Consider two processes on a system as shown in figure. The volume… - Vidyadip

MCQ

Consider two processes on a system as shown in figure. The volumes in the initial states are the same in the two processes and the volumes in the final states are also the same. Let $\Delta\text{W}_1$ and $\Delta\text{W}_2$ be the work done by the system in the processes $A$ and $B$ respectively.

A
$\Delta\text{W}_1>\Delta\text{W}_2.$
B
$\Delta\text{W}_1=\Delta\text{W}_2.$
✓
$\Delta\text{W}_1<\Delta\text{W}_2.$
D
Nothing can be said about the relation between $\Delta\text{W}_1$ and $\Delta\text{W}_2.$

✓

Answer

Correct option: C.

$\Delta\text{W}_1<\Delta\text{W}_2.$

Work done by the system, $\Delta\text{W}=\text{P}\Delta\text{V}$
Here,
$P =$ Pressure in the process
$\Delta\text{V} =$ Change in volume during the process
Let $V_i$ and $V_f$ be the volumes in the initial states and final states for processes $A$ and $B,$ respectively. Then,
$\Delta\text{W}_1=\text{P}_1\Delta\text{V}_1$
$\Delta\text{W}_2=\text{P}_2\Delta\text{V}_2$
But $\Delta\text{V}_2=\Delta\text{V}_1,$
$\big[(\text{V}_{\text{f}_1}-\text{V}_{\text{i}_1})=(\text{V}_{\text{f}_2}-\text{V}_{\text{i}_2})\big]$
$\Rightarrow\frac{\Delta\text{W}_1}{\Delta\text{W}_2}=\frac{\text{P}_1}{\text{P}_2}$
$\Rightarrow\Delta\text{W}_1<\Delta\text{W}_2\ [\because\text{P}_2>\text{P}_1]$

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