A quantity of 1 g of water on evaporation at atmospheric pressure forms of steam. Heat of vaporizati — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

Heat input $q_p = \Delta H = 540 \text{ cal}$. The expansion volume $\Delta V = 1671 \text{ cm}^3 - 1 \text{ cm}^3 = 1670 \text{ cm}^3 = 1.67 \text{ L}$. Work done $W_{by} = P \Delta V = 1 \text{ atm} \times 1.67 \text{ L} = 1.67 \text{ L-atm}$. Converting to calories ($1 \text{ L-atm} \approx 24.2 \text{ cal}$), $W_{by} \approx 40.4 \text{ cal}$. The increase in internal energy $\Delta U = \Delta H - W_{by} = 540 - 40.4 = 499.6 \text{ cal} \approx 500 \text{ cal}$.