An ideal gaseous sample at initial state () is allowed to expand to volume using two different proce — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

Integrating for work: $W_1 = \int_{V_o}^{2V_o} \frac{P_o V_o^2}{V^2} dV = 0.5 P_o V_o$. $W_2 = \int_{V_o}^{2V_o} \frac{P_o V_o}{V} dV = P_o V_o \ln 2 \approx 0.693 P_o V_o$. The magnitude of work in the second process ($0.693 P_o V_o$) is greater than in the first ($0.5 P_o V_o$).