Question
Assuming that water vapour is an ideal gas, the internal energy change(∆U) when 1 mol of water is vapourised at 1 bar pressure and 100°C, (Given: Molar enthalpy of vapourisation of water at 1 bar and 373 K = 41 kJ mol-1 and R = 8.3 J mol–1K–1 will be) –
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4.100 kJ mol–1
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3.7904 kJ mol–1
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37.904 kJ mol–1
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41.00 kJ mol–1
Solution
C.
37.904 kJ mol–1
∆ng =1−0 =1
∆H =∆U +∆ngRT
∆U =∆H −∆ngRT
= 41 – 8.3 × 10-3 × 373
= 37.9 kJ mol-1
