For water Δ_vap H = 41 kJ mol⁻¹ 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 nearly___________ kJ mol⁻¹

[Use : R = 8.3 J mol⁻¹ K⁻¹]

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41kJ mol⁻¹

38 kJ/mol

44kJ mol⁻¹

Zero kJ mol⁻¹

answer is B.

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Detailed Solution

H₂O(l) → H₂O(g) : ΔH = 41 kJ/mol

$\Rightarrow$ From the relation : ΔH = ΔU + Δn_gRT

⇒ $\Rightarrow$ 41kJ/mol = ΔU + (1) × $\frac{8.3}{1000}$ × 373

ΔU = 41 − 3.0959 = 38 kJ/mol

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