The molar entropy of a constant volume sample of neon at 500 K if it is 46.2 cal/K-mol at 250 K, is — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

$\Delta S = C_V \ln\left(\frac{T_2}{T_1}\right)$. For neon (monoatomic), $C_V = \frac{3R}{2} \approx 3 \text{ cal/K-mol}$. $\Delta S = 3 \ln\left(\frac{500}{250}\right) = 3 \ln 2 = 3(0.7) = 2.1 \text{ cal/K-mol}$. Final entropy $S_{500} = S_{250} + \Delta S = 46.2 + 2.1 = 48.3 \text{ cal/K-mol}$.