The work done by one mole of an ideal gas in the reversible process: , from (1 atm, 300 K) to is — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

Process $PV^3 = \text{constant} \implies n = 3$. $P_1 V_1^3 = P_2 V_2^3$. Using $V \propto T/P$, this is $T^3 P^{-2} = \text{constant}$. $300^3 / 1^2 = T_2^3 / (2\sqrt{2})^2 \implies T_2^3 = 8 \times 300^3 \implies T_2 = 600 \text{ K}$. Work done by the gas $W_{by} = \frac{R(T_1 - T_2)}{n - 1} = \frac{R(300 - 600)}{3 - 1} = -150 R$. The magnitude (amount of work) is $150 R$.