ABC26PH0210 · Statistical Thermodynamics

Subject: Physical Chemistry · Chapter: Statistical Thermodynamics · Topic: Statistical Thermodynamics – General · Exam: CSIR-NET DEC 2019 · Marks: 2 · Difficulty: Medium

The molar residual entropy (in $\mathrm{J\,K^{-1}}$) of solid OCS would be closest to
(a)0
(b)2.9
(c)5.8
(d)8.7
Answer
SELF-PRACTICE — the source book printed no answer.

Nothing is invented here, so this question has no answer on record.

Explanation
Residual entropy, $S=k_{B}\ln\left(W^{N_{A}}\right)$, where $W$ = number of arrangements. Arrangement for solid OCS: O=C=S, S=C=O; so $W=2$. $S=k_{B}\ln(2^{N_{A}})=N_{A}k_{B}\ln 2=R\ln 2$. $S=8.314\times0.693=5.7628\,\mathrm{J\,K^{-1}}$.

Open in whiteboard · Browse this chapter in the app