Question

One mole of 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 of these

Answer 1

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

This question can be solved by maxima

and minima.

Since the graph shown here is a straight

line, we can write it's equation in

the form of $:y=mx+c$

here $y=P,x=V,m=\frac{P_2-P_1}{V_2-V_1}=\frac{\frac{P_0}{2}-P_0}{2V_0-V_0}$

$\Rightarrow m=\frac{-P_0}{2V_0}$

$c$ is the intercept and $c=\frac{3P_0}{2}$

We can find $c$ by extra polating the graph

Now $P=\frac{-P_0}{2V_0}V+\frac{3P_0}{2}$ ...(1)

Also, from equation of state, we know the

$PV=nRT$

for $1mole$

$PV=RT$

or $P=\frac{RT}{V}$

Substituting this in equation (1)

$\frac{RT}{V}=\frac{-P_0}{2V_0}V+\frac{3P_0}{2}$

$T=\frac{-P_0}{2R{V}_0}.V^2+\frac{3P_0}{2R}V$

To find the minimum temperature, we

will differentiate the above equation

i.e., $\frac{dT}{dV}=-\frac{2P_0}{2R{V}_0}.V+\frac{3P_0}{2R}$

for $\frac{dT}{dV}=0\Rightarrow \frac{P_0}{R{V}_0}.V=\frac{3P_0}{2R}$

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

This is the critical point.

$\frac{d^{2}T}{d{V}^2}=-\frac{P_0}{R{V}_0}$

which is negative since

pressure and volume can't

be negative.

$\Rightarrow$ pt. of maximum

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

maximum temperature $=T_{max}=\frac{-P_0}{2R{V}_0}\frac{9V_0^2}{4}+\frac{3P_0}{2R}.\frac{3V_0}{2}$

$\Rightarrow T_{max}=\frac{9P_0{V}_0}{8R}$

So, option (c) is correct option.