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Q. image
In the given figure, $1$ represents isobaric, $2$ represents isothermal and $3$ represents adiabatic processes of an ideal gas. If $\Delta U_{1}, \Delta U_{2}, \Delta U_{3}$ be the changes in internal energy in these processes respectively, then

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Solution:

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$\Delta T_{1}=T_{0}$
$\Delta T_{2}=0$
$\Delta T_{3}=-v e$
$\Rightarrow \Delta U_{1}>\Delta U_{2}>\Delta U_{3}$