Adds in the factor of number of moles of gas “n”. The Ideal Gas Law PV = nRT Adds in the factor of number of moles of gas “n”. P = Pressure (atm) V = Volume (Liters) T = Temperature (Kelvin) n = number of moles Remember when dealing with moles: # moles (n) = grams of substance Gram Formula Mass

What is “R”? R is a constant, called the “Universal Gas Constant” It is derived by plugging in the values for 1 mole of gas at STP. (22.4 Liters, 273K, 1 atm) R = 0.0821 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. L • atm Mol • K

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

Example # 1 Honors Packet A 9.81 L cylinder contains 23.5 moles of nitrogen at 23° C. What pressure is exerted by the gas?

Example # 2 Honors Packet A pressure of 850 mmHg is exerted by 28.6 grams of sulfur dioxide at a temp of 40 °C. Calculate the volume of the vessel holding the gas.

Density is Hidden in this Formula Density = Mass (g) Volume # moles = Mass (g) Gram formula mass Can you rework PV = nRT to solve for Density?

Example # 7 in Honors Packet What is the density of neon at 40 °C and 1.23 atmospheres?

Dalton’s Law of Partial Pressures Ptotal = P1+P2+…. Total pressure of a mixture of gases in a container is the sum of the individual pressures (partial pressures) of each gas, as if each took up the total space alone. This is often useful when gases are collected “over water”

Collecting Gas over Water Pressureatm = Pressure (O2) + Pressure (H2O vapor) http://www.kentchemistry.com/links/GasLaws/dalton.htm Crash Course: Partial Pressure and Vapor Pressure http://www.youtube.com/watch?v=JbqtqCunYzA&safe=active

Look up water-vapor pressure on for 22.5°C. Dalton’s Law H2 gas is collected over water at 22.5°C. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa. GIVEN: PH2 = ? Ptotal = 94.4 kPa PH2O = 2.72 kPa WORK: Ptotal = PH2 + PH2O 94.4 kPa = PH2 + 2.72 kPa PH2 = 91.7 kPa Look up water-vapor pressure on for 22.5°C.

Look up water-vapor pressure for 35.0°C. Dalton’s Law A gas is collected over water at a temp of 35.0°C when the barometric pressure is 742.0 torr. What is the partial pressure of the dry gas? DALTON’S LAW GIVEN: Pgas = ? Ptotal = 742.0 torr PH2O = 42.2 torr WORK: Ptotal = Pgas + PH2O 742.0 torr = PH2 + 42.2 torr Pgas = 699.8 torr Look up water-vapor pressure for 35.0°C.

Examples of Vapor Pressure Tables

You can even use Table H to find VP of Water

Using Mole Fraction (Honors) Moles gas (X) x P (total) = P (X) Total Moles Gas

Example # 4 Honors Packet A mixture of 2.00 moles H2, 3.00 moles NH3, 4.00 moles CO2, and 5.00 moles N2 exert a total pressure of 800 mmHg. What is the partial pressure of each gas?

Graham’s Law Diffusion Effusion Spreading of gas molecules throughout a container until evenly distributed. Effusion Passing of gas molecules through a tiny opening in a container Smaller and lighter gas particles do this faster! https://www.youtube.com/watch?feature=player_embedded&v=H7QsDs8ZRMI https://www.youtube.com/watch?feature=player_embedded&v=L41KhBPBymA Crash Course: Grahams Law http://www.youtube.com/watch?v=TLRZAFU_9Kg&safe=active

Graham’s Law Speed of diffusion/effusion At the same temp & KE, heavier molecules move more slowly. Ex: Which of the following gases will diffuse most rapidly? A.) N2 B.) CO2 C.) CH4

Graham’s Law Formula Graham’s Law 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

Graham’s Law Kr diffuses 1.381 times faster than Br2. Determine the relative rate of diffusion for krypton and bromine. The first gas is “Gas A” and the second gas is “Gas B”. Relative rate mean find the ratio “vA/vB”. Kr diffuses 1.381 times faster than Br2.

Put the gas with the unknown speed as “Gas A”. Graham’s Law 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? Put the gas with the unknown speed as “Gas A”.

Graham’s Law An unknown gas diffuses 4.0 times faster than O2. Find its molar mass. The first gas is “Gas A” and the second gas is “Gas B”. The ratio “vA/vB” is 4.0. Square both sides to get rid of the square root sign.

Example # 1 Honors Packet Under the same conditions of temp and pressure, how many times faster will hydrogen effuse compared to carbon dioxide?

Example #7 Honors Packet If CO2 diffuses from a flask in 30 seconds and an unknown gas diffuses from the same flask in 5 minutes, calculate the molecular weight of this unknown gas.

Avogadro’s Principle Volume of a gas is directly proportional to the number of moles of gas particles present. V n

Equal volumes of gases contain equal numbers of moles of particles at constant temp & pressure true for any gas

Avogadros Law: https://www. youtube. com/watch Molar Volume of a Gas: https://www.youtube.com/watch?feature=player_embedded&v=4b852VIEkHQ

Practice Questions At the same temperature and pressure, 1.0 liter of CO(g) and 1.0 liter of CO2(g) have equal masses and the same number of molecules B. different masses and a different number of molecules C. equal volumes and the same number of molecules D. different volumes and a different number of molecules

Which rigid cylinder contains the same number of gas molecules at STP as a 2.0-liter rigid cylinder containing H2(g) at STP? (1) 1.0-L cylinder of O2(g) (2) 2.0-L cylinder of CH4(g) (3) 1.5-L cylinder of NH3(g) (4) 4.0-L cylinder of He(g

Which two samples of gas at STP contain the same total number of molecules? (1) 1 L of CO(g) and 0.5 L of N2(g) (2) 2 L of CO(g) and 0.5 L of NH3(g) (3) 1 L of H2(g) and 2 L of Cl2(g) (4) 2 L of H2(g) and 2 L of Cl2(g)

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

Gas Stoichiometry Problem What volume of CO2 forms from 5.25 g of CaCO3 at 1.017atm & 25ºC? CaCO3  CaO + CO2 5.25 g ? L non-STP Looking for liters: Start with stoich and calculate moles of CO2. 5.25 g CaCO3 1 mol CaCO3 100.09g 1 mol CO2 CaCO3 = 1.26 mol CO2 Plug this into the Ideal Gas Law to find liters.

Gas Stoichiometry Problem What volume of CO2 forms from 5.25 g of CaCO3 at 1.017atm & 25ºC? GIVEN: P = 1.017atm n = 1.26 mol T = 25°C = 298 K R = .0821 Latm/molK WORK: PV = nRT (1.017atm)V =(1mol)(.0821 Latm/molK)(298K) V = 1.26 L CO2

Gas Stoichiometry Problem How many grams of Al2O3 are formed from 15.0 L of O2 at 0.96 atm & 21°C? 4 Al + 3 O2  2 Al2O3 15.0 L non-STP ? g GIVEN: P = 0.96 atm V = 15.0 L n = ? T = 21°C = 294 K R = .0821 Latm/molK WORK: PV = nRT (0.96 atm ) (15.0 L) = n (.0821 Latm/molK) (294K) n = 0.597 mol O2 Given liters: Start with Ideal Gas Law and calculate moles of O2. NEXT 

Gas Stoichiometry Problem How many grams of Al2O3 are formed from 15.0 L of O2 at 0.96 atm & 21°C? 4 Al + 3 O2  2 Al2O3 15.0L non-STP ? g Use stoich to convert moles of O2 to grams Al2O3. 0.597 mol O2 2 mol Al2O3 3 mol O2 101.96 g Al2O3 1 mol Al2O3 = 40.6 g Al2O3

Example # 10 Honors Multiple Choice Ni(CO)4 (l) → Ni (s) + 4CO (g) What volume of CO is formed from the complete decomposition of 444g of Ni(CO)4 at 752 torr and 22.0 °C Do stoich for 444g to find moles of CO. Plug the moles into Ideal Gas Law to get V at non standard conditions.

Some Cool Videos Crash Course: Ideal Gas Laws http://www.youtube.com/watch?v=BxUS1K7xu30&safe=active Crash Course: Ideal Gas Law Problems http://www.youtube.com/watch?v=8SRAkXMu3d0 Crash Course: Real Gases http://www.youtube.com/watch?v=GIPrsWuSkQc&safe=active