Question

If two gases \(CH_4\) and \(SO_2\), are allowed to enter from the two ends of a 1 km long vacuum tube at the same time, where will the gases meet from the \(CH_4\) end?

• A. \(500\ \text{m}\)
• B. \(620\ \text{m}\)
• C. \(667\ \text{m}\)
• D. \(720\ \text{m}\)

Hint:

According to Graham’s law, \[ \text{Rate of diffusion} \propto \frac{1}{\sqrt{\text{Molar mass}}} \] Lighter gases diffuse faster.

Answer 1

The Correct Option is C

Step 1: Use Graham’s law of diffusion.
According to Graham’s law, \[ \frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}} \] Since distance travelled in same time is proportional to rate of diffusion, \[ \frac{d_{CH_4}}{d_{SO_2}} = \sqrt{\frac{M_{SO_2}}{M_{CH_4}}} \]

Step 2: Calculate molar masses.
For methane, \[ M_{CH_4}=16 \] For sulphur dioxide, \[ M_{SO_2}=64 \] Thus, \[ \frac{d_{CH_4}}{d_{SO_2}} = \sqrt{\frac{64}{16}} = \sqrt{4} = 2 \] Hence, \[ d_{CH_4}:d_{SO_2}=2:1 \]

Step 3: Use total length of tube.
Total length of tube: \[ 1\ \text{km}=1000\ \text{m} \] Let \[ d_{SO_2}=x \] Then, \[ d_{CH_4}=2x \] Therefore, \[ 2x+x=1000 \] \[ 3x=1000 \] \[ x=\frac{1000}{3} \] Hence, \[ d_{CH_4}=\frac{2000}{3} \approx 667\ \text{m} \]

Step 4: Final conclusion.
Therefore, the gases meet at \[ \boxed{667\ \text{m}} \] from the \(CH_4\) end.