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

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?

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.

To solve this problem, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

Let's start solving the problem step by step:

a) To calculate the partial pressure and mole fractions of argon and nitrogen after the valve is opened and the gases are allowed to mix, we can use Dalton's Law of Partial Pressures. According to Dalton's law, the total pressure of the mixture is the sum of the partial pressures of each gas.

Partial pressure of argon (PAr) = (nAr / nTotal) * PTotal
Partial pressure of nitrogen (PN2) = (nN2 / nTotal) * PTotal

To calculate the number of moles of each gas, we can use the ideal gas law rearranged to solve for n:

n = PV / RT

Given:
Volume of Ar ball (VAr) = 1.0 L
Pressure of Ar ball (PAr initial) = 5 atm
Volume of N2 ball (VN2) = 10.0 L
Pressure of N2 ball (PN2 initial) = 2 atm

First, we need to find the number of moles of argon and nitrogen in each ball:

nAr = PAr initial * VAr / RT
nN2 = PN2 initial * VN2 / RT

Substituting the values:
nAr = (5 atm * 1.0 L) / (0.0821 atm L/mol K * 298 K) ≈ 0.199 mol
nN2 = (2 atm * 10.0 L) / (0.0821 atm L/mol K * 298 K) ≈ 0.810 mol

The total number of moles (nTotal) is the sum of the moles of argon and nitrogen:
nTotal = nAr + nN2 ≈ 1.009 mol

Now we can calculate the partial pressure and mole fractions:

PAr = (nAr / nTotal) * PTotal
PN2 = (nN2 / nTotal) * PTotal

Since the gases mix and fill both balls, the total pressure (PTotal) is the same as the sum of the initial pressures of argon and nitrogen before mixing:
PTotal = PAr initial + PN2 initial

Substituting the values:
PAr = (0.199 mol / 1.009 mol) * (5 atm + 2 atm) ≈ 1.252 atm
PN2 = (0.810 mol / 1.009 mol) * (5 atm + 2 atm) ≈ 5.748 atm

The mole fractions can be calculated as the ratio of the number of moles of each gas to the total number of moles:

Mole fraction of argon (XAr) = nAr / nTotal
Mole fraction of nitrogen (XN2) = nN2 / nTotal

Substituting the values:
XAr = 0.199 mol / 1.009 mol ≈ 0.197
XN2 = 0.810 mol / 1.009 mol ≈ 0.803

So, after the valve is opened and the gases mix, the partial pressure of argon is approximately 1.252 atm, the partial pressure of nitrogen is approximately 5.748 atm, the mole fraction of argon is approximately 0.197, and the mole fraction of nitrogen is approximately 0.803.

b) To calculate the total pressure of the mixture, we can use Dalton's Law again, which states that the total pressure is the sum of the individual partial pressures:

PTotal mixture = PAr + PN2

Substituting the values:
PTotal mixture = 1.252 atm + 5.748 atm ≈ 7.0 atm

Therefore, the total pressure of the mixture is approximately 7.0 atm.

c) If we added enough argon to raise its mole fraction to 0.5 in the mixture, we can calculate the new total pressure using the same principle as in part b.

First, we need to find the new number of moles of argon and nitrogen to satisfy the given mole fraction:

nAr_new = XAr_new * nTotal
nN2_new = (1 - XAr_new) * nTotal

Given:
XAr_new = 0.5

Substituting the values:
nAr_new = 0.5 * 1.009 mol ≈ 0.505 mol
nN2_new = (1 - 0.5) * 1.009 mol ≈ 0.504 mol

Now we can calculate the new total pressure:

PTotal_new = (nAr_new / nTotal) * PTotal mixture

Substituting the values:
PTotal_new = (0.505 mol / 1.009 mol) * 7.0 atm ≈ 3.49 atm

Therefore, if we added enough argon to raise its mole fraction to 0.5 in the mixture, the new total pressure would be approximately 3.49 atm.