The expansion process of 2.0 moles of argon proceeds so that the gas pressure increases in direct pr — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

Process $P \propto V \implies PV^{-1} = \text{const}$, so $n = -1$. $C = C_V + \frac{R}{1-n} = 1.5R + 0.5R = 2R$. Since $T \propto PV \propto V^2$, if volume doubles, temperature quadruples ($T_2/T_1 = 4$). $\Delta S = n_{\text{mol}} C \ln\left(\frac{T_2}{T_1}\right) = 2(2R)\ln(4) = 4R(2\ln 2) = 8R\ln 2 = 8(2)(0.7) = 11.2 \text{ cal/K}$.