JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Chemical Thermodynamics

(A) Freezing of water at 273 K and 1 atm. ${\mathrm{H}}_2\mathrm{O}(l)\to {\mathrm{H}}_2\mathrm{O}(s)$ at the normal freezing point is an equilibrium process, so $\Delta G=0$ $\to$ (T). The liquid becomes an ordered solid, so $\Delta S_{\text{sys}}<0$ $\to$ (R). Freezing is exothermic ($q\ne 0$) and water expands on freezing ($w\ne 0$), so $\Delta U\ne 0$ and neither (P), (Q) nor (S) applies.

(B) Free expansion of 1 mol of an ideal gas into a vacuum, isolated. Isolated, so no heat crosses the boundary: $q=0$ $\to$ (P). The gas expands against zero external pressure: $w=0$ $\to$ (Q). Hence $\Delta U=q+w=0$ $\to$ (S). The gas occupies a larger volume, so $\Delta S_{\text{sys}}>0$ and (R) does not apply.

(C) Mixing equal volumes of two ideal gases at constant T and P in an isolated container. Ideal gases do not interact, so mixing costs no heat and no work: $q=0$ $\to$ (P), $w=0$ $\to$ (Q), $\Delta U=0$ $\to$ (S). Mixing raises the entropy, $\Delta S_{\text{sys}}>0$, so (R) does not apply.

**(D) Reversible heating 300 K $\to$ 600 K at 1 atm, then reversible cooling back to 300 K at 1 atm.** The gas returns to its initial state, so every state function is unchanged over the cycle: $\Delta U=0$ $\to$ (S), $\Delta G=0$ $\to$ (T), and $\Delta S_{\text{sys}}=0$, which is not $<0$, so (R) does not apply. The path is reversed step for step, so the heat absorbed while heating is given back while cooling and the work done by the gas is returned to it: over the whole process $q=0$ $\to$ (P) and $w=0$ $\to$ (Q).

Answer: (A) $\to$ (R), (T); (B) $\to$ (P), (Q), (S); (C) $\to$ (P), (Q), (S); (D) $\to$ (P), (Q), (S), (T).