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Pressure and Temperature

An ideal gas ...

Question

An ideal 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 processes respectively, 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 $Δ U_{1} > Δ U_{2} > Δ U_{3}$
Internal energy for an ideal gas, $Δ U = n C_{V} Δ T$

Process $1$ is an isobaric ( $P =$ constant) expansion.
Using ideal gas equation, $P V = n R T$ ,
$\Rightarrow V \propto T$
Since the volume of the gas increases, therefore temperature of the gas will increase,
$Δ U_{1}$ = positive

Process $2$ is an isothermal process, thus $Δ T = 0$
$∴ Δ U_{2} = 0$

Process $3$ is an adiabatic expansion, therefore gas will do work at the expense of internal energy.
$Δ U = W$ $(Q = 0)$
Hence, internal energy of the gas decreases,
$Δ U_{3}$ = negative

Therefore, $Δ U_{1} > Δ U_{2} > Δ U_{3}$ is the correct option.

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An ideal gas is expanded from volume V0 to 2V0 under three different processes. Process 1 is isobaric process, process 2 is isothermal and process 3 is adiabatic. Let ΔU1, ΔU2 and ΔU3 be the change in internal energy of the gas in these three processes respectively, then