JEE Advanced 2023 Paper 2 · Q15 · Thermodynamics

Correct answer:300

Solution

At the transition temperature $T_\text{tr} = 600$ K, the two phases are in equilibrium, so $\Delta G_\text{tr} = 0$ and $\Delta H_\text{tr} = T_\text{tr} \Delta S_\text{tr} = 600 \times 1 = 600$ J mol$^{-1}$. Kirchhoff's law: $\dfrac{d(\Delta H)}{dT} = \Delta C_p = C_{p,\beta} - C_{p,\alpha} = 1$ J mol$^{-1}$ K$^{-1}$ (constant). Integrating from $600$ K down to $300$ K: $\Delta H(300) - \Delta H(600) = \Delta C_p (300 - 600) = 1 \times (-300) = -300$ J mol$^{-1}$. Hence $\Delta H(300) = H_\beta(300) - H_\alpha(300) = 600 - 300 = 300$ J mol$^{-1}$.

Want to drill deeper? Open this question inside SolveKar AI and tap "Why?" on any step — the AI Mentor reads your full solution context and explains until it clicks. Download SolveKar AI →