Heat And Thermodynamics Ques 132
- $n$ moles of an ideal gas undergoes a process $A$ and $B$ as shown in the figure. The maximum temperature of the gas during the process will be
(2016 Main)
(a) $\frac{9}{4} \frac{p _0 v _0}{n R}$
(b) $\frac{3}{2} \frac{p _0 v _0}{n R}$
(c) $\frac{9}{2} \frac{p _0 v _0}{n R}$
(d) $\frac{9 p _0 v _0}{n R}$
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Answer:
Correct Answer: 132.(a)
Solution:
Formula:
- $p-V$ equation for path $A B$
$$ p=-\left(\frac{p _0}{V _0}\right) V+3 p _0 $$
$\Rightarrow \quad p V=3 p _0 V-\frac{p _0}{V _0} V^{2}$
or
$$ T=\frac{p V}{n R}=\frac{1}{n R}\left(3 p _0 V-\frac{p _0}{V _0} V^{2}\right) $$
For maximum temperature,
$$ \begin{aligned} \frac{d T}{d V} & =0 \Rightarrow 3 p _0-\frac{2 p _0 V}{V _0}=0 \\ \Rightarrow \quad V & =\frac{3}{2} V _0 \text { and } p=3 p _0-\frac{p _0}{V _0}=\frac{3 p _0}{2} \end{aligned} $$
Therefore, at these values :
$$ \therefore \quad T _{\max }=\frac{\left(\frac{3 p _0}{2}\right)\left(\frac{3 V _0}{2}\right)}{n R}=\frac{9 p _0 V _0}{4 n R} $$
BETA
