Question

$n$ moles of an ideal gas undergoes a process $A\to B$ as shown in the figure. The maximum temperature of the gas during the process will be:

• A. $\frac{3P_0{V}_0}{2nR}$
• B. $\frac{9P_0{V}_0}{4nR}$
• C. $\frac{9P_0{V}_0}{2nR}$
• D. $\frac{9P_0{V}_0}{nR}$

Answer 1

The correct option is B $\frac{9P_0{V}_0}{4nR}$

Since $P-V$ graph of the process is a straight line and two point $(V_0,2P_0)$ and $(2V_0,P_0)$ are known, its equation will be
$(P-P_0)=(2P_0-P_0)/(V_0-2V_0)(V-2V_0)=\left(P_0\right)/\left(V_0\right)(2V_0-V)$
$:.P=3P_0-\left(P_{0}V\right)/\left(V_0\right)$

According to equation for ideal gas,
$T=\left(pV\right)/\left(nR\right)$
$\left(3P_0-\frac{P_{0}V}{V_0}\right)\frac{V}{nR}$
$=(3P_0{V}_{0}V-P_0{V}^2)/\left(nR{V}_0\right)$

For $T$ to be maximum, we need $\left(dT\right)/\left(dV\right)=0$ and $\left(d^{2}T\right)/\left(d{V}^2\right)<0$

$3P_0{V}_0-2P_{0}V=0$
or $V=\left(3V_0\right)/\left(2\right)$

So we get,
$T_{max}=\frac{3P_0{V}_0\left(\frac{3V_0}{2}\right)-P_0\left(\frac{9{V_0}^2}{4}\right)}{nR{V}_0}=\frac{9P_0{V}_0}{4nR}$