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_{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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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 B $Q_{1} > Q_{2} > Q_{3} and Δ U_{1} = Δ U_{2} = Δ U_{3}$
Initial and final condition is same for all process,
$Δ U_{1} = Δ U_{2} = Δ U_{3}$
And $Δ Q = Δ U + Δ W$

Work done $=$ Area of $P - V$ graph

$Δ W_{1} > Δ W_{2} > Δ W_{3}$

So, $Q_{1} > Q_{2} > Q_{3}$

Hence, option $B$ is the correct answer.

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An ideal gas goes from state A to state B via three different processes as indicated in the P−V diagram. If Q1, Q2, Q3 indicate the heat absorbed by the gas along the three processes and ΔU1, ΔU2, ΔU3 indicate the change in internal energy along the three processes respectively, then

The correct option is B Q1>Q2>Q3 and ΔU1=ΔU2=ΔU3 Initial and final condition is same for all process, ΔU1=ΔU2=ΔU3 And ΔQ=ΔU+ΔW Work done =Area of P−V graph ΔW1>ΔW2>ΔW3 So, Q1>Q2>Q3 Hence, option B is the correct answer.