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. $\frac{6P_0{V}_0}{nR}$
• B. $\frac{9P_0{V}_0}{4nR}$
• C. $\frac{3P_0{V}_0}{2nR}$
• D. $\frac{7P_0{V}_0}{nR}$

Answer 1

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

The equation of straight line = $y-y_1=m(x-x_1)$
where m is slope of the line.
Slope of the line = $\frac{2P_o-P_o}{V_o-2V_o}$
$=-\frac{P_o}{V_o}$
Equation of line
$P-2P_o=-\frac{P_o}{V_o}(V-V_o)$
$P=-\frac{P_{o}V}{V_o}+3P_o$
Since the given gas is an ideal gas hence we can use PV= nRT
$\frac{nRT}{V}=-\frac{P_{o}V}{V_o}+3P_o$ ....(i)
differentiating with respect to V we get:
$nR\frac{dT}{dV}=-\frac{P_o}{V_o}\left(2V\right)+3P_o$
For maximum temperature put $\frac{dT}{dV}=0$
$\frac{dT}{dV}=3P_o{V}_o-2P_{o}V$
For maximum temperature
$3P_o{V}_o-2P_{o}V=0$
$\Rightarrow V=\frac{3V_o}{2}$...(ii)

putting (ii) value in eq. (i):

$T_{max}=(-\frac{3P_o{V}_o}{2V_o}+3P_o)\frac{3V_o}{2nR}$
$T_{max}=\frac{3P_o{V}_o}{2V_o}\times \frac{3V_o}{2nR}$
$T_{max}=\frac{9P_0{V}_0}{4nR}$