Question

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A vessel contains equal no of moles of helium and methane . Due to a hole in the vessel , half of the gaseous mixture effused out. What is the ratio of the number of moles of helium and methane remaining in vessel?​

Answer 1

HEYA!!!!!!

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We have , PV = nRT

Now, at constant P and T, we get, V∝ n ( no. of moles)

V1 / V2 = n1 / n2

As half of the gaseous mixture effused out, half is still remaining in the vessel, so

Since, V2 = V1 /2 so,

Substituting values we get:

V1/V1/2 =n1/n2

n1 / n2 = 2

Hence, the ratio of no. of moles of helium and methane remaining is = 1:2

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Answer 2

Answer: 1:2

Explanation:

We know that,

At a same time rate of diffusion of quantity of diffusion of a gas is inversely proportional to square root of its molecular mass. It is know  as Graham's law of diffusion. If the  diffused moles of a gas  is r and molecular mass is M. Then,

r ∝ 1/√M

So,

Molecular mass of He(M₁) = 4

Molecular mass of Methane CH₄ (M₂) = 16

If there diffused moles are r₁ and r₂ respectively,

Then,

$\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}\\\;\\\frac{r_1}{r_2}=\sqrt{\frac{16}{4}}\\\;\\\frac{r_1}{r_2}=\sqrt{4}\\\;\\\frac{r_1}{r_2}=2$

From this, We can say that helium diffused 2 times more than the methane

And it is mentioned that in starting both have name number of moles but after diffusion helium diffused 2 times more.

It means that now  in the vessel there must be less helium in comparison of methane.

So there methane will  remain twice  more than helium

Hence, Ratio of Helium and methane will be  1:2