'x' cm\(^3\) of CH\(_4\) gas diffused through a porous membrane in 25 min, Under the same conditions 'y' cm\(^3\) of another gas of molar mass 64 g mol\(^{-1}\) diffused in 20 min. The ratio of x and y is:
Show Hint
- \(1:2\)
- \(2:1\)
- \(2:5\)
- \(5:2\)
The Correct Option is D
Solution and Explanation
Step 1: Apply Graham's Law of Effusion. Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
Step 2: Calculate the relative rates of diffusion. Given that:
Molar mass of CH\(_4\) = 16 g/mol.
Molar mass of the other gas = 64 g/mol.
\[ \frac{{Rate of CH}_4}{{Rate of other gas}} = \frac{\sqrt{64}}{\sqrt{16}} = \frac{8}{4} = 2 \]
Step 3: Relate the rates to the volumes diffused over time. Using the given times for diffusion: \[ \frac{x}{25} = 2 \times \frac{y}{20} \]
Step 4: Solve for the ratio \(\frac{x}{y}\). \[ \frac{x}{y} = 2 \times \frac{25}{20} = 2.5 \] Since the ratio \(2.5\) is equivalent to \(5:2\), the correct answer is confirmed to be option (4).
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