In the figure, an ideal gas is expanded from volume $V_{0}$ to $2 V_{0}$ 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 the internal energy of the gas is 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 A $Δ U_{1} > Δ U_{2} > Δ U_{3}$
According to first law of thermodynamic $d Q = d U + P △ V,$

Isobaric Isothermal Adiabatic
Pressure is constant. $△ T = 0$ $d Q = 0$
So, $d Q - P △ V = △ V_{1}$ $△ U_{2} = 0$ $- P △ V = △ U_{3}$

So, $△ U_{1} > △ U_{2} > △ U_{3}$
Hence, the answer is $△ U_{1} > △ U_{2} > △ U_{3} .$

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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: