Question

At what temperature will the rate of diffusion of $N_2$ be 1.6 times the rate of diffusion of $S{O}_2$ at ${27}^0$C.

• A. ${336}^0$C
• B. ${50}^0$C
• C. 300 K
• D. ${63}^0$C

Answer 1

The correct option is D ${63}^0$C

$\text{Rate of diffusion}\propto \sqrt{\frac{T}{M}}$

$\therefore \frac{R_{N_2}}{R_{S{O}_2}}=\sqrt{\frac{T_{N_2}}{M_{N_2}}\times \frac{M_{S{O}_2}}{T_{S{O}_2}}}$

$1.6=\sqrt{\frac{T_{N_2}}{28}\times \frac{64}{300}}$

$\frac{T_{N_2}\times 64}{8400}=2.56$

$T_{N_2}=336K=63°C$