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IDEAL gas law

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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 mole Additional KEY Terms Molar volumeMolar mass

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Avogadro (1776-1856) Any sample of any gas at the same temperature and pressure will contain the same number of particles.

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* Experiment conducted at constant temperature and pressure

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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.

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mol · K Experiments with n, V, P, and T give Ideal gas law: R = experimentally determined ideal gas constant. Value of R depends on the units used for pressure. L · kPa 8.314 L · atm mol · K 0.0821 L · mmHg 62.4 R = P1P1 P2P2 T2T2 T1T1 V1V1 V2V2 k n

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A steel container with a volume of 20.0 L is filled with nitrogen gas to a final pressure of 2000.0 kPa at 27.0 0 C. How many moles of gas was used? = PV nTR = 2000(20.0) n (300) 8.314 L · kPa mol · K 8.314 27°C + 273 = 300 K 16.0 moles

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The molecular mass of water (H 2 O) is 18.0 µ... 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 µ The mass of 1 mole of Al atoms = 27.0 g So...the molar mass of water is 18.0 g/mol. The mass of 1 mole is the molar mass - (g/mol)

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Molar mass of lead (II) chloride, PbCl 2 ? 1 particle of PbCl 2 - 1 atom of Pb, 2 atoms of Cl 1 mole of PbCl 2 - 1 mole of Pb, 2 moles of Cl PbCl 2 = 207.2 g/mol + 2(35.5 g/mol) = 278.2 g/mol The molar mass of lead (II) chloride is 278.2 g/mol.

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What pressure is exerted by 640.0 g of methane (CH 4 ) gas in a sealed 5.35L container at 27 ºC? 640.0 g 16.0 g 1 mol = 40.0 moles = PVnTR 27°C + 273 = 300 K L · kPa mol · K 8.314 (5.35) (300)8.314 1.9 x 10 4 kPa = Molar mass of CH 4 = 16.0 g/mol (40.0)

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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 37.0 0 C? 0.086 mol 1 mol 32.0 g = 2.75 g of O 2 = PV nTR 37°C + 273 = 310 K L · kPa mol · K 8.314 = 101.3(2.2) (310) 8.314 0.086 moles Molar mass of O 2 = 32.0 g/mol

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CAN 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 mole Additional KEY Terms Molar volumeMolar mass

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