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(A) The degree of freedom $(f)$ for diatomic nitrogen $(N_2)$ is $5$.The degree of freedom $(f)$ for monoatomic nitrogen $(N)$ is $3$.Let the initial

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

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(A) The degree of freedom $(f)$ for diatomic nitrogen $(N_2)$ is $5$.The degree of freedom $(f)$ for monoatomic nitrogen $(N)$ is $3$.Let the initial number of moles be $n$. If $\alpha$ fraction dissociates,the number of moles of $N_2$ remaining is $n(1-\alpha)$.Since each mole of $N_2$ dissociates into $2$ moles of $N$,the number of moles of $N$ formed is $2n\alpha$.Since the container is insulated,the total internal energy remains constant $(U_i = U_f)$.Initial internal energy: $U_i = n 	imes \frac{5}{2} RT$.Final internal energy: $U_f = n(1-\alpha) 	imes \frac{5}{2} RT_2 + 2n\alpha 	imes \frac{3}{2} RT_2$.Equating $U_i = U_f$:$\frac{5}{2} nRT = nRT_2 [\frac{5}{2}(1-\alpha) + 3\alpha] = nRT_2 [\frac{5 - 5\alpha + 6\alpha}{2}] = nRT_2 [\frac{5+\alpha}{2}]$.Solving for $T_2$: $T_2 = \frac{5T}{5+\alpha}$.The change in temperature is $\Delta T = T_2 - T = \frac{5T}{5+\alpha} - T = \frac{5T - 5T - \alpha T}{5+\alpha} = \frac{-\alpha T}{5+\alpha}$.The fractional change in temperature is $\frac{\Delta T}{T} = \frac{-\alpha}{5+\alpha}$.

Nitrogen gas is filled in an insulated container. If $\alpha$ fraction of moles dissociates without exchange of any energy,then the fractional change in its temperature is ..............

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