A gas is expanded from volume $V_{0}$ to $2 V_{0}$ under three different processes. Process 1 is isobaric process, 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 process. Then

Δ U_{1} > Δ U_{2} > Δ U_{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Δ U_{1} < Δ U_{2} < Δ U_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Δ U_{2} < Δ U_{1} < Δ U_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Δ U_{2} < Δ U_{3} < Δ U_{1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $Δ U_{1} > Δ U_{2} > Δ U_{3}$
From first law of thermodynamics,
Heat supplied ( $Δ Q$ )= work done by system( $Δ W$ ) + increase in internal energy( $Δ U$ )
All 3 processes being expansions,
$Δ W > 0$ , for each of them
For isothermal process, $Δ U = 0$
For adiabatic process, $Δ Q = 0$
For isobaric expansion, $Δ U = n C_{p} Δ T > 0$
Thus, for process 1, $Δ U_{1} > 0$
for process 2, $Δ U_{2} = 0$
for process 3, $Δ U_{3} = - Δ W_{3} < 0$
Thus, $Δ U_{1} > Δ U_{2} > Δ U_{3}$ is the answer.

Suggest Corrections

Similar questions

Q. A gas is expanded from volume

V_{0}

2 V_{0}

under three different processes. 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

A gas is expanded from volume $V_{0} t o 2 V_{0}$ under three different processes, as shown in the figure. Process 1 is isobaric process, process 2 is isothermal and process 3 is adiabatic. Let $Δ U_{1}, Δ U_{2}$ and $Δ U_{3}$ be the change in the internal energy of the gas in these three processes, then: