An ideal gas undergoes isothermal expansion from (10 atm, 1 L) to (1 atm, 10 L) either by path–I (in — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

Path I (reversible): $W_I = P_1V_1 \ln(V_2/V_1) = 10 \ln(10) = 23.03\text{ L-atm}$. Path II (irreversible steps): Expands to $V = 10/5 = 2\text{ L}$ against $5\text{ atm}$, work = $5(2-1) = 5\text{ L-atm}$. Expands to $10\text{ L}$ against $1\text{ atm}$, work = $1(10-2) = 8\text{ L-atm}$. Total $W_{II} = 13\text{ L-atm}$. For isothermal ideal gas $\Delta U=0$, so heat $q = W$. Ratio is $23.03 / 13 = 2.303 / 1.3$.