Two gases are mixed in a 10.0L fixed volume flask: 8.0 L of O2 at 2.00 atm and 2.0 L of N2 at 3.00 atm.
A) Calculate the partial pressure for each gas and the total pressure.
I got ...
PO2 = 1.60 atm
PN2= 0.600 atm
B) How much (volume in liters) argon at 5.0 atm must be added to the 10.0 L flask to lower the mole fraction of oxygen to 1/2 ( i.e., to get XO2=0.5)?
Hint: maybe it is easier to do part C first?
C) What is the total pressure from part B (Ar, O2, N2)?
A is ok but you didn't calculate the total pressure. Of course, it's 2.2 atm.
B. You want XO2 to be 0.5
(pO2/Ptotal) = 0.5
What must total pressure be.
(1.6/y) = 0.5
Solve for y and you get 3.2 for the total pressure; i.e., that way, 1.6/3.2 = 0.5
So what pAr is needed.
Ptotal = pO2 + pN2 + pAr
Solve for pAr.
Then pAr x (z/10) = 5 x (z/10) = volume Ar in L.
To solve this problem, we will use Dalton's Law of Partial Pressures, which states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of the individual gases.
A) Calculate the partial pressure for each gas and the total pressure:
To calculate the partial pressure for each gas, use the formula:
Partial Pressure = (Volume of Gas / Total Volume) * Total Pressure
For O2:
Partial Pressure of O2 = (8.0 L / 10.0 L) * 2.00 atm
Partial Pressure of O2 = 1.60 atm
For N2:
Partial Pressure of N2 = (2.0 L / 10.0 L) * 3.00 atm
Partial Pressure of N2 = 0.600 atm
The total pressure is simply the sum of the partial pressures:
Total Pressure = Partial Pressure of O2 + Partial Pressure of N2
Total Pressure = 1.60 atm + 0.600 atm
Total Pressure = 2.20 atm
B) To calculate the volume of argon at 5.0 atm that must be added to the flask, we need to find the new mole fraction of oxygen (XO2). The mole fraction (X) of a gas is given by the equation:
X = Moles of Gas / Total Moles of all Gases
Let's assign variables:
VO2 = initial volume of O2 (8.0 L)
VN2 = initial volume of N2 (2.0 L)
PT = total pressure (2.20 atm)
VO2+A = final volume of O2 and argon
VA = volume of argon to be added
PO2+A = partial pressure of O2 after adding argon (1/2 of the original)
We are given that XO2 = 0.5 (since we want to lower the mole fraction of oxygen to 1/2).
Using the equation for mole fraction, we can solve for the total moles of all gases:
XO2 = MO2 / (MO2 + MN2)
0.5 = MO2 / (MO2 + MN2)
MO2 + MN2 = 2 * MO2
MN2 = MO2
Since nitrogen and oxygen have the same number of moles, we can write the equation for total moles of all gases:
(VO2 + VA) / 22.4 L (1 mole) = (8.0 L / 22.4 L) + (2.0 L / 22.4 L) + (VA / 22.4 L)
VO2 + VA = 8.0 + 2.0 + VA
VO2 = 10.0 L
Now that we have the new volume of O2 and argon, we can use Dalton's Law of Partial Pressures again to find the partial pressure of O2 after adding argon.
PO2+A = (VO2+A / (VO2+A + VN2)) * PT
1/2 * PT = (VO2+A / (VO2+A + 2.0)) * 2.20 atm
Simplifying the equation, we have:
VO2+A / (VO2+A + 2.0) = 1/4
Solving for VO2+A:
VO2 + 2 = 4 VO2 + 4
3 VO2 = 2
VO2 = 2/3 L
Finally, to find the volume of argon that needs to be added:
VA = VO2+A - VO2
VA = (2/3) L - 8.0 L
VA = -7.33 L (Ignoring this negative value since physical volumes cannot be negative)
C) Now, we can find the total pressure after adding the argon. The total pressure will be the sum of the partial pressures of all gases.
Total Pressure = Partial Pressure of O2 + Partial Pressure of Ar + Partial Pressure of N2
Since we already have the partial pressure of O2 (PO2) and the partial pressure of N2 (PN2) from part A, we just need to find the partial pressure of argon (Partial Pressure of Ar). Since the volume of argon (VA) is negative, indicating that no argon should be added, the partial pressure of argon is 0.
Total Pressure = PO2 + Partial Pressure of Ar + PN2
Total Pressure = 1.60 atm + 0 atm + 0.600 atm
Total Pressure = 2.20 atm