For water H = 41 kJ mol at 373 K and 1 bar pressure. Assuming that water vapour is an ideal gas that — Thermodynamics and Thermochemistry Chemistry Question
Question
For water H = 41 kJ mol at 373 K and 1 bar pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is _____ kJ mol [Use: R = 8.3 J mol K ] vap –1 –1 –1 -1
💡 Solution & Explanation
**Step 1: Identify the relationship between enthalpy and internal energy** For phase changes: ΔH = ΔU + Δ(PV) Since vapor is an ideal gas and occupies much larger volume than liquid: ΔH = ΔU + Δ(nRT) **Step 2: Simplify for one mole of water** For 1 mole: ΔH = ΔU + RT Rearranging: ΔU = ΔH - RT **Step 3: Convert R to consistent units** R = 8.3 J mol⁻¹ K⁻¹ = 0.0083 kJ mol⁻¹ K⁻¹ **Step 4: Calculate RT** RT = 0.0083 kJ mol⁻¹ K⁻¹ × 373 K RT = 3.0959 kJ mol⁻¹ ≈ 3.10 kJ mol⁻¹ **Step 5: Calculate ΔU** ΔU = ΔH - RT ΔU = 41 kJ mol⁻¹ - 3.10 kJ mol⁻¹ ΔU = 37.9 kJ mol⁻¹ ≈ 38.0 kJ mol⁻¹ Therefore, the answer is 38.00.