Molar enthalpy change for vapourisation of 1 mole of water at 1 bar and 100°c is 41KJ mol-1. Calculate the internal energy change, when 1 mole of water is vaporised at1bar pressure and 100°c.

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Detailed Solution

H2O(l)→H2O(g) ∆H=∆U+∆ngRT ∆U=∆H-∆ngRT Given, ∆H=41×103 J mol-1 , T=100°+273=373K ∆U =41×103-1(8.3)373 =41×103-3.096×103=37.904×103J mol-1

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