# Dalton’s Law. Dalton’s Law of Partial Pressure The total pressure of a mixture of gases equals the SUM of the partial pressures for each gas in the mixture.

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Dalton’s Law

Dalton’s Law of Partial Pressure The total pressure of a mixture of gases equals the SUM of the partial pressures for each gas in the mixture. Partial pressure – the pressure exerted by an individual gas in a mixture of gases.

Example The total pressure of a mixture of hydrogen, helium, and argon is 99.3 kPa. If the partial pressure of helium is 42.7 kPa and the partial pressure of argon is 54.7 kPa, what is the partial pressure of hydrogen?

Example The total pressure of a mixture of hydrogen, helium, and argon is 99.3 kPa. If the partial pressure of helium is 42.7 kPa and the partial pressure of argon is 54.7 kPa, what is the partial pressure of hydrogen? P T = 99.3 kPa P He = 42.7 kPa P Ar = 54.7 kPa

Example The total pressure of a mixture of hydrogen, helium, and argon is 99.3 kPa. If the partial pressure of helium is 42.7 kPa and the partial pressure of argon is 54.7 kPa, what is the partial pressure of hydrogen? P T = 99.3 kPa P He = 42.7 kPa P Ar = 54.7 kPa P H 2 = 99.3 – (42.7 + 54.7) = 1.9 kPa

Another Useful Relationship P A =n A x P T n T P A is the partial pressure of a gas in a mixture P T is the total pressure of the mixture of gases n A is the number of moles of a gas in the mixture n T is the total number of moles of gas in the mixture

Example A mixture of 6.0 g of Ar gas and 8.0 g of O 2 gas has a total pressure of 66 kPa. Calculate the partial pressure exerted by each gas.

Example A mixture of 6.0 g of Ar gas and 8.0 g of O 2 gas has a total pressure of 66 kPa. Calculate the partial pressure exerted by each gas. n Ar = 6.0 g x 1 mol/39.95g = 0.152 mol Ar n O 2 = 8.0 g x 1 mol/32.00g = 0.25 mol O 2

Example A mixture of 6.0 g of Ar gas and 8.0 g of O 2 gas has a total pressure of 66 kPa. Calculate the partial pressure exerted by each gas. n Ar = 6.0 g x 1 mol/39.95g = 0.150 mol Ar n O2 = 8.0 g x 1 mol/32.00g = 0.25 mol O 2 n T = 0.152+ 0.25 = 0.400 mol

Example A mixture of 6.0 g of Ar gas and 8.0 g of O 2 gas has a total pressure of 66 kPa. Calculate the partial pressure exerted by each gas. n Ar = 6.0 g x 1 mol/39.95g = 0.150 mol Ar n O2 = 8.0 g x 1 mol/32.00g = 0.25 mol O 2 n T = 0.150+ 0.25 = 0.400 mol P Ar = 0.150/0.400 x 66 kPa = 25 kPa P O2 = 0.25/0.400 x 66 kPa = 41 kPa Check?

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