GASES Chapter 14

From last chapter… Kinetic Molecular Theory Particles in an ideal gas… have no volume. have elastic collisions. are in constant, random, straight-line motion. don’t attract or repel each other. have an avg. KE directly related to Kelvin temperature.

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

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

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

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

Standard Temperature & Pressure STP Standard Temperature & Pressure 0°C 273 K 1 atm 101.325 kPa -OR-

The Gas Laws 14.2 P V T

Boyle’s Law The pressure and volume of a gas are inversely related at constant mass & temp P1V1 = P2V2 P V

Gas Law Problem BOYLE’S LAW P V A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. BOYLE’S LAW GIVEN: V1 = 100. mL P1 = 150. kPa V2 = ? P2 = 200. kPa P V WORK:

Charles’ Law The volume and absolute temperature (K) of a gas are directly related at constant mass & pressure V T

Gas Law Problem CHARLES’ LAW T V A gas occupies 473 cm3 at 36°C. Find its volume at 94°C. CHARLES’ LAW GIVEN: V1 = 473 cm3 T1 = 36°C = 309K V2 = ? T2 = 94°C = 367K T V WORK:

Gay-Lussac’s Law The pressure and absolute temperature (K) of a gas are directly related at constant mass & volume P T

Gas Law Problem GAY-LUSSAC’S LAW P T A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr? GAY-LUSSAC’S LAW GIVEN: P1 = 765 torr T1 = 23°C = 296K P2 = 560. torr T2 = ? P T WORK:

Combined Gas Law P1V1 T1 = P2V2 T2

Gas Law Problem COMBINED GAS LAW P T V V1 = 7.84 cm3 P1 = 71.8 kPa A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP. COMBINED GAS LAW GIVEN: V1 = 7.84 cm3 P1 = 71.8 kPa T1 = 25°C = 298 K V2 = ? P2 = 101.325 kPa T2 = 273 K P T V WORK:

Avogadro’s Law The volume and number of moles of a gas are directly related at constant temperature & pressure V n

Gas Law Problem AVOGADRO’S LAW n V GIVEN: V1 = 36.7 L n1 = 1.5 mol Consider two sample of N2 gas. Sample 1 contains 1.5 mol of N2 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 N2 in Sample 2. AVOGADRO’S LAW GIVEN: V1 = 36.7 L n1 = 1.5 mol V2 = 16.5 L n2 = ? n V WORK:

UNIVERSAL GAS CONSTANT 14.3 - Ideal Gas Law PV=nRT UNIVERSAL GAS CONSTANT R=0.0821 Latm/molK R=8.315 dm3kPa/molK

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

14.4 - Dalton’s 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. Dalton’s 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, Ptotal = P1 + P2 + P3 +… Note: you can calculate the partial pressures of the gases if they behave ideally using the ideal gas law (P = nRT/V)

Dalton’s 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 N2. Calculate the partial pressure of oxygen and the moles of oxygen present.

Graham’s 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

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

Graham’s Law Problem Calculate the ratio of the velocity of hydrogen molecules (H2) to the velocity of carbon dioxide (CO2) molecules at the same temperature.

Gas Stoichiometry Yes, we are going back to those 3 step problems… 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…

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 CaCO3 (s)  CaO (s) + CO2 (g)