States of Matter

States of Matter

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) –

  • 4.100 kJ mol–1

  • 3.7904 kJ mol–1

  • 37.904 kJ mol–1

  • 41.00 kJ mol–1

Answer

C.

37.904 kJ mol–1

straight H subscript 2 straight O space left parenthesis straight l right parenthesis space rightwards arrow with vaporisation on top space straight H subscript 2 straight O space left parenthesis straight g right parenthesis
∆ng =1−0 =1
∆H =∆U +∆ngRT
∆U =∆H −∆ngRT
= 41 – 8.3 × 10-3 × 373
= 37.9 kJ mol-1

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