Total and Partial Pressure  Dalton's Law of Partial Pressures
How to calculate total pressure and partial pressures for gas mixtures from Ideal Gas Law.
The term partial pressure is used when we have a mixture of two or several gases in the same volume, and it expresses the pressure that is caused by each of the induvidual gases in the mixture.
The total pressure of the gas mixture is the sum of the partial pressure of the component gases:
P _{ tot } = ∑P _{ i } = P_{1} + P_{2}+ P_{3} ...
Where
P _{ tot } = the total pressure
P _{ i } = the pressure of component i ( i can vary from 1,2,3.....up to the number of different gases in the mixture)
From the Ideal Gas Law we have:
PV = nRT or P = nRT / V
Then,
P _{ tot } = n _{ tot } RT/V and P _{ i } = n _{ i } RT/V
where
n _{ i } = the number of moles of component i
n _{ tot } = the total number of moles in the gas mixture, which is the sum of all n _{ i }.
R = the gas constant = 8.3145 [J/mol K] = 0.08206 [L atm/mol K] = 62.37 [L torr /mol K]
T = absolute temperature [K]
V = volume [m3] or [L]
For a gas mixture, the temperature and the volume is the same for all gases, and the gas constant is always the same, and then we get:
P _{ i } /P _{ tot } = (n _{ i } RT/V)/(n _{ tot } RT/V) = n _{ i } /n _{ tot }
We can express the concentration of one gas in the gas mixture as the mole fraction, X _{ i } :
X _{ i } = n _{ i } /n _{ tot }
and then
P _{ i } /P _{ tot } = X _{ i } or P _{ i } = X _{ i } P _{ tot }
See also Nonideal gas  Van der Waal's equation and constants
Example 1.
Dry air consists mainly of nitrogen (78.09vol% or 75.47wt%),oxygen (20.95vol% or 23.20 wt%), argon (0.93vol% or 1.28wt%) and carbondioxide (0.03vol% or 0.046wt%).
If you have 100 g of dry air in a 50 liter closed container, what will the partial pressure of each gas be, and what will the total pressure be at 120°C?
First, we must find how many moles of each gas, using the weight fraction of each gas and molweight of the gases :
n _{ N2 } = 100[g] * 0.7547 /28.02 [g/mol] = 2.693 mol N _{ 2 }
n _{ O2 } = 100[g] * 0.2320 /32.00 [g/mol] = 0.725 mol O _{ 2 }
nAr = 100[g] * 0.0128 /39.95 [g/mol] = 0.032 mol Ar
n _{ CO2 } = 100[g] * 0.00046 /44.01 [g/mol] = 0.001 mol CO _{ 2 }
n _{ tot } = n _{ N2 } + n _{ O2 } + n _{ Ar } + n _{ CO2 } = 3.451 mol gas
Then, assuming the gas mixture behaves ideally, we have:
The total pressure, P _{ tot } = n _{ tot } RT/V = 3.451 [mol]* 0.08206 [L atm/mol K]* (273+120) [K] / 50 [L] = 2.226 atm
P _{ N2 } = X _{ N2 } *P _{ tot } = n _{ N2 } /n _{ tot } *P _{ tot } = (2.693[mol]/3.451[mol])*2.226 atm = 1.737 atm
P _{ O2 } = X _{ O2 } *P _{ tot } = n _{ O2 } /n _{ tot } *P _{ tot } = (0.725[mol]/3.451[mol])*2.226 atm = 0.468 atm
P _{ Ar } = X _{ Ar } *P _{ tot } = n _{ Ar } /n _{ tot } *P _{ tot } = (0.032[mol]/3.451[mol])*2.226 atm = 0.021 atm
P _{ CO2 } = X _{ CO2 } *P _{ tot } = n _{ CO2 } /n _{ tot } *P _{ tot } =(0.001[mol]/3.451[mol])*2.226 atm = 0.0006 atm
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