Q147 · CSIR-NET Chemistry, December 2019

Paper: CSIR-NET December 2019 · Subject: Physical Chemistry · Chapter: Statistical Thermodynamics · Topic: Partition Functions · Marks: 2 · Difficulty: Medium

The entropy in terms of internal energy $\{\mathrm{U}-\mathrm{U}(0)\}$, and canonical partition function (Q) is given by $\mathrm{S}=\frac{\mathrm{U}-\mathrm{U}(0)}{\mathrm{T}}+\mathrm{k} \ln \mathrm{Q}$. Assuming that T the atoms are distinguishable, the corresponding expression for a monatomic perfect gas is [Here $\mathrm{n}, \mathrm{m}, \mathrm{N}_{\mathrm{A}}, \mathrm{k}$ and h represent number of moles, mass of atom, Avogadro constant,Boltzmann constant and Planck constant, respectively. $\mathrm{U}(0)$ is internal energy at $\mathrm{T}=0]$. Options:
(a)$\mathrm{S}=\frac{5}{2} \mathrm{nR}+\mathrm{nR} \ln \left\{\left.\mathrm{V}(2 \pi \mathrm{mkT})^{\frac{3}{2}} \right\rvert\, \mathrm{nN}_{\mathrm{A}} \mathrm{h}^{3}\right\}$
(b)$\mathrm{S}=\frac{3}{2} \mathrm{nR}+\mathrm{nR} \ln \left\{\left.\mathrm{V}(2 \pi \mathrm{mkT})^{\frac{3}{2}} \right\rvert\, \mathrm{h}^{3}\right\}$
(c)$\mathrm{S}=\frac{5}{2} \mathrm{nR}+\mathrm{nR} \ln \left\{\left.\mathrm{V}(2 \pi \mathrm{mkT})^{\frac{3}{2}} \right\rvert\, \mathrm{h}^{3}\right\}$
(d)$\mathrm{S}=\frac{3}{2} \mathrm{nR}+\mathrm{nR} \ln \left\{\left.\mathrm{V}(2 \pi \mathrm{mkT})^{\frac{3}{2}} \right\rvert\, \mathrm{nN}_{\mathrm{A}} \mathrm{h}^{3}\right\}$
Answer
Answer: B ✓ checked by 4AB · confidence high
Explanation
For distinguishable atoms Q = qᴺ, with no 1/N! term. S = U/T + k ln Q = (3/2)nR + nR ln[V(2πmkT)^(3/2)/h³] (b).

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