The `DeltaH` for vaporization of a liquid is 20kJ/mol. Assuming ideal behaviour, the change in internal energy for the vaporization of 1 mol of the li
The `DeltaH` for vaporization of a liquid is 20kJ/mol. Assuming ideal behaviour, the change in internal energy for the vaporization of 1 mol of the liquid at `60^(@)C` and l bar is close to -
A. `13.2kJ//mol`
B. `17.2kJ//mol`
C. `19.5kJ//mol`
D. `20.0kJ//mol`
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Correct Answer - B
`DeltaH=DeltaE+Deltan_(g)RT`
`20=DeltaE+8.314xx10^(-3)xx333`
`DeltaE=17.2kJ//mol`
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