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A gas is expanded from volume V(0) to 2V...

A gas is expanded from volume `V_(0)` to `2V_(0)` under three different processes. Process 1 is isobaric process 2 is isothermal and process 3 is adiabatic. Let `DeltaU_(1), DeltaU_(2)` and `DeltaU_(3)` be the change in internal energy of the gas in these three processes. Then

A

`DeltaU_(1) gt DeltaU_(2) gt DeltaU_(3)`

B

`DeltaU_(1) lt DeltaU_(2) lt DeltaU_(3)`

C

`DeltaU_(2) lt DeltaU_(1) lt DeltaU_(3)`

D

`DeltaU_(2) lt DeltaU_(3) lt DeltaU_(1)`

Text Solution

Verified by Experts

The correct Answer is:
A
`DeltaU_(1) = +ve`
`DeltaU_(2) = 0`
`DeltaU_(3) = -ve`
`therefore DeltaU_(1) gt DeltaU_(2) gt DeltaU_(3)`
As volume increases , `DeltaW = +ve`.
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