2Solve quantitative problems involving the relationships among the pressure, temperature, and volume of a gas using dimensional analysis.Include: Avogadro, Ideal Gas Law and the moleAdditional KEY TermsMolar volume Molar mass
3Avogadro ( )Any sample of any gas at the same temperature and pressure will contain the same number of particles.
4* Experiment conducted at constant temperature and pressure
5Particles are now counted in moles (n). Defined 22.4 L as the molar volume for any gas.Particles are now counted in moles (n).The number of moles is directly proportional to pressure, temperature, and volume.
6P1 V1 = P2 R k V2 n T1 T2 L · kPa 8.314 mol · K Experiments with n, V, P, and T give Ideal gas law:P1V1=P2RkV2nT1T2R = experimentally determined ideal gas constant.Value of R depends on the units used for pressure.L · atmL · kPa0.08218.314mol · Kmol · KL · mmHg62.4mol · K
7A steel container with a volume of 20 A steel container with a volume of 20.0 L is filled with nitrogen gas to a final pressure of kPa at 27.00C.How many moles of gas was used?L · kPamol · K8.314=PVnRT27°C = 300 K=2000(20.0)16.0 molesn8.314(300)
8The mass of 1 mole is the molar mass - (g/mol) One mole of any particle has a mass equal to its total formula mass – IN GRAMS.The mass of 1 atom of Al = 27.0 µThe mass of 1 mole of Al atoms = 27.0 gThe molecular mass of water (H2O) is 18.0 µ ...So...the molar mass of water is 18.0 g/mol.
9Molar mass of lead (II) chloride, PbCl2 ? 1 particle of PbCl2 - 1 atom of Pb, 2 atoms of Cl1 mole of PbCl2 - 1 mole of Pb, 2 moles of ClPbCl2 = g/mol + 2(35.5 g/mol)= g/molThe molar mass of lead (II) chloride is g/mol.
10What pressure is exerted by 640.0 g of methane (CH4) gas in a sealed 5.35L container at 27 ºC? Molar mass of CH4 = 16.0 g/mol640.0 g16.0 g1 mol= molesL · kPamol · K8.314PV=nRT27°C = 300 K(40.0)8.314(300)1.9 x 104 kPa =(5.35)
11= 2.75 g of O2 = P V n R T = 101.3 (2.2) 0.086 moles 8.314 (310) A child lung capacity is 2.2 L. How many grams of oxygen gas do lungs hold at a pressure of 1.00 atm and a normal body temperature of 37.00C?L · kPamol · K8.314=PVnRT37°C = 310 K=101.3(2.2)0.086 moles8.314(310)Molar mass of O2 = 32.0 g/mol1 mol32.0 g0.086 mol= g of O2
13CAN YOU / HAVE YOU?Solve quantitative problems involving the relationships among the pressure, temperature, and volume of a gas using dimensional analysis.Include: Avogadro, Ideal Gas Law and the moleAdditional KEY TermsMolar volume Molar mass