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An ideal gas undergoes a circular cycle centred at $4 \,atm , 4 L$ as shown in the diagram. The maximum temperature attained in this process is close to
$\frac{30}{R}$
$\frac{36}{R}$
$\frac{24}{R}$
$\frac{16}{R}$
Solution
(a)
From given cyclic process,
Process equation is
$(p-4)^2+(V-4)^2=4 \quad \dots(i)$
Now, from $p V=n R T$
We can say that $T$ is maximum when $p V$ is maximum.
Now, for given cyclic process, $p V$ maximum occur when $p^2 V^2$ is maximum.
Now,
$p^2 V^2=p^2\left(4-(p-4)^2\right)$ [from Eq. $(i)$]
Now, $p^2 V^2$ is maximum when $\frac{d}{d p} p^2 V^2=0$
$\Rightarrow \quad \frac{d}{d p}\left(p^2 \cdot\left(p^2 \cdot\left(4-(p-4)^2\right)=0\right.\right.$
$\Rightarrow \quad p^2-8 p+14=0$
$=4 \pm \sqrt{2}$
and from Eq. $(i)$, we get
$p=4 \pm \sqrt{2}, V=4 \pm \sqrt{2}$
Taking positive values, we have
$(p V)_{\max } \Rightarrow p=4+\sqrt{2}$
and $V=4+\sqrt{2}$
So, by gas equation, we have
$T_{\max } =\frac{(p V)_{\max }}{R} \quad \text { [for } 1 \,mol \text { of gas] }$
$=\frac{(4+\sqrt{2})(4+\sqrt{2})}{R}$
$=\frac{16+2+2 \times 4 \times \sqrt{2}}{R}$
$=\frac{29.32}{R}=\frac{30}{R}$
