1. Use the ideal gas law

For an ideal gas, $PV=nRT \,\Rightarrow\, T=\frac{PV}{nR}.$

So, along the given process on the $P$-$V$ diagram, the temperature is proportional to the product $PV$.

2. Equation of the path $A \to B$

From the graph, the straight line joins the intercepts $A:(V_0,3P_0), \qquad B:(3V_0,0).$

Hence its equation is obtained from two-point form.

Slope: $m=\frac{0-3P_0}{3V_0-V_0}=\frac{-3P_0}{2V_0}.$

Therefore, $P-3P_0=\frac{-3P_0}{2V_0}(V-V_0).$

Simplifying, $P=\frac{9P_0}{2}-\frac{3P_0}{2V_0}V.$

3. Write temperature as a function of volume

Using $T=\frac{PV}{nR},$ we get $T(V)=\frac{1}{nR}\left(\frac{9P_0}{2}V-\frac{3P_0}{2V_0}V^2\right).$

So, $T(V)=\frac{3P_0}{2nR}\left(3V-\frac{V^2}{V_0}\right).$

4. Maximize $T(V)$

Differentiate with respect to $V$: $\frac{dT}{dV}=\frac{1}{nR}\left(\frac{9P_0}{2}-\frac{3P_0}{V_0}V\right).$

Set it equal to zero: $\frac{9P_0}{2}-\frac{3P_0}{V_0}V=0.$

$\frac{9}{2}=3\frac{V}{V_0}$

$\frac{V}{V_0}=\frac{3}{2}.$

So the temperature is maximum at $V=\frac{3V_0}{2}.$

5. Find corresponding pressure

Substitute into the line equation: $P=\frac{9P_0}{2}-\frac{3P_0}{2V_0}\cdot \frac{3V_0}{2}$

$P=\frac{9P_0}{2}-\frac{9P_0}{4}=\frac{9P_0}{4}.$

6. Compute maximum temperature

$T_{\max}=\frac{PV}{nR}=\frac{1}{nR}\left(\frac{9P_0}{4}\right)\left(\frac{3V_0}{2}\right)$

$T_{\max}=\frac{27P_0V_0}{8nR}.$

7. Compare with options

Given options are:

• A: $\dfrac{9P_0V_0}{2nR}$
• B: $\dfrac{9P_0V_0}{nR}$
• C: $\dfrac{9P_0V_0}{4nR}$
• D: $\dfrac{3P_0V_0}{2nR}$

Our derived value is $\boxed{\frac{27P_0V_0}{8nR}}.$ This does not match any option.

So the stored correct answer $\text{C}$ is not consistent with the straight-line graph described above. If the figure indeed has endpoints $A(V_0,3P_0)$ and $B(3V_0,0)$, then the correct maximum temperature should be $\boxed{\frac{27P_0V_0}{8nR}}.$