Question

At what temperature, the rate of effusion of $N_2$ would be $1.625$ times than that of $S{O}_2$ at $50^{\circ }C$ ?

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

Answer 1

Answer: (C)

The rate of effusion is directly proportional to the square root of temperature and inversely proportional to the square root of molecular weight.
$r\propto U;U=\sqrt{\frac{3RT}{M}}$
$r\propto \sqrt{\frac{T}{M}}$
$\therefore \frac{r_{N_2}}{r_{S{O}_2}}=\sqrt{\frac{T_{N_2{M}_{S{O}_2}}}{T_{S{O}_2{M}_{N_2}}}}$
The rate of effusion of $N_2$ is $1.625$ times than that of $S{O}_2$ at $50^{\circ }C$. or
$\frac{r_{N_2}}{r_{S{O}_2}}=\sqrt{\frac{T_1\times 64}{(50+273)\times 28}}=1.625$
$T_2=373K$
At $373K$, the rate of effusion of $N_2$ would be $1.625$ times than that of $S{O}_2$ at $50^{\circ }C$.