GATE 2025 CY – Question 62

Physical Chemistry · Statistical Thermodynamics: Micro-canonical, Canonical and Grand canonical ensembles, Boltzmann distribution, Partition functions and thermodynamic properties. Statistical mechanics of non-interacting systems, ideal monoatomic, diatomic gases, translational, rotational, vibrational and electronic partition functions. · 2 marks · Numerical answer

The mean energy of a molecule having two available energy states at $\epsilon=0$ J and $\epsilon=4.14\times10^{-21}$ J at 300 K is _____ $\times10^{-21}$ J (rounded off to two decimal places). [Given: Boltzmann constant $k_B=1.38\times10^{-23}\ \mathrm{J\,K^{-1}}$]

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Correct answer: 1.10 to 1.12

Explanation

For a two-level system with energies $0$ and $\epsilon$ (no degeneracy), the Boltzmann populations are
$$p_0=\frac{1}{1+e^{-\epsilon/k_BT}},\qquad p_1=\frac{e^{-\epsilon/k_BT}}{1+e^{-\epsilon/k_BT}}.$$

**The exponent.**
$$\frac{\epsilon}{k_BT}=\frac{4.14\times10^{-21}}{1.38\times10^{-23}\times300}=\frac{4.14\times10^{-21}}{4.14\times10^{-21}}=1 .$$

**Mean energy:**
$$\langle\epsilon\rangle=0\cdot p_0+\epsilon\,p_1=\epsilon\,\frac{e^{-1}}{1+e^{-1}}=\frac{\epsilon}{1+e}=\frac{4.14\times10^{-21}}{3.718}=1.113\times10^{-21}\text{ J}.$$

The mean energy is **1.11 × 10⁻²¹ J**.