Q83 · CSIR-NET Chemistry, December 2017
Paper: CSIR-NET December 2017 · Subject: Physical Chemistry · Chapter: Thermodynamics · Topic: Thermodynamics – General · Marks: 2 · Difficulty: Medium
In stretching of a rubber band, $\mathrm{dG}=\mathrm{Vdp}-\mathrm{SdT}+\mathrm{fdL}$. Which of the following relations in true?
(a)$\left(\frac{\partial \mathrm{S}}{\partial \mathrm{L}}\right)_{\mathrm{p}, \mathrm{T}}=-\left(\frac{\partial \mathrm{f}}{\partial \mathrm{T}}\right)_{\mathrm{p}, \mathrm{L}}$
(b)$\left(\frac{\partial \mathrm{S}}{\partial \mathrm{L}}\right)_{\mathrm{p}, \mathrm{T}}=-\left(\frac{\partial \mathrm{f}}{\partial \mathrm{V}}\right)_{\mathrm{p}, \mathrm{L}}$
(c)$\left(\frac{\partial \mathrm{S}}{\partial \mathrm{L}}\right)_{\mathrm{p}, \mathrm{T}}=-\left(\frac{\partial \mathrm{V}}{\partial \mathrm{T}}\right)_{\mathrm{p}, \mathrm{L}}$
(d)$\left(\frac{\partial \mathrm{S}}{\partial \mathrm{L}}\right)_{\mathrm{p}, \mathrm{T}}=-\left(\frac{\partial \mathrm{f}}{\partial \mathrm{p}}\right)_{\mathrm{T}}$
Answer
Answer: A ✓ checked by 4AB · confidence high
Explanation
dG = Vdp − SdT + f dL is an exact differential. The cross-derivative gives (∂S/∂L)_{p,T} = −(∂f/∂T)_{p,L} (a).
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