A quantity of 1 g of water on evaporation at atmospheric pressure forms 1671 cm^3 of steam. Heat of — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

Δ H = 540 cal. W = P * Δ V = 1.013 * 10^5 N/m^2 * 1671 * 10^-6 m^3 = 169.3 Joules = 169.3 / 4.184 cal = 40.5 cal. Δ U = Δ H - W = 540 - 40.5 = 499.5 cal approx 500 cal.