A gas is expanded from 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 process 3 is adiabatic. Let `Delta U_(1), Delta U_(2)` and `DeltaU_(3)` be the change in internal energy of the gas in these three processes. Then

Options

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

Answer

A

Explanation

Process 1 is isobaric expansion (P = constant)
Hence temperature of gas will increase, `Delta U_(1)` = positive
Process 2 is an isothermal process, `Delta U_(2) = 0`
Process 3 is an adiabatic expansion.
Hence temperature of gas will fall.
`:. Delta U_(3)` = negative `:. Delta U_(1) gt Delta U_(2) gt Delta U_(3)`