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_{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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Solution

The correct option is D $Q_{1} > Q_{2} > Q_{3}$ and $Δ U_{1} = Δ U_{2} = Δ U_{3}$
For all process $1, 2$ and $3$ .
$Δ U = U_{B} - U_{A}$ is same
Therefore, $Δ U_{1} = Δ U_{2} = Δ U_{3}$
Now, $Δ Q = Δ U + Δ W$
Now, $Δ W =$ work done by the gas.
Therefore $Δ Q_{1} > Δ Q_{2} > Δ Q_{3}$

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Q. 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

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