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Chapter 11 Gases 2006, Prentice Hall

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2 CHAPTER OUTLINE Properties of Gases Pressure & Its Measurement The Gas Laws Vapor Pressure & Boiling Point Combined Gas Law Avogadros Law STP & Molar Volume Ideal Gas Law Partial Pressures

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3 PROPERTIES OF GASES Gases are the least dense and most mobile of the three phases of matter. Particles of matter in the gas phase are spaced far apart from one another and move rapidly and collide with each other often. Gases occupy much greater space than the same amount of liquid or solid. This is because the gas particles are spaced apart from one another and are therefore compressible. Solid or liquid particles are spaced much closer and cannot be compressed.

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4 PROPERTIES OF GASES Gases are characterized by four properties. These are: 1. Pressure(P) 2. Volume(V) 3. Temperature(T) 4. Amount(n)

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5 KINETIC-MOLECULAR THEORY Scientists use the kinetic-molecular theory (KMT) to describe the behavior of gases. The KMT consists of several postulates: 1. Gases consist of small particles (atoms or molecules) that move randomly with rapid velocities. 3. The distance between the particles is large compared to their size. Therefore the volume occupied by gas molecules is small compared to the volume of the gas. 2. Gas particles have little attraction for one another. Therefore attractive forces between gas molecules can be ignored.

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6 KINETIC-MOLECULAR THEORY 5. The average kinetic energy of gas molecules is directly proportional to the absolute temperature (Kelvin). Animation 4. Gas particles move in straight lines and collide with each other and the container frequently. The force of collision of the gas particles with the walls of the container causes pressure.

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Tro's Introductory Chemistry, Chapter 7 Kinetic Molecular Theory

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8 PRESSURE & ITS MEASUREMENT Pressure is the result of collision of gas particles with the sides of the container. Pressure is defined as the force per unit area. Pressure is measured in units of atmosphere (atm) or mmHg or torr. The SI unit of pressure is pascal (Pa) or kilopascal (kPa). 1 atm= 760 mmHg= kPa 1 mmHg= 1 torr

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Tro's Introductory Chemistry, Chapter 9 Common Units of Pressure UnitAverage Air Pressure at Sea Level pascal (Pa)101,325 kilopascal (kPa) atmosphere (atm)1 (exactly) millimeters of mercury (mmHg)760 (exactly) inches of mercury (inHg)29.92 torr (torr)760 (exactly) pounds per square inch (psi, lbs./in 2 )14.7

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10 PRESSURE & ITS MEASUREMENT Atmospheric pressure can be measured with the use of a barometer. Mercury is used in a barometer due to its high density. At sea level, the mercury stands at 760 mm above its base.

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Tro's Introductory Chemistry, Chapter 11 Atmospheric Pressure & Altitude the higher up in the atmosphere you go, the lower the atmospheric pressure is around you at the surface the atmospheric pressure is 14.7 psi, but at 10,000 ft is is only 10.0 psi rapid changes in atmospheric pressure may cause your ears to pop due to an imbalance in pressure on either side of your ear drum

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Tro's Introductory Chemistry, Chapter 12 Pressure Imbalance in Ear If there is a difference in pressure across the eardrum membrane, the membrane will be pushed out – what we commonly call a popped eardrum.

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13 Example 1: The atmospheric pressure at Walnut, CA is 740. mmHg. Calculate this pressure in torr and atm. 1 mmHg= 1 torr 740. mmHg= 740. torr 740. mmHg = atm

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14 Example 2: The barometer at a location reads 1.12 atm. Calculate the pressure in mmHg and torr. 1 mmHg= 1 torr 851 mmHg= 851 torr 1.12 atm = 851 mmHg

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15 PRESSURE & MOLES OF A GAS The pressure of a gas is directly proportional to the number of particles (moles) present. The greater the moles of the gas, the greater the pressure

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16 BOYLES LAW The relationship of pressure and volume in gases is called Boyles Law. At constant temperature, the volume of a fixed amount of gas is inversely proportional to its pressure. As pressure increases, the volume of the gas decreases Pressure and volume of a gas are inversely proportional

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17 BOYLES LAW Boyles Law can be mathematically expressed as P 1 x V 1 = P 2 x V 2 P and V under initial condition P and V under final condition

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18 Example 1: A sample of gas has a volume of 12-L and a pressure of 4500 mmHg. What is the volume of the gas when the pressure is reduced to 750 mmHg? V 2 = 72 L P 1 = 4500 mmHg P 2 = 750 mmHg V 1 = 12 L V 2 = ??? P 1 x V 1 = P 2 x V 2 Pressure decreases Volume increases

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19 Example 2: A sample of hydrogen gas occupies 4.0 L at 650 mmHg. What volume would it occupy at 2.0 atm? V 2 = 1.7 L P 1 = 650 mmHg P 2 = 2.0 atm V 1 = 4.0 L V 2 = ??? P 1 x V 1 = P 2 x V 2 Pressures must be in the same unit P 2 = 2.0 atm= 1520 mmHg Volume decreases Pressure increases

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20 CHARLESS LAW The relationship of temperature and volume in gases is called Charles Law. At constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature. As temperature increases, the volume of the gas increases Temperature and volume of a gas are directly proportional

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21 CHARLESS LAW Charles Law can be mathematically expressed as V and T under initial condition V and T under final condition T must be in units of K

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22 Example 1: A 2.0-L sample of a gas is cooled from 298K to 278 K, at constant pressure. What is the new volume of the gas? V 2 = 1.9 L V 1 = 2.0 L V 2 = ??? T 1 = 298 K T 2 = 278 K Volume decreases Temperature decreases

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23 Example 2: If 20.0 L of oxygen is cooled from 100ºC to 0ºC, what is the new volume? V 2 = 14.6 L V 1 = 20.0 L V 2 = ??? T 1 = 100ºC T 2 = 0ºC T 1 = 100ºC + 273= 373 K T 2 = 0ºC + 273= 273 K Temperature s must be in K Volume decreases Temperature decreases

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24 GAY-LUSSACS LAW The relationship of temperature and pressure in gases is called Gay-Lussacs Law. At constant volume, the pressure of a fixed amount of gas is directly proportional to its absolute temperature. As temperature increases, the pressure of the gas increases Temperature and pressure of a gas are directly proportional

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25 GAY-LUSSACS LAW Gay-Lussacs Law can be mathematically expressed as P and T under initial condition P and T under final condition T must be in units of K

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26 Example 1: An aerosol spray can has a pressure of 4.0 atm at 25 C. What pressure will the can have if it is placed in a fire and reaches temperature of 400 C ? P 1 = 4.0 atm P 2 = ??? T 1 = 25ºC T 2 = 400ºC T 1 = 25ºC + 273= 298 K T 2 = 400ºC + 273= 673 K Temperatures must be in K

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27 Example 1: An aerosol spray can has a pressure of 4.0 atm at 25 C. What pressure will the can have if it is placed in a fire and reaches temperature of 400 C ? P 2 = 9.0 atm P 1 = 4.0 atm P 2 = ??? T 1 = 298 K T 2 = 673 K Pressure increases Temperature increases

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28 Example 2: A cylinder of gas with a volume of 15.0-L and a pressure of 965 mmHg is stored at a temperature of 55 C. To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg? T 2 = 289 K P 1 = 965 mmHg P 2 = 850 mmHg T 1 = 55ºC T 2 = ??? T 1 = 55ºC + 273= 328 K T 2 = 16ºC Temperature decreases Pressure decreases

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29 VAPOR PRESSURE & BOILING POINT In an open container, liquid molecules at the surface that possess sufficient energy, can break away from the surface and become gas particles or vapor. In a closed container, these gas particles can accumulate and create pressure called vapor pressure. Vapor pressure is defined as the pressure above a liquid at a given temperature. Vapor pressure varies with each liquid and increases with temperature.

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30 VAPOR PRESSURE & BOILING POINT Listed below is the vapor pressure of water at various temperatures.

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31 VAPOR PRESSURE & BOILING POINT A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure). For example, at sea level, water reaches its boiling point at 100 C since its vapor pressure is 760 mmHg at this temperature.

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32 VAPOR PRESSURE & BOILING POINT At higher altitudes, where atmospheric pressure is lower, water reaches boiling point at temperatures lower than 100 C. For example, in Denver, where atmospheric pressure is 630 mmHg, water boils at 95 C, since its vapor pressure is 630 mmHg at this temperature.

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33 COMBINED GAS LAW All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law. Initial condition This law is useful for studying the effect of changes in two variables. Final condition

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34 COMBINED GAS LAW The individual gas laws studied previously are embodied in the combined gas law. Boyles Law Charless Law Gay-Lussacs Law

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35 Example 1: A 25.0-mL sample of gas has a pressure of 4.00 atm at a temperature of 10 C. What is the volume of the gas at a pressure of 1.00 atm and a temperature of 18 C? V 2 = 103 mL P 1 = 4.00 atm V 1 = 25.0 mL T 1 = 283 K P 2 = 1.00 atm V 2 = ??? T 2 = 291 K Both pressure and temp. change Must use combined gas law

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36 AVOGADROS LAW The relationship of moles and volume in gases is called Avogadros Law. At constant temperature and pressure, the volume of a fixed amount of gas is directly proportional to the number of moles. As number of moles increases, the volume of the gas increases Number of moles and volume of a gas are directly proportional

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Tro's Introductory Chemistry, Chapter 37 Avogadros Law

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38 AVOGADROS LAW As a result of Avogadros Law, equal volumes of different gases at the same temp. and pressure contain equal number of moles (molecules). 2 Tanks of gas of equal volume at the same T & P contain the same number of molecules

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39 AVOGADROS LAW For example: 2 H 2 (g) + 1 O 2 (g) 2 H 2 O (g) 2 molecules1 molecule2 molecules 2 moles1 mole2 moles 2 Liters1 Liter2 Liters Avogadros Law also allows chemists to relate volumes and moles of a gas in a chemical reaction. This relationship is only valid for gaseous substances

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40 Example 1: A sample of helium gas with a mass of 18.0 g occupies 1.6 Liters at a particular temperature and pressure. What mass of oxygen would occupy1.6 L at the same temperature and pressure? mol of He= 4.50 mol mol of O 2 = mol of He= 4.50 mol Same Temp. & Pressure mass of O 2 = 144 g

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41 Example 2: How many Liters of NH 3 can be produced from reaction of 1.8 L of H 2 with excess N 2, as shown below? 1.8 L H 2 = 1.2 L NH 3 N 2 (g) + 3 H 2 (g) 2 NH 3 (g) 3 2

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42 STP & MOLAR VOLUME To better understand the factors that affect gas behavior, a set of standard conditions have been chosen for use, and are referred to as Standard Temperature and Pressure (STP). STP = 1 atm (760 mmHg) 0ºC (273 K)

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43 STP & MOLAR VOLUME At STP conditions, one mole of any gas is observed to occupy a volume of 22.4 L. V = 22.4 L Molar Volume at STP

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44 Example 1: If 2.00 L of a gas at STP has a mass of 3.23 g, what is the molar mass of the gas? mol of gas = = mol Molar mass == 36.2 g/mol Molar volume at STP

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45 Example 2: A sample of gas has a volume of 2.50 L at 730 mmHg and 20 C. What is the volume of this gas at STP? V 2 = 2.24 L P 1 = 730 mmHg V 1 = 2.50 L T 1 = 293 K P 2 = 760 mmHg V 2 = ??? T 2 = 273 K

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46 IDEAL GAS LAW Combining all the laws that describe the behavior of gases, one can obtain a useful relationship that relates the volume of a gas to the temperature, pressure and number of moles. Universal gas constant R = L atm/mol K

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47 IDEAL GAS LAW This relationship is called the Ideal Gas Law, and commonly written as: P V = n R T Pressure in atm Volume in Liters Number of moles Temp. in K

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48 Example 1: A sample of H 2 gas has a volume of 8.56 L at a temp. of 0 C and pressure of 1.5 atm. Calculate the moles of gas present. = 0.57 mol P = 1.5 atm V = 8.56 L T = 273 K n = ??? R = PV = nRT

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49 Example 2: What volume does 40.0 g of N 2 gas occupy at 10 C and 750 mmHg? P =750 mmHg V = ??? T = 283 K n = not given m = 40.0 g R = mol of N 2 == 1.43 mol P = 750 mmHg Moles must be calculated from mass Pressure must be converted to atm P = atm

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50 Example 2: What volume does 40.0 g of N 2 gas occupy at 10 C and 750 mmHg? V = 33.7 L P =0.987 atm V = ??? T = 283 K n = 1.43 R =

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Tro's Introductory Chemistry, Chapter 51 Ideal vs. Real Gases Real gases often do not behave like ideal gases at high pressure or low temperature Ideal gas laws assume 1)no attractions between gas molecules 2)gas molecules do not take up space based on the Kinetic-Molecular Theory at low temperatures and high pressures these assumptions are not valid

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Tro's Introductory Chemistry, Chapter 52 Ideal vs. Real Gases

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53 PARTIAL PRESSURES Many gas samples are mixture of gases. For example, the air we breathe is a mixture of mostly oxygen and nitrogen gases. Since gas particles have no attractions towards one another, each gas in a mixture behaves as if it is present by itself, and is not affected by the other gases present in the mixture. In a mixture, each gas exerts a pressure as if it was the only gas present in the container. This pressure is called partial pressure of the gas.

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54 DALTONS LAW In a mixture, the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture. This is called Daltons law of partial pressures. Total pressure of a gas mixture P total = P 1 + P 2 + P 3 + ··· Sum of the partial pressures of the gases in the mixture =

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55 PARTIAL PRESSURES P He = 2.0 atm + P Ar = 4.0 atmP Total = ???P He + P Ar Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture = 6.0 atm

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56 PARTIAL PRESSURES The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture. For example, in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas, the partial pressure of each gas is one-half of the total pressure in the container.

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57 PARTIAL PRESSURES P He = 2.0 atm + P Ar = 4.0 atmP Total = 6.0 atm The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture Twice as many moles of Ar compared to He P Ar = 2 P He

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58 Example 1: A scuba tank contains a mixture of oxygen and helium gases with total pressure of 7.00 atm. If the partial pressure of oxygen in the tank is 1140 mmHg, what is the partial pressure of helium in the tank? P total = P oxygen + P helium P oxygen = 1140 mmHg x 1 atm 760 mmHg = 1.50 atm P helium = P Total - P oxygen = 7.00 atm – 1.50 atm= 5.50 atm

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59 Example 2: A mixture of gases contains 2.0 mol of O 2 gas and 4.0 mol of N 2 gas with total pressure of 3.0 atm. What is the partial pressure of each gas in the mixture? P total = P oxygen + P nitrogen P nitrogen = 2 P oxygen Twice as many moles of N 2 compared to O 2 P nitrogen =2/3 (P total )= 2.0 atm P oxygen =1/3 (P total )= 1.0 atm

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60 THE END

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