Avogadro’s Law The Ideal Gas Law Combined Gas Laws STP

Volume and Moles How does adding more molecules of a gas change the volume of the air in your glove? If the glove has a leak, how does the loss of air molecules change the volume?

Avogadro’s Law Remember you must convert units to moles and liters. We can calculate volume or number of moles using this equation. Volume of gas is Directly proportional to Number of moles (n) of gas V = an (at constant T and P) V = a n (V) is volume of gas in Liters (a) is a constant V1 = V2 (n) is the number of moles n1 n2 Initial Final   Remember you must convert units to moles and liters.

Learning Check #1 True (1) or False(2) 1.___The Pressure exerted by a gas at constant Volume is not affected by the Temp. of the gas. 2.___ At constant Pressure, the Volume of a gas is directly proportional to the absolute Temp. 3.___ At constant Temperature, doubling the Pressure will cause the Volume of a gas sample to decrease to one-half its original V.

Solution #1 True (1) or False(2) 1. (2)The P exerted by a gas at constant V is not affected by the T of the gas. 2. (1) At constant P, the V of a gas is directly proportional to the absolute T 3. (1) At constant T, doubling the P will cause the V of a gas sample to decrease to one-half its original V.

Learning Check #2 If 3.20 g of O2 gas occupies a volume of 2.24 L at 0°C and 1.0 atm, what volume would be occupied by 32.0 g of O2 gas under same conditions? Remember you must convert units to moles and liters. Prepare a data table and convert units: DATA TABLE Initial conditions Final conditions V1 = 2.24 L V2 = ? L n1 = 3.20 g O2 (1 mol/ 32g) = .1 mol O2 n2 = 32.0 g (1 mol/ 32g) = 1 mol O2 V1 = V2 V2 = V1 n2 n1 n2 n1 Initial Final

Solution #2 V2 = 2.24 L 1.0 mol O2 22.4 L O2 or 2.24 x 101 L O2 DATA TABLE Initial conditions Final conditions V1 = 2.24 L V2 = ? L n1 = 3.20 g O2 (1 mol/ 32g) = .1 mol O2 n2 = 32.0 g (1 mol/ 32g) = 1 mol O2 V1 = V2 V2 = V1 n2 n1 n2 n1 Initial Final V2 = 2.24 L 1.0 mol O2 22.4 L O2 or 2.24 x 101 L O2 .1 mol O2

Avogadro’s Law The Ideal Gas Law Combined Gas Laws STP

Ideal Gas Law Can be combined to form PV = nRT The equality for the 3 variables involved in Boyle’s, Charles’, and Avogadro’s law. Can be combined to form PV = nRT You must convert all units to moles, liters, Kelvin and atm. (P) is the Pressure in atm (V) is volume in Liters (n) is the number of particles in moles (T) is Temperature in Kelvin (R) is the universal gas constant; R = 0.08206 L atm / mol K

Ideal Gases PV = nRT R is known as the Universal gas constant STP Describes how most real gases behave at STP 273 K (0°C) and pressures of 1 atm 1mol of any gas at STP takes up 22.4 L of volume Using STP conditions: (P) (V) R = PV = (1.00 atm) (22.4 L) = 0.08206 L- atm nT (1mol) (273K) mol - K (n) (T)

Learning Check #1 Dinitrogen monoxide (N2O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23.0°C, what is the pressure (atm) in the tank in the dentist office? Rearrange Ideal Gas Law Equation to solve for P: PV = nRT P = nRT V Remember to convert all units, Kelvin, liters, mol, and atm. Prepare a data table: DATA TABLE P = ? atm V = 20.0 L n = 2.86 mol R = .08206 L atm T = 23 °C + 273= 296 K mol K

Solution #1 V Substitute values of n, R, T and V and solve for P n R T Dinitrogen monoxide (N2O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23.0°C, what is the pressure (atm) in the tank in the dentist office? DATA TABLE P = ? atm V = 20.0 L n = 2.86 mol R = .08206 L atm T = 23 °C + 273= 296 K K mol Rearrange ideal gas law for unknown P: P = nRT V Substitute values of n, R, T and V and solve for P n R T P = 2.86 mol .08206 L atm 296 K 20.0 L K-mol = 3.47 atm

Learning Check #2 A 5.00 L cylinder contains oxygen gas at 20.0°C and 735 mm Hg. How many grams of oxygen are in the cylinder? Rearrange Ideal Gas Law Equation to solve for n: PV = nRT n = PV RT Remember to convert all units, Kelvin, liters, mol, and atm. Prepare a data table: DATA TABLE P = 735 mm Hg (1 atm/760 mm Hg) = .967 atm V = 5.00 L R = .08206 L atm T = 20°C + 273 = 293 K n = ? Mol O2 mol K

Solution #2 DATA TABLE P = 735 mm Hg (1 atm/760 mm Hg) = .967 atm V = 5.00 L R = .08206 L atm T = 20°C + 273= 293 K n = ? mol O2 mol K Solve ideal gas equation for n (moles) n = PV RT = (.967 atm) (5.00 L) (mol K) = .201 mol (293 K) (.08206 L atm) = 0. 201 mol O2 x 32.0 g O2 = 6.43 g O2 1 mol O2

Avogadro’s Law The Ideal Gas Law Combined Gas Laws STP

Combined Gas Laws Boyle’s - P1V1 = P2V2 Charles’ - V1 = V2 T1 T2 Combined Gas Law P1V1 = P2V2 T1 T2 You can rearrange the equation to solve for pressure, volume or temperature. Example: Rearrange the combined gas law to solve for V2 P1V1T2 = P2V2T1 V2 = P1V1T2 P2T1

Combined Gas Laws Problem #1 A sample of He2 has a volume of 0.180 L, pressure of 0.800 atm and a temperature of 29.0°C. What is the new temperature (K) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm? Rearrange Combined Gas Law Equation to solve for T2: P1V1T2 = P2V2T1 T2 = P2V2T1 P1V1 Remember to convert all units, Kelvin, liters, mol, and atm: DATA TABLE P1 = 0.800 atm P2 = 3.20 atm V1 = 0.180 L V2 = 90.0 mL = .090 L T2 = ? K T1 = 29°C + 273 = 302 K

Solution #1 DATA TABLE P1 = 0.800 atm P2 = 3.20 atm V1 = 0.180 L V2 = 90.0 mL = .090 L T2 = ? K T1 = 29°C + 273 = 302 K Solve for T2 Enter data T2 = P2V2T1 T2 = 3.20 atm x .090 L x 302 K = 604 K P1V1 0.800 atm .180 L T2 = 604 K or 6.04 x 102 K

Learning Check #2 A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temp. in K when the gas has a volume of 0.315 L and a pressure of 802 mm Hg? Rearrange Combined Gas Law Equation to solve for T2: P1V1T2 = P2V2T1 T2 = P2V2T1 P1V1 Remember to convert all units, Kelvin, liters, mol, and atm: DATA TABLE P1 = ____atm P2 = ____atm V1 = ____L V2 = ______L T2 = ? K T1 = _____K

Solution #2 T2 = 180 K DATA TABLE P1 = 0.850 atm P2 = 802 mm Hg (1atm/760 mm Hg) = 1.06 atm V1 = 0.675 L V2 = .315 L T2 = ? K T1 = 35°C + 273 = 308 K Solve for T2 Enter data T2 = P2V2T1 T2 = 1.06 atm x .315 L x 308 K = 179 K P1V1 0.850 atm .675 L T2 = 180 K

Gas Laws Avogadro’s Law The Ideal Gas Law Combined Gas Laws STP and Molar Volume

STP and Molar volume The volumes of gases can be compared when they have the same temperature and pressure (STP). Standard temperature 0°C or 273 K Standard pressure 1 atm (760 mm Hg)

P1 = _____ V1 = _____ T1 = _____ K Learning Check #1 A sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 atm and –25°C? Data Table: P1 = _____ V1 = _____ T1 = _____ K P2 = _____ V2 = ?? T2 = _____K P1V1T2 = P2V2T1 V2 = P1V1T2 P2T1 V2 = 15 L x atm x K = _____L atm K

Solution #1 Data Table: P1 = 1.0 atm V1 = 15 L T1 = 273 K P2 = 2.0 atm V2 = ?? T2 = 248 K V2 = 15 L x 1.0 atm x 248 K = 6.8 L 2.0 atm 273 K

Molar Volume At STP 2.0 g H2 16.0 g CH4 44.0 g CO2 1 mole 1 mole 1mole (STP) (STP) (STP) V = 22.4 L V = 22.4 L V = 22.4 L It does not matter what gas you are dealing with, if you have exactly one mol of that gas you will have 22. 4 L of that gas. 1 mole of a gas at STP = 22.4 L 22.4 L and 1 mole 1 mole 22.4 L

Learning Check #1 A. What is the volume at STP of 4.00 g of CH4? 1) 5.60 L 2) 11.2 L 3) 44.8 L 4.00 g CH4 L CH4 B. How many grams of He2 are present in 8.00 L of gas at STP? 1) 25.6 g 2) 0.357 g 3) 1.43 g 8.00 L He2 g He2

Solution #1 A. What is the volume at STP of 4.00 g of CH4? 4.00 g CH4 x 1 mole CH4 x 22.4 L CH4(STP) = 5.60 L CH4 16.0 g CH4 1 mole CH4 B. How many grams of He2 are present in 8.00 L of gas at STP? 8.00 L He2 x 1 mole He2 x 8.00 g He2 = 1.43 g He2 22.4 L He2 1 mole He2