A swimmer coming out from a pool is covered with a film of water weighing about 18 g. calculte the internal energy of vaporisation at `100^(@)`, `[Delta_("vap")H^(@) for water at 373K = 40. 66 kJ mol^(-1)]`A. `35.67 kJmol ^(-1)`B. `35.67 kJmol ^(-1)`C. `36.57kJmol ^(-1)`D. `38.75kJmol ^(-1)`

A swimmer coming out from a pool is covered with a film of water weigh

1.	A swimmer coming out from a pool is covered with a film of water weighing about 18 g. calculte the internal energy of vaporisation at `100^(@)`, `[Delta_("vap")H^(@) for water at 373K = 40. 66 kJ mol^(-1)]`A. `35.67 kJmol ^(-1)`B. `35.67 kJmol ^(-1)`C. `36.57kJmol ^(-1)`D. `38.75kJmol ^(-1)`
Answer» Correct Answer - b we can repesent the process of evaporation as `18gH_(2)O(l)overset("veporistion")to 18 g H_(2)O(g)` number of moles in ` 18 g H_(2)O(l)` `(18g)/(18g mol^(-1))=1 mol` `Delta_(vap)U=Delta_(vap)H^(-)-pDeltaV` `Delta_(vap)H^(-)-Deltan_(g)RT` assume steam behave as an ideal ga `Delta_(vap)U=(40.66)-(1)(8.314xx10^(-3))(373) ` 40.66 - 3.10 ` 37.56 KJ mol^(-1)`

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A swimmer coming out from a pool is covered with a film of water weighing about 18 g. calculte the internal energy of vaporisation at `100^(@)`, `[Delta_("vap")H^(@) for water at 373K = 40. 66 kJ mol^(-1)]`A. `35.67 kJmol ^(-1)`B. `35.67 kJmol ^(-1)`C. `36.57kJmol ^(-1)`D. `38.75kJmol ^(-1)`

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