A 1.0 L ball containing Ar at 5 atm is connected to a 10.0 L ball containing N2 at 2 atm.

Question

a) Calculate the partial pressure and mole fractions of argon and nitrogen after the valve is opened and the gases are allowed to mix (they fill the balls on both sides).

b) What is the total pressure of the mixture?

c) If we added enough argon to raise its mole fraction to 0.5 in the mixture, what would the new total pressure be?

Answer 1

1.
p1v1 = p2v2
p1= 5 atm
v1 = 1 L
p2 = ?
v2 = 11
Solve for p2 = partial pressure Ar in the combined system.

Do the same and calculate pN2.

Ptotal = pN2 + pAr

pN2 = XN2*Ptotal. Substitute and solve for XN2.

pAr = XAr*Ptotal. Solve for XAr.

3.
If XAr = 0.5, then XN2 must be 0.5 since XN2 + XAr must = 1.00
pN2 = XN2*Ptotal. Partial P N2 must still be 1.82 since adding something else may change the partial pressure of that something else but the partial pressure N2 is the partial pressure of N2. That allows you to calculate Ptotal.