In the reversible adiabatic expansion of an ideal monoatomic gas, the final volume is 8 times the in — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

For a reversible adiabatic expansion, $T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$. For a monoatomic ideal gas, $\gamma = 5/3$, so $\gamma - 1 = 2/3$. The ratio $\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{2/3} = \left(\frac{1}{8}\right)^{2/3} = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.