2Molar Volume Avogadro’s Principle - equal volumes of gases at same T and P contain equal numbers of particles (equal number of moles)
3one mole of any gas at STP has a volume of 22.4 L 1 mole H2 = 22.4 L at STP1 mole CO2 = 22.4 L at STP1 mole CH4 = 22.4 L at STPCO2CH4H2
4No!!!!!!! Molar mass is equal to mass on periodic table. Would the mass of this 22.4 L of gas be the same for all gases?No!!!!!!!Molar mass is equal to mass on periodic table.22.4 L = 1 mole = mass (g) on periodic table
5one mole of any gas at STP has a volume of 22.4 L 1 mole H2 = 22.4L at STP1 mole CO2 = 22.4 L at STP1 mole CH4 = 22.4 L at STPH2CO2CH42 g44 g16 g
6Volume Ratios = Molar ratios in balanced equation N H2 2NH31 mole nitrogen reacts with 3 moles hydrogen to produce 2 moles ammoniaOr1 liter nitrogen reacts with 3 liters hydrogen to produce 2 liters ammonia
7Remember Combined Gas law P1V1 = P2V2T T2Since number of molecules is directly proportional to volume, it too goes on bottom of equationP1V1 = P2V2T1n1 T2n2
8P1V1 = constant T1n1 (1atm)(22.4L) = constant (273K)(1mol) Plug in standard temperature,pressure, volume, & # of moles to find constant(1atm)(22.4L) = constant (273K)(1mol).0821 atm . L = constantK . Mol
9PV = nRT P = pressure (atm or kPa) V = volume (L or dm3) n = number of molesR = universal gas constantT = temperature (K)
10Universal Gas Constant R = L-atm mole-KR = 8.31 L-kPa mole-K(1 dm3 = 1 L)Units in equation must all match
11Sample ProblemA rigid steel cylinder with a volume of 20.0 L is filled with nitrogen gas to a final pressure of atm at 27oC. How many moles of gas does the cylinder hold?
12PV = nRT V = 20.0 L N2 P = 200.0 atm T = 27oC +273 = 300 K R= n = ? L-atm Or L-kPa mole-K mole-K
13n = 162 mol N2 PV= nRT (200 atm)(20 L N2) = n (.0821 atm-L)(300 K) K-mol(200 atm)(20 L N2) = n(.0821 atm-L)(300 K)n = 162 mol N2
14How many grams N2 is 162 mol? 162 mol ( 28 g ) 1 mol N2 = 4,536 g N2 Periodic table mass of nitrogen = 14 gN2 = 28 g162 mol ( 28 g )1 mol N2= 4,536 g N2
15Sample Problem 2A deep underground cavern contains 2.24 x 106 L of methane gas (CH4) at a pressure of 15.0 atm and a temperature of 42oC. How many grams of methane does this natural gas deposit contain?
16V = 2.24 x 106 L CH4P = 15.0 atmT = 42oC = 315 KR = atmxL/molxK? grams
17PV = nRT or PV = n RT (15.0 atm)(2.24 x 106 L CH4) = n (315K)(.0821 atm-L/K-mol)n = 1.30 x 106 mol CH4CH4 = 16g16g1 mol CH4= 2.08 x 107 g
18A really, really difficult one now! How many liters of ammonia (NH3) gas would be produced at STP if 5.00 g of hydrogen reacts with excess nitrogen?N H2 2NH3
19PV = nRT P = 1.00 atm V = ? R = .0821 atm-L/K-mol T = 273 K At STP volume of NH3 produced?P = 1.00 atmV = ?R = atm-L/K-molT = 273 Kn = 5g H2 ___mol NH3?
20N 2 + 3H2 2NH3 5 g H2 (1 mol H2)(2 mol NH3) 2 g 3 mol H2
21P = 1.00 atm, V = ? n = 1.67 mol NH3 R = .0821 atm-L/K-mol T = 273 K V = nRTP(1.67 mol)(.0821 atm-L/K-mol )(273K)(1.00 atm)V = 37.4 L