3 mole of gas " X " and 2 moles of gas " Y enters from end " P " and " Q " of the cylinder respectively. The cylinder has the area of cross-section A, shown as under - The length of the cylinder is 15

Question

3 mole of gas " X " and 2 moles of gas " Y enters from end " P " and " Q " of the cylinder respectively. The cylinder has the area of cross-section A, shown as under -

The length of the cylinder is 150 cm. The gas "X" intermixes with gas "Y" at the point A. If the molecular weight of the gases X and Y is 20 and 80 respectively, then what will be the distance of point A from Q?

TestHub · Accepted Answer

According to Graham's law, the rate of diffusion (and distance traveled $x$ ) is proportional to the number of moles ( $n$ ) and inversely proportional to the square root of molecular weight ( $M$ ):

$\frac{x _{X}}{x _{Y}} = \frac{n _{X}}{n _{Y}} \frac{M _{Y}}{M _{X}}$

Substituting the values gives the ratio of distances from ends P ( $x_{X}$ ) and Q ( $x_{Y}$ ):

$\frac{x _{X}}{x _{Y}} = \frac{3}{2} \times \frac{80}{20} = \frac{3}{2} \times 2 = 3$

So, $x_{X} = 3 x_{Y}$ .

Since the total length of the cylinder is 150 cm, we can set up the equation:

$x_{X} + x_{Y} = 3 x_{Y} + x_{Y} = 150 c m$

Solving for $x_{Y}$ gives $4 x_{Y} = 150 c m ⟹ x_{Y} = 37.5 c m$ .

Chemistry - Ideal Gas Question with Solution | TestHub

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