One mole of a gas expands obeying the relation as shown in the $P/V$ diagram. The maximum temperature in this process is equal to
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This question can be solved by maxima
and minima.
since the graph shown here is a straight line, we can write it's equation in the form of $:$
$y=m x+c$
here $y=P, x=V, m=\frac{P_{2}-P_{1}}{V_{2}-V_{1}}=\frac{\frac{P_{0}}{2}-P_{0}}{2 V_{0}-V_{0}}$
$\Rightarrow m=\frac{-P_{0}}{2 V_{0}}$
$c$ is the intercept and $c=\frac{3 P_{0}}{2}$
We can find $c$ by extra polating the graph Now
$P=\frac{-P_{0}}{2 V_{0}} V+\frac{3 P_{0}}{2}$ $...(1)$
Also, from equation of state, we know the $P V=n R T$
for $1$ mole $P V=R T$
or $P=\frac{R T}{V}$
substituting this in equation $( 1)$
$\frac{R T}{V}=\frac{-P_{0}}{2 V_{0}} V+\frac{3 P_{0}}{2}$
$T=\frac{-\bar{P}_{0}}{2 R V_{0}} \cdot V^{2}+\frac{3 P_{0}}{2 R} V$
To find the minimum temperature, we will differentiate the above equation
i.e., $\frac{d T}{d V}=-\frac{2 P_{0}}{2 R V_{0}} \cdot V+\frac{3 P_{0}}{2 R}$
for $\frac{d T}{d V}=0 \Rightarrow \frac{P_{0}}{R V_{0}} \cdot V=\frac{3 P_{0}}{2 R}$
$\Rightarrow V=\frac{3 V_{0}}{2}$
This is the critical point.
$\frac{d^{2} T}{d V^{2}}=-\frac{P_{0}}{R V_{0}}$
which is negative since pressure and volume can't be negative.
$\Rightarrow$ pt of maximum
Now at $V=\frac{3 V_{0}}{2}$
maximum temperature $=T_{\max }=\frac{-P_{0}}{2 R V_{0}} \frac{9 V_{0}^{2}}{4}+\frac{3 P_{0}}{2 R} \cdot \frac{3 V_{0}}{2}$
$\Rightarrow T_{\max }=\frac{9 P_{0} V_{0}}{8 R}$
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