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As per your syllabus, for vaporisation of 1 mol of water at 100 °C and 1 bar, given ΔH = 41 kJ/mol, what is ΔU? (R = 8.3 J/mol K)
A≈ 41.0 kJ/mol
B≈ 44.0 kJ/mol
C≈ 0 kJ/mol
D≈ 37.9 kJ/mol
Answer & Solution
Correct answer: D. ≈ 37.9 kJ/mol
As per your syllabus, ΔU = ΔH − Δn_g RT = 41 − (1)(8.3)(373)/1000 = 41 − 3.096 = 37.9 kJ/mol.
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