{5}{2}R\right)\Delta T = \frac{11}{2}R\Delta T$$ $$-(3)$$

Substitute $$\Delta T$$ from $$(1)$$ into $$(3)$$:
$$\Delta U = \frac{11}{2}R \left(\frac{22}{R}\right) = \frac{11 \times 22}{2} = 121 \text{ J}$$

Hence, the change in internal energy of the gas mixture is 121 Joule.

Total heat absorbed in a multi-step process is the sum of heat transfers, where $$Q_{A \rightarrow B} = 0$$ for an adiabatic process and $$Q_{B \rightarrow C} = nRT\ln\left(\frac{V_f}{V_i}\right)$$ for an isothermal process.

Given: $$n = 1$$, $$V_1 = 10\text{ m}^3$$, $$T_1 = 600\text{ K}$$, $$T_2 = 60\text{ K}$$, $$C_{p,m} = \frac{5}{2}R \implies \gamma = \frac{5}{3}$$

Along adiabatic process $$A \rightarrow B$$:

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

$$600 \times 10^{2/3} = 60 \times V_2^{2/3} \implies 10 \times 10^{2/3} = V_2^{2/3} \implies V_2 = 10^{5/2}\text{ m}^3$$

$$Q_{A \rightarrow B} = 0$$

Along isothermal process $$B \rightarrow C$$:

$$Q_{B \rightarrow C} = nRT_2\ln\left(\frac{V_3}{V_2}\right) = RT_2\ln\left(\frac{V_3}{10^{5/2}}\right)$$

Equating to total heat absorbed:

$$RT_2\ln\left(\frac{V_3}{10^{5/2}}\right) = RT_2\ln 10 \implies \frac{V_3}{10^{5/2}} = 10 \implies V_3 = 10^{7/2}\text{ m}^3$$

$$\log V_3 = \log\left(10^{7/2}\right) = \frac{7}{2}$$

$$2\log V_3 = 2 \times \frac{7}{2} = 7$$

We use the first law of thermodynamics: $$\Delta U = Q - W$$.

The heat absorbed during vaporisation is:

$$Q = mL = 1 \times 2257 = 2257 \text{ kJ}$$

Now, the work done during expansion at constant pressure is:

$$W = P\Delta V = P(V_2 - V_1)$$

$$W = 1 \times 10^5 \times (1.671 - 1.00 \times 10^{-3})$$

$$W = 1 \times 10^5 \times 1.670 = 1.670 \times 10^5 \text{ J} = 167 \text{ kJ}$$

So the change in internal energy is:

$$\Delta U = Q - W = 2257 - 167 = 2090 \text{ kJ}$$

Hence, the answer is $$+2090$$ kJ.

The gas expands under a constant pressure of $$P = 3 \times 10^5$$ Pa and its volume increases by $$\Delta V = 1600$$ cm³, which equals $$1600 \times 10^{-6}$$ m³, with 10% of the heat supplied being used to perform expansion work.

The work done by the gas during expansion is given by the product of pressure and volume change:

$$W = P\Delta V = 3 \times 10^5 \times 1600 \times 10^{-6} = 480 \text{ J}$$

Since this expansion work corresponds to 10% of the total heat input $$Q$$, it follows that:

$$Q = \frac{480}{0.10} = 4800 \text{ J}$$

By applying the first law of thermodynamics, the increase in internal energy is:

$$\Delta U = Q - W = 4800 - 480 = 4320 \text{ J}$$

Therefore, the internal energy of the gas increases by 4320 J.

$$\frac{\theta_1 - \theta_2}{t} = K \left( \frac{\theta_1 + \theta_2}{2} - \theta_s \right)$$

Case 1: Cooling from $$80^\circ\text{C}$$ to $$60^\circ\text{C}$$

$$\frac{80 - 60}{300} = K \left( \frac{80 + 60}{2} - 20 \right)$$

$$\frac{20}{300} = K (70 - 20)$$

$$\frac{1}{15} = K(50) \implies K = \frac{1}{750}\text{ s}^{-1}$$

Case 2: Cooling from $$60^\circ\text{C}$$ to $$40^\circ\text{C}$$

$$\frac{60 - 40}{t_2} = \frac{1}{750} \left( \frac{60 + 40}{2} - 20 \right)$$

$$\frac{20}{t_2} = \frac{1}{750} (50 - 20)$$

$$\frac{20}{t_2} = \frac{1}{750} (30)$$

$$t_2 = 20 \times 25 = 500\text{ s}$$

$$\frac{dT}{dt} = -k(T - T_s)$$

For small temperature changes: $$\frac{T_1 - T_2}{t} = k \left( \frac{T_1 + T_2}{2} - T_s \right)$$

$$\frac{12}{2} = k(92 - 22)$$

$$6 = k(70) \implies k = \frac{6}{70} = \frac{3}{35}$$

$$\frac{6}{t'} = k(72 - 22)$$

$$\frac{6}{t'} = \frac{150}{35} = \frac{30}{7}$$

$$t' = \frac{6 \times 7}{30} = \frac{42}{30} = 1.4 \text{ minutes}$$

We need to identify which statement is NOT true for adiabatic compression of a gas. An adiabatic process is one in which no heat is exchanged between the system and its surroundings, meaning $$Q = 0$$. The First Law of Thermodynamics states:
$$\Delta U = Q - W$$

For an adiabatic process ($$Q = 0$$):
$$\Delta U = -W$$

Option 1: "There is no heat supplied to the system." This is TRUE. By definition, an adiabatic process has $$Q = 0$$, meaning no heat enters or leaves the system.

Option 2: "There is no change in the internal energy." This is NOT TRUE. From the First Law with $$Q = 0$$, we have $$\Delta U = -W$$. During compression, work is done ON the gas, so the gas does negative work ($$W < 0$$). Therefore:
$$\Delta U = -W = -(\text{negative value}) > 0$$
The internal energy INCREASES during adiabatic compression. The statement $$\Delta U = 0$$ would only hold if $$W = 0$$ (no work done), which contradicts the fact that the gas is being compressed. Note: $$\Delta U = 0$$ is characteristic of an isothermal process for an ideal gas, not an adiabatic one.

Option 3: "The temperature of the gas increases." This is TRUE. Since $$\Delta U > 0$$, and for an ideal gas the internal energy depends only on temperature via $$U = nC_vT$$, an increase in internal energy means an increase in temperature. This is why adiabatic compression heats up a gas, as seen in diesel engines where adiabatic compression raises the air temperature enough to ignite fuel.

Option 4: "The change in internal energy is equal to the work done on the gas." This is TRUE. From $$\Delta U = -W$$, and noting that work done ON the gas is $$W_{on} = -W$$, we get $$\Delta U = W_{on}$$. Since no heat escapes ($$Q = 0$$), all the work done on the gas goes entirely into increasing its internal energy.

The statement that is NOT true is Option 2.
The correct answer is Option 2.

Solution :

According to law of equipartition of energy, each degree of freedom contributes :

$$\frac{1}{2}R$$

to molar specific heat at constant volume.

For a diatomic gas :

• Translational degrees of freedom = 3

Contribution :

$$\frac{3}{2}R$$

• Rotational degrees of freedom = 2

Contribution :

$$R$$

• One vibrational mode contributes two degrees of freedom (kinetic + potential)

Contribution :

$$R$$

Therefore,

Total molar specific heat at constant volume :

$$C_V = \frac{3}{2}R + R + R$$

$$C_V = \frac{7}{2}R$$

Final Answer :

$$\frac{7}{2}R$$

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.

If $$dQ$$ and $$dW$$ represent the heat supplied to the system and the work done on the system respectively, then according to the first law of thermodynamics $$dQ = dU - dW$$. The first law states that the change in internal energy equals heat added plus work done on the system: $$dU = dQ + dW_{\text{on}}$$ where $$dW_{\text{on}}$$ is the work done on the system (IUPAC convention). Rearranging gives $$dQ = dU - dW_{\text{on}}$$, matching the assertion since $$dW$$ denotes the work done on the system. Assertion A is correct.

The first law of thermodynamics is indeed a statement of the law of conservation of energy applied to thermodynamic systems — energy can neither be created nor destroyed, only converted from one form to another. Reason R is correct.

The mathematical form $$dQ = dU - dW$$ follows directly from energy conservation: the heat supplied equals the change in internal energy minus the work done on the system, so R provides the fundamental basis for A. R is the correct explanation of A.

Both A and R are correct, and R is the correct explanation of A.

We are given an Assertion-Reason question about the efficiency of a reversible (Carnot) heat engine.

Assertion A: Efficiency of a reversible heat engine will be highest at $$-273°C$$ temperature of cold reservoir.

Reason R: The efficiency of a Carnot engine depends on both the hot and cold reservoir temperatures, given by $$\eta = 1 - \dfrac{T_2}{T_1}$$.

The efficiency of a Carnot engine is $$\eta = 1 - \dfrac{T_2}{T_1}$$, where $$T_2$$ and $$T_1$$ are the cold and hot reservoir temperatures in Kelvin. When the cold reservoir is at $$-273°C = 0$$ K, we get $$\eta = 1 - \dfrac{0}{T_1} = 1 = 100\%$$, which is the maximum possible efficiency. So Assertion A is true.

The Reason states the correct formula and correctly notes that efficiency depends on both reservoir temperatures. So Reason R is also true.

Now, since the formula in R directly explains why the efficiency is maximum when $$T_2 = 0$$ K (substituting gives $$\eta = 1$$), R is the correct explanation of A.

Hence, the correct answer is Option D.

Heat energy $$Q = 735$$ J given to a diatomic gas at constant pressure. The gas molecules rotate but do not oscillate.

Diatomic gas with rotation, no oscillation: $$f = 5$$ (3 translational + 2 rotational).

At constant pressure: $$Q = nC_p\Delta T$$ and $$\Delta U = nC_v\Delta T$$.

The answer is Option A: 525 J.

For an ideal gas, the internal energy depends only on temperature: $$U = nC_vT$$.

In an isothermal process, the temperature remains constant, and therefore the internal energy remains constant ($$\Delta U = 0$$).

By the first law of thermodynamics: $$\Delta Q = \Delta U + \Delta W = 0 + \Delta W = \Delta W$$

So in an isothermal process, all the heat supplied to the system is used to do work, while the internal energy stays constant.

Note: In an isochoric process, volume is constant ($$W = 0$$, but $$\Delta U$$ can change). In an adiabatic process, $$Q = 0$$ but $$U$$ changes. In an isobaric process, pressure is constant but $$U$$ can change.

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.

A body cools from $$60°\text{C}$$ to $$40°\text{C}$$ in $$7$$ minutes and the surrounding temperature remains at $$10°\text{C}$$, so we wish to determine its temperature after the next $$7$$ minutes.

According to Newton’s Law of Cooling in exponential form, the temperature at time $$t$$ satisfies the relation $$T(t) - T_s = (T_0 - T_s)\,e^{-kt}$$ where $$T_s = 10°\text{C}$$ is the ambient temperature.

By setting $$t = 7$$ minutes and using the initial drop from $$60°\text{C}$$ to $$40°\text{C}$$, we substitute into the formula to obtain $$40 - 10 = (60 - 10)\,e^{-7k}$$, which simplifies to $$30 = 50\cdot e^{-7k}$$ and hence $$e^{-7k} = \dfrac{3}{5}$$.

For the subsequent $$7$$ minutes, the temperature $$T$$ starting from $$40°\text{C}$$ satisfies $$T - 10 = (40 - 10)\,e^{-7k}$$. Substituting the value $$e^{-7k} = \dfrac{3}{5}$$ yields $$T - 10 = 30 \times \dfrac{3}{5} = 18$$, so $$T = 18 + 10 = 28°\text{C}$$.

Therefore, the temperature after the next $$7$$ minutes is 28°$$\text{C}$$.

A Carnot engine operates between two reservoirs with efficiency $$\frac{1}{3}$$. When the cold reservoir temperature is raised by $$x$$, the efficiency decreases to $$\frac{1}{6}$$. The hot reservoir temperature is $$99°$$C.
In Kelvin this gives $$T_H = 99 + 273 = 372$$ K.

The initial efficiency satisfies the relation $$\eta = 1 - \frac{T_C}{T_H}$$. Substituting $$\frac{1}{3} = 1 - \frac{T_C}{372}$$ leads to $$\frac{T_C}{372} = \frac{2}{3}$$ and hence $$T_C = 248$$ K.

After increasing the cold reservoir temperature by $$x$$, the efficiency becomes $$\frac{1}{6} = 1 - \frac{T_C + x}{372}$$. Rearranging gives $$\frac{T_C + x}{372} = \frac{5}{6}$$, so $$T_C + x = 310$$.

Solving for $$x$$ yields $$x = 310 - 248 = 62$$ K, so the answer is Option D: 62 K.

A Carnot engine has 50% efficiency at a source temperature of 600 K. Find the new source temperature needed to achieve 70% efficiency while keeping the sink temperature unchanged.

Find the sink temperature.

Carnot efficiency: $$\eta = 1 - \frac{T_{\text{cold}}}{T_{\text{hot}}}$$

$$0.50 = 1 - \frac{T_{\text{cold}}}{600}$$

$$T_{\text{cold}} = 600 \times 0.50 = 300 \text{ K}$$

Find the new source temperature.

$$0.70 = 1 - \frac{300}{T_{\text{new}}}$$

$$\frac{300}{T_{\text{new}}} = 0.30$$

$$T_{\text{new}} = \frac{300}{0.30} = 1000 \text{ K}$$

The correct answer is 1000 K.

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 need to find $$\gamma = \frac{C_p}{C_v}$$ for an adiabatic process.

Adiabatic process relation. For an adiabatic process: $$PV^\gamma = \text{constant}$$

Therefore: $$\frac{P_2}{P_1} = \left(\frac{V_1}{V_2}\right)^\gamma$$

Substitute given values: $$V_1 = 8$$ litres, $$V_2 = 27$$ litres, $$\frac{P_2}{P_1} = \frac{16}{81}$$

$$\frac{16}{81} = \left(\frac{8}{27}\right)^\gamma$$

Express both sides as powers: $$\frac{8}{27} = \left(\frac{2}{3}\right)^3$$ and $$\frac{16}{81} = \left(\frac{2}{3}\right)^4$$

$$\left(\frac{2}{3}\right)^4 = \left[\left(\frac{2}{3}\right)^3\right]^\gamma = \left(\frac{2}{3}\right)^{3\gamma}$$

Equate exponents: $$3\gamma = 4 \implies \gamma = \frac{4}{3}$$

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

We have a gas at temperature $$T$$ that is adiabatically expanded to double its volume, with $$\gamma = \frac{3}{2}$$.

For an adiabatic process, $$TV^{\gamma - 1} = \text{constant}$$. With $$\gamma - 1 = \frac{1}{2}$$:

$$T_1 V_1^{1/2} = T_2 V_2^{1/2}$$

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

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

Now the work done by the gas in an adiabatic process (for 1 mole) is

$$W = \frac{R(T_1 - T_2)}{\gamma - 1}$$

Substituting:

$$W = \frac{R\left(T - \frac{T}{\sqrt{2}}\right)}{\frac{1}{2}} = 2RT\left(1 - \frac{1}{\sqrt{2}}\right) = 2RT \cdot \frac{\sqrt{2} - 1}{\sqrt{2}}$$

Simplifying (by multiplying numerator and denominator by $$\sqrt{2}$$):

$$W = \frac{2RT(\sqrt{2} - 1) \cdot \sqrt{2}}{2} = RT(\sqrt{2})(\sqrt{2} - 1) = RT(2 - \sqrt{2})$$

Hence, the correct answer is $$W = RT(2 - \sqrt{2})$$.

Work Done = $$\text{Area of } \triangle CDE = \frac{1}{2} \times \text{Base} \times \text{Height}$$

$$W = \frac{1}{2} \times 2 \text{ m}^3 \times 300 \text{ Pa}$$

$$W = 300 \text{ J}$$

Since the cycle $$C \to D \to E \to C$$ is clockwise, the net work done is positive. 

The absolute temperatures for the boiling point ($$T_H$$) and freezing point ($$T_C$$) of water are:

$$T_H = 100^\circ\text{C} = 373\text{ K}$$

$$T_C = 0^\circ\text{C} = 273\text{ K}$$

Maximum limiting Carnot efficiency:   $$\eta_{\text{Carnot}} = 1 - \frac{273}{373} = \frac{100}{373} \approx 0.268 \text{ or } 26.8\%$$

Since any actual operating engine has an efficiency strictly lower than the Carnot limit ($$\eta < 26.8\%$$), its efficiency is automatically less than $$27\%$$.

Given: Both gases start at pressure P, volume V. Monoatomic gas ($$\gamma = 5/3$$).

Container A (isothermal compression):

$$P_A V_A = PV$$, with $$V_A = V/8$$:

$$P_A = 8P$$

Container B (adiabatic compression):

$$P_B V_B^\gamma = PV^\gamma$$, with $$V_B = V/8$$:

$$P_B (V/8)^{5/3} = PV^{5/3}$$

$$P_B = P \times 8^{5/3} = P \times (2^3)^{5/3} = P \times 2^5 = 32P$$

Ratio:

$$\frac{P_B}{P_A} = \frac{32P}{8P} = 4$$

The correct answer is Option 2: 4.

Statement I: At $$T_1 = -73°C = 200$$ K and $$T_2 = 527°C = 800$$ K.

Since $$v_{rms} \propto \sqrt{T}$$: $$\frac{v_2}{v_1} = \sqrt{\frac{800}{200}} = \sqrt{4} = 2$$

The rms speed is doubled. TRUE.

Statement II: PV = nRT and translational KE = $$\frac{3}{2}nRT$$.

So $$PV = \frac{2}{3} \times KE_{translational}$$, NOT equal to KE.

FALSE.

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

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.

Solution :

Given :

Mass of ice,

$$m = 600\text{ g} = 0.6\text{ kg}$$

Initial temperature,

$$-12^\circ\text{C}$$

Heat supplied,

$$Q = 184\text{ kJ}$$

Specific heat of ice,

$$s = 2222.3\text{ J kg}^{-1}\ ^\circ\text{C}^{-1}$$

Latent heat of fusion,

$$L = 336\text{ kJ kg}^{-1}$$

Heat required to raise temperature of ice from $$-12^\circ\text{C}$$ to $$0^\circ\text{C}$$ :

$$Q_1 = ms\Delta T$$

$$= 0.6 \times 2222.3 \times 12$$

$$\approx 16000\text{ J}$$

$$= 16\text{ kJ}$$

Remaining heat :

$$Q_2 = 184 - 16$$

$$= 168\text{ kJ}$$

Heat required to completely melt the ice :

$$Q_m = mL$$

$$= 0.6 \times 336$$

$$= 201.6\text{ kJ}$$

Since available heat $$168\text{ kJ}$$ is less than $$201.6\text{ kJ}$$, all the ice will not melt.

Mass of ice melted :

$$m' = \frac{168}{336}$$

$$= 0.5\text{ kg}$$

Remaining ice :

$$0.6 - 0.5 = 0.1\text{ kg}$$

Water formed :

$$0.5\text{ kg}$$

Therefore,

Ice : Water

$$= 0.1 : 0.5$$

$$= 1 : 5$$

Also, final temperature remains $$0^\circ\text{C}$$.

Hence,

(A) Correct

(D) Correct

Final Answer :

A and D

We need to find the final pressure when an ideal gas is suddenly compressed from volume $$V_0$$ to $$V_0/4$$.

First, identify the type of process.

The key word is "suddenly" compressed. A sudden compression means the process happens so quickly that there is no time for heat exchange between the gas and its surroundings. This is an adiabatic process ($$Q = 0$$).

Now state the adiabatic relation for an ideal gas.

For a reversible adiabatic process involving an ideal gas, pressure and volume are related by:

$$ PV^\gamma = \text{constant} $$

where $$\gamma = C_p/C_v$$ is the ratio of specific heats. This relation is derived from the first law of thermodynamics ($$dU = -PdV$$ for adiabatic process) combined with the ideal gas law and the relation $$dU = nC_v dT$$.

Now apply the adiabatic relation to initial and final states.

Initial state: pressure $$P_0$$, volume $$V_0$$

Final state: pressure $$P_f$$, volume $$V_0/4$$

Using $$P_i V_i^\gamma = P_f V_f^\gamma$$:

$$ P_0 V_0^\gamma = P_f \left(\frac{V_0}{4}\right)^\gamma $$

Next, solve for $$P_f$$.

$$ P_f = P_0 \cdot \frac{V_0^\gamma}{\left(\frac{V_0}{4}\right)^\gamma} = P_0 \cdot \frac{V_0^\gamma}{\frac{V_0^\gamma}{4^\gamma}} = P_0 \cdot 4^\gamma $$

Therefore, the final pressure is:

$$ P_f = P_0 (4)^\gamma $$

Note that since $$\gamma > 1$$ for all gases (for a monoatomic gas $$\gamma = 5/3$$, for diatomic $$\gamma = 7/5$$), the factor $$4^\gamma > 4$$. This means the pressure increase in an adiabatic compression is greater than what would occur in an isothermal compression (where $$P_f$$ would be simply $$4P_0$$). This is because during adiabatic compression, the temperature also rises (no heat is removed), which further increases the pressure.

The correct answer is Option 1: $$P_0(4)^\gamma$$.

Solution :

From first law of thermodynamics :

$$\Delta Q = \Delta U + W$$

Given, no heat is supplied or extracted :

$$\Delta Q = 0$$

Therefore,

$$\Delta U = -W$$

From graph, process occurs from A to B.

Coordinates :

At \(A\),

$$P_A = 10\text{ kPa}$$

$$V_A = 200\text{ cc}$$

At \(B\),

$$P_B = 50\text{ kPa}$$

$$V_B = 50\text{ cc}$$

Work done by gas is area under \(PV\) graph :

$$W = \frac{(P_A+P_B)}{2}(V_B-V_A)$$

$$= \frac{(10+50)}{2}(50-200)$$

$$= 30 \times (-150)$$

$$= -4500\text{ kPa-cc}$$

Now,

$$1\text{ kPa-cc} = 10^{-3}\text{ J}$$

Therefore,

$$W = -4500 \times 10^{-3}$$

$$= -4.5\text{ J}$$

Hence,

$$\Delta U = -(-4.5)$$

$$= 4.5\text{ J}$$

Final Answer :

$$4.5\text{ J}$$

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:

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

Given: Under isothermal conditions, $$P = aV^{-3}$$ where $$a$$ is a constant.

The bulk modulus at constant temperature is defined as:

$$B = -V \frac{dP}{dV}$$

Differentiate P with respect to V.

$$\frac{dP}{dV} = a \cdot (-3) \cdot V^{-4} = -3aV^{-4}$$

Compute the bulk modulus.

$$B = -V \cdot (-3aV^{-4}) = 3aV^{-3} = 3P$$

Therefore, the correct answer is Option A: $$\mathbf{3P}$$.

We have a Carnot engine operating between a hot reservoir at $$T_H = 127°C = 400$$ K and a cold reservoir at $$T_C = 27°C = 300$$ K, producing work $$W = 2$$ kJ.

The efficiency of the Carnot engine is:

$$\eta = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{400} = 1 - 0.75 = 0.25$$

Now, since $$\eta = \frac{W}{Q_H}$$, the heat absorbed from the hot reservoir is:

$$Q_H = \frac{W}{\eta} = \frac{2}{0.25} = 8 \text{ kJ}$$

Hence, the correct answer is Option 1.

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.

By the first law of thermodynamics, the rate of heat supply equals the rate of change of internal energy plus the rate of work done by the system:

$$\frac{dQ}{dt} = \frac{dU}{dt} + \frac{dW}{dt}$$

We are given $$\frac{dQ}{dt} = 1000\;\text{W}$$ and $$\frac{dW}{dt} = 200\;\text{W}$$. Substituting:

$$1000 = \frac{dU}{dt} + 200$$

So $$\frac{dU}{dt} = 1000 - 200 = 800\;\text{W}$$. Hence, the rate at which internal energy increases is 800 W.

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 $$28$$.

We need to identify the correct statements about heat given to an ideal gas in an isothermal process.

In an isothermal process:

Temperature remains constant, i.e., $$\Delta T = 0$$.

Statement A: Internal energy will decrease.

For an ideal gas, internal energy depends only on temperature: $$U = nC_vT$$. Since $$\Delta T = 0$$, we have $$\Delta U = 0$$. Statement A is INCORRECT.

Statement B: Internal energy will increase.

Since $$\Delta U = 0$$, Statement B is INCORRECT.

Statement C: Internal energy will not change.

Since $$\Delta U = 0$$, Statement C is CORRECT. ✓

Statement D: The gas will do positive work.

By first law: $$Q = \Delta U + W$$. Since $$\Delta U = 0$$ and $$Q > 0$$ (heat is given), we get $$W = Q > 0$$.

The gas does positive work (it expands). Statement D is CORRECT. ✓

Statement E: The gas will do negative work.

Since $$W > 0$$, Statement E is INCORRECT.

The correct answer is Option 4: C and D only.

For an isothermal process involving an ideal gas, we have $$PV = nRT = \text{constant}$$.

This gives the relation $$P = \frac{nRT}{V}$$, which is a rectangular hyperbola in the P-V plane.

Key properties of isothermal curves (isotherms) on a P-V diagram:

1. Each isotherm is a rectangular hyperbola.

2. For a higher temperature, the hyperbola is farther from the origin.

3. At any given volume, a higher temperature gives a higher pressure (since $$P = \frac{nRT}{V}$$).

Since $$T_3 > T_2 > T_1$$, the isotherm for $$T_3$$ should be the outermost curve (farthest from origin), $$T_2$$ in the middle, and $$T_1$$ closest to the origin.

The correct graph shows rectangular hyperbolas that do not intersect, with $$T_3$$ being the highest curve.

The answer is Option A.

Matching thermodynamic processes with their characteristics:

A. Isothermal Process: Temperature is constant, so internal energy doesn't change (for ideal gas). ΔU = 0, Q = W → II. No change in internal energy

B. Adiabatic Process: No heat exchange (Q = 0), so W = -ΔU. Work done by gas decreases internal energy → I. Work done by the gas decreases internal energy

C. Isochoric Process: Constant volume, so no work done (W = 0). All heat goes to change internal energy → IV. No work is done on or by the gas

D. Isobaric Process: Constant pressure, heat absorbed goes partly to increase internal energy and partly to do work (Q = ΔU + W) → III

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

The correct answer is Option 2.

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): $$\boxed{\gamma \propto T^0}$$.

The mean free path is given by:

$$\lambda = \frac{k_BT}{\sqrt{2}\pi d^2 P}$$

At constant pressure, $$\lambda \propto T$$.

At STP (273 K): $$\lambda_1 = 1500d$$

At 373 K (same pressure):

$$\lambda_2 = \lambda_1 \times \frac{T_2}{T_1} = 1500d \times \frac{373}{273}$$

$$\lambda_2 = 1500d \times 1.366 \approx 2049d$$

This matches option 3: 2049d.

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.