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. $\frac{9P_0{V}_0}{nR}$
• D. $n=\frac{C_P}{C_V}$

Answer 1

Answer: (B)

$\frac{9P_0{V}_0}{4nR}$

$\text{Solution}$

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

$(P-P_0)=\frac{(2P_0-P_0)}{(V_0-2V_0)}(V-2V_0)=\frac{P_0}{V_0}(2V_0-V)$

$\therefore P=3P_0-\frac{P_{0}V}{V_0}$

According to equation for ideal gas,

$T=\frac{pV}{nR}=(3P_0-\frac{P_{0}V}{V_0})\frac{V}{nR}=\frac{3P_0{V}_{0}V-P_0{V}^2}{nR{V}_0}$

For T to be maximum,

$\frac{dT}{dV}=03P_0{V}_0-2P_{0}V=0$ or

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

Putting the value in Eq. (i) we get

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

$\text{Hence B is the correct option}$