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# ch5 - Chapter 5 Gases Chapter Goals Understand the basis of...

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Chapter 5 Gases Chapter Goals Understand the basis of the gas laws and know how to use each (Boyle‟s law, Charles‟ law, Avogadro‟s hypothesis, Dalton‟s law). Use the (model) ideal gas law. Apply the gas laws to stoichiometric calculations. Understand kinetic-molecular theory of gases, especially the distribution of molecular speeds (energies). Recognize why real gases do not behave like ideal gases. Solids, Liquids, and Gases Density < solids or liquids. Liquids & gases are fluids; they flow easily Density (g/mL) Solid Liquid Gas H 2 O 0.917 0.998 0.000588 CCl 4 1.70 1.59 0.00503 Solid to liquid to gas, via heating. Gases may be liquefied by cooling and compressing General Properties of Gases There is a lot of “free” space in a gas. Gases can be expanded infinitely; compressed. Gases occupy containers uniformly and completely. Gases diffuse and mix rapidly. Gases exert pressure on their surroundings Importance of Gases Airbags fill with N 2 gas in an accident. Gas is generated by the decomposition of sodium azide, NaN 3 . 2 NaN 3 (s) 2 Na(s) + 3 N 2 (g) Physical Properties of Gases Gas properties can be modeled mathematically. Model depends on four quantities (parameters): V = volume of the gas (L) T = temperature (K) n = amount (moles) P = pressure (atmospheres)

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(arises from molecular collision) Pressure is force per unit area. force F Pressure = ──── P = ── area A N SI unit: pascal, 1 Pa = ─── = 1kg m −1 s −2 m 2 Pressure Atmospheric pressure (pressure of the atmosphere) is measured with a barometer , invented by Torricelli (1643). Definitions of standard pressure 76 cm Hg = = 760 mm Hg = 760 torr = = 1 atmosphere = 1 atm 1 atm = 101.3 kPa = 1.013 10 5 Pa 1 bar = 1 10 5 Pa = 0.987 atm H 2 O density ~ 1 g/mL Hg density = 13.6 g/mL IDEAL GAS LAW Brings together gas properties. Can be derived from experiment and theory. P V = n R T Boyle‟s Law: Calculations involving P-V changes PV = k V 1 or V= k ── or V 1/P P 1/P P 1 V 1 = k 1 for one sample of a gas. P 2 V 2 = k 2 for a second sample of a gas. k 1 = k 2 for the same sample of a gas (same number of moles at the same T.) Thus we can write Boyle‟s Law mathematically as P 1 V 1 = P 2 V 2 for constant n and T Boyle‟s Law: The Volume-Pressure Relationship Example: At 25 o C a sample of He has a volume of 4.00 10 2 mL under a pressure of 7.60 10 2 torr. What volume would it occupy under a pressure of 2.00 atm at the same T? Firstly , we need to convert to same unit of P: Then, using Boyle’s law  torr 1520 mL 400 torr 760 V 2 atm 1 torr 760 atm 2.00 = 1520 torr 2 1 1 2 P V P V 2 2 1 1 V P V P Charles‟ Law: The Volume-Temperature Relationship; The Absolute Temperature Scale 0 5 10 15 20 25 30 35 0 50 100 150 200 250 300 350 400 Volume (L) vs. Temperature (K)
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ch5 - Chapter 5 Gases Chapter Goals Understand the basis of...

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