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 changed in internal energy of the gas in these three processes. Then

Δ U_{1} > Δ U_{2} > Δ U_{3}

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Δ U_{1} < Δ U_{2} < Δ U_{3}

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Δ U_{2} < Δ U_{1} < Δ U_{3}

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Δ U_{2} < Δ U_{3} < Δ U_{1}

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Solution

The correct option is C $Δ U_{1} > Δ U_{2} > Δ U_{3}$
For process (1): Isobaric

$Δ W = P Δ V$

$⟹ P V = n R T$

$⟹ P Δ V = n R Δ T$

$⟹ Δ T = \frac{Δ W}{n R}$

$⟹ Δ U = \frac{n R}{(r - 1)}$

$Δ T = \frac{n R}{r - 1} . \frac{Δ W}{n R}$

$⟹ U_{1} = \frac{Δ T}{r - 1}$

For process (2): Isothermal: $Δ T = 0$

$⟹ δ U = 0$

For process (3): Adiabatic: $Δ Q = 0$

$⟹ Δ W = - Δ U$

$⟹ U_{3} = - Δ W$

$⟹ \frac{n R}{r - 1} Δ T = Δ W$

$⟹ Δ T = \frac{Δ W (r - 1)}{n R}$

So,

$U_{1} > U_{2} > U_{3}$

Hence, the answer is $Δ U_{1} > Δ U_{2} > Δ U_{3}$

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