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

Engineering

Physics

Thermodynamics

Question

$\frac{3 P_{0} V_{0}}{2 nR}$

$\frac{9 P_{0} V_{0}}{4 nR}$

$\frac{9 P_{0} V_{0}}{2 nR}$

$\frac{9 P_{0} V_{0}}{nR}$

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$P = - \frac{P_{0}}{V_{0}} V + 3 P_{0} = \frac{nRT}{V}$

$T = \frac{1}{nR} [- \frac{P_{0}}{V_{0}} V^{2} + 3 P_{0} V]$

$\frac{dT}{dV} = 0 V = \frac{3 V_{0}}{2}$

$T = \frac{9 P_{0} V_{0}}{4 nR}$