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 $-\frac{\alpha}{5+\alpha}$