Additional Gas Laws

Graham’s Law b Diffusion Spreading of gas molecules throughout a container until evenly distributed. b Effusion Passing of gas molecules through a tiny opening in a container

C. Johannesson Graham’s Law KE = ½mv 2 b Speed of diffusion/effusion Kinetic energy is determined by the temperature of the gas. At the same temp & KE, heavier molecules move more slowly.  Larger m  smaller v

C. Johannesson Graham’s Law b Graham’s Law Rate of diffusion of a gas is inversely related to the square root of its molar mass. The equation shows the ratio of Gas A’s speed to Gas B’s speed.

C. Johannesson b Determine the relative rate of diffusion for krypton and bromine. Kr diffuses times faster than Br 2. Graham’s Law The first gas is “Gas A” and the second gas is “Gas B”. Relative rate mean find the ratio “v A /v B ”.

C. Johannesson b A molecule of oxygen gas has an average speed of 12.3 m/s at a given temp and pressure. What is the average speed of hydrogen molecules at the same conditions? Graham’s Law Put the gas with the unknown speed as “Gas A”.

C. Johannesson b An unknown gas diffuses 4.0 times faster than O 2. Find its molar mass. Graham’s Law The first gas is “Gas A” and the second gas is “Gas B”. The ratio “v A /v B ” is 4.0. Square both sides to get rid of the square root sign.

Dalton’s Law b The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases. P total = P 1 + P

When a H 2 gas is collected by water displacement, the gas in the collection bottle is actually a mixture of H 2 and water vapor.

Dalton’s law of partial pressures and a table of known vapor pressures of water can be used to determine the pressure of dry gas that has been collected.

C. Johannesson GIVEN: P H2 = ? P total = 94.4 kPa P H2O = 2.72 kPa WORK: P total = P H2 + P H2O 94.4 kPa = P H kPa P H2 = 91.7 kPa Dalton’s Law b Hydrogen gas is collected over water at 22.5°C. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa. Look up water-vapor pressure on chart for 22.5°C. Sig Figs: Round to least number of decimal places. The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H 2 and water vapor.

C. Johannesson GIVEN: P gas = ? P total = torr P H2O = 42.2 torr WORK: P total = P gas + P H2O torr = P H torr P gas = torr b A gas is collected over water at a temp of 35.0°C when the barometric pressure is torr. What is the partial pressure of the dry gas? Look up water-vapor pressure on table for 35.0°C and convert to torr. Sig Figs: Round to least number of decimal places. Dalton’s Law The total pressure in the collection bottle is equal to barometric pressure and is a mixture of the “gas” and water vapor.

Sample Problem #1 Mixtures of helium and oxygen are used in the air tanks of underwater divers for deep dives. For a particular dive, 12 L of O 2 at 25 o C and 1.0 atm and 46 L of He at 25 o C and 1.0 atm were both pumped into a 5.0-L tank. A)Calculate the partial pressure of each gas B)Calculate the total pressure in the tank at 25 o C (298 K). 14 *Because the partial pressure of each gas depends on the total moles present, we must first calculate n for each gas using the ideal gas law. PV = nRT n = PV ÷ RT n O 2 = (1.0 x 12) ÷ ( x 298) = 0.49 mol n He = (1.0 x 46) ÷ ( x 298) = 1.9 mol P = nRT ÷ V P O 2 = (0.49 x x 298) ÷ 5.0 = 2.4 atm P He = (1.9 x x 298) ÷ 5.0 = 9.3 atm *The tank has a volume of 5.0-L and temperature of 298 K, so we can figure out the partial pressure of each gas Determine the total pressure: P total = P O 2 + P He = 2.4 atm atm = 11.7 atm

Sample Problem #2 A sample of solid potassium chlorate, KClO 3, was heated in a test tube and decomposed according to the reaction: 2KClO 3 (s)  2KCl (s) + 3O 2 (g) The oxygen produced was collected by displacement of water at 22 o C. The resulting mixture of O 2 and H 2 O vapor had a pressure of 754 torr and a volume of L. Calculate the partial pressure of O 2 in the gas collected and the number of moles of O 2 present. The vapor pressure of water (P H 2 O ) at 22 o C is 21 torr. 15

Sample Problem #2 Solution We know: P total = 754 torr P H 2 O = 21 torr Next, we solve the ideal gas law for n O 2 n O 2 = (P O 2 V) ÷ (RT) Then solve: P = atm V = L T = 22 o C = = 295 K R = L atm/K mol 16 So we can determine: P total = P H 2 O + P O 2 OR P total - P H 2 O = P O 2 SO… 754 torr – 21 torr = 733 torr P O 2 = 733 torr Convert pressure from torr to atm: 733 torr x (1 atm/760 torr) = atm n O 2 = (0.964 x 0.650) ( x 295) n O 2 = mol O 2

Dalton’s Law Practice Problems 1.If a gaseous mixture is made of 2.41 g of He and 2.79 g of Ne in 1.04-L container at 25 o C, what will be the partial pressure of each gas and the total pressure in the container? 2.How many moles of helium gas would be required to fill a 2.41-L container to a pressure of 759 mm Hg at 25 o C? 3.A sample of oxygen gas (O 2 ) is saturated with water vapor (H 2 O) at 27 o C. The total pressure of the mixture is 772 torr, and the vapor pressure of water is 26.7 torr at 27 o C. What is the partial pressure of the oxygen gas? 4.Suppose a gaseous mixture of 1.15 g helium and 2.91 g argon is placed in a 5.25-L container at 273 o C. What pressure would exist in the container? 5.A tank contains a mixture of 3.0 mol of N2, 2.0 mol of O2, and 1.0 mol of CO2 at 25 o C and a total pressure of 10.0 atm. Calculate the partial pressure (in torr) of each gas in the mixture. 17