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If water vapour is assumed to be a perfe...

If water vapour is assumed to be a perfect gas, molar enthalpy change for vaporization of 1 mol of water at 1 bar and `100^(@)C` is `51 kJ mol ^(-1)`. Calculate the internal energy, when 1 mol of water is vapourised at one bar pressure and `100^(@)C`.

Text Solution

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The change `H_(2)O(I) rightarrow H_(2)O(g)`
`Delta H = Delta U + n_(g)RT`
or `Delta U = Delta H - Delta n_(g)RT `
`= 51 kJ mol^(-1) -1 xx8.314 J mol^(-1) K^(-1) xx373 K`
`= 51 kJ mol^(-1) -3101 J mol^(-1)`
`= 51 kJ mol^(-1) - 3.101 kJ mol^(-1)`
`= 47.9 kJ mol^(-1)`.
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