A sample of gas expands from${\mathit{V}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}{\mathbf{\text{m}}}^{\mathbf{\text{3}}}$ and${\mathit{p}}_{\mathbf{1}}\mathbf{=}\mathbf{40}\mathbf{ }\mathbf{\text{Pa}}$ to${\mathit{V}}_{\mathbf{2}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}{\mathbf{\text{m}}}^{\mathbf{\text{3}}}$ and${\mathit{p}}_{\mathbf{2}}\mathbf{=}\mathbf{10}\mathbf{}\mathbf{\text{Pa}}$along path Bin the p-Vdiagram in Fig. 18-58. It is then compressed back to ${\mathit{V}}_{\mathbf{1}}$ along either path Aor path C. Compute the net work done by the gas for the complete cycle along (a) path BAand (b) path BC.

The graph depicts pressure vs. volume.

Initial volume,$V_1=1 {\text{m}}^{\text{3}}$.

Initial pressure,$p_1=40\text{Pa}$.

Final volume,$V_2=4.0{\text{m}}^{\text{3}}$

Final pressure, $p_2=10 \text{Pa}$.

By using the given graph and values for the points on the graph, we can find the work done by the gas. It can be calculated using the area under the curve.

Formula:

Work done by the gas under a thermodynamic cycle

$$\begin{gathered} W=P\Delta V \\ =\frac{P_i+P_f}{2}\Delta V \end{gathered}$$ …(i)

a.

We know that the area under the curve is the work done. When we consider the two paths BA and BC, then the work done for BA cycle is the area under the curve of the horizontal line and the tilted straight line. The work done for BC cycle is area under the curve of the horizontal and tilted lines.

The area under the curve of the vertical straight line is zero.

We have to find the work done by the gas along two paths which can be calculated by the area under the curve.

Now, we find the work done for BC cycle, tilted in a straight line, $\left(1.0m^3,40Pa\right)to\left(4.0m^3,10Pa\right)$using equation (i) is given by

$$\begin{gathered} W_1=\frac{40 \text{Pa}+10 \text{Pa}}{2}\left(4.0-1.0\right){\text{m}}^{\text{3}} \\ =75\text{J} \end{gathered}$$.

Work done for BC cycle, along horizontal line,$\left(4.0m^3,10Pa\right)to\left(1.0m^3,10Pa\right)$ using equation (i) is given by

$$\begin{gathered} W_2=\frac{10 \text{Pa}+10 \text{Pa}}{2}\left(1.0-4.0\right) {\text{m}}^{\text{3}} \\ =-30.0\text{J} \end{gathered}$$.

Work done for BC cycle vertical line,$\left(1.0m^3,10Pa\right)to\left(1.0m^3,40Pa\right)$using equation (i) is given by

$$\begin{gathered} W_3=\frac{10 \text{Pa}+40 \text{Pa}}{2}\left(4.0-4.0\right){\text{m}}^{\text{3}} \\ =0\text{J} \end{gathered}$$.

Work done for BA cycle, horizontal line $\left(4.0m^3,40Pa\right)to\left(1.0m^3,40Pa\right)$ using equation (i) is given by

$$\begin{gathered} W_1^{\text{'}}=\frac{40 \text{Pa}+40 \text{Pa}}{2}\left(1.0-4.0\right){\text{m}}^{\text{3}} \\ =-120.0\text{J} \end{gathered}$$

For work done during BA cycle, the tilted part is the same as the BC cycle, i.e., $W_1=75.0\text{J}$and the main difference is the horizontal part.

So, the total work done during BA cycle is given by

$$\begin{gathered} W^{\text{'}}=W_1^{\text{'}}+W_1 \\ =\left(-120 \text{J}+75 \text{J}\right) \\ =-45\text{J} \end{gathered}$$.

Hence, the total work done along path BA is .

b.

Now, the total work done during BC cycle is given by

$$\begin{gathered} W=W_1+W_2+W_3 \\ =\left(75 \text{J}-30 \text{J}+0 \text{J}\right) \\ =45 \text{J} \end{gathered}$$

Hence, the work done along path BC is $45 \text{J}$.