A gas is expanded from volume V 0 to 2 V 0 under three different process. Process 1 is isobaric, process 2 is isothermal and process 3 is adiabatic. Let Δ U 1 , Δ U 2 and Δ U 3 be the change

Question

A gas is expanded from volume V 0 ​ to 2 V 0 ​ under three different process. Process 1 is isobaric, process 2 is isothermal and process 3 is adiabatic. Let Δ U 1 ​ , Δ U 2 ​ and Δ U 3 ​ be the change in internal energy of the gas in these three processes. Then :

TestHub · Accepted Answer

The internal energy change for an ideal gas is given by $Δ U = n C_{v} Δ T$ .

1. Isobaric expansion: $P$ constant, $V$ increases, so $T$ increases ( $P V = n RT$ ). Thus, $Δ T > 0$ , so $Δ U_{1} > 0$ .

2. Isothermal expansion: $T$ constant. Thus, $Δ T = 0$ , so $Δ U_{2} = 0$ .

3. Adiabatic expansion: $Q = 0$ . Gas does work, so its internal energy decreases. Thus, $Δ T < 0$ , so $Δ U_{3} < 0$ .

Therefore, $Δ U_{1} > Δ U_{2} > Δ U_{3}$ .

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JEE Main 2027 (Class 12+)