Each of the vessels 1 and 2 contain 1.2 moles of gaseous helium. The ratio of the vessels volumes is — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

$S_2 - S_1 = n C_V \ln\left(\frac{T_2}{T_1}\right) + n R \ln\left(\frac{V_2}{V_1}\right)$. He is monoatomic, $C_V = 1.5R$. Given $\frac{T_2}{T_1} = 0.5$ and $\frac{V_2}{V_1} = 2$. $S_2 - S_1 = 1.2(1.5R) \ln(0.5) + 1.2R \ln(2) = -1.8R \ln 2 + 1.2R \ln 2 = -0.6R \ln 2$. Using $R=2 \text{ cal/mol-K}$, $\Delta S = -0.6(2)(0.7) = -0.84 \text{ cal/K}$.