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Chapter 11 Gases 2006, Prentice Hall

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**CHAPTER OUTLINE Properties of Gases Pressure & Its Measurement**

The Gas Laws Vapor Pressure & Boiling Point Combined Gas Law Avogadro’s Law STP & Molar Volume Ideal Gas Law Partial Pressures

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PROPERTIES OF GASES 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. 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.

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**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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**KINETIC-MOLECULAR THEORY**

The KMT consists of several postulates: Scientists use the kinetic-molecular theory (KMT) to describe the behavior of gases. 1. Gases consist of small particles (atoms or molecules) that move randomly with rapid velocities. 2. Gas particles have little attraction for one another. Therefore attractive forces between gas molecules can be ignored. 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.

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**KINETIC-MOLECULAR THEORY**

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. 5. The average kinetic energy of gas molecules is directly proportional to the absolute temperature (Kelvin). Animation

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**Kinetic Molecular Theory**

Tro's Introductory Chemistry, Chapter

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**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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**Common Units of Pressure**

Average 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) pounds per square inch (psi, lbs./in2) 14.7 Tro's Introductory Chemistry, Chapter

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**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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**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 Tro's Introductory Chemistry, Chapter

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**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.” Tro's Introductory Chemistry, Chapter

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**Example 1: 1 mmHg = 1 torr 740. mmHg = 740. torr 760 1 740. mmHg**

The atmospheric pressure at Walnut, CA is 740. mmHg. Calculate this pressure in torr and atm. 1 mmHg = 1 torr 740. mmHg = 740. torr 760 1 740. mmHg = atm

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**Example 2: 1 760 1.12 atm = 851 mmHg 1 mmHg = 1 torr 851 mmHg**

The barometer at a location reads 1.12 atm. Calculate the pressure in mmHg and torr. 1 760 1.12 atm = 851 mmHg 1 mmHg = 1 torr 851 mmHg = 851 torr

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**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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BOYLE’S LAW At constant temperature, the volume of a fixed amount of gas is inversely proportional to its pressure. The relationship of pressure and volume in gases is called Boyle’s Law. As pressure increases, the volume of the gas decreases Pressure and volume of a gas are inversely proportional

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**P and V under initial condition P and V under final condition**

BOYLE’S LAW Boyle’s Law can be mathematically expressed as P1 x V1 = P2 x V2 P and V under initial condition P and V under final condition

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**Example 1: Pressure decreases Volume increases P1= 4500 mmHg**

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? Pressure decreases Volume increases P1= 4500 mmHg P2= 750 mmHg V1 = 12 L V2 = ??? P1 x V1 = P2 x V2 V2 = 72 L

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**Pressures must be in the same unit**

Example 2: A sample of hydrogen gas occupies 4.0 L at 650 mmHg. What volume would it occupy at 2.0 atm? Pressure increases Volume decreases P2 = 2.0 atm = 1520 mmHg P1= 650 mmHg P2= 2.0 atm V1 = 4.0 L V2 = ??? P1 x V1 = P2 x V2 Pressures must be in the same unit V2 = 1.7 L

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

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**V and T under initial condition V and T under final condition**

CHARLES’S LAW Charles’ Law can be mathematically expressed as T must be in units of K V and T under initial condition V and T under final condition

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**Temperature decreases**

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? Temperature decreases Volume decreases V1 = 2.0 L V2 = ??? T1 = 298 K T2 = 278 K V2 = 1.9 L

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**Temperatures must be in K Temperature decreases**

Example 2: If 20.0 L of oxygen is cooled from 100ºC to 0ºC, what is the new volume? Temperatures must be in K Volume decreases Temperature decreases T1= 100ºC + 273 = 373 K V1 = 20.0 L V2 = ??? T1 = 100ºC T2 = 0ºC T2= 0ºC + 273 = 273 K V2 = 14.6 L

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

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**P and T under final condition P and T under initial condition**

GAY-LUSSAC’S LAW Gay-Lussac’s Law can be mathematically expressed as T must be in units of K P and T under final condition P and T under initial condition

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**Temperatures must be in K**

Example 1: An aerosol spray can has a pressure of 4.0 atm at 25C. What pressure will the can have if it is placed in a fire and reaches temperature of 400C ? Temperatures must be in K P1 = 4.0 atm P2 = ??? T1 = 25ºC T2 = 400ºC T1= 25ºC + 273 = 298 K T2= 400ºC + 273 = 673 K

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**Temperature increases**

Example 1: An aerosol spray can has a pressure of 4.0 atm at 25C. What pressure will the can have if it is placed in a fire and reaches temperature of 400C ? Temperature increases Pressure increases P1 = 4.0 atm P2 = ??? T1 = 298 K T2 = 673 K P2 = 9.0 atm

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**Temperature decreases**

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 55C. To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg? Pressure decreases Temperature decreases P1 = 965 mmHg P2 = 850 mmHg T1 = 55ºC T2 = ??? T1= 55ºC + 273 = 328 K T2 = 289 K T2 = 16ºC

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**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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**VAPOR PRESSURE & BOILING POINT**

Listed below is the vapor pressure of water at various temperatures.

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**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 100C since its vapor pressure is 760 mmHg at this temperature.

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**VAPOR PRESSURE & BOILING POINT**

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

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

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COMBINED GAS LAW The individual gas laws studied previously are embodied in the combined gas law. Charles’s Law Gay-Lussac’s Law Boyle’s Law

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**Both pressure and temp. change Must use combined gas law**

Example 1: A 25.0-mL sample of gas has a pressure of 4.00 atm at a temperature of 10C. What is the volume of the gas at a pressure of 1.00 atm and a temperature of 18C? Both pressure and temp. change P1= 4.00 atm V1= 25.0 mL T1= 283 K P2= 1.00 atm V2= ??? T2 = 291 K Must use combined gas law V2 = 103 mL

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AVOGADRO’S LAW At constant temperature and pressure, the volume of a fixed amount of gas is directly proportional to the number of moles. The relationship of moles and volume in gases is called Avogadro’s Law. 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**

Avogadro’s Law Tro's Introductory Chemistry, Chapter

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AVOGADRO’S LAW As a result of Avogadro’s 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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**This relationship is only valid for gaseous substances**

AVOGADRO’S LAW Avogadro’s Law also allows chemists to relate volumes and moles of a gas in a chemical reaction. This relationship is only valid for gaseous substances For example: 2 H2 (g) O2 (g) 2 H2O (g) 2 molecules 1 molecule 2 molecules 2 moles 1 mole 2 moles 2 Liters 1 Liter 2 Liters

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**Example 1: mol of He = 4.50 mol mol of O2 = mol of He = 4.50 mol**

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? Same Temp. & Pressure mol of He = 4.50 mol mol of O2 = mol of He = 4.50 mol mass of O2 = 144 g

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**Example 2: N2 (g) + 3 H2 (g) 2 NH3 (g) 3 2 1.8 L H2 = 1.2 L NH3**

How many Liters of NH3 can be produced from reaction of 1.8 L of H2 with excess N2, as shown below? N2 (g) + 3 H2 (g) 2 NH3 (g) 3 2 1.8 L H2 = 1.2 L NH3

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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 = 0ºC (273 K) 1 atm (760 mmHg)

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

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**Example 1: 22.4 1 mol of gas = = 0.0893 mol Molar mass = = 36.2 g/mol**

If 2.00 L of a gas at STP has a mass of 3.23 g, what is the molar mass of the gas? 22.4 1 mol of gas = = mol Molar volume at STP Molar mass = = 36.2 g/mol

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Example 2: A sample of gas has a volume of 2.50 L at 730 mmHg and 20C. What is the volume of this gas at STP? P1= 730 mmHg V1= 2.50 L T1= 293 K P2= 760 mmHg V2= ??? T2 = 273 K V2 = 2.24 L

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**Universal gas constant**

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

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Example 1: A sample of H2 gas has a volume of 8.56 L at a temp. of 0C and pressure of 1.5 atm. Calculate the moles of gas present. PV = nRT P = 1.5 atm V = 8.56 L T = 273 K n = ??? R = = 0.57 mol

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**Moles must be calculated from mass Pressure must be converted to atm**

Example 2: What volume does 40.0 g of N2 gas occupy at 10C and 750 mmHg? Moles must be calculated from mass mol of N2 = = 1.43 mol P =750 mmHg V = ??? T = 283 K n = not given m = 40.0 g R = Pressure must be converted to atm P = 750 mmHg P = atm

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**Example 2: P =0.987 atm V = ??? T = 283 K n = 1.43 R = 0.0821**

What volume does 40.0 g of N2 gas occupy at 10C and 750 mmHg? P =0.987 atm V = ??? T = 283 K n = R = V = 33.7 L

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**Tro's Introductory Chemistry, Chapter**

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

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**Tro's Introductory Chemistry, Chapter**

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

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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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**DALTON’S LAW Ptotal = P1 + P2 + P3 + ···**

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 Dalton’s law of partial pressures. Ptotal = P1 + P2 + P3 + ··· Total pressure of a gas mixture Sum of the partial pressures of the gases in the mixture =

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PARTIAL PRESSURES + PHe = 2.0 atm PAr = 4.0 atm PTotal = ??? PHe + PAr = 6.0 atm Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture

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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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**Twice as many moles of Ar compared to He**

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

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**Example 1: Ptotal = Poxygen + Phelium**

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? Ptotal = Poxygen + Phelium 1 atm 760 mmHg Poxygen = 1140 mmHg x = 1.50 atm Phelium = 7.00 atm – 1.50 atm = PTotal - Poxygen = 5.50 atm

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**Twice as many moles of N2 compared to O2**

Example 2: A mixture of gases contains 2.0 mol of O2 gas and 4.0 mol of N2 gas with total pressure of 3.0 atm. What is the partial pressure of each gas in the mixture? Twice as many moles of N2 compared to O2 Ptotal = Poxygen + Pnitrogen Pnitrogen = 2 Poxygen Pnitrogen = 2/3 (Ptotal ) = 2.0 atm Poxygen = 1/3 (Ptotal ) = 1.0 atm

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THE END

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