Chapter 13 Gases and Pressure. Pressure and Force. Pressure is the force per unit area on a surface. Force Area. Pressure =

Transcription

1 Chapter 13 Gas Laws

2 Chapter 13 Gases and Pressure Pressure and Force Pressure is the force per unit area on a surface. Pressure = Force Area

3 Chapter 13 Gases and Pressure Gases in the Atmosphere The atmosphere of Earth is a layer of gases surrounding the planet that is retained by Earth's gravity. By volume, dry air is 78% nitrogen, 21% oxygen, 0.9% argon, 0.04% CO 2, and small amounts of other gases.

4 Chapter 13 Gases and Pressure Atmospheric Pressure Atmospheric pressure is the force per unit area exerted on a surface by the weight of the gases that make up the atmosphere above it.

5 Chapter 13 Gases and Pressure Measuring Pressure A common unit of pressure is millimeters of mercury (mm Hg). 1 mm Hg is also called 1 torr in honor of Evangelista Torricelli who invented the barometer (used to measure atmospheric pressure). The average atmospheric pressure at sea level at 0 C is 760 mm Hg, so one atmosphere (atm) of pressure is 760 mm Hg.

6 Chapter 13 Gases and Pressure Measuring Pressure (continued) Pressure can also be measured in pascals (Pa): 1 Pa = 1 N/m 2. One pascal is very small, so usually kilopascals (kpa) are used instead. One atm is equal to kpa. 1 atm = 760 mm Hg (Torr) = kpa

7 Chapter 13 Gases and Pressure Units of Pressure

8 Converting Pressure Sample Problem The average atmospheric pressure in Denver, CO is atm. Express this pressure in: a. millimeters of mercury (mm Hg) atm b. kilopascals (kpa) Chapter 13 Gases and Pressure x 760 mm Hg = 1 atm 631 mm Hg atm x kpa = 1 atm 84.1 kpa

9 Chapter 13 Gases and Pressure Dalton s Law of Partial Pressures Dalton s law of partial pressures - the total pressure of a gas mixture is the sum of the partial pressures of the component gases. P T = P 1 + P 2 + P 3

10 Dalton s Law of Partial Pressures Sample Problem A container holds a mixture of gases A, B & C. Gas A has a pressure of 0.5 atm, Gas B has a pressure of 0.7 atm, and Gas C has a pressure of 1.2 atm. a. What is the total pressure of this system? P T = P 1 + P 2 + P 3 P T = 0.5 atm atm atm = 2.4 atm b. What is the total pressure in mm Hg? 2.4 atm x Chapter 13 Gases and Pressure mm Hg atm = 1800 mm Hg

11 Chapter 13 Gases and Pressure Dalton s Law of Partial Pressures #3 Gases and Pressure WS Sample Problem Explain how to calculate the partial pressure of a dry gas that is collected over water when the total pressure is atmospheric pressure. Because water molecules at a liquid surface evaporates gas collected is not pure! The gas collected is a mixture of dry gas and water vapor Use Dalton s Law to solve for partial pressure: p dry gas = P atm p water

12 Chapter 13 The Gas Laws Gases and Pressure Gas pressure is caused by collisions of the gas molecules with each other and with the walls of their container. The greater the number of collisions, the higher the pressure will be.

13 Chapter 13 The Gas Laws Pressure Volume Relationship When the volume of a gas is decreased, more collisions will occur. Pressure is caused by collisions. Therefore, pressure will increase. This relationship between pressure and volume is inversely proportional.

14 Chapter 13 The Gas Laws Boyle s Law Boyle s Law The volume of a fixed mass of gas varies inversely with the pressure at a constant temperature. P 1 V 1 = P 2 V 2 P 1 and V 1 represent initial conditions, and P 2 and V 2 represent another set of conditions.

15 Boyle s Law Sample Problem A sample of oxygen gas has a volume of ml when its pressure is atm. What will the volume of the gas be at a pressure of atm if the temperature remains constant? Solution: Chapter 13 The Gas Laws P 1 V 1 = P 2 V 2 (0.947 atm) (150.0 ml) = (0.987 atm) V 2 (0.947 atm) (150.0 ml) V 2 = = 144 ml (0.987 atm)

16 Chapter 13 The Gas Laws Volume Temperature Relationship the pressure of gas inside and outside the balloon are the same. at low temperatures, the gas molecules don t move as much therefore the volume is small. at high temperatures, the gas molecules move more causing the volume to become larger.

17 Chapter 13 The Gas Laws Charles s Law Charles s Law The volume of a fixed mass of gas at constant pressure varies directly with the Kelvin temperature. V 1 V 2 = T 1 T 2 V 1 and T 1 represent initial conditions, and V 2 and T 2 represent another set of conditions.

18 Chapter 13 The Gas Laws The Kelvin Temperature Scale Absolute zero The theoretical lowest possible temperature where all molecular motion stops. The Kelvin temperature scale starts at absolute zero (-273 o C.) This gives the following relationship between the two temperature scales: K = o C + 273

19 Charles s Law Sample Problem A sample of neon gas occupies a volume of 752 ml at 25 C. What volume will the gas occupy at 50 C if the pressure remains constant? Solution: V 1 V = 2 T 1 T ml V = K 323 K 752 ml V 2 = x 298 K Chapter 13 The Gas Laws 323 K K = o C T 1 = = 298 T 2 = = 323 = 815 ml

20 Chapter 13 The Gas Laws Pressure Temperature Relationship Increasing temperature means increasing kinetic energy of the particles. The energy and frequency of collisions depend on the average kinetic energy of the molecules. Therefore, if volume is kept constant, the pressure of a gas increases with increasing temperature.

21 Chapter 13 The Gas Laws Gay-Lussac s Law Gay-Lussac s Law The pressure of a fixed mass of gas varies directly with the Kelvin temperature. P 1 P 2 = T 1 T 2 P 1 and T 1 represent initial conditions. P 2 and T 2 represent another set of conditions.

22 Chapter 13 The Gas Laws The Combined Gas Law The combined gas law is written as follows: P 1 V 1 P 2 V 2 = T 1 T 2 Each of the other gas laws can be obtained from the combined gas law when the proper variable is kept constant.

23 Chapter 13 The Gas Laws The Combined Gas Law Sample Problem A helium-filled balloon has a volume of 50.0 L at 25 C and 1.08 atm. What volume will it have at atm and 10.0 C? K = o C Solution: T 1 = = 298 P 1 V 1 P 2 V = 2 T 2 = = 283 T 1 T 2 (1.08 atm) (50.0 L) (0.855 atm) V = K 283 K (1.08 atm) (50.0 L) V (283 K) 2 = = 60.0 L (298 K) (0.855 atm)

24 Chapter 13 The Gas Laws Avogadro s Law In 1811, Amedeo Avogadro discovered that the volume of a gas is proportional to the number of molecules (or number of moles.) Avogadro s Law - equal volumes of gases at the same temperature and pressure contain equal numbers of molecules, or: V 1 V 2 = n 1 n 2

25 Chapter 13 Gas Volumes and the Ideal Gas Law Standard Molar Volume Standard Temperature and Pressure (STP) is 0 o C and 1 atm. The Standard Molar Volume of a gas is the volume occupied by one mole of a gas at STP. It has been found to be 22.4 L.

26 Chapter 13 Gas Volumes and the Ideal Gas Law Molar Volume Conversion Factor Standard Molar Volume can be used as a conversion factor to convert from the number of moles of a gas at STP to volume (L), or vice versa.

27 Molar Volume Conversion Sample Problem a. What quantity of gas, in moles, is contained in 5.00 L at STP? 5.00 L Chapter 13 Gas Volumes and the Ideal Gas Law x mol L = mol b. What volume does moles of a gas occupy at STP? mol x 22.4 L 1 mol = 17.2 L

28 Chapter 13 Gas Volumes and the Ideal Gas Law Volume Ratios You can use the volume ratios as conversion factors just like you would use mole ratios. 2 CO(g) + O 2 (g) 2 CO 2 (g) 2 molecules 1 molecule 2 molecules 2 mole 1 mole 2 mol 2 volumes 1 volume 2 volumes Example: What volume of O 2 is needed to react completely with L of CO to form CO 2? L CO x 1 L O 2 = L O 2 2 L CO

29 Chapter 13 Gas Volumes and the Ideal Gas Law The Mole Map You can now convert between number of particles, mass (g), and volume (L) by going through moles.

30 Gas Stoichiometry Sample Problem Assume that 5.61 L H 2 at STP reacts with excess CuO according to the following equation: CuO(s) + H 2 (g) Cu(s) + H 2 O(g) a. How many moles of H 2 react? 5.61 L H 2 x 1 mol H 2 = mol H L H 2 b. How many grams of Cu are produced? 5.61 L H 2 x 1 mol H L H 2 Chapter 13 Gas Volumes and the Ideal Gas Law x 1 mol Cu 1 mol H 2 x 63.5 g Cu 1 mol Cu = 15.9 g Cu

31 Chapter 13 Gas Volumes and the Ideal Gas Law The Ideal Gas Law All of the gas laws you have learned so far can be combined into a single equation, the ideal gas law: PV = nrt R represents the ideal gas constant which has a value of (L atm)/(mol K).

32 The Ideal Gas Law Sample Problem What is the pressure in atmospheres exerted by a mol sample of nitrogen gas in a 10.0 L container at 298 K? Solution: PV = nrt Chapter 13 Gas Volumes and the Ideal Gas Law P (10.0 L) = (0.500 mol) ( L atm/mol K) (298 K) P (0.500 mol) ( L atm/mol K) (298 K) = (10.0 L) = 1.22 atm

33 Chapter 13 Diffusion and Effusion Diffusion and Effusion Diffusion is the gradual mixing of two or more gases due to their spontaneous, random motion. Effusion is the process whereby the molecules of a gas confined in a container randomly pass through a tiny opening in the container.

34 Chapter 13 Diffusion and Effusion Graham s Law of Effusion Light molecules move faster than heavy ones. Graham s law of effusion says the greater the molar mass of a gas, the slower it will effuse.