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$n$ moles of an ideal gas undergo a process $A \rightarrow B$ as shown in the figure. Maximum temperature of the gas during the process is
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Verified Answer
The correct answer is:
$\frac{9 P_0 V_0}{4 n R}$
$\frac{9 P_0 V_0}{4 n R}$
For given graph, equation of $P$ - $V$ line is
$$
P-2 P_0=\frac{2 P_0-P_0}{V_0-2 V_0}\left(V-V_0\right)
$$
So, $P=3 P_0-\frac{P_0}{V_0} V$ as $P V=n R T$
$$
\Rightarrow\left(3 P_0-\frac{P_0}{V_0} V\right) V=n R T ;
$$
For maximum temperature $\frac{d T}{d V}=0$
$$
\Rightarrow \frac{d T}{d V}=3 P_0-\frac{2 P_0}{V_0} V=0 \Rightarrow V=\frac{3}{2} V_0
$$
Also $P=\frac{3}{2} P_0$
So, $T=\frac{P V}{n R}=\frac{1}{n R} \times \frac{3}{2} P_0 \times \frac{3}{2} V_0=\frac{9 P_0 V_0}{4 n R}$
$$
P-2 P_0=\frac{2 P_0-P_0}{V_0-2 V_0}\left(V-V_0\right)
$$
So, $P=3 P_0-\frac{P_0}{V_0} V$ as $P V=n R T$
$$
\Rightarrow\left(3 P_0-\frac{P_0}{V_0} V\right) V=n R T ;
$$
For maximum temperature $\frac{d T}{d V}=0$
$$
\Rightarrow \frac{d T}{d V}=3 P_0-\frac{2 P_0}{V_0} V=0 \Rightarrow V=\frac{3}{2} V_0
$$
Also $P=\frac{3}{2} P_0$
So, $T=\frac{P V}{n R}=\frac{1}{n R} \times \frac{3}{2} P_0 \times \frac{3}{2} V_0=\frac{9 P_0 V_0}{4 n R}$
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