A liquid has molar ∆vapH = 30.0 kJ/mol at 300 K. Treating the vapour as ideal, ∆vapU is about:
A29.0 kJ/mol
B32.5 kJ/mol
C27.5 kJ/mol
D30.0 kJ/mol
Answer & Solution
Correct answer: C. 27.5 kJ/mol
∆U = ∆H - ∆ngRT = 30.0 - (1 x 8.314 x 300)/1000 = 30.0 - 2.49 = 27.5 kJ/mol.
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