Question

${}^{\prime }{n}^{\prime }$ 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. $\frac{9P_0{V}_0}{2nR}$
• C. $\frac{9P_0{V}_0}{nR}$
• D. $\frac{9P_0{V}_0}{4nR}$

Answer 1

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

$\begin{array}{c}y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)\\ P-P_0=\frac{2p_0-p_0}{v_{0}2v_0}\left(V-2V_0\right)\\ =\frac{-P_0}{V_0}\left(V-2V_0\right)\\ P=\frac{-P_0}{V_0}V+3p_0\\ PV=\frac{-P_0}{V_0}V+3p_{0}V\\ nRT=\frac{-P_0}{V_0}V+3p_{0}V\\ T=\frac{1}{nR}\left(\frac{-P_0}{V_0}{V}^2+3p_{0}V\right)\\ \frac{dT}{dV}=0\left(formaxtemperature\right)\\ \frac{-P_0}{V_0}2V+3p_0=0\\ \frac{-P_0}{V_0}2V=-3p_0\\ V=\frac{3}{2}{V}_0\left(condtionformaxtemperature\right)\\ T_{max}=\frac{1}{nR}\left(\frac{-P_0}{V_0}\times \frac{9}{4}{V}_0^2+3p_0\times \frac{3}{2}{V}_0\right)\\ T_{max}=\frac{1}{nR}\left(-\frac{9}{4}{P}_0{V}_0+\frac{9}{2}{P}_0{V}_0\right)\\ =\frac{9}{4}\frac{P_0{V}_0}{nR}\end{array}$