Step 1: State Graham's Law of Diffusion.
At the same temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass: \[ r \propto \frac{1}{\sqrt{M}} \] Since both gases are at the same conditions and enter simultaneously, the distance each gas travels in the same time is proportional to its rate of diffusion.
Step 2: Write the distance ratio.
\[ \frac{d_{CH_4}}{d_{SO_2}} = \sqrt{\frac{M_{SO_2}}{M_{CH_4}}} \] This ratio holds because the gas with lower molar mass (lighter gas) diffuses faster and covers more distance.
Step 3: Calculate molar masses and the ratio.
$ M_{CH_4} = 12 + 4(1) = 16\text{ g mol}^{-1} $. $ M_{SO_2} = 32 + 2(16) = 64\text{ g mol}^{-1} $. \[ \frac{d_{CH_4}}{d_{SO_2}} = \sqrt{\frac{64}{16}} = \sqrt{4} = 2 \] So $ CH_4 $ travels twice as far as $ SO_2 $ in the same time.
Step 4: Set up the distance equation.
Total tube length = 1000 m. Let $ d_{SO_2} = x $, then $ d_{CH_4} = 2x $. The gases meet when their combined distances equal the tube length: \[ 2x + x = 1000 \implies 3x = 1000 \implies x = \frac{1000}{3}\text{ m} \]
Step 5: Find the meeting point from the CH4 end.
\[ d_{CH_4} = 2x = \frac{2000}{3} \approx 666.7\text{ m} \approx 667\text{ m} \] from the $ CH_4 $ end. This makes sense: the lighter $ CH_4 $ travels further.
Step 6: Final answer.
\[ \boxed{667\text{ m from the }CH_4\text{ end}} \]