Question

One mole of a ideal gas expands obeying the relation as shown in the $P/V$ diagram. The maximum temperature in this process is equal to

• A. $\frac{P_0{V}_0}{R}$
• B. $\frac{3P_0{V}_0}{R}$
• C. $\frac{9P_0{V}_0}{8R}$
• D. None

Answer 1

The correct option is C $\frac{9P_0{V}_0}{8R}$

From the graph shown the equation of line is
$P-P_0=\left(\frac{\frac{P_0}{2}-P_0}{2V_0-V_0}\right)(V-V_0)$
$P-P_0=\frac{-P_0}{2V_0}(V-V_0)$
$P=\frac{-P_{0}V}{2V_0}+\frac{3P_0}{2}$
Now we know $PV=nRT$
$\Rightarrow (\frac{3}{2}{P}_0-\frac{P_{0}V}{2V_0})V=nRT$
For maximum temperature
$\frac{dT}{dV}=0\Rightarrow \frac{3}{2}{P}_0-\frac{P_{0}V}{V_0}=0$
$V=\frac{3}{2}{V}_0$
$T_{max}=(\frac{3}{2}{P}_0-\frac{P_0}{2V_0}.\frac{3}{2}{V}_0)\frac{3}{2}{V}_0.\frac{1}{nR}$
$=\frac{3}{4}{P}_0.\frac{3}{2}{V}_0.\frac{1}{R}=\frac{9P_0{V}_0}{8R}$