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GASES Chapter 14

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From last chapter… Kinetic Molecular Theory Particles in an ideal gas… – have no volume. – have elastic collisions. – are in constant, random, straight-line motion. – dont attract or repel each other. – have an avg. KE directly related to Kelvin temperature.

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Real Gases Particles in a REAL gas… – have their own volume – attract each other Gas behavior is most ideal… – at low pressures – at high temperatures – in nonpolar atoms/molecules

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Properties of Gases Compressibility – gases are easily compressed because of the space between the particles in a gas Gases expand to take the shape and volume of their container

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Factors Affecting Gas Pressure Amount of gas – more particles have more collisions with the container walls and thus create more pressure Volume – if you reduce the volume of the container, the particles are more compressed and exert a greater pressure on the walls of the container Temperature – increasing temperature increases the kinetic energy of the particles, which then strike the walls of the container with more energy

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Remember? Units of Pressure KEY UNITS AT SEA LEVEL 101.325 kPa (kilopascal) 1 atm 760 mm Hg 760 torr 14.7 psi *These are all equivalent amounts of pressure

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Standard Temperature & Pressure 0°C 273 K 1 atm101.325 kPa -OR- STP

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The Gas Laws 14.2

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Boyles Law The pressure and volume of a gas are inversely related – at constant mass & temp P V P 1 V 1 = P 2 V 2

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GIVEN: V 1 = 100. mL P 1 = 150. kPa V 2 = ? P 2 = 200. kPa WORK: Gas Law Problem A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. BOYLES LAW P V

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V T Charles Law The volume and absolute temperature (K) of a gas are directly related – at constant mass & pressure

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GIVEN: V 1 = 473 cm 3 T 1 = 36°C = 309K V 2 = ? T 2 = 94°C = 367K WORK: Gas Law Problem A gas occupies 473 cm 3 at 36°C. Find its volume at 94°C. CHARLES LAW T V

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P T Gay-Lussacs Law The pressure and absolute temperature (K) of a gas are directly related – at constant mass & volume

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GIVEN: P 1 = 765 torr T 1 = 23°C = 296K P 2 = 560. torr T 2 = ? WORK: Gas Law Problem A gas pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr? GAY-LUSSACS LAW P T

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Combined Gas Law P1V1T1P1V1T1 = P2V2T2P2V2T2

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GIVEN: V 1 = 7.84 cm 3 P 1 = 71.8 kPa T 1 = 25°C = 298 K V2 = ?V2 = ? P 2 = 101.325 kPa T 2 = 273 K WORK: Gas Law Problem A gas occupies 7.84 cm 3 at 71.8 kPa & 25°C. Find its volume at STP. P T V COMBINED GAS LAW

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V n Avogadros Law The volume and number of moles of a gas are directly related – at constant temperature & pressure

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GIVEN: V 1 = 36.7 L n 1 = 1.5 mol V 2 = 16.5 L n 2 = ? WORK: Gas Law Problem Consider two sample of N 2 gas. Sample 1 contains 1.5 mol of N 2 and has a volume of 36.7 L at 25°C and 1 atm. Sample 2 has a volume of 16.5 L at 25°C and 1 atm. Calculate the number of moles of N 2 in Sample 2. AVOGADROS LAW n V

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14.3 - Ideal Gas Law UNIVERSAL GAS CONSTANT R=0.0821 L atm/mol K R=8.315 dm 3 kPa/mol K PV=nRT

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GIVEN: P = ? atm n = 0.412 mol T = 16°C = 289 K V = 3.25 L R = 0.0821 L atm/mol K WORK: Ideal Gas Law Problem Calculate the pressure in atmospheres of 0.412 mol of He at 16°C & occupying 3.25 L.

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14.4 - Daltons Law of Partial Pressures The partial pressure of a gas is the pressure that the gas would exert if it were alone in the container. Daltons Law of Partial Pressures says that the total pressure of a mixture of gas is equal to the sum of the partial pressures of all gases in the mixture. Or, P total = P 1 + P 2 + P 3 +… Note: you can calculate the partial pressures of the gases if they behave ideally using the ideal gas law (P = nRT/V)

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Daltons Law Example A 2.0 L flask contains a mixture of nitrogen gas and oxygen gas at 25°C. The total pressure of the gaseous mixture is 0.91 atm, and the mixture is known to contain 0.050 mol of N 2. Calculate the partial pressure of oxygen and the moles of oxygen present.

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Grahams Law of Effusion Diffusion is the tendency of molecules to move toward areas of lower concentration until the concentration is uniform throughout Effusion is when a gas escapes through a tiny hole in its container Gases of lower molar mass diffuse and effuse faster than gases of higher molar mass

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Grahams Law of Effusion The rate of effusion of a gas is inversely proportional to the square root of the gass molar mass. This equation compares effusion rates for two gases

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Grahams Law Problem Calculate the ratio of the velocity of hydrogen molecules (H 2 ) to the velocity of carbon dioxide (CO 2 ) molecules at the same temperature.

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Gas Stoichiometry Molar volume of a gas is the volume that is occupied by 1 mol of an ideal gas at STP. – 1 mol of gas occupies 22.4 L Yes, we are going back to those 3 step problems…

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Gas Stoichiometry Problem Quicklime, CaO, is produced by heating calcium carbonate. Calculate the volume of carbon dioxide produced at STP from the decomposition of 152 g of calcium carbonate according to the reaction CaCO 3 (s) CaO (s) + CO 2 (g)

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