Answer:
(a) 
(b) 
Explanation:
Hello,
In this case, since the both gases behave ideally, with the given information we can compute the moles of He in A:
Thus, since the final pressure is 3.60 bar, we can write:
The moles of helium could be computed via solver as:
Or algebraically:
In such a way, the volume of the compartment B is:
Finally, he mole fraction of He is:
Regards.
Since the container of the gas is rigid, the volume of the gas will remain constant. Therefore, when the number of particles were decreased in half then the pressure will also be half of the original given they both are subjected to the same temperature.
PV = nRT
V, T and R are constants so they can be lumped together to a constant k.
P/n = k
P1/n1 = P2/n2
since n2 = n1/2
P1/n1 = P2/<span>n1/2</span>
P2 = P1/2
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