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The molar enthalpy change for H(2)O(l)hA...

The molar enthalpy change for `H_(2)O(l)hArrH_(2)O(g)` at 373 K and 1 atm is 41 kJ/mol. Assume ideal behaviour, the internal energy change for vaporization of 1 mol of water at 373 K and 1 atm in KJ `"mol"^(-1)` is :

A

`30.2`

B

`41.0`

C

`48.1`

D

`37.9`

Text Solution

Verified by Experts

The correct Answer is:
D
`W=-nRT=-(1xx8.314xx10^(-3)xx373)kJ`
=-3.10kJ
`q=DeltaH=41kJ`
`&DeltaE=q+w=(41-3.1)~= 37.9kJ`
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