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

A gas is expanded from volume V_(0) to 2V_(0) under three different pr

1.	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
Answer» `Delta U_(1) gt Delta U_(2) gt Delta U_(3)` `Delta U_(1) lt Delta U_(2) lt Delta U_(3)` `Delta U_(2) lt Delta U_(1) lt Delta U_(3)` `Delta U_(2) lt Delta U_(3) lt Delta U_(1)` Solution :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)`