Two moles of an ideal gas () was allowed to expand reversibly and adiabatically from 1 L, to 32 L. T — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

$T_1 = 527^\circ\text{C} = 800 \text{ K}$. For reversible adiabatic expansion, $T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \implies 800 \times (1)^{0.4} = T_2 \times (32)^{0.4} \implies 800 = T_2 \times (2^5)^{2/5} \implies 800 = 4 T_2 \implies T_2 = 200 \text{ K}$. Molar enthalpy change $\Delta H_m = C_{p,m} \Delta T$. With $\gamma = 1.4$, $C_{p,m} = \frac{7}{2} R$. $\Delta H_m = \frac{7}{2} R (200 - 800) = \frac{7}{2} R (-600) = -2100 R$.