91影视

In the first process the pressure changes where as the volume is constant at $V_1=0.2$. Therefore, the first process is drawn with red line. In the second process the pressure is constant while the volume changes from $V_1$to $V_2$, it is drawn with green line.

To calculate the work done during both the process. We will calculate the work done for each process and then combine the two, work done to get the total work done.

In the first process:

the volume is constant therefore the work done is zero.

In the second process,

$W_1=0$

The pressure is kept constant at$p=5脳{10}^5$ , and the volume changes from$V_1$to$V_2$and the work done is given by$W_2$ .

$$\begin{gathered} W_2=p\left(V2-V1\right) \\ =5脳{10}^{5}pax\left(0.12-0.2\right) \\ =-4脳{10}^{4}J \end{gathered}$$

Now the total work done is:

$$\begin{gathered} W_t=W_1+W_2 \\ =0+W1+W2 \\ =0+\left(-4脳{10}^{4}J\right) \\ =-4脳{10}^{4}J \end{gathered}$$

Therefore, the total work done by the processes is$-4脳{10}^{4}J$