JEE Thermodynamics Questions

$$\eta_{\text{net}} = \frac{211}{243}$$

Using individual Carnot engine relations: $$Q_1 = Q_0(1-\eta)$$

$$Q_2 = Q_1(1-\eta) = Q_0(1-\eta)^2$$

$$Q_5 = Q_0(1-\eta)^5$$

From net efficiency definition: $$\eta_{\text{net}} = 1 - \frac{Q_5}{Q_0}$$

$$\frac{211}{243} = 1 - (1-\eta)^5$$

$$(1-\eta)^5 = 1 - \frac{211}{243} = \frac{32}{243}$$

$$1-\eta = \left(\frac{2^5}{3^5}\right)^{1/5} = \frac{2}{3}$$

$$\eta = 1 - \frac{2}{3} = \frac{1}{3} \approx 0.33$$

Using heat exchange relations:

$$dQ = -C_{v1} dT_1 = C_{p2} dT_2$$

$$dT_1 = -\frac{2 dQ}{3R},\quad dT_2 = \frac{2 dQ}{5R}$$

$$d(\Delta T) = dT_1 - dT_2 = -dQ\left(\frac{2}{3R} + \frac{2}{5R}\right) = -\frac{16}{15R} dQ$$

Substituting $$dQ = \frac{KA}{x}\Delta T dt$$:

$$\frac{d(\Delta T)}{\Delta T} = -\frac{16KA}{15xR} dt$$

$$\int_{\Delta T_0}^{\Delta T_0/2} \frac{d(\Delta T)}{\Delta T} = -\frac{16KA}{15xR} \int_{0}^{t} dt$$

$$\ln 2 = \frac{16KA}{15xR} t \implies t = \frac{15\ln 2}{16}\frac{xR}{KA}$$

$$n = \frac{15 \times 0.7}{16} = 0.65625 \approx 0.66$$

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

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:

$$(P_0,V_0)$$

For one complete cycle,

$$ΔU=0$$

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:

$$W_2=P\Delta V=\frac{P_0}{4}(V_0-4V_0)$$$$=-\frac{3}{4}P_0V_0$$

1. Isochoric heating

$$W_3=0$$

Net work:

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

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.

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.

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.

Mass of $$CO_{2}$$ given is 50 g. Molar mass of $$CO_{2}$$ is $$12 + 2 \times 16 = 44\;\mathrm{g\,mol^{-1}}$$.

Number of moles, $$n = \frac{\text{mass}}{\text{molar mass}} = \frac{50}{44} = 1.13636\;\mathrm{mol}\,.$$

Temperature, $$T = 17^\circ\mathrm{C} = 17 + 273 = 290\;\mathrm{K}\,.$$

Formula: For an ideal gas, total translational kinetic energy is given by $$E_{\mathrm{trans}} = \frac{3}{2}\,n\,R\,T\,.$$

Substituting values into $$E_{\mathrm{trans}}$$: $$E_{\mathrm{trans}} = \frac{3}{2} \times 1.13636 \times 8.314\;\mathrm{J\,mol^{-1}K^{-1}} \times 290\;\mathrm{K}\quad-(1)$$

Compute step by step:
$$1.13636 \times 8.314 = 9.449\quad(\text{approx})$$
$$9.449 \times 290 = 2740.21\quad(\text{approx})$$
$$\frac{3}{2} \times 2740.21 = 4110.315\quad(\text{approx})$$

The result is $$\approx 4.11\times10^3\;\mathrm{J}$$. The closest given option is Option B: $$4102.8\;\mathrm{J}\,.$$

Answer: Option B, $$4102.8\;\mathrm{J}\,.$$

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 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:

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

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

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.

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.

Case 1: JK (isobaric at $$P_0$$)

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 $$W_{JK} = P_0\,(V_K-V_J) = P_0 \left(\frac{2RT_0}{P_0}\right)=2RT_0.$$

Case 2: KL (isothermal at $$T = 3T_0$$)

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 $$Q_{KL}=W_{KL}=-3RT_0\ln2.$$

Case 3: LM (isobaric at $$2P_0$$)

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 $$W_{LM}=2P_0\left(-\frac{RT_0}{P_0}\right)=-2RT_0.$$

Case 4: MJ (isothermal at $$T=T_0$$)

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 $$Q_{MJ}=W_{MJ}=RT_0\ln2.$$

Total work for the cycle

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

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: $$2 : 1$$

For an isobaric process (constant pressure), the work done by the gas is given by:

$$W = P \Delta V$$

Using the ideal gas law, $$PV = nRT$$, we can express the work as:

$$W = P \Delta V = nR \Delta T$$

Given that the work done by the gas is 200 J:

$$nR \Delta T = 200 \text{ J} \quad \text{(Equation 1)}$$

The heat supplied to the gas in an isobaric process is:

$$Q = n C_p \Delta T$$

where $$C_p$$ is the molar specific heat at constant pressure.

For a diatomic gas with $$\gamma = 1.4$$, the ratio of specific heats is $$\gamma = \frac{C_p}{C_v} = \frac{7}{5}$$. The molar specific heat at constant volume is $$C_v = \frac{5}{2}R$$ (since diatomic gases have 5 degrees of freedom). Therefore:

$$C_p = \gamma C_v = \frac{7}{5} \times \frac{5}{2}R = \frac{7}{2}R$$

Substituting into the expression for Q:

$$Q = n \left( \frac{7}{2}R \right) \Delta T = \frac{7}{2} (nR \Delta T)$$

Using Equation 1, where $$nR \Delta T = 200 \text{ J}$$:

$$Q = \frac{7}{2} \times 200 = 7 \times 100 = 700 \text{ J}$$

Thus, the heat given to the gas is 700 J.

Verification using the first law of thermodynamics:

The change in internal energy is:

$$\Delta U = n C_v \Delta T = n \left( \frac{5}{2}R \right) \Delta T = \frac{5}{2} (nR \Delta T) = \frac{5}{2} \times 200 = 500 \text{ J}$$

From the first law, $$\Delta U = Q - W$$:

$$500 = Q - 200 \implies Q = 700 \text{ J}$$

This confirms the result.

The correct option is D. 700 J.

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,

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

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

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

Substituting: $$P\left(\frac{nRT}{P}\right)^\gamma = \text{const}$$

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

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

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

$$P \propto T^{\gamma/(\gamma-1)}$$

Given $$P \propto T^3$$:

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

$$\gamma = 3\gamma - 3$$

$$2\gamma = 3$$

$$\gamma = \frac{3}{2}$$

The answer is $$\frac{C_p}{C_v} = \frac{3}{2}$$, which corresponds to Option (2).

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.

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: $$\frac{3}{T}$$.

The rms speed and average speed of gas molecules are given by:

$$v_{rms} = \sqrt{\frac{3RT}{M}}, \quad v_{avg} = \sqrt{\frac{8RT}{\pi M}}$$

The ratio:

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

Given: $$v_{rms} = \left(1 + \frac{5}{x}\right)^{1/2} \cdot v_{avg}$$

Squaring both sides:

$$\frac{v_{rms}^2}{v_{avg}^2} = 1 + \frac{5}{x} = \frac{3\pi}{8}$$

$$\frac{5}{x} = \frac{3\pi}{8} - 1 = \frac{3\pi - 8}{8}$$

$$x = \frac{40}{3\pi - 8}$$

Using $$\pi = \frac{22}{7}$$:

$$3\pi - 8 = \frac{66}{7} - \frac{56}{7} = \frac{10}{7}$$

$$x = \frac{40}{\frac{10}{7}} = \frac{40 \times 7}{10} = 28$$

Therefore, the correct answer is Option C: $$\mathbf{28}$$.

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.

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$$^{-1}$$.

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.

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:

$$v_{rms} = \sqrt{\frac{3RT}{M_{mix}}}, \quad v_{sound} = \sqrt{\frac{\gamma_{mix} RT}{M_{mix}}}$$

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

$$\frac{3RT}{M_{mix}} = 2 \times \frac{\gamma_{mix} RT}{M_{mix}}$$

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

$$3 = 2\gamma_{mix}$$

$$\gamma_{mix} = \frac{3}{2}$$

Determine the specific heats of the individual gases.

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

$$C_{v,He} = \frac{3}{2}R, \quad C_{p,He} = \frac{5}{2}R$$

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

$$C_{v,H_2} = \frac{5}{2}R, \quad C_{p,H_2} = \frac{7}{2}R$$

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:

$$C_{v,mix} = \frac{n_{He} \cdot C_{v,He} + n_{H_2} \cdot C_{v,H_2}}{n_{He} + n_{H_2}} = \frac{2 \times \frac{3}{2}R + n \times \frac{5}{2}R}{2 + n} = \frac{3R + \frac{5nR}{2}}{2 + n} = \frac{R(6 + 5n)}{2(2 + n)}$$

$$C_{p,mix} = \frac{n_{He} \cdot C_{p,He} + n_{H_2} \cdot C_{p,H_2}}{n_{He} + n_{H_2}} = \frac{2 \times \frac{5}{2}R + n \times \frac{7}{2}R}{2 + n} = \frac{5R + \frac{7nR}{2}}{2 + n} = \frac{R(10 + 7n)}{2(2 + n)}$$

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

$$\gamma_{mix} = \frac{C_{p,mix}}{C_{v,mix}} = \frac{10 + 7n}{6 + 5n} = \frac{3}{2}$$

Cross-multiplying:

$$2(10 + 7n) = 3(6 + 5n)$$

$$20 + 14n = 18 + 15n$$

$$20 - 18 = 15n - 14n$$

$$n = 2$$

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 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 minimum temperature of the hot reservoir for a heat engine.

Since the cold reservoir temperature is $$T_C = 324$$ K, the heat absorbed from the hot reservoir is $$Q_H = 300$$ J, and the heat delivered to the cold reservoir is $$Q_C = 180$$ J, the engine must operate at maximum (Carnot) efficiency to achieve the minimum hot reservoir temperature.

For a Carnot engine, the relationship between heat exchanged and reservoir temperatures is given by

$$\frac{Q_C}{Q_H} = \frac{T_C}{T_H}$$

Substituting the given values into this expression, we have

$$\frac{180}{300} = \frac{324}{T_H}$$

From this,

$$\frac{3}{5} = \frac{324}{T_H}$$

This gives

$$T_H = \frac{324 \times 5}{3}$$

Next, simplifying the right-hand side yields

$$T_H = \frac{1620}{3} = 540$$ K

The minimum temperature of the hot reservoir is 540 K.

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.

Work done in an adiabatic process is given by $$W = \frac{P_A V_A - P_B V_B}{\gamma - 1}$$, and for an isothermal process it is given by $$W = P_B V_B \ln\left(\frac{V_C}{V_B}\right)$$.

Given: $$P_A = 100\text{ kPa}$$, $$V_A = 0.80\text{ m}^3$$, $$P_B = 300\text{ kPa}$$, $$\gamma = \frac{5}{3}$$, $$\left(\frac{1}{3}\right)^{0.6} = 0.5$$, $$\ln 2 \approx 0.7$$

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

$$P_A V_A^\gamma = P_B V_B^\gamma \implies V_B = V_A \left(\frac{P_A}{P_B}\right)^{1/\gamma} = 0.80 \left(\frac{1}{3}\right)^{0.6} = 0.80 \times 0.5 = 0.40\text{ m}^3$$

$$W_{A \rightarrow B} = \frac{100 \times 0.80 - 300 \times 0.40}{\frac{5}{3} - 1} = \frac{80 - 120}{\frac{2}{3}} = -60\text{ kJ} \implies \vert{}W_{A \rightarrow B}\vert{} = 60\text{ kJ}$$

Along isothermal expansion $$B \rightarrow C$$ (where $$V_C = V_A = 0.80\text{ m}^3$$):

$$W_{B \rightarrow C} = P_B V_B \ln\left(\frac{V_C}{V_B}\right) = 300 \times 0.40 \ln\left(\frac{0.80}{0.40}\right) = 120 \ln 2 = 120 \times 0.7 = 84\text{ kJ}$$

Along isochoric path $$C \rightarrow A$$:

$$W_{C \rightarrow A} = 0$$

Total work done $$A \rightarrow B \rightarrow C$$:

$$W_{\text{total}} = W_{A \rightarrow B} + W_{B \rightarrow C} = -60 + 84 = 24\text{ kJ}$$

Process I - Water ⟶ Steam at 100 °C, constant pressure $$P = 10^{5}\;{\rm Pa}$$

Heat supplied equals the enthalpy change: $$Q = \Delta H = mL$$.
Given mass $$m = 10^{-3}\;{\rm kg}$$ and latent heat $$L = 2250\;{\rm kJ\,kg^{-1}}$$,

$$\Delta H = 10^{-3}\times 2250\;{\rm kJ}=2.25\;{\rm kJ}$$

The work done on the surroundings is $$W = P\Delta V$$ with
$$\Delta V = 10^{-3}-10^{-6}=9.99\times 10^{-4}\;{\rm m^{3}}$$, hence

$$W = 10^{5}\times 9.99\times 10^{-4}\;{\rm J}\approx 1.0\times 10^{2}\;{\rm J}=0.10\;{\rm kJ}$$

By the first law, $$\Delta U = \Delta H - P\Delta V = 2.25-0.10 = 2.15\;{\rm kJ}\approx 2\;{\rm kJ}$$.
Thus I → (P).

Process II - 0.2 mol rigid diatomic gas, isobaric, $$V\;{\rm to}\;3V$$ at $$T_i = 500\;{\rm K}$$

At constant pressure, $$\dfrac{T_f}{T_i} = \dfrac{V_f}{V_i}=3\; \Rightarrow\; T_f = 1500\;{\rm K}$$, so $$\Delta T = 1000\;{\rm K}$$.

For a rigid diatomic gas, $$C_v = \dfrac{5}{2}R = 20\;{\rm J\,mol^{-1}K^{-1}}$$.
Hence

$$\Delta U = nC_v\Delta T = 0.2\times 20\times 1000 = 4000\;{\rm J}=4\;{\rm kJ}$$

Therefore II → (R).

Process III - 1 mol monatomic gas, adiabatic compression from $$V_i=\dfrac13\;{\rm m^{3}}$$, $$P_i = 2\;{\rm kPa}$$ to $$V_f = \dfrac{V_i}{8}$$

Initial temperature: $$T_i=\dfrac{P_iV_i}{R}= \dfrac{2000\times\dfrac13}{8}=83.3\;{\rm K}$$.

For a monatomic gas, $$\gamma=\dfrac53$$ and $$TV^{\gamma-1}={\rm const}$$.
Thus $$T_f = T_i\left(\dfrac{V_i}{V_f}\right)^{\gamma-1}=83.3\times 8^{2/3}=83.3\times 4 = 333\;{\rm K}$$.

$$\Delta T = 333-83.3 = 250\;{\rm K}$$, and $$C_v = \dfrac32 R = 12\;{\rm J\,mol^{-1}K^{-1}}$$, so

$$\Delta U = nC_v\Delta T = 1\times 12\times 250 = 3000\;{\rm J}=3\;{\rm kJ}$$

Hence III → (T).

Process IV - 3 mol vibrating diatomic gas, isobaric; heat supplied $$Q = 9\;{\rm kJ}$$

For a diatomic gas with vibrations active:
$$C_v = \dfrac{7}{2}R,\quad C_p=C_v+R=\dfrac{9}{2}R$$

At constant pressure, $$Q = nC_p\Delta T \Rightarrow \Delta T = \dfrac{Q}{nC_p}$$.
Internal-energy change: $$\Delta U = nC_v\Delta T = \dfrac{C_v}{C_p}\,Q = \dfrac{7/2}{9/2}\,Q = \dfrac79\,Q$$

$$\Delta U = \dfrac79 \times 9\;{\rm kJ}=7\;{\rm kJ}$$

Thus IV → (Q).

Collecting the matches:
I → P, II → R, III → T, IV → Q

Hence the correct option is Option C.

The total surface area of the cube is $$24$$ m$$^2$$.

A cube has 6 faces, so the area of each face is:

$$\frac{24}{6} = 4 \text{ m}^2$$

The side length of the cube is:

$$a = \sqrt{4} = 2 \text{ m}$$

The original volume of the cube is:

$$V = a^3 = 2^3 = 8 \text{ m}^3$$

The coefficient of volume expansion is:

$$\gamma = 3\alpha = 3 \times 5.0 \times 10^{-4} = 1.5 \times 10^{-3} \text{ °C}^{-1}$$

The increase in volume is:

$$\Delta V = V \gamma \Delta T = 8 \times 1.5 \times 10^{-3} \times 10$$

$$\Delta V = 0.12 \text{ m}^3$$

Converting to cm$$^3$$ ($$1 \text{ m}^3 = 10^6 \text{ cm}^3$$):

$$\Delta V = 0.12 \times 10^6 = 1.2 \times 10^5 \text{ cm}^3$$

Hence, the correct answer is Option B.

We need to find the temperature of the source of a Carnot engine. The efficiency of a Carnot engine is given by $$\eta = 1 - \frac{T_2}{T_1}$$ and it is stated that initially $$\eta = 50\% = 0.5$$. Therefore $$0.5 = 1 - \frac{T_2}{T_1}$$ implies $$\frac{T_2}{T_1} = 0.5$$ and hence $$T_2 = 0.5 \, T_1 \quad \cdots (1)$$.

When the sink temperature is reduced by 40°C (equal to 40 K), the efficiency increases by 30%. If this increase were additive, the new efficiency would be $$0.5 + 0.3 = 0.8$$ so that $$0.8 = 1 - \frac{T_2 - 40}{T_1}$$ giving $$\frac{T_2 - 40}{T_1} = 0.2$$. Substituting from (1) yields $$\frac{0.5 \, T_1 - 40}{T_1} = 0.2$$, leading to $$0.5 \, T_1 - 40 = 0.2 \, T_1$$ and hence $$0.3 \, T_1 = 40$$ so that $$T_1 = \frac{40}{0.3} = 133.3 \text{ K}$$, which does not match any of the given options.

Interpreting the 30% increase as a relative increase instead gives a new efficiency of $$0.5 \times 1.3 = 0.65$$. In that case $$0.65 = 1 - \frac{T_2 - 40}{T_1}$$ implies $$\frac{T_2 - 40}{T_1} = 0.35$$. Using $$T_2 = 0.5 \, T_1$$ again, we have $$\frac{0.5 \, T_1 - 40}{T_1} = 0.35$$, so $$0.5 \, T_1 - 40 = 0.35 \, T_1$$, giving $$0.15 \, T_1 = 40$$ and hence $$T_1 = \frac{40}{0.15} = 266.67 \text{ K} \approx 266.7 \text{ K}$$.

Therefore, the correct answer is Option C: 266.7 K.

We need to find the steady-state junction temperature T between a steel rod and a copper rod connected in series, with the steel end at 0°C and the copper end at 50°C.

Since the system is in steady state, the rate of heat flow through both rods is equal, so $$\frac{K_1 A_1 (T - 0)}{L_1} = \frac{K_2 A_2 (50 - T)}{L_2}$$ where subscript 1 refers to steel and subscript 2 refers to copper.

We are given that $$\frac{K_2}{K_1} = 9$$, $$\frac{A_1}{A_2} = 2$$, and $$\frac{L_1}{L_2} = 2$$.

Substituting these ratios into the heat balance equation gives $$\frac{K_1 A_1}{L_1} \cdot T = \frac{K_2 A_2}{L_2} \cdot (50 - T)\,.$$

From the above, the ratio of the coefficients is $$\frac{K_1 A_1 / L_1}{K_2 A_2 / L_2} = \frac{K_1}{K_2} \cdot \frac{A_1}{A_2} \cdot \frac{L_2}{L_1} = \frac{1}{9} \times 2 \times \frac{1}{2} = \frac{1}{9}\,.$$

Substituting this result into the previous relation yields $$\frac{1}{9} \cdot T = (50 - T)\,, $$ which can be written as $$T = 9(50 - T) = 450 - 9T\,, $$ leading to $$10T = 450$$ and hence $$T = 45°C\,. $$

Therefore, the correct answer is Option C: $$45°C$$.

We need to find the efficiency of a Carnot engine operating between the steam point and the ice point. The steam point corresponds to the hot reservoir at $$T_H = 100°C = 373$$ K, and the ice point corresponds to the cold reservoir at $$T_C = 0°C = 273$$ K.

The efficiency of a Carnot engine is given by $$\eta = 1 - \frac{T_C}{T_H}$$. Substituting the temperatures yields $$\eta = 1 - \frac{273}{373} = \frac{373 - 273}{373} = \frac{100}{373}$$.

Evaluating this result gives $$\eta = \frac{100}{373} = 0.2681 = 26.81\%$$. Therefore, 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 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.

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.

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 $$f = \frac{2}{\gamma - 1}$$.

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 $$25$$.

We have three points whose temperatures are fixed: $$A \;(200^{\circ}\text{C}),\; B \;(100^{\circ}\text{C}),\; D \;(125^{\circ}\text{C})$$. An identical rod $$AB$$ has total thermal resistance $$10\;\text{K W}^{-1}$$, and rod $$CD$$ also has thermal resistance $$10\;\text{K W}^{-1}$$. Because the connection to rod $$CD$$ is made at the mid-point $$C$$ of $$AB$$, rod $$AB$$ is effectively split into two equal halves: $$AC$$ and $$CB$$.

The resistance of a portion is directly proportional to its length. Since each half is half the full length, each half has half the resistance: $$R_{AC}=R_{CB}=\dfrac{10}{2}=5\;\text{K W}^{-1}.$$

Let the steady-state temperature of the junction $$C$$ be $$T\;({}^{\circ}\text{C}).$$ For any rod the heat current $$Q$$ is given by the basic conduction relation $$Q=\dfrac{\Delta T}{R},$$ where $$\Delta T$$ is the temperature difference across the rod and $$R$$ is its thermal resistance.

The heat current from $$A$$ to $$C$$ is therefore $$Q_{AC}=\dfrac{200-T}{5}.$$

The heat current from $$C$$ to $$B$$ is $$Q_{CB}=\dfrac{T-100}{5}.$$

The heat current from $$C$$ to $$D$$ is $$Q_{CD}=\dfrac{T-125}{10}.$$

At junction $$C$$ the system is in steady state, so the heat arriving at $$C$$ from $$A$$ must equal the heat leaving $$C$$ towards $$B$$ and $$D$$: $$Q_{AC}=Q_{CB}+Q_{CD}.$$ Substituting the expressions, we get $$\dfrac{200-T}{5}=\dfrac{T-100}{5}+\dfrac{T-125}{10}.$$

To clear denominators, multiply the entire equation by $$10$$: $$2(200-T)=2(T-100)+(T-125).$$

Simplifying each side, we obtain $$400-2T=2T-200+T-125.$$ Combining like terms on the right, $$400-2T=3T-325.$$

Now move all the $$T$$ terms to one side and the constants to the other: $$400+325=3T+2T.$$ $$725=5T.$$ So, $$T=\dfrac{725}{5}=145^{\circ}\text{C}.$$

The required heat current in rod $$CD$$ is $$Q_{CD}=\dfrac{T-125}{10}=\dfrac{145-125}{10}=\dfrac{20}{10}=2\;\text{W}.$$

Hence, the heat current in $$CD$$ is $$2\;\text{W}$$, so the value of $$P$$ is $$2$$.

So, the answer is $$2$$.

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 $$400$$.

$$\text{Given polytropic process equation: } P V^{1/2} = \text{constant}$$

$$\text{From the ideal gas equation: } P = \frac{nRT}{V} \implies P \propto \frac{T}{V}$$

$$\text{Substituting } P \text{ into the process equation: } \left(\frac{T}{V}\right) V^{1/2} = \text{constant} \implies T V^{-1/2} = \text{constant}$$

$$\text{Setting up the ratio between state 1 and state 2: } T_1 V_1^{-1/2} = T_2 V_2^{-1/2}$$

$$\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{-1/2} = \left(\frac{V_2}{V_1}\right)^{1/2}$$

$$\text{Given that } V_2 = 2V_1 \implies \frac{V_2}{V_1} = 2:$$ $$\frac{T_2}{T_1} = (2)^{1/2} = \sqrt{2}$$

For the complete cycle,

$$W_{\text{net}}=W_{AB}+W_{BC}+W_{CA}$$

For A→B, expansion is isothermal.

For one mole,

Since at A,

and

$$V_B=2V_1$$

so

$$W_{AB}=P_1V_1\ln2$$

For B→CB \to CB→C, volume is constant, so

$$W_{BC}=0$$

For C→A, process is adiabatic compression.

For adiabatic process,

Q=0

and

For compression from C to A,

$$W_{CA}=\frac{P_CV_C-P_AV_A}{\gamma-1}$$

Now

and

$$P_AV_A=P_1V_1$$

Thus

So net work done is

$$W_{\text{net}}=RT\ln2-\frac{P_1V_1}{2(\gamma-1)}$$

We need to determine what happens to a bimetallic strip made of metals $$A$$ and $$B$$ when it is placed in a cold bath, given that metal $$A$$ has a higher coefficient of linear expansion ($$\alpha_A > \alpha_B$$) than metal $$B$$.

1. Understanding Thermal Expansion and Contraction

The change in length ($$\Delta L$$) of a metallic rod due to a change in temperature ($$\Delta T$$) is governed by the formula:

$$\Delta L = L_0 \alpha \Delta T$$

Where:

• $$L_0$$ is the initial length.
• $$\alpha$$ is the coefficient of linear expansion.
• $$\Delta T$$ is the change in temperature.

2. Behavior in a Cold Bath (Cooling Phase)

When the strip is placed in a cold bath, the temperature decreases ($$\Delta T < 0$$), causing both metals to contract (shrink).

• Since metal $$A$$ has a higher coefficient of expansion ($$\alpha_A > \alpha_B$$), it is more sensitive to temperature variations.
• This means that upon cooling, metal $$A$$ will contract more (become shorter) than metal $$B$$.

3. Direction of Bending

Because the two metals are bonded together rigidly, they cannot slide past one another. To accommodate the unequal shortening while remaining joined:

• The metal that contracts more (metal $$A$$) must form the inner curve of the bend.
• The metal that contracts less (metal $$B$$) must form the outer curve of the bend.

Based on the standard mounting arrangement where metal $$A$$ is placed on the left and metal $$B$$ is placed on the right, metal $$A$$ shortening more forces the strip to curve inward toward its own side.

Therefore, the bimetallic strip will bend towards the left.

Correct Answer: Option D: Bend towards the left

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.

According to Newton’s Law of Cooling, the rate at which the temperature $$T$$ of a body changes is proportional to the excess temperature over that of the surroundings. If $$T_s$$ is the constant temperature of the surroundings and $$k$$ is the proportionality constant, the law is written as

Integrating between an initial temperature $$T_1$$ and a final temperature $$T_2$$ over a time interval $$t$$ gives the well-known working formula

We first use the given cooling from 61 °C to 59 °C in 4 min to determine the constant $$k$$. Here

Substituting in the formula, we have

Solving for $$k$$,

Now we need the time required for the body to cool from 51 °C to 49 °C. For this second stage we put

Using the same surroundings temperature $$T_s = 30^\circ\text{C}$$ and the same $$k$$, the time $$t'$$ for this cooling is

We substitute the value of $$k$$ obtained earlier:

Evaluating the logarithms (natural logarithms):

So

The calculated time is practically 6 minutes.

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.

Thermal stress in a material arises when the material is constrained and prevented from expanding or contracting due to temperature changes. The stress is given by $$\sigma = Y \alpha \Delta T$$, where $$Y$$ is Young's modulus, $$\alpha$$ is the coefficient of linear expansion, and $$\Delta T$$ is the temperature change.

When a rod is lying freely and is heated, it is free to expand without any constraint. Since there is no external force or support preventing the expansion, no thermal stress develops in the rod. Therefore, Assertion A is true.

Reason R states that on heating, the length of the rod increases. This is indeed true for most materials, as they undergo thermal expansion when heated.

However, R is not the correct explanation of A. The fact that the rod's length increases on heating is not why thermal stress is absent. Thermal stress is absent because the rod is free to expand. If the rod were fixed at both ends, it would still try to expand on heating, but the constraint would cause thermal stress. The key factor is the freedom to expand, not the expansion itself.

Therefore, both A and R are true but R is NOT the correct explanation of A, which corresponds to Option (3).

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.

For two identical wires (same length $$l$$ and cross-sectional area $$A$$) connected in series, the total thermal resistance is the sum of individual resistances. The thermal resistance of each wire is $$R_i = \frac{l}{K_i A}$$, so the total resistance is $$R = \frac{l}{K_1 A} + \frac{l}{K_2 A}$$.

The effective conductivity $$K_{\text{eff}}$$ of the combination (total length $$2l$$, same area $$A$$) satisfies $$R = \frac{2l}{K_{\text{eff}} A}$$. Therefore $$\frac{2l}{K_{\text{eff}} A} = \frac{l}{K_1 A} + \frac{l}{K_2 A}$$.

Simplifying: $$\frac{2}{K_{\text{eff}}} = \frac{1}{K_1} + \frac{1}{K_2} = \frac{K_1 + K_2}{K_1 K_2}$$, which gives $$K_{\text{eff}} = \frac{2K_1 K_2}{K_1 + K_2}$$.

The correct answer is option 1: $$\frac{2K_1 K_2}{K_1 + K_2}$$.

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 completely reversible (Carnot) heat engine, the efficiency $$\eta$$ is related to the source temperature $$T_1$$ and the sink (cold reservoir) temperature $$T_2$$ by the well-known Carnot relation:

$$\eta \;=\;1-\dfrac{T_2}{T_1}$$

We are first told that the efficiency is $$\dfrac14$$. Substituting this value, we have

$$\dfrac14 \;=\;1-\dfrac{T_2}{T_1}$$

Now we isolate the fraction $$\dfrac{T_2}{T_1}$$. Moving terms carefully,

$$\dfrac{T_2}{T_1}=1-\dfrac14=\dfrac34$$

Multiplying both sides by $$T_1$$ gives

$$T_2=\dfrac34\,T_1$$

Next, the problem says that the sink temperature is lowered by 58 °C, and the efficiency becomes double. Because temperature differences are identical in the Celsius and Kelvin scales, we can use the same numerical value in kelvin. Let the new sink temperature be $$T_2'$$ and the (unchanged) source temperature remain $$T_1$$. We therefore have

$$T_2'=T_2-58$$

The efficiency is now doubled, becoming $$2\times\dfrac14=\dfrac12$$. Using the Carnot formula again,

$$\dfrac12 \;=\;1-\dfrac{T_2'}{T_1}$$

Solving for $$T_2'$$,

$$\dfrac{T_2'}{T_1}=1-\dfrac12=\dfrac12\quad\Longrightarrow\quad T_2'=\dfrac12\,T_1$$

But we also have $$T_2' = T_2 - 58$$. Substituting $$T_2=\dfrac34\,T_1$$ obtained earlier,

$$\dfrac12\,T_1 \;=\;\dfrac34\,T_1 - 58$$

Bringing the terms containing $$T_1$$ to one side,

$$\dfrac34\,T_1 - \dfrac12\,T_1 = 58$$

The left-hand side simplifies since $$\dfrac34-\dfrac12=\dfrac14$$, giving

$$\dfrac14\,T_1 = 58$$

Multiplying both sides by 4, we find the source temperature:

$$T_1 = 4 \times 58 = 232\text{ K}$$

Finally, we determine the original sink temperature from $$T_2=\dfrac34\,T_1$$:

$$T_2=\dfrac34 \times 232 = \dfrac{696}{4}=174\text{ K}$$

Hence, the correct answer is Option C.

For an adiabatic process involving an ideal gas, we have $$PV^\gamma = \text{constant}$$.

Taking the differential of both sides: $$V^\gamma \, dP + P \cdot \gamma V^{\gamma - 1} \, dV = 0$$.

Dividing throughout by $$PV^\gamma$$: $$\frac{dP}{P} + \gamma \frac{dV}{V} = 0$$.

Therefore, the fractional change in pressure is $$\frac{dP}{P} = -\gamma \frac{dV}{V}$$.