An ideal gas goes from state $A$ to state $B$ via three different processes as indicated in the P-V diagram-
If $Q_{1}, Q_{2}, Q_{3}$ indicate the heat absorbed by the gas along the three processes and $Δ U_{1}, Δ U_{2}, Δ U_{3}$ indicate the change in internal energy along the three processes respectively, then-

Q_{3} > Q_{2} > Q_{1}

and

Δ U_{1} = Δ U_{2} = Δ U_{3}

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Q_{1} = Q_{2} = Q_{3}

and

Δ U_{1} > Δ U_{2} > Δ U_{3}

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Q_{3} > Q_{2} > Q_{1}

and

Δ U_{1} > Δ U_{2} > Δ U_{3}

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Q_{1} > Q_{2} > Q_{3}

and

Δ U_{1} = Δ U_{2} = Δ U_{3}

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Solution

The correct option is B $Q_{1} > Q_{2} > Q_{3}$ and $Δ U_{1} = Δ U_{2} = Δ U_{3}$
$d U_{1} = d U_{2} = d U_{3}$
$Q_{1} > Q_{2} > Q_{3}$
Because
$Δ Q = Δ W + d U$
Change in internal energy independent of path and work depends on path and the heat required increases as the work done increases.
$∴$ $Δ Q_{1} > Δ Q_{2} > Δ Q_{3}$

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