CK-12-Chemistry Intermediate

14.4. Gas Mixtures and Molecular Speeds http://www.ck12.org

Another 20.0 liter vessel contains 3.0 mol of helium at a pressure of 1800 mmHg. These two gases are mixed
together in an identical 20.0 liter vessel. Because each will exert its own pressure according to Dalton’s law, we can
express the partial pressures as follows:

PH 2 =XH 2 ×PTotaland PHe=XHe×PTotal

The partial pressure of a gas in a mixture is equal to its mole fraction multiplied by the total pressure. For our
mixture of hydrogen and helium:

XH 2 =

1 .0 mol

1 .0 mol+ 3 .0 mol

= 0 .25 and XHe=

3 .0 mol

1 .0 mol+ 3 .0 mol

= 0. 75

The total pressure according to Dalton’s law is 600 mmHg + 1800 mmHg = 2400 mmHg. So, each partial pressure
will be:

PH 2 = 0. 25 ×2400 mmHg=600 mmHg

PHe= 0. 75 ×2400 mmHg=1800 mmHg

The partial pressures of each gas in the mixture do not change, since they were mixed into the same size vessel and
the temperature was not altered.

Sample Problem 14.8: Dalton’s Law

A flask contains a mixture of 1.24 moles of hydrogen gas and 2.91 moles of oxygen gas. If the total pressure is 104
kPa, what is the partial pressure of each gas?

Step 1: List the known quantities and plan the problem.

Known

• 1.24 mol H 2


• 2.91 mol O 2


• PTotal= 104 kPa

Unknown

• PH 2 =? kPa


• PO 2 =? kPa

First, the mole fraction of each gas can be determined. Then, the partial pressure can be calculated by multiplying
the mole fraction by the total pressure.

Step 2: Solve.

XH 2 =

1 .24 mol

1 .24 mol+ 2 .91 mol

= 0. 299

XO 2 =

2 .91 mol

1 .24 mol+ 2 .91 mol

= 0. 701

PH 2 = 0. 299 ×104 kPa= 31 .1 kPa

PO 2 = 0. 701 ×104 kPa= 72 .9 kPa

Step 3: Think about your result.

The hydrogen is slightly less than one third of the mixture, so it exerts slightly less than one third of the total
pressure.