JEE Main 2026 — Thermodynamics (C) Question with Solution

At the transition temperature and $1$ atm pressure, the two states $A$ and $B$ are in equilibrium.

For a phase transition of a pure substance at $1$ atm, both states are in their standard states, so the reaction quotient $Q=1$.

Using the relation $\Delta G=\Delta G^0+RT\ln Q$, we get $\Delta G=\Delta G^0$.

At equilibrium, $\Delta G=0$, which implies $\Delta G^0=0$.

Given $\Delta G^0=105-35\log T$, we can equate it to zero:

$105-35\log T=0$

$35\log T=105$

$\log T=3$

$T=10^3=1000$ K

Converting the temperature to Celsius:

$T{(}^{\circ }\text{C})=1000-273=727^{\circ }\text{C}$

Answer: $727$