If two gases $\mathrm{CH}_4$ and $\mathrm{SO}_2$, are allowed to enter from the two ends of a $1…
If two gases $\mathrm{CH}_4$ and $\mathrm{SO}_2$, are allowed to enter from the two ends of a $1 \mathrm{~km}$ long vacuum tube at the same time, where will the gases meet from the $\mathrm{CH}_4$ end?
$500 \mathrm{~m}$
$620 \mathrm{~m}$
$667 \mathrm{~m}$
$720 \mathrm{~m}$
Solution
$\mathrm{M}_{\mathrm{CH}_4}=16 ;\left(\mathrm{C}_{\mathrm{CH}_4}\right)_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{16}}=\frac{\sqrt{3 \mathrm{RT}}}{4} \mathrm{~m} / \mathrm{s}$
$
\begin{aligned}
& \mathrm{M}_{\mathrm{SO}_2}=64 ;\left(\mathrm{C}_{\mathrm{SO}_2}\right)_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{64}}=\frac{\sqrt{3 \mathrm{RT}}}{8} \mathrm{~m} / \mathrm{s} \\
& \therefore\left(\mathrm{C}_{\mathrm{CH}_4}\right)_{\mathrm{rms}}>\left(\mathrm{C}_{\mathrm{SO}_2}\right)_{\mathrm{rms}}
\end{aligned}
$
Suppose, at time ' $\mathrm{t}$ ', both the gas will meet.
At time $\mathrm{t}, \mathrm{CH}_4$ molecules travel $\frac{\sqrt{3 \mathrm{RT}}}{4} \mathrm{tm}$ and At time $\mathrm{SO}_2$ molecules travel $\frac{\sqrt{3 \mathrm{RT}}}{8} \mathrm{tm}$.
$\therefore \frac{\sqrt{3 \mathrm{RT}} \mathrm{t}}{4}+\frac{\sqrt{3 \mathrm{RT}} \mathrm{t}}{8}=1000$
or $\frac{\sqrt{3 \mathrm{RT}} \mathrm{t}}{4}\left(1+\frac{1}{2}\right)=1000$
or $\frac{\sqrt{3 \mathrm{RT}} \mathrm{t}}{4} \times \frac{3}{2}=1000$
or $\quad t=\frac{8000}{3 \sqrt{3 R T}}$
$\therefore$ Distance travelled by time ' $\mathrm{t}$ '
$
=\frac{\sqrt{3 \mathrm{RT}}}{4} \times \frac{8000}{3 \sqrt{3 \mathrm{RT}}}=667 \mathrm{~m}
$