Suppose that the container of the volumo-V\'is filled at foom temperature and at

Question

Suppose that the container of the volumo-V'is filled at foom temperature and at total pressure p with a mixture of H gas of 70% and He gas of 30% in the atomic concentration. Let us consider situation when the hole of thesi is suffciently small so that molecules emerge through the hole by effusion if a vacuum is maintained container is filled with only He gas at the room temperature and pressure p, the pressure will decrease to the half of its original value after 30 min. What will be the atomic ratio of two gas mixtures remained in the container after one hour? ze of radius r is made and outside the container. On the other hand, it is known that when the

Explanation / Answer

The two gases are present as 70% H2 and 30% He atomic concentrations.

He is a monatomic gas whereas H2 is a diatomic gas so in terms of actual mole% concentration they will be

H2 :He = 35% : 30% mole%

Since the hole permits molecules to leave by effusion we can assume that after the same time the same number of molecules will leave from the mixture too.

Now with pure Helium gas since P becomes half in the equation

PV=nRT

V, R and T being constant n which is number of moles will become Half after 30 min

In a similar analogy in the mixture also n will become 1/2 in 30 min and another 1/2 of the total in 30 min more

which is 1/2 x 1/2 = 1/4 of the original pressure

Since the mixture is random the H2 will become 1/4 and He will become 1/4

so by mole% we will have 8.75% H2 and 7.5% He of the original in mole% compared to the original.

In terms of atomic ratio we will have 8.75 x 2 atoms of H2 which is 17.5 atoms% versus 7.5% He atoms

In ratio it is 70% hydrogen and 30% Helium.

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