__Gas Mixtures and
Partial Pressure__

So far we have only been dealing with pure gases. How do we apply the gas laws to mixtures of two or more gases?

__Dalton’s Law of
Partial Pressures__ ® The total pressure of a mixture of
gases equals the sum of the pressures that each gas would exert
if it were alone.

Mathematically Dalton’s Law of Partial Pressures can be written as:

P_{tot}
= P_{1} + P_{2} + P_{3} + …

This property greatly simplifies our life (actually it probably doesn’t simplify your life that much, but it makes working the homework, quiz and exam problems a lot easier).

When we apply the ideal gas law to mixtures of gases each component gas will have its own P and n, but all of the component gases will have the same T and V.

__Example__

What is
the total pressure exerted by a mixture of 2.00 g of H_{2}
(MW = 2.016 g/mol), 8.00 g of N_{2} (MW = 28.01 g/mol)
and 12.0 g of Ar (AW = 39.95 g/mol) at 273 K in a 10.0 L vessel?

First we need to calculate the number of moles of each gas:

2.00 g H_{2}
´ [1 mol/2.016 g] =
0.992 mol

8.00 g N_{2}
´ [1 mol/28.01 g] =
0.286 mol

12.0 g Ar ´ [1 mol/39.95 g] = 0.300 mol

Now we can calculate the partial pressure of each gas:

P(H_{2})
= nRT/V = [(0.992)(0.08206)(273)]/(10.0) = 2.22 atm

P(N_{2})
= nRT/V = [(0.286)(0.08206)(273)]/(10.0) = 0.641 atm

P(Ar) = nRT/V = [(0.300)(0.08206)(273)]/(10.0) = 0.672 atm

We can now use the law of partial pressures to calculate the total pressure:

P_{tot}
= P(H_{2}) + P(N_{2}) + P(Ar) = 2.22 + 0.641 +
0.672 = 3.53 atm

A somewhat shorter approach to the same answer would have been to total the number of moles before applying the ideal gas law:

n_{tot}
= n(H_{2}) + n(N_{2}) + n(Ar) = 0.922 + 0.286 +
0.300 = 1.578 mol

P_{tot}
= nRT/V = [(1.578)(0.08206)(273)]/(10.0) = 3.53 atm

Another way we can use Dalton’s Law of Partial Pressures is to calculate the partial pressure of any gas in a mixture if we know the total pressure and the number of moles of each gas (and vice versa).

P_{1}/P_{tot}
= n_{1}/n_{tot}

__Example__

If a gas
which is 5 mol% H_{2} and 95 mol% N_{2} is
present in a container at 200 atm, what is the partial pressure
of each gas?

P(H_{2})
= P_{tot}n_{H2}/n_{tot} = [(200
atm)(0.05n_{tot})]/( n_{tot}) = 10 atm

P(N_{2})
= P_{tot}n_{N2}/n_{tot} = [(200
atm)(0.95n_{tot})]/( n_{tot}) = 190 atm