Q103 · CSIR-NET Chemistry, June 2016
Paper: CSIR-NET June 2016 · Subject: Physical Chemistry · Chapter: Thermodynamics · Topic: Phase Equilibria · Marks: 2 · Difficulty: Medium
Phase diagram of a compound is shown below 
The slopes of the lines $\mathrm{OA}, \mathrm{AC}$ and AB are $\tan \frac{\pi}{4}, \tan \frac{\pi}{6}$ and $\tan \frac{\pi}{3}$ respectively. If melting point and $\Delta \mathrm{H}$ of melting are 300 K and 3 kJ mole ${ }^{-1}$ respectively, the change in the volume on melting is
The slopes of the lines $\mathrm{OA}, \mathrm{AC}$ and AB are $\tan \frac{\pi}{4}, \tan \frac{\pi}{6}$ and $\tan \frac{\pi}{3}$ respectively. If melting point and $\Delta \mathrm{H}$ of melting are 300 K and 3 kJ mole ${ }^{-1}$ respectively, the change in the volume on melting is(a)$10 \tan \frac{\pi}{3}$
(b)$10 \tan \frac{\pi}{4}$
(c)$10 \cot \frac{\pi}{3}$
(d)$10 \cot \frac{\pi}{4}$
Answer
Answer: C ✓ checked by 4AB · confidence high
Explanation
AB is the melting line, so its slope is dP/dT = tan(π/3) = ΔH/(TΔV). ΔV = 3000/(300 tan π/3) = 10 cot(π/3) (c).
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