Degree of freedom of diatomic nitrogen \(=5\)

Degree of freedom of monoatomic nitrogen \(=3\)

Let initial number of moles be \(n\) and \(\alpha\) fraction dissociated.

So fraction dissociated \(=n \alpha\) fraction remaining \(=n-n \alpha\).

\(n \alpha\) break into two so new atoms formed is actually \(2 n \alpha\).

Initial energy is given by \(=n \times \frac{f}{2} \times R T=n \times \frac{5}{2} \times R T\)

Final energy \(=(n-n \alpha) \frac{5}{2} R T_2+2 n \alpha \times \frac{3}{2} R T_2\)

\(=\frac{5}{2} n R T_2-\frac{5}{2} n \alpha R T_2+n \alpha 3 R T_2\)

\(=\frac{5}{2} n R T_2+\frac{n \alpha R T_2}{2}\)

\(=\frac{(5+2) n R T_2}{2}\)

Change in energy is given on zero.

\(\frac{5 n R T}{2}=\frac{(5+\alpha) n R T_2}{2}\)

\(\frac{5 T}{5+\alpha}=T_2\)

\(\Delta T=T_2-T\)

or \(\Delta T=\frac{5 T}{5+\alpha}-T=\frac{-\alpha}{5+\alpha} T\)

Fractional change in temperature \(=\frac{\Delta T}{T}\) or \(-\frac{\alpha}{5+\alpha}\)