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 gs in these three processes then

Options

A. `DeltaU_(1) gt DeltaU_(2) gt DeltaU_(2)`
B. `DeltaU_(1) lt DeltaU_(2) lt DeltaU_(2)`
C. `DeltaU_(2) lt DeltaU_(1) lt DeltaU_(3)`
D. `DeltaU_(2) lt DeltaU_(3) lt DeltaU_(1)`

Answer

A

Explanation

Process 2 is an isothermal process, Hence, `Delta U_(2) = 0`
Process 1 is an isobatic (P=constant) expansion.
or , `Delta U_(1)` = positiveProcess 3 is an adiabatic expansion. Hence, temperature will decrease.
or `DeltaU_(3) = negative`
therefore, `DeltaU_(1) gt DeltaU_(2) gt DeltaU_(3)` is the correct option.