A sample of perfect gas that initially occupies 15.0 L at 300 K and 1.0 bar is compressed isothermal — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

$\Delta S = n R \ln\left(\frac{V_2}{V_1}\right)$. We know $nR = \frac{P_1 V_1}{T_1} = \frac{10^5 \text{ Pa} \times 15 \times 10^{-3} \text{ m}^3}{300 \text{ K}} = \frac{1500}{300} = 5 \text{ J/K}$. We want $\Delta S = -5.0 \text{ J/K}$. $-5.0 = 5 \ln\left(\frac{V_2}{15}\right) \implies \ln\left(\frac{V_2}{15}\right) = -1.0$. Given $\ln 0.36 = -1.0$, $V_2 = 15 \times 0.36 = 5.4 \text{ L}$.