12. THERMODYNAMICS (HM)
Explanation:
(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
\(T_{\max } =\frac{(p V)_{\max }}{R} \quad \text { [for } 1 \,mol \text { of gas] }\)
\(=\frac{(4+\sqrt{2})(4+\sqrt{2})}{R}\)
\(=\frac{16+2+2 \times 4 \times \sqrt{2}}{R}\)
\(=\frac{29.32}{R}=\frac{30}{R}\)
