10 g of argon gas is compressed isothermally and reversibly at a temperature of from 10 L to 5 L. Ca — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

For an isothermal process involving an ideal gas, temperature is constant ($\Delta T = 0$), so $\Delta H = 0$ and $\Delta U = 0$. By the First Law, $q = -w$. Reversible work done $w = -2.303 nRT \log(V_2/V_1) = -2.303 \times (\frac{10}{40}) \times 2 \text{ cal K}^{-1}\text{mol}^{-1} \times 300 \text{ K} \times \log(\frac{5}{10}) = -2.303 \times 0.25 \times 2 \times 300 \times (-0.3010) = +103.99 \text{ cal}$. Thus, heat exchanged is $q = -w = -103.99 \text{ cal}$.