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An ideal gas ...

Question

An ideal gas of mass $m$ in a state A goes to another state B via three different processes as shown in the fig. If $Q_{1}$ , $Q_{2}$ and $Q_{3}$ denote heats absorbed by the gas along three paths respectively then

Q_{1} < Q_{2} < Q_{3}

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

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

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

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Solution

The correct option is A $Q_{1} < Q_{2} < Q_{3}$
From 1st law of thermodynamics
$Δ Q = Δ U + Δ W$
where, $Δ Q$ is the heat supplied to the system, $Δ U$ is the change in internal energy, $Δ W$ is the work done by the system

In the P-V diagram work done is area under the curve.
Process 3 lies above processes 1 and 2 and has maximum area. Process 1 has the least area.
Hence, $Δ W_{3}$ > $Δ W_{2}$ > $Δ W_{1}$

Since initial states for all three processes are same and final states for all three process are same,
$Δ U_{1} = Δ U_{2} = Δ U_{3}$
$∴ Δ Q_{3}$ > $Δ Q_{2}$ > $Δ Q_{1}$

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