EOC Problem 6.046
A mixture of gases contains 0.65 mol of N2, 0.70 mol of O2, and 0.90 mol of He.
What is the partial pressure of each gas (in atm and in torr) in a 25.0 L cylinder at 340 K?
What is total pressure?
Use mols each gas in PV = nRT and solve for p.
Then Ptotal = pN2 + pO2 + pHe
To find the partial pressure of each gas, we can use the formula:
Partial pressure (P) = moles of gas (n) * gas constant (R) * temperature (T) / volume (V)
Given:
Moles of N2 (n1) = 0.65 mol
Moles of O2 (n2) = 0.70 mol
Moles of He (n3) = 0.90 mol
Volume (V) = 25.0 L
Temperature (T) = 340 K
First, let's find the partial pressure of each gas in atm.
Using the ideal gas law equation PV = nRT, we can rearrange it to solve for P:
P = nRT / V
For N2:
Partial pressure of N2 (P1) = n1RT / V
P1 = (0.65 mol) * (0.0821 L*atm/(K*mol)) * (340 K) / (25.0 L)
P1 = 1.011 atm
For O2:
Partial pressure of O2 (P2) = n2RT / V
P2 = (0.70 mol) * (0.0821 L*atm/(K*mol)) * (340 K) / (25.0 L)
P2 = 1.203 atm
For He:
Partial pressure of He (P3) = n3RT / V
P3 = (0.90 mol) * (0.0821 L*atm/(K*mol)) * (340 K) / (25.0 L)
P3 = 1.460 atm
Now let's convert the partial pressure of each gas to torr.
To convert from atm to torr, we multiply by the conversion factor 1 atm = 760 torr.
Partial pressure of N2 in torr = 1.011 atm * 760 torr/atm = 768.76 torr
Partial pressure of O2 in torr = 1.203 atm * 760 torr/atm = 915.28 torr
Partial pressure of He in torr = 1.460 atm * 760 torr/atm = 1110.56 torr
Finally, let's find the total pressure of the mixture.
Total pressure (Pt) = P1 + P2 + P3
Pt = 1.011 atm + 1.203 atm + 1.460 atm
Pt = 3.674 atm
To convert the total pressure to torr:
Total pressure (Pt) in torr = 3.674 atm * 760 torr/atm = 2793.44 torr
Therefore, the partial pressure of each gas is as follows:
N2: 1.011 atm or 768.76 torr
O2: 1.203 atm or 915.28 torr
He: 1.460 atm or 1110.56 torr
The total pressure is 3.674 atm or 2793.44 torr.