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Unit mass of liquid of volume V(1) compl...

Unit mass of liquid of volume `V_(1)` completely turns into a gas of volume `V_(2)` at constant atmospheric pressure P and temperature T. The latent heat of vaporization is "L". Then the change in internal energy of the gas is

A

zero

B

`P(V_(2)-V_(1))`

C

`L-P(V_(2)-V_(1))`

D

L

Text Solution

Verified by Experts

The correct Answer is:
C
`Q = mL = 1 xx L = L, W = P(V_(2)-V_(1))`
Now `Q =DeltaU +W`
or `L = DeltaU + P(V_(2)-V_(1))`
`therefore DeltaU = L - P(V_(2)-V_(1))`
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