Chapter 11 The Gaseous State by Christopher Hamaker © 2011 Pearson Education, Inc. Chapter 11

Gases assume the volume and shape of their containers. Physical Characteristics of Gases Gases assume the volume and shape of their containers. Gases are the most compressible state of matter. Gases will mix evenly and completely when confined to the same container. Gases have much lower densities than liquids and solids.

Properties of Gases There are five important properties of gases: Gases have an indefinite shape. Gases have low densities. Gases can compress. Gases can expand. Gases mix completely with other gases in the same container. Let’s take a closer look at these properties. Chapter 11

Detailed Gas Properties Gases have an indefinite shape. A gas takes the shape of its container, filling it completely. If the container changes shape, the gas also changes shape. Gases have low densities: The density of air is about 0.001 g/mL compared to a density of 1.0 g/mL for water. Air is about 1000 times less dense than water. Chapter 11

Detailed Gas Properties, Continued 3. Gases can compress. The volume of a gas decreases when the volume of its container decreases. If the volume is reduced enough, the gas will liquefy. 4. Gases can expand. A gas constantly expands to fill a sealed container. The volume of a gas increases if there is an increase in the volume of the container. Chapter 11

Detailed Gas Properties, Continued 5. Gases mix completely with other gases in the same container. Air is an example of a mixture of gases. When automobiles emit nitrogen oxide gases into the atmosphere, they mix with the other atmospheric gases. A mixture of gases in a sealed container will mix to form a homogeneous mixture. Chapter 11

Elements that exist as gases at 250C and 1 atmosphere

The Greenhouse Effect Several gases contribute to the greenhouse effect. High energy radiation strikes Earth’s surface, and is converted to heat. This heat is radiated from the surface as infrared radiation. This infrared radiation is absorbed by the gases, and released as heat in all directions, heating the atmosphere. Chapter 11

Gas Pressure Gas pressure is the result of constantly moving gas molecules striking the inside surface of their container. The more often the molecules collide with the sides of the container, the higher the pressure. The higher the temperature, the faster the gas molecules move. Chapter 11

Example Video Gas volume and pressure (1:04 min) http://www.youtube.com/watch?v=t-Iz414g-ro © 2011 Pearson Education, Inc. Chapter 4

Atmospheric Pressure Atmospheric pressure is a result of the air molecules in the environment. Evangelista Torricelli invented the barometer in 1643 to measure atmospheric pressure. Atmospheric pressure is 29.9 inches of mercury or 760 torr at sea level. Chapter 11

Units of Pressure Standard pressure is the atmospheric pressure at sea level, 29.9 inches of mercury. Here is standard pressure expressed in other units. Chapter 11

Pressure P = Force/unit area Force exerted per unit area of surface by molecules in motion. P = Force/unit area 1 atmosphere = 14.7 psi 1 atmosphere = 760 mm Hg 1 atmosphere = 101,325 Pascals 1 Pascal = 1 kg/m.s2 2

Gas Pressure Conversions The barometric pressure is 27.5 in. Hg. What is the barometric pressure in atmospheres? We want atm; we have in. Hg. Use 1 atm = 29.9 in. Hg: = 0.920 atm 27.5 in. Hg x 1 atm 29.9 in Hg Chapter 11

Worked Example 5.1

Variables Affecting Gas Pressure There are three variables that affect gas pressure: The volume of the container. The temperature of the gas. The number of molecules of gas in the container. Chapter 11

Volume Versus Pressure When volume decreases, the gas molecules collide with the container more often, so pressure increases. When volume increases, the gas molecules collide with the container less often, so pressure decreases. Chapter 11

Temperature Versus Pressure When temperature decreases, the gas molecules move slower and collide with the container less often, so pressure decreases. When temperature increases, the gas molecules move faster and collide with the container more often, so pressure increases. Chapter 11

Molecules Versus Pressure When the number of molecules decreases, there are fewer gas molecules colliding with the side of the container, so pressure decreases. When the number of molecules increases, there are more gas molecules colliding with the side of the container, so pressure increases. Chapter 11

Boyle’s Gas Experiment Robert Boyle trapped air in a J-tube using liquid mercury. He found that the volume of air decreased as he added more mercury. When he halved the volume, the pressure doubled. Chapter 11

Boyle’s Law Boyle’s law states that the volume of a gas is inversely proportional to the pressure at constant temperature. Inversely proportional means two variables have a reciprocal relationship. Mathematically, we write: V ∝ 1 . P Chapter 11

Boyle’s Law, Continued If we introduce a proportionality constant, k, we can write Boyle’s law as follows: We can also rearrange it to PV = k. Let’s take a sample of gas at P1 and V1, and change the conditions to P2 and V2. Because the product of pressure and volume is constant, we can write: P1V1 = k = P2V2 V = k x 1 . P Chapter 11

Boyle’s Law Boyle’s Law - volume of a gas is inversely proportional to pressure if the temperature and number of moles is held constant. PV = k1 or PiVi = PfVf

Applying Boyle’s Law To find the new pressure after a change in volume: To find the new volume after a change in pressure: V1 x P1 P2 = V2 Pfactor P1 x V1 V2 = P2 Vfactor Chapter 11

Boyle’s Law Problem A 1.50 L sample of methane gas exerts a pressure of 1650 mm Hg. What is the final pressure if the volume changes to 7.00 L? The volume increased and the pressure decreased as we expected. P1 x V1 V2 = P2 1650 mm Hg x 1.50 L 7.00 L = 354 mm Hg Chapter 11

A Problem to Consider A sample of chlorine gas has a volume of 1.8 L at 1.0 atm. If the pressure increases to 4.0 atm (at constant temperature), what would be the new volume? 7

A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P x V = constant P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL V2 = 154 mL P1 x V1 V2 726 mmHg x 946 mL 154 mL = P2 = = 4460 mmHg

Examples using Boyle's Law 1. A 5.0 L sample of a gas at 25oC and 3.0 atm is compressed at constant temperature to a volume of 1.0 L. What is the new pressure? 2. A 3.5 L sample of a gas at 1.0 atm is expanded at constant temperature until the pressure is 0.10 atm. What is the volume of the gas?

Example Video Boyle's Law Demonstration (0:36 min) http://www.youtube.com/watch?v=62Hsaag2Yzk © 2011 Pearson Education, Inc. Chapter 4

Charles’s Law In 1783, Jacques Charles discovered that the volume of a gas is directly proportional to the temperature in Kelvin. This is Charles’s law. V ∝ T at constant pressure. Notice that Charles’s law gives a straight line graph. Chapter 11

Charles’s Law, Continued We can write Charles’s law as an equation using a proportionality constant, k. V = k T or = k Again, let’s consider a sample of gas at V1 and T1, and change the volume and temperature to V2 and T2. Because the ratio of volume to temperature is constant, we can write: V T V1 T1 V2 T2 = k = Chapter 11

Illustration of Charles’s Law Below is an illustration of Charles’s law. As a balloon is cooled from room temperature with liquid nitrogen (–196 C), its volume decreases. Chapter 11

Applying Charles’s Law To find the new volume after a change in temperature: To find the new temperature after a change in volume: V1 x T2 T1 = V2 Tfactor T1 x V2 V1 = T2 Vfactor Chapter 11

Charles’ Law Charles’ Law - volume of a gas varies directly with the absolute temperature (K) if pressure and number of moles of gas are constant. or

Charles’s Law Problem A 275 L helium balloon is heated from 20 C to 40 C. What is the final volume at constant P? We first have to convert the temp from C to K: 20 C + 273 = 293 K 40 C + 273 = 313 K V1 x T2 T1 = V2 275 L x 313 K 293 K = 294 L Chapter 11

A Problem to Consider A sample of methane gas that has a volume of 3.8 L at 5.0 oC is heated to 86.0 oC at constant pressure. Calculate its new volume. 7

A sample of carbon monoxide gas occupies 3. 20 L at 125 0C A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V1 /T1 = V2 /T2 V1 = 3.20 L V2 = 1.54 L T1 = 398.15 K T2 = ? T1 = 125 (0C) + 273.15 (K) = 398.15 K V2 x T1 V1 1.54 L x 398.15 K 3.20 L = T2 = = 192 K

Examples Using Charles' Law 1. A 2.5 L sample of gas at 25oC is heated to 50oC at constant pressure. Will the volume double? 2. What would be the volume in question 1? 3. What temperature would be required to double the volume in question 1?

Example Video Charles' Law Demonstration (1:30 min) http://www.youtube.com/watch?v=GcCmalmLTiU © 2011 Pearson Education, Inc. Chapter 4

Gay-Lussac’s Law In 1802, Joseph Gay-Lussac discovered that the pressure of a gas is directly proportional to the temperature in Kelvin. This is Gay-Lussac’s Law. P ∝ T at constant temperature. Notice that Gay-Lussac’s law gives a straight line graph. Chapter 11

Gay-Lussac’s Law, Continued We can write Gay-Lussac’s law as an equation using a proportionality constant, k. P = k T or = k Let’s consider a sample of gas at P1 and T1, and change the volume and temperature to P2 and T2. Because the ratio of pressure to temperature is constant, we can write: P T P1 T1 P2 T2 = k = Chapter 11

Illustration of Gay-Lussac’s Law Here is an illustration of Gay- Lussac’s law. As the temperature of a gas in a steel cylinder increases, the pressure increases. Chapter 11

Applying Gay-Lussac’s Law To find the new volume after a change in temperature: To find the new temperature after a change in volume: P1 × T2 T1 = P2 Tfactor T1 × P2 P1 = T2 Pfactor Chapter 11

The Empirical Gas Laws Gay-Lussac’s Law: The pressure exerted by a gas at constant volume is directly proportional to its absolute temperature. P a Tabs (constant moles and V) or 3

Gay-Lussac’s Law Problem A steel container of nitrous oxide at 15.0 atm is cooled from 25 C to –40 C. What is the final volume at constant V? We first have to convert the temp from C to K: 25 C + 273 = 298 K –40 C + 273 = 233 K P1 x T2 T1 = P2 15.0 atm x 298 K 233 K = 11.7 atm Chapter 11

A Problem to Consider An aerosol can has a pressure of 1.4 atm at 25 oC. What pressure would it attain at 1200 oC, assuming the volume remained constant? 7

Example Video Gay Lussac's Law Demo (1:23 min) http://www.youtube.com/watch?v=wODW_9MRHWk © 2011 Pearson Education, Inc. Chapter 4

Combined Gas Law When we introduced Boyle’s, Charles’s, and Gay-Lussac’s laws, we assumed that one of the variables remained constant. Experimentally, all three (temperature, pressure, and volume) usually change. By combining all three laws, we obtain the combined gas law: P1V1 T1 P2V2 T2 = Chapter 11

Applying the Combined Gas Law To find a new volume when P and T change: To find a new pressure when V and T change: To find a new temperature when P and V change: T2 T1 V2 = V1 x P1 P2 x Pfactor Tfactor T2 T1 P2 = P1 x V1 V2 x Vfactor Tfactor V2 V1 T2 = T1 x P2 P1 x Pfactor Vfactor Chapter 11

Combined Gas Law Problem In a combined gas law problem, there are three variables: P, V, and T. Let’s apply the combined gas law to 10.0 L of carbon dioxide gas at 300 K and1.00 atm. If the volume and Kelvin temperature double, what is the new pressure? Conditions P V T initial 1.00 atm 10.0 L 300 K final P2 20.0 L 600 K Chapter 11

Combined Gas Law 1 This law is used when a sample of gas undergoes change involving volume, pressure, and temperature simultaneously.

Combined Gas Law Problem, Continued P2 = P1 x V1 V2 x 600 K 300 K P2 = 1.00 atm x 10.0 L 20.0 L x P2 = 1.00 atm Chapter 11

A Problem to Consider A sample of carbon dioxide occupies 4.5 L at 30 oC and 650 mm Hg. What volume would it occupy at 800 mm Hg and 200 oC? 7

Example Using the Combined 2 Example Using the Combined Law Calculate the temperature when a 0.50 L sample of gas at 1.0 atm and 25oC is compressed to 0.05 L of gas at 5.0 atm.

Example Video Combined Gas Law Demonstration... (0:27 min) http://www.youtube.com/watch?v=3mH3guEe0_Q © 2011 Pearson Education, Inc. Chapter 4

Vapor Pressure Vapor pressure is the pressure exerted by the gaseous vapor above a liquid when the rates of evaporation and condensation are equal. Vapor pressure increases as temperature increases. Chapter 11

Example Video Vapor Pressure (18:04 min) http://www.youtube.com/watch?v=hA5jddDYcyg © 2011 Pearson Education, Inc. Chapter 4

Dalton’s Law Dalton’s law of partial pressures states that the total pressure of a gaseous mixture is equal to the sum of the individual pressures of each gas. P1 + P2 + P3 + … = Ptotal The pressure exerted by each gas in a mixture is its partial pressure, Pn. Chapter 11

Dalton’s Law Calculation An atmospheric sample contains nitrogen, oxygen, and argon. If the partial pressure of nitrogen is 587 mm Hg, oxygen is 158 mm Hg, and argon is 7 mm Hg, what is the barometric pressure? Ptotal = Pnitrogen + Poxygen + Pargon Ptotal = 587 mm Hg + 158 mm Hg + 7 mm Hg Ptotal = 752 mm Hg Chapter 11

A sample of natural gas contains 8. 24 moles of CH4, 0 A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)? Pi = Xi PT PT = 1.37 atm 0.116 8.24 + 0.421 + 0.116 Xpropane = = 0.0132 Ppropane = 0.0132 x 1.37 atm = 0.0181 atm

A Problem to Consider Given a mixture of gases in the atmosphere at 760 torr, what is the partial pressure of N2 (c = 0 .7808) at 25 oC? 5

Collecting a Gas Over Water We can measure the volume of a gas by displacement. By collecting the gas in a graduated cylinder, we can measure the amount of gas produced. The gas collected is referred to as “wet” gas since it also contains water vapor. Chapter 11

Example Video Dalton's Law (2:00 min) http://www.youtube.com/watch?v=tTGehRESHmA © 2011 Pearson Education, Inc. Chapter 4

Kinetic Molecular Theory of Gases 3 Kinetic Molecular Theory of Gases Provides an explanation of the behavior of gases that we have studied in this chapter. Summary follows: 1. Gases are made up of small atoms or molecules that are in constant and random motion. 2. The distance of separation is very large compared to the size of the atoms or molecules. the gas is mostly empty space.

3. All gas particles behave independently. No attractive or repulsive forces exist between them. 4. Gas particles collide with each other and with the walls of the container without losing energy. The energy is transferred from one atom or molecule to another. 5. The average kinetic energy of the atoms or molecules is proportional to absolute temperature. K.E. = 1/2mv2 so as temperature goes up, the speed of the particles goes up.

4 How does the Kinetic Molecular Theory of Gases explain the following statements? Gases are easily compressible. Gases will expand to fill any available volume. Gases have low density.

Ideal Gas Behavior An ideal gas is a gas that behaves in a predictable and consistent manner. Ideal gases have the following properties: Gases are made up of very tiny molecules. Gas molecules demonstrate rapid motion in straight lines and in random directions. Gas molecules have no attraction for one another. Gas molecules undergo elastic collisions. The average kinetic energy of gas molecules is proportional to the Kelvin temperature, KE ∝ T. Chapter 11

Absolute Zero The temperature where the pressure and volume of a gas theoretically reaches zero is absolute zero. If we extrapolate T versus P or T versus V graphs to zero pressure or volume, the temperature is 0 Kelvin, or –273 C. Chapter 11

Ideal Gas Law Recall that the pressure of a gas is inversely proportional to volume and directly proportional to temperature and the number of molecules (or moles): If we introduce a proportionality constant, R, we can write the equation: P ∝ nT . V P = RnT . V Chapter 11

Ideal Gas Law, Continued We can rearrange the equation to read: PV = nRT This is the ideal gas law. The constant R is the ideal gas constant, and has a value of 0.0821 atmL/molK. Chapter 11

The Gaseous State Ideal Gas Concept Ideal gas - a model of the way that particles of a gas behave at the microscopic level. We can measure the following of a gas: temperature, volume, pressure and quantity (mass) We can systematically change one of the properties and see the effect on each of the others.

The Ideal Gas Law Combining Boyle’s Law, Charles’ Law and Avogadro’s Law gives the Ideal Gas Law. PV=nRT R (ideal gas constant) = 0.08206 L.Atm/mol.K

The conditions 0 0C and 1 atm are called standard temperature and pressure (STP). Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L. PV = nRT R = PV nT = (1 atm)(22.414L) (1 mol)(273.15 K) R = 0.082057 L • atm / (mol • K)

Let’s calculate the molar volume at STP using the ideal gas law: PV = nRT What would be the pressure? 1 atm What would be the temperature? 273 K What would be the number of moles? 1 mol 22.4 L

Ideal Gas Law Problem How many mole of hydrogen gas occupy 0.500 L at STP? At STP, T = 273K and P = 1 atm. Rearrange the ideal gas equation to solve for moles: n = PV . RT n = (1 atm)(0.500 L) . (0.0821 atmL/molK)(273K) n = 0.0223 moles Chapter 11

A Problem to Consider An experiment calls for 3.50 moles of chlorine, Cl2. What volume would this be if the gas volume is measured at 34 oC and 2.45 atm? 5

What is the volume (in liters) occupied by 49.8 g of HCl at STP? T = 0 0C = 273.15 K P = 1 atm PV = nRT n = 49.8 g x 1 mol HCl 36.45 g HCl = 1.37 mol V = nRT P V = 1 atm 1.37 mol x 0.0821 x 273.15 K L•atm mol•K V = 30.6 L

Worked Example 5.3

Worked Example 5.4

Examples using The Ideal Gas Law 1. What is the volume of gas occupied by 5.0 g CH4 at 25oC and 1 atm? 2. What is the mass of N2 required to occupy 3.0 L at 100oC and 700 mmHg?

Example Video Ideal Gas Equation: PV=nRT (9:22 min) http://www.youtube.com/watch?v=WScwPIPqZa0 © 2011 Pearson Education, Inc. Chapter 4

Chapter Summary Gases have variable shape and volume. The pressure of a gas is directly proportional to the temperature and the number of moles present. The pressure of a gas is inversely proportional to the volume it occupies. Standard temperature and pressure are exactly 1 atmosphere and 0 C (273 K). Chapter 11

Chapter Summary, Continued Boyle’s law is: P1V1 = P2V2. Charles’s law is: Gay-Lussac’s law is: The combined gas law is: V1 T1 V2 T2 = . P1 T1 P2 T2 = . P1V1 T1 P2V2 T2 = . Chapter 11

Chapter Summary, Continued Dalton’s law of partial pressures is: P1 + P2 + P3 + … = Ptotal. The ideal gas law is: PV = nRT. R is the ideal gas constant: 0.0821 atmL/molK. Chapter 11