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What is `Delta U` when `2.0` mole of liquid water vaporises at `100^(@)C`? The heat of vaporisation `(Delta H_("vap".))` of water at `100^(@)C` is `40

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What is `Delta U` when `2.0` mole of liquid water vaporises at `100^(@)C`? The heat of vaporisation `(Delta H_("vap".))` of water at `100^(@)C` is `40.66 KJmol^(-1)`.
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`Delta U= DeltaH-(P_(2)V_(2)-P_(1)V_(1)) ( because V_(2) gt gt V_(1))`
`Delta H-(P_(2)V_(2)) rArrDeltaH-nRT`
`Delta U= 40.66xx2-(2xx8.314xx373)/(1000)=75.12 KJ`
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