Gas Laws Compressible They have no constant volume. *They fill the container they are placed in. They have mass. They have high Kinetic energy (KE = ½ mv 2 ) Movement is random and rapid. Gases exert pressure. *They undergo thousands of collisions with each other and with the container walls. Real Gas Properties

1.All gases consist of very small particles with very small (point) masses. 2.Large distances separate gas particles. 1.The volume of the gas particles themselves is assumed to be zero because it is negligible compared with the total volume of the container. 3.Gas particles are in constant, rapid, linear motion. 4.Collisions within the container are perfectly elastic. 5.The average KE depends only on the temperature of the gas. 6.Gas particles exert no force on one another. 1.Attractive forces between gas particles are so weak that the model assumes them to be zero. Kinetic Molecular Theory

Gas Forms Monatomic (Nobel Gases) Diatomic (N 2, O 2, H 2, F 2,Cl 2 ) Molecules (Compounds such as CO 2, H 2 O and NO 2 ) Ideal gases are described by the kinetic-molecular theory postulates. Ideal gases are considered point masses with no volume. No attractive forces exist between gas particles. Ideal Gas Properties

Measuring Gases Temperature (T): The average kinetic energy of a system. Units: Celcius (°C) and convert to Kelvin (K) K = °C The Kelvin temperature scales is based on absolute zero. “0” K is equal to –273 °C. Absolute zero suggests that matter has no energy and all motion has stopped. Amount of Gas (n ): “n” is the letter designated to indicate moles of gas. Recall: 1 mole of gas = 22.4 L at STP STP = 0 °C and 1 atm

Volume (v): The volume of a gas varies with container size. Recall: 1 mL = 1 cm 3 1 L = 1 dm 3 Pressure (P): Force per unit area. Particles strike the container with a force relative to their velocity (v) and rebound with an equal force in the opposite direction. With an abundance of collisions, the force per unit area becomes constant. Atmospheric pressure results by air colliding with objects on Earth and Earth itself. Standard Atmospheric pressure (at sea-level) is: 1 atm = 760 torr = 760 mm Hg = kPa = 1 N/m 2

Barometers measure atmospheric pressure. Mercury Barometers measure pressure by the changes that occur in the height of a column of Hg supported by the atmosphere. Aneroid Barometers exist in a metal can (aneroid) that contains a near vacuum. The pressure is measured relative to the changes in the aneroid size.

An instrument used to measure the pressure of enclosed gases relative to the atmospheric pressure in a closed container. A U-shaped glass tube is filled with mercury. One end of the tube is open into the container in which the gas pressure is to be measure while the other end of the tube is open to the surrounding atmoshpere. If the gas pressure is greater it will push the mercury away from it. If the atmospheric pressure is greater, it will push the mercury down.

P gas = P atm - hP gas = P atm + h At Sea-Level!

You have a closed container attached to a U-tube. The mercury in the open-ended side of the tube is higher, the difference between the heights of the mercury columns in 27 mm. You have measured the atmospheric pressure as 755 mm Hg. What is the pressure of the gas in the container in atmospheres? Manometer Sample problems

A balloon is attached to an open-ended manometer. The mercury level in the manometer is 13 mm lower on the side attached to the balloon than on the side open to the atmosphere. The pressure of the atmosphere is measured to be 755 mm Hg. What is the pressure of the balloon in torr and kPa?

Boyle’s Law Robert Boyle measured the volume (v) of the air at different pressures (p) while keeping the temperature constant. PV = k b where k is a constant With this relationship, we can measure the pressure and volume in the laboratory and then mathematically determine a second set of data for the same substances. P 1 V 1 = k b P 1 V 1 = P 2 V 2 = P 2 V 2

P 1 V 1 = P 2 V 2 This suggests that pressure and volume are inversely proportional. Boyle’s Law is valid for real gases except at very low temperatures and very high pressures. Boyle’s Law The curve is a hyperbola indicating an inverse relationship.

Gas Law Practice problems #1

Jacque Charles measured the volume (v) of the gases at different temperature (T) while keeping the pressures constant. V = k c T where k c is a constant With this relationship, we can measure the temperature and volume in the laboratory and then mathematically determine a second set of data for the same substances. V 1 = k c T 1 = V 2 T 2 V 1 = T 1 V 2 T 2 Charles Law

V 1 = V 2 T 1 T 2 This suggests that Temperature and volume are directly proportions. Charles Law was used to estimate the point of absolute zero. Charles Law You must always put your temperature in Kelvin for any gas law calculation. (Remember, you can’t divide by 0!)

Gas Law Practice problems #3

Avogadro’s Law Equal volumes of gases at the same temperature and pressure contain an equal number of particles. V = k a n or V = k a n Where V = volume, k 3 = a constant, and n = mols of gas This law suggests:  All gases show the same physical behavior.  Gases with a larger volume must consist of a greater number of particles.  Volume is directly proportional to mols of gas Remember!!!! 1 mol of gas at STP = 22.4 L

Dalton’s Law of Partial Pressure The sum of the partial pressures of all the components in a gas mixture is equal to the total pressure of the gas mixture. P T = P a + P b + P c +…….P n What is the atmospheric pressure if the partial pressure of nitrogen, oxygen, and argon are mm Hg, mm Hg and 0.5 mm Hg respectively? P T = P a + P b + P c P T = mm Hg mm Hg mm Hg P T = mm Hg

Gas Law Practice problems #7

Combining Boyle’s & Charles’ Laws P 1 V 1 = P 2 V 2 T 1 T 2 The combined gas law relates temperature, pressure and volume of a gas. Temperature and volume are directly proportional Temperature and Pressure are directly proportional Pressure and volume are inversely proportional

Gas Law Practice problems #6

Ideal Gas Law PV = nRT Where: P = pressure in “atm”, V = volume in “L”, n = mols of gas T = temperature in Kelvin. R = Ideal Gas Law Constant = atmL/molK How many moles of gas at 100.0°C does it take to fill a 1.00 L flask to a pressure of 1.50 atm? PV = nRT PV = n RT n = (1.50 atm) (1.00 L) = mols ( atmL)(373 K) molK T = 100.0°C = K P = 1.50 atm V =1.00 L

A camping stove uses a 5.00 L propane tank that holds 3.00 kg of liquid C 3 H 8. How large a container would be needed to hold the same amount of propane as a gas at 25.0 °C and a pressure of kPa? Givens: V = ? T = 25.0 °C P = kPa m = 3.00 kg = 298 K = 3.00 atm = 3000 g 1 mol C 3 H 8 = 68.1 mols C 3 H g c 3 H 8 PV = nRTV = nRT P V = (68.1 mols)(( atmL)(298 K) molK 3.00 atm = 555 L

Variations of the Ideal Gas Law PV = nRT Moles (n) = mass of given FM or MM of given So make a substitution PV = mRT (FM) We also know, density = mass volume P = mRT V (FM) Substitute d for m V P = dRT (FM) Thus, you have three forms of the Ideal Gas law!

Gas Diffusion  The movement of a gas throughout an area.  Diffusion is also dependent on how big the gas particle is. Lighter gas particles diffuse faster than heavier particles because the velocity of lighter particles is greater than those of heavier particles. 1.Explain how gas density is related to molar mass and temperature. Since there are the same number of gas particles in equal volumes at the same temp and pressure (1 mol gas = 22.4 L at STP), the density of a gas is directly proportional to the mass of its particles. Also since temperature increase causes gas volume to increase, fewer particles in a unit of volume will exist thus causing the density to decrease. 2. Describe two ways in which a balloon can get lift in air. To achieve lift, the gas inside the balloon must be less dense than the air outside. This can be done by either using a gas whose particles have a smaller mass than the weighted average mass of the particle is air, or using hot gases that have expanded enough to make them less dense than the surrounding air.

Gas Effusion  The movement of a gas through a hole so tiny that they do not stream through but instead pass through one particle at a time.  Effusion is dependent on how big the gas particle is. Lighter gas particles effuse faster than heavier particles. Ex… CO 2, O 3, C 2 H 6, and SO 3. Rank the gases in order of increasing effusion rate. **First find the molecular masses of each gas. CO 2 =44 g/mol, O 3 =48 g/mol, C 2 H 6 =30 g/mol, and SO 3 =80 g/mol Thus, ranking from slow to fast…SO 3, O 3, CO 2,and C 2 H 6