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. $\frac{9\, P_{0}\, V_{0}}{4nR}$
• B. $\frac{3\, P_{0} \,V_{0}}{2 nR}$
• C. $\frac{9\, P_{0} \,V_{0}}{2 nR}$
• D. $\frac{9\, P_{0} \,V_{0}}{nR}$

Answer 1

Answer: (A)

Equation of line $AB$
$y -y_{1} = \frac{y_{2} -y_{1}}{x_{2} -x_{1}}\left(x-x_{1}\right) $
$P - P_{0} = \frac{2P_{0} - P_{0}}{V_{0} - 2V_{0}}$
$\left(V - 2V_{0}\right) = - \frac{P_{0}}{V_{0}} \left(V - 2V_{0}\right) $
$P = - \frac{P_{0}}{V_{0}} V + 3P_{0} $
$PV = - \frac{P_{0}}{V_{0}} V^{2} + 3P_{0}V$
$ nRT = - \frac{P_{0}}{V_{0}}V^{2} + 3P_{0}V $
$T = \frac{1}{nR}\left(\frac{P_{0}}{V_{0} }V^{2} + 3P_{0}V\right)$
$ \frac{dT}{dV} = 0 $ (For maximum temperature)
$ \frac{P_{0}}{V_{0}} 2V + 3P_{0} = 0 $
$\frac{P_{0}}{V_{0}}2V = - 3 P_{0} $
$V = \frac{3}{2}V_{0} $ (Condition for maximum temperature)
$T_{max } = \frac{1}{nR} \left( - \frac{P_{0}}{V_{0} } \times \frac{9}{4} V_{0}^{2}+3P_{0} \times\frac{3}{2}V_{0}\right) $
$= \frac{1}{nR} \left( - \frac{9}{4} P_{0}V_{0} + \frac{9}{2} P_{0}V_{0}0\right) = \frac{9}{4} \frac{P_{0}V_{0}}{nR} $0