Using PV = nRT (Honors) P = Pressure V = Volume T = Temperature N = number of moles R is a constant, called the Ideal Gas Constant Instead of learning a different value for R for all the possible unit combinations, we can just memorize one value and convert the units to match R. R = L atm Mol K

Ideal Gas Law: P V = n R T P = pressure in atm V = volume measured in Liters n =# of moles T = temperature K R=Universal gas constant = L atm/(mol K) = L atm/(mol K)

GIVEN: P = ? atm n = mol T = 16°C = 289 K V = 3.25 L R = L  atm/mol  K WORK: PV = nRT P(3.25)=(0.412)(0.0821)(289) L mol L  atm/mol  K K L mol L  atm/mol  K K P = 3.01 atm Ideal Gas Law (Honors) Calculate the pressure in atmospheres of mol of He at 16°C & occupying 3.25 L. Calculate the pressure in atmospheres of mol of He at 16°C & occupying 3.25 L. IDEAL GAS LAW

Some Cool Videos (Honors) Crash Course: Ideal Gas Laws Crash Course: Ideal Gas Laws Crash Course: Ideal Gas Law Problems Crash Course: Ideal Gas Law Problems Crash Course: Real Gases Crash Course: Real Gases Crash Course: Grahams Law Crash Course: Grahams Law

Dalton’s Law of Partial Pressures Total pressure of a mixture of gases in a container equals the sum of the individual partial pressures of each gas. Total pressure of a mixture of gases in a container equals the sum of the individual partial pressures of each gas. P total = P 1 + P P atm = P H2 + P H2O

Crash Course: Partial Pressure and Vapor Pressure v=JbqtqCunYzA&safe=active v=JbqtqCunYzA&safe=active This is often useful when gases are collected “over water”

You can use Table H

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 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. 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 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.

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 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? 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? DALTON’S LAW Look up water-vapor pressure for 35.0°C. 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.

Using Mole Fraction (Honors)

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

Graham’s Law Speed of diffusion/effusion Speed of diffusion/effusion Kinetic energy is determined by the temperature of the gas. Kinetic energy is determined by the temperature of the gas. At the same temp & KE, heavier molecules move more slowly. At the same temp & KE, heavier molecules move more slowly.

Graham’s Law Formula (Honors) Graham’s Law Graham’s Law Rate of diffusion of a gas is inversely related to the square root of its molar mass. Rate of diffusion of a gas is inversely related to the square root of its molar mass. Ratio of gas A’s speed to gas B’s speed

Determine the relative rate of diffusion for krypton and bromine. 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 ”.

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? 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”.

An unknown gas diffuses 4.0 times faster than O 2. Find its molar mass. 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.

V n Avogadro’s Principle Equal volumes of gases contain equal numbers of moles Equal volumes of gases contain equal numbers of moles at constant temp & pressure at constant temp & pressure true for any gas true for any gas Equal volumes of gases at the same T and P have the same number of molecules.

Gas Stoichiometry (Honors) Moles  Liters of a Gas Moles  Liters of a Gas STP - use 22.4 L/mol STP - use 22.4 L/mol Non-STP - use ideal gas law Non-STP - use ideal gas law Non- STP Problems Non- STP Problems Given liters of gas? Given liters of gas? start with ideal gas law start with ideal gas law Looking for liters of gas? Looking for liters of gas? start with stoichiometry conv. start with stoichiometry conv.

1 mol CaCO g Gas Stoichiometry Problem What volume of CO 2 forms from 5.25 g of CaCO 3 at 103 kPa & 25ºC? What volume of CO 2 forms from 5.25 g of CaCO 3 at 103 kPa & 25ºC? 5.25 g CaCO 3 = 1.26 mol CO 2 CaCO 3  CaO + CO 2 1 mol CO 2 1 mol CaCO g ? L non-STP Looking for liters: Start with stoich and calculate moles of CO 2. Plug this into the Ideal Gas Law to find liters.

WORK: PV = nRT (103 kPa)V =(1mol)(8.315 dm 3  kPa/mol  K )(298K) V = 1.26 dm 3 CO 2 Gas Stoichiometry Problem What volume of CO 2 forms from 5.25 g of CaCO 3 at 103 kPa & 25ºC? What volume of CO 2 forms from 5.25 g of CaCO 3 at 103 kPa & 25ºC? GIVEN: P = 103 kPa V = ? n = 1.26 mol T = 25°C = 298 K R = dm 3  kPa/mol  K

WORK: PV = nRT (97.3 kPa) (15.0 L) = n (8.315 dm 3  kPa/mol  K ) (294K) n = mol O 2 Gas Stoichiometry Problem How many grams of Al 2 O 3 are formed from 15.0 L of O 2 at 97.3 kPa & 21°C? How many grams of Al 2 O 3 are formed from 15.0 L of O 2 at 97.3 kPa & 21°C? GIVEN: P = 97.3 kPa V = 15.0 L n = ?n = ?n = ?n = ? T = 21°C = 294 K R = dm 3  kPa/mol  K 4 Al + 3 O 2  2 Al 2 O L non-STP ? g Given liters: Start with Ideal Gas Law and calculate moles of O 2. NEXT 

2 mol Al 2 O 3 3 mol O 2 Gas Stoichiometry Problem How many grams of Al 2 O 3 are formed from 15.0 L of O 2 at 97.3 kPa & 21°C? How many grams of Al 2 O 3 are formed from 15.0 L of O 2 at 97.3 kPa & 21°C? mol O 2 = 40.6 g Al 2 O 3 4 Al + 3 O 2  2 Al 2 O g Al 2 O 3 1 mol Al 2 O Lnon-STP ? g Use stoich to convert moles of O 2 to grams Al 2 O 3.