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A gas takes part in two processes in which it is heated from the same initial state 1 to the same final temperature. The process are shown on the P-V diagram by the straight line 1-2 and 1-3. 2 and 3 are the points on the same isothermal curve. $Q_{1}$ and $Q_{2}$ are the heat transfer along the two processes. Then in which case will the heat transfer be more ?

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Solution

Let the volumes and temperatures of gas at points $1$ , $2$ and $3$ are $V_{1}, T_{1}$ , $V_{2}, T_{2}$ and $V_{3}, T_{3}$ respectively.
Given : $T_{2} = T_{3}$
Thus $Δ U_{1} = n C_{v} (T_{2} - T_{1})$
Also $Δ U_{2} = n C_{v} (T_{3} - T_{1})$
But $T_{2} = T_{3}$ $⟹ Δ U_{1} = Δ U_{2}$
Work done by the gas in process 1, $W_{1} = n R T l n \frac{V_{2}}{V_{1}}$

Also work done by gas in process 2, $W_{2} = n R T l n \frac{V_{3}}{V_{1}}$
$V_{3} > V_{2}$ $⟹ W_{2} > W_{1}$
From 1st law : $Q = Δ U + W$
$∴$ $Q_{1} = Δ U_{1} + W_{1}$
Also $Q_{2} = Δ U_{2} + W_{2}$
$∴$ $Q_{2} - Q_{1} = W_{2} - W_{1} > 0$ $(∵ Δ U_{1} = Δ U_{2})$
$⟹$ $Q_{2} > Q_{1}$

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