Question

At what temperature, the rate of effusion of $N_2$ would be $1.625$ times than that of $SO_2$ at $50^{\circ} C$ ?

• A. 110 K
• B. 173 K
• C. 373 K
• D. 273 K

Answer 1

Answer: (C)

373 K

Answer 2

The rate of effusion is directly proportional to the square root of temperature and inversely proportional to the square root of molecular weight.
$r \propto U ; U =\sqrt{\frac{3 RT }{ M }}$
$r \propto \sqrt{\frac{ T }{ M }}$
$\therefore \frac{ r _{ N _{2}}}{ r _{ SO _{2}}}=\sqrt{\frac{ T _{ N _{2} M _{ SO _{2}}}}{ T _{ SO _{2} M _{ N _{2}}}}}$
The rate of effusion of $N _{2}$ is $1.625$ times than that of $SO _{2}$ at $50^{\circ} C$. or
$\frac{ r _{ N _{2}}}{ r _{ SO _{2}}}=\sqrt{\frac{ T _{1} \times 64}{(50+273) \times 28}}=1.625$
$T _{2}=373\, K$
At $373\, K$, the rate of effusion of $N _{2}$ would be $1.625$ times than that of $SO _{2}$ at $50^{\circ} C$.