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Relating Translational KE and T

A gaseous mix...

Question

A gaseous mixture containing He,CH4 and SO2 was allowed to effuse through a fine hole then find what the molar ratio of gases coming out initially?(Given mixture contains He,CH4 and SO2 in 1:2:3 mole ratio).

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Solution

Effusion is process in which gas is escaped from container having a tiny hole into an evacuated space. According to kinetic theory of gases, lesser is the mass, more will be the speed of gas because most probable speed of its molecules is higher.
Graham's Law of Diffusion states that the rate of effusion is directly proportional to the speed of the gaseous molecules and inversely proportional to the square root of molar mass.
$R a t e o f E f f u s i o n α s p e e d α \frac{1}{\sqrt{M}} M = m o l a r m a s s$
In the given question, mole ratio of He : CH₄ : SO₂ = 1 :2 : 3
Molar mass of He = 4 g
Molar mass of CH₄ = 16 g
Molar mass of SO₂= 64 g
$N u m b e r o f m o l e s = \frac{n u m b e r o f m o l e s o f t h e c o m p o n e n t}{t o t a l n u m b e r o f m o l e s} M o l e r a t i o o f H e : C H_{4} : S O_{2} = 1 : 2 : 3 M o l a r m a s s r a t i o o f H e : C H_{4} : S O_{2} = 4 : 16 : 64 = 1 : 4 : 16 \sqrt{\frac{m a s s o f H e}{m a s s o f C H_{4}}} = \sqrt{\frac{4}{16}} = \frac{1}{2} \frac{E f f u s i o n r a t e o f H e}{E f f u s i o n r a t e o f C H_{4}} = \frac{1}{\frac{1}{2}} = 2 \sqrt{\frac{m a s s o f C H_{4}}{m a s s o f S O_{2}}} = \sqrt{\frac{16}{64}} = \frac{1}{2} \frac{E f f u s i o n r a t e o f C H_{4}}{E f f u s i o n r a t e o f S O_{2}} = \frac{1}{\frac{1}{2}} = 2 S o, t h e r a t i o o f e f f u s i o n r a t e o f H e : C H_{4} : S O_{2} = 4 : 2 : 1$

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