Gas Mixtures and Partial Pressures

Gas Mixtures and Partial Pressures

How do we deal with gases composed of a mixture of two or more different substances?

John Dalton (1766-1844) - (gave us Dalton's atomic theory)

The total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone

The partial pressure of a gas:

Dalton's Law of Partial Pressures:

Pt = P1 + P2 + P3 + ...

If each of the gases behaves independently of the others then we can apply the ideal gas law to each gas component in the sample:

And so on for all components. Therefore, the total pressure Pt will be equal to:

At constant temperature and volume, the total pressure of a gas sample is determined by the total number of moles of gas present, whether this represents a single substance, or a mixture


A gaseous mixture made from 10 g of oxygen and 5 g of methane is placed in a 10 L vessel at 25C. What is the partial pressure of each gas, and what is the total pressure in the vessel?

(10 g O2)(1 mol/32 g) = 0.313 mol O2

(10 g CH4)(1 mol/16 g) = 0.616 mol CH4

V=10 L


Pt = PO2 + PCH4 = 0.702 atm + 1.403 atm = 2.105 atm

Partial Pressures and Mole Fractions

The ratio of the partial pressure of one component of a gas to the total pressure is:


The ratio of the partial pressure to the total pressure is equal to the mole fraction of the component gas

The partial pressure of a gas is equal to its mole fraction times the total pressure


a) A synthetic atmosphere is created by blending 2 mol percent CO2, 20 mol percent O2 and 78 mol percent N2. If the total pressure is 750 torr, calculate the partial pressure of the oxygen component.

Mole fraction of oxygen is (20/100) = 0.2

Therefore, partial pressure of oxygen = (0.2)(750 torr) = 150 torr

b) If 25 liters of this atmosphere, at 37C, have to be produced, how many moles of O2 are needed?

PO2 = 150 torr (1 atm/760 torr) = 0.197 atm

V = 25 L

T = (273+37K)=310K

R=0.0821 L atm/mol K

PV = nRT

n = (PV)/(RT) = (0.197 atm * 25 L)/(0.0821 L atm/mol K * 310K)

n = 0.194 mol

1996 Michael Blaber