Presentation on theme: "GASES Chapter 14. From last chapter… Kinetic Molecular Theory Particles in an ideal gas… – have no volume. – have elastic collisions. – are in constant,"— Presentation transcript:
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.
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
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 0°C 273 K 1 atm101.325 kPa -OR- STP
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
V n Avogadros Law The volume and number of moles of a gas are directly related – at constant temperature & pressure
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
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
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.
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)
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.
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
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
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.
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…
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)