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The latent heat of vaporisation of liqui...

The latent heat of vaporisation of liquid is `10kcal mol^(-1)` at `1 atm` and `227^(@)C`. What will be the change in internal energy of `3 mol es ` of the liquid at same condition ?

A

`14kcal`

B

`-14kcal`

C

`27kcal`

D

`-27kcal`

Text Solution

Verified by Experts

The correct Answer is:
C
`3H_(2)O_((l))rarr 3H_(2)O_((g)), Deltan_(g)=3`
`:' DeltaH=DeltaU+Deltan_((g))RT`
`:. DeltaU=DeltaH-Deltan_((g))RT`
`=10xx3-3xx2xx10^(-3)xx500`
`( :' R=2xx10^(-3)kcal K^(-1)mol^(-1))`
`=27kcal`
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