JEE Advanced 2023 Paper 2 · Q14 · Thermodynamics

Question

The value of entropy change, $S_\beta - S_\alpha$ (in J mol$^{-1}$ K$^{-1}$), at $300$ K is ___. [Use: $\ln 2 = 0.69$. Given: $S_\beta - S_\alpha = 0$ at $0$ K.]

SolveKar AI · Accepted Answer

At the transition temperature $T_	ext{tr} = 600$ K, the plot gives $\Delta S_	ext{tr} = S_\beta(600) - S_\alpha(600) = 6 - 5 = 1$ J mol$^{-1}$ K$^{-1}$. Cooling each phase from $600$ K to $300$ K at constant pressure gives $S_\beta(300) - S_\beta(600) = \int_{600}^{300} (C_{p,\beta}/T)\,dT = C_{p,\beta} \ln(300/600)$, and similarly for $\alpha$. Subtracting: $\Delta S(300) - \Delta S(600) = (C_{p,\beta} - C_{p,\alpha}) \ln(300/600) = (1)(-\ln 2) = -0.69$ J mol$^{-1}$ K$^{-1}$. Hence $\Delta S(300) = 1 - 0.69 = 0.31 \approx 0.30$ J mol$^{-1}$ K$^{-1}$. (Using $\ln 2 = 0.69$ as given, the answer truncated/rounded to two decimal places is $0.30$.)

JEE Advanced 2023 Paper 2 · Q14 · Thermodynamics

Solution