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Calculate the work done when 1.0 mol of ...

Calculate the work done when `1.0` mol of water at `373K` vaporises against an atmosheric pressure of `1.0atm`. Assume ideal gas behaviour.

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The correct Answer is:
3100J
`PV=nRT` or`V= (nRT)/(P) = (1mol xx 0.0821 L atmK^(-1) mol^(-1) xx 373 K)/(1atm) =30.6 L`
Taking volume of liquid water to be negligible, `Delta V =V_("vapour") - V_(H_(2)O(l))=30.6 L`.
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