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Question 16 options:

A mixture of equal moles of He and Xe are in a plastic container at 25ºC and 874 mm pressure. If only He leaks from the container and the pressure decreases to 600 mm at 25º C, what is the mole fraction of

(1) He and

(2) Xe in the container after the leak?

2 answers

PV=nRT

V,R, and T remain the same before and after the leak.

Let the combined moles of He and Xe be n1 before the leak, and n2 after the leak.

Before the leak:
n1=874V/RT
Because n1 is equally split between He and Xe, the mole fraction of n1 that is He is (1/2)(874V/RT)=437V, as is the mole fraction of n1 that is Xe.

After the leak:
n2=600V/RT

n2 is not equally split because we lost some He. However, we still have the same amount of Xe, so the mole fraction of Xe is now:
(437V/RT) / (600V/RT) = 437/600

The mole fraction of He is then 1 minus this fraction.
Correction: under "Before the leak:" , 437V should be 437V/RT