Question

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

• A. $\frac{-\alpha}{5+\alpha}$
• B. $\frac{\alpha}{3+\alpha}$
• C. $\frac{-3 \alpha}{2+\alpha}$
• D. $\frac{5 \alpha}{2+3 \alpha}$

Answer 1

Answer: (A)

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