Dalton’s Law of Dalton’s Law of Partial PressurePartial Pressure

Dalton’s Law of Partial Dalton’s Law of Partial PressurePressure

John DaltonJohn Dalton responsible for atomic theoryresponsible for atomic theory also studied gas mixturesalso studied gas mixtures

the P of gas mixture is the sum of the the P of gas mixture is the sum of the individual pressures of each gas individual pressures of each gas alonealone

the P that each gas exerts in the the P that each gas exerts in the mixture is independent of the P that mixture is independent of the P that are exerted by other gasesare exerted by other gases

Dalton’s Law of Partial Dalton’s Law of Partial PressurePressure

The total pressure in a container is The total pressure in a container is the sum of the partial pressures of all the sum of the partial pressures of all the gases in the container. the gases in the container.

PPTT = P = P11 + P + P22 + P + P33 + …P + …Pnn Where PWhere P11, P, P22, and P, and Pnn are the partial pressures of the are the partial pressures of the

gases involvedgases involved

Partial Pressure-Partial Pressure- P of each gas in P of each gas in mixturemixture

Why?Why?

the particles of each gas in a the particles of each gas in a mixture have an equal chance mixture have an equal chance to hit the wallsto hit the walls

so each gas exerts P so each gas exerts P independent of that exerted independent of that exerted by other gasesby other gases

total P is result of the total # total P is result of the total # of collisions per unit of wall of collisions per unit of wall areaarea

Ex. A mixture of oxygen, hydrogen Ex. A mixture of oxygen, hydrogen and nitrogen gases exerts a total and nitrogen gases exerts a total pressure of 278 kPa.  If the partial pressure of 278 kPa.  If the partial pressures of the oxygen and the pressures of the oxygen and the hydrogen are 112 kPa and 101 kPa hydrogen are 112 kPa and 101 kPa respectively, what would be the respectively, what would be the partial pressure exerted by the partial pressure exerted by the nitrogennitrogen..PT= 278 kPa

PO= 112 kPa

PH= 101 kPa

PN= ?

PT = PO + PH + PN

278 kPa = 112 kPa + 101 kPa + PN

PN = 65.0 kPa

Vapour Pressure DefinedVapour Pressure DefinedVapour pressure is the pressure exerted by a Vapour pressure is the pressure exerted by a

vapour. Ex. the Hvapour. Ex. the H22O(g) in a sealed containerO(g) in a sealed container. .

• Yet, molecules both leave and join the surface, so vapour pressure also pushes molecules up.

Eventually the air above the water is filled with vapour pushing down. As temperature , more molecules fill the air, and vapour pressure .

• To measure vapour pressure we can heat a sample of liquid on top of a column of Hg and see the pressure it exerts at different °C.

Gas Collected Over WaterGas Collected Over Water

Production of oxygen by thermal Production of oxygen by thermal decomposition of KCIO3.decomposition of KCIO3.

Water DisplacementWater Displacement gas produced is less dense than water gas produced is less dense than water

so it replaces the water in the bottleso it replaces the water in the bottle gas collected is not pure because it gas collected is not pure because it

contains vapor from the watercontains vapor from the water

PPTT = P = Pgasgas + P + Pwaterwaterequal to atmospheric pressure

set for a certain T

ExampleExample Oxygen gas from decomposition of KClOOxygen gas from decomposition of KClO33 was was

collected by water displacement. The barometric collected by water displacement. The barometric pressure and the temperature during the experiment pressure and the temperature during the experiment were 731.0 mm Hg and 20.0were 731.0 mm Hg and 20.0°°C respectively. If the C respectively. If the partial pressure of water vapor is 17.5 mm Hg at partial pressure of water vapor is 17.5 mm Hg at 20.020.0°°C. What was the partial pressure of oxygen C. What was the partial pressure of oxygen collected?collected?

PPTT = P = PDGDG + P + PH2OH2O

731.0 mm Hg = P731.0 mm Hg = PO2O2 + 17.5 mm Hg + 17.5 mm Hg

PPO2O2 = 713.5 mm Hg = 713.5 mm Hg

ExampleExample

Find the partial pressure by 2 gases Find the partial pressure by 2 gases (A and B) mixed if the overall (A and B) mixed if the overall pressure is 790 mmHg. The percent pressure is 790 mmHg. The percent by volume is A: 20% and B: 80%.by volume is A: 20% and B: 80%.

PPTT = P = PAA + P + PB B = 790 mmHg= 790 mmHg A: 0.20 x 790 = 158 mmHgA: 0.20 x 790 = 158 mmHg B: 0.80 x 790 = 632 mmHgB: 0.80 x 790 = 632 mmHg

PPTT = P = PDGDG + P + PH2OH2O

PPTT = 113.0 kPa = 113.0 kPa

PPH2OH2O = 1.6 kPa = 1.6 kPa

PPDGDG = ? = ? PPDGDG = 113.0 kPa – 1.6 kPa = 113.0 kPa – 1.6 kPa

PPDGDG = 111 kPa

Ex. A sample of hydrogen gas is Ex. A sample of hydrogen gas is collected over water at 14.0 collected over water at 14.0 ooC.  C.  The pressure of the resultant The pressure of the resultant mixture is 113.0 kPa.  What is the mixture is 113.0 kPa.  What is the pressure that is exerted by the pressure that is exerted by the dry hydrogen alone?dry hydrogen alone?

If the above gas is 2.3 L, what is the new volume at standard pressure?

PP11 = 111.4 kPa = 111.4 kPa

VV11 = 2.3 L = 2.3 L

PP22 = 101.325 kPa = 101.325 kPa

VV22 = ? = ?

PP11 V V11 = P = P22 V V22

1 12

2

P VV =

P

2 (111.4 kPa)(2.3L)

V = 101.325 kPa

V2 = 2.53 L