The entropy versus temperature plot for phases α and β at 1 bar pressure is given. ST and S0 are entropies of the phases at temperatures T and 0 K, respectively.
(see diagram)
The transition temperature for α to β phase change is 600 K and CP,β−CP,α=1Jmol−1K−1. Assume (CP,β−CP,α) is independent of temperature in the range of 200 to 700K.CP,α and CP,β are heat capacities of α and β phases, respectively.
The value of entropy change, Sβ−Sα (in Jmol−1K−1 ), at 300 K is ____. [Use : ln2=0.69 Given : Sβ−Sα=0 at 0 K]
Answer: 0.31
Solution
The two phases have equal entropy at 0 K, and CP,β−CP,α=1Jmol−1K−1 is constant, so the entropy gap varies as dTd(Sβ−Sα)=TCP,β−CP,α=T1. At the transition temperature 600 K the diagram gives Sβ−Sα=1. Integrating between 300 K and 600 K, (Sβ−Sα)600−(Sβ−Sα)300=ln300600=ln2=0.69, so (Sβ−Sα)300=1−0.69=0.31Jmol−1K−1.
JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Numerical answer