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25/09/2021 5:09 pm
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Assuming that water vapour is an ideal gas, the internal energy change (ΔU) 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-1 K-1) will be
(a) 41.00 kJ mol-1
(b) 4.100 kJ mol-1
(c) 3.7904 kJ mol-1
(d) 37.904 kJ mol-1
1 Answer
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25/09/2021 5:16 pm
Correct answer: (d) 37.904 kJ mol-1
Explanation:
Given ΔH = 41 kJ mol-1 = 41000 J mol-1
T = 100°C = 273 + 100 = 373 K
Δn = 1
ΔU = ΔH – ΔnRT = 41000 - (1 × 8.314 × 373)
= 37898.88 J mol-1 ≃ 37.9 kJmol-1