91Ó°ÊÓ

There are four process that is given $1→3,3→2,2→4$and $4→1$.

Work done by the first process $1→3$at constant pressure $p_1$and change of volume from $V_1$to $V_2$is given by:$W_1=p_1\left(V_1-V_2\right)$

Work done by the second process $3→2$, here volume is constant therefore no work done:

$$\begin{gathered} W_t=W_1+W_2+W_4 \\ =p_1\left(V_2+V_1\right)+0+p_2\left(V_1-V_2\right)+0 \\ =\left(p_1-p_2\right)\left(V_2+V_1\right) \end{gathered}$$

For the third process $2→4$, pressure is constant and volume changes from $V_2$to $V_1$and hence the work done is:$W_3=p_2\left(V_1-V_2\right)$

For the fourth process $4→1$ the volume is constant therefore no work is done:

$W_4=0$

To calculate the total work done we can simply add all the work done by the four processes:

$$\begin{gathered} W_t=W_1+W_2+W_4 \\ =p_1\left(V_2+V_1\right)+0+p_2\left(V_1-V_2\right)+0 \\ =\left(p_1-p_2\right)\left(V_2+V_1\right) \end{gathered}$$

Here the$\left(p_1-p_2\right)$ represents the area enclosed by the loop and$\left(V_2-V_1\right)$ is the dimension of the rectangular shape in the pV diagram. Therefore, the work done is equals to the area enclosed by the loop.

If the loop is to be traversed in the opposite direction, the value of the work done will remain same but will have opposite sign.

Therefore, the direction will be -W

$W_t=-\left(p_1-p_2\right)\left(V_2-V_1\right)$

Hence, the work done by if the loop is traversed in the opposite direction is given by$W_t=-\left(p_1-p_2\right)\left(V_2-V_1\right)$