Question

12.00 g of a gaseous mixture of He and methane was taken in a container and to the mixture 8.00g of oxygen gas was added at same temperature. The pressure inside the container increased to a factor of 7/6. What was the weight percentage of methane in the original mixture?

Comments

pls solve answer is 66.67 percentage

Answer 1

Let mass of methane in original mixture be x g

So total number of moles in original mixture n₁ = (12-x)/4+x/16

When 8g O₂ is added to it total number of moles become n₂= (12-x)/4+x/16+ 8/32

Temperature and volume remaining constant pressure is proportional to number of moles as we know PV=nRT.

So P1V=n1RT...(1)

P2V=n2RT...(2)

Dividing (2) by (1) we get

 n₂/n₁=P₂/P₁

_{But by the problem P2/P1=7/6}

Hence n2/n1=7/6

=>[ (12-x)/4+x/16+ 8/32]/[(12-x)/4+x/16]=7/6

=> x= 8g

so mass of original mixture is 12g

Hence the amount of Methane in original mixture =8/12*100%=66.67%

Comments

I am not able to understand how n2/n1 became equal to 7/6 pls specify it briefly...

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Is it OK know?

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what is the value of p1 and p2 pls specify

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Check it now

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it is increased by a factor 7/6 so we should not add 7/6 to p i.e p2=p1+7/6

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Increased by a factor of 7/6 means , changed value P2 is 7/6 times of initial value P1.

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how temperature and pressure are constant

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Not pressure, the temperature and volume of the container are constant as stated in the question.

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Thanks you are great sir