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Question

If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourization of 1 mol of water at 1 bar and 1000C is 41 kJ.mol1. Calculate the internal energy change, when 1 mol of water is vaporized at 1 bar pressure and 1000C.

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Solution

i) The reaction is: H2O(l)H2O(g)
We know that,
ΔH=ΔU+ΔngRTΔU=ΔHΔngRT
So on substituting the given values we get,
ΔU=41kJmol18.3×3731000
ΔU=37.904 kJ.mol1

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