A gas is expanded form volume `V_(0) to 2V_(0)` under three different processes as shown in the figure . Process 1 is isobaric process process 2 is isothermal and 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

Options

A. `DeltaU_(1)gtDeltaU_(2)gtDeltaU_(3)`
B. `DeltaU_(1)ltDeltaU_(2)ltDeltaU_(3)`
C. `DeltaU_(1)ltDeltaU_(2)ltDeltaU_(3)`
D. `DeltaU_(1)ltDeltaU_(2)ltDeltaU_(3)`

Answer

A

Explanation

Process 1 is isobaric expansion (P=constant)Hence temperature of gas will increase.
`DeltaU_(1)` =positive
process 2 is an adiabatic expansion
Gence temperature of gas will fall
`DeltaU_(3)`=negative
`DeltaU_(1)gt DeltaU_(2)gt DeltaU_(3)`