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)]

Answer»

`35.67 kJmol ^(-1)`
`35.67 kJmol ^(-1)`
`36.57kJmol ^(-1)`
`38.75kJmol ^(-1)`

SOLUTION :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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