Thermodynamics

JEE Main level, single correct, medium

173 of 190

Question 173 of 190

nn moles of an ideal gas undergo a process A→BA \rightarrow B as shown in the figure. Maximum temperature of the gas during the process is

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Answer: (D) 9P0V04nR\frac{9 P_0 V_0}{4 n R}

For given graph, equation of PP - VV line is
P−2P0=2P0−P0V0−2V0(V−V0)P-2 P_0=\frac{2 P_0-P_0}{V_0-2 V_0}\left(V-V_0\right)
So, P=3P0−P0V0VP=3 P_0-\frac{P_0}{V_0} V as PV=nRTP V=n R T
⇒(3P0−P0V0V)V=nRT;\Rightarrow\left(3 P_0-\frac{P_0}{V_0} V\right) V=n R T ;
For maximum temperature dTdV=0\frac{d T}{d V}=0
⇒dTdV=3P0−2P0V0V=0⇒V=32V0\Rightarrow \frac{d T}{d V}=3 P_0-\frac{2 P_0}{V_0} V=0 \Rightarrow V=\frac{3}{2} V_0
Also P=32P0P=\frac{3}{2} P_0
So, T=PVnR=1nR×32P0×32V0=9P0V04nRT=\frac{P V}{n R}=\frac{1}{n R} \times \frac{3}{2} P_0 \times \frac{3}{2} V_0=\frac{9 P_0 V_0}{4 n R}

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