Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved
Why study gases? An understanding of real world phenomena. An understanding of how science “works.” Copyright © Cengage Learning. All rights reserved

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. Gases exert pressure on their surroundings

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 most important gas laws involve the relationship between
Measurement of Gases The most important gas laws involve the relationship between number of moles (n) of gas volume (V) temperature (T) pressure (P) Pressure - force per unit area. Gas pressure is a result of force exerted by the collision of particles with the walls of the container.

Pressure Force exerted per unit area of surface by molecules in motion. 1 atmosphere = 14.7 psi 1 atmosphere = 760 mm Hg = 760 torr 1 atmosphere = 101,325 Pascals 1 Pascal = 1 kg/m.s2 = Newton/meter2 2

Barometer - measures atmospheric pressure.
Invented by Evangelista Torricelli A commonly used unit of pressure is the atmosphere (atm). 1 atm is equal to: 760 mmHg 760 torr 76 cmHg

Worked Example 5.1

Pressure Conversions: An Example
The pressure of a gas is measured as 2.5 atm. Represent this pressure in both torr and pascals. Copyright © Cengage Learning. All rights reserved

Elements that exist as gases at 250C and 1 atmosphere

Liquid Nitrogen and a Balloon
Copyright © Cengage Learning. All rights reserved

Gas laws can be deduced from observations like these. Mathematical relationships among the properties of a gas (Pressure, Volume, Temperature and Moles) can be discovered. Copyright © Cengage Learning. All rights reserved

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

Boyle’s law To play movie you must be in Slide Show Mode
PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

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

EXERCISE! A sample of helium gas occupies 12.4 L at 23°C and atm. What volume will it occupy at 1.20 atm assuming that the temperature stays constant? 9.88 L (0.956 atm)(12.4 L) = (1.20 atm)(V2) The new volume is 9.88 L. Copyright © Cengage Learning. All rights reserved

Examples using Boyle's Law
2 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?

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 To play movie you must be in Slide Show Mode

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 = K T2 = ? T1 = 125 (0C) (K) = K V2 x T1 V1 1.54 L x K 3.20 L = T2 = = 192 K

EXERCISE! Suppose a balloon containing 1.30 L of air at 24.7°C is placed into a beaker containing liquid nitrogen at –78.5°C. What will the volume of the sample of air become (at constant pressure)? L The new volume will become L. Remember to convert the temperatures to Kelvin. (1.30 L) / ( ) = V2 / ( ) Copyright © Cengage Learning. All rights reserved

Examples Using Charles' Law
2 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?

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

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

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

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.

Avogadro’s Law 1 Avogadro’s Law - equal volumes of an ideal gas contain the same number of moles if measured under the same conditions of temperature and pressure. or

EXERCISE! If 2.45 mol of argon gas occupies a volume of 89.0 L, what volume will 2.10 mol of argon occupy under the same conditions of temperature and pressure? 76.3 L The new volume is 76.3 L. (2.45 mol) / (89.0 L) = (2.10 mol) / (V2) Copyright © Cengage Learning. All rights reserved

Example Using Avogaro's Law
2 Assuming no change in temperature and pressure, how many moles of gas would be needed to double the volume occupied by 0.50 moles of gas?

Molar Volume - the volume occupied by 1 mol of any gas.
Molar Volume of a Gas Molar Volume - the volume occupied by 1 mol of any gas. STP - Standard Temperature and Pressure T = 273 K (or 0oC) P = 1 atm At STP the molar volume of a gas is 22.4 L/mol We will learn to calculate the volume later.

The molar volume of a gas. Photo courtesy of James Scherer.

Figure 5.11

Gas Densities We know: density = mass/volume Let’s calculate the density of H2. What is the mass of 1 mol of H2? 2.0 g What is the volume of 1 mol of H2? 22.4 L Density = 2.0 g/22.4 L = g/L

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) = L.Atm/mol.K

Ideal Gas Equation 1 Boyle’s law: V a (at constant n and T) P
Charles’ law: V a T (at constant n and P) Avogadro’s law: V a n (at constant P and T) V a nT P V = constant x = R nT P R is the gas constant PV = nRT

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 L. PV = nRT R = PV nT = (1 atm)(22.414L) (1 mol)( K) R = 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

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 = 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 x K L•atm mol•K V = 30.6 L

Worked Example 5.3

Worked Example 5.4

EXERCISE! An automobile tire at 23°C with an internal volume of 25.0 L is filled with air to a total pressure of 3.18 atm. Determine the number of moles of air in the tire mol The number of moles of air is 3.27 mol. (3.18 atm)(25.0 L) = n( )(23+273) Copyright © Cengage Learning. All rights reserved

EXERCISE! What is the pressure in a L tank that contains kg of helium at 25°C? 114 atm The pressure is 114 atm. Remember to convert kg of helium into moles of helium before using the ideal gas law. (P)(304.0) = ([5.670×1000]/4.003)(0.0821)(25+273) Copyright © Cengage Learning. All rights reserved

EXERCISE! At what temperature (in °C) does 121 mL of CO2 at 27°C and 1.05 atm occupy a volume of 293 mL at a pressure of 1.40 atm? 696°C The new temperature is 696°C. (1.05 atm)(121 mL) / (27+273) = (1.40 atm)(293 mL) / (T ) Copyright © Cengage Learning. All rights reserved

EXERCISE! A sample of oxygen gas has a volume of 2.50 L at STP. How many grams of O2 are present? 3.57 g 3.57 g of O2 are present. (1.00 atm)(2.50 L) = n( )(273) n = mol O2 × g/mol = 3.57 g O2 Copyright © Cengage Learning. All rights reserved

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?

Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? PV = nRT n, V and R are constant nR V = P T = constant P1 = 1.20 atm T1 = 291 K P2 = ? T2 = 358 K P1 T1 P2 T2 = P2 = P1 x T2 T1 = 1.20 atm x 358 K 291 K = 1.48 atm

Worked Example 5.5

Worked Example 5.6

Worked Example 5.7

Molecular Weight Determination
The relationship between moles and mass. or 12

Molecular Weight Determination
If we substitute this in the ideal gas equation, we obtain If we solve this equation for the molecular mass, we obtain 12

A Problem to Consider A 15.5 gram sample of an unknown gas occupied a volume of 5.75 L at 25 oC and a pressure of 1.08 atm. Calculate its molecular mass. 5

A 2. 10-L vessel contains 4. 65 g of a gas at 1. 00 atm and 27. 0 0C
A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and C. What is the molar mass of the gas? dRT P M = d = m V 4.65 g 2.10 L = = 2.21 g L 2.21 g L 1 atm x x K L•atm mol•K M = M = 54.6 g/mol

Density Determination
If we look again at our derivation of the molecular mass equation, we can solve for m/V, which represents density. 12

A Problem to Consider Calculate the density of ozone, O3 (Mm = 48.0g/mol), at 50 oC and 1.75 atm of pressure. 5

Density (d) Calculations
m is the mass of the gas in g m V = PM RT d = M is the molar mass of the gas Molar Mass (M ) of a Gaseous Substance dRT P M = d is the density of the gas in g/L

Worked Example 5.8

Gas Stoichiometry g C6H12O6 mol C6H12O6 mol CO2 V CO2
What is the volume of CO2 produced at 37 0C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C6H12O6 (s) + 6O2 (g) CO2 (g) + 6H2O (l) g C6H12O mol C6H12O mol CO V CO2 1 mol C6H12O6 180 g C6H12O6 x 6 mol CO2 1 mol C6H12O6 x 5.60 g C6H12O6 = mol CO2 0.187 mol x x K L•atm mol•K 1.00 atm = nRT P V = = 4.76 L

Stoichiometry Problems Involving Gas Volumes
Consider the following reaction, which is often used to generate small quantities of oxygen. Suppose you heat mol of potassium chlorate, KClO3, in a test tube. How many liters of oxygen can you produce at 298 K and 1.02 atm? 13

First we must determine the number of moles of oxygen produced by the reaction. 13

Now we can use the ideal gas equation to calculate the volume of oxygen under the conditions given. 13

1 Dalton’s Law of Partial Pressures Dalton’s Law - a mixture of gases exerts a pressure that is the sum of the pressures that each gas would exert if it were present alone under the same conditions. Pt=p1+p2+p3+... For example, the total pressure of our atmosphere is equal to the sum of the pressures of N2 and O2.

EXERCISE! Consider the following apparatus containing helium in both sides at 45°C. Initially the valve is closed. After the valve is opened, what is the pressure of the helium gas? The pressure is 2.25 atm. There are two ways to solve the problem: 1) Find the new pressure of the gas on the left side after the valve is opened (P1V1=P2V2). The pressure becomes 1.50 atm. (2.00 atm)(9.00 L) = (P2)(9.00 L L) P2 = 1.50 atm Find the new pressure of the gas on the right side after the valve is opened (P1V1=P2V2). The pressure becomes 0.75 atm. (3.00 atm)(3.00 L) = (P2)(9.00 L L) P2 = 0.75 atm The total pressure is therefore = 2.25 atm. 2) Find the number of moles in each chamber separately (using PV = nRT), add the number of moles and volumes on each side, then use PV = nRT to solve for the new pressure. 3.00 atm 3.00 L 2.00 atm 9.00 L Copyright © Cengage Learning. All rights reserved

EXERCISE! 27.4 L of oxygen gas at 25.0°C and 1.30 atm, and 8.50 L of helium gas at 25.0°C and 2.00 atm were pumped into a tank with a volume of 5.81 L at 25°C. Calculate the new partial pressure of oxygen. 6.13 atm Calculate the new partial pressure of helium. 2.93 atm Calculate the new total pressure of both gases. 9.06 atm For oxygen: P1V1=P2V2 (1.30 atm)(27.4 L) = (P2)(5.81 L) P2 = 6.13 atm For helium: (2.00 atm)(8.50 L) = (P2)(5.81 L) P2 = 2.93 atm The total pressure is therefore = 9.06 atm. Copyright © Cengage Learning. All rights reserved

Partial Pressures of Gas Mixtures
The partial pressure of a component gas, “A”, is then defined as Applying this concept to the ideal gas equation, we find that each gas can be treated independently. 13

Partial Pressures of Gas Mixtures
The composition of a gas mixture is often described in terms of its mole fraction. The mole fraction, c , of a component gas is the fraction of moles of that component in the total moles of gas mixture. 13

A sample of natural gas contains 8. 24 moles of CH4, 0
A sample of natural gas contains 8.24 moles of CH4, moles of C2H6, and 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 Xpropane = = Ppropane = x 1.37 atm = atm

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

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.

Kinetic Molecular Theory
To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! He H2 You are holding two balloons of the same volume. One contains helium, and one contains hydrogen. Complete each of the following statements with “different” or “the same” and be prepared to justify your answer. Copyright © Cengage Learning. All rights reserved

Molecular View of Boyle’s Law

Molecular View of Charles’s Law

Molecular View of the Ideal Gas Law

CONCEPT CHECK! You have a sample of nitrogen gas (N2) in a container fitted with a piston that maintains a pressure of 6.00 atm. Initially, the gas is at 45°C in a volume of 6.00 L. You then cool the gas sample. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Which best explains the final result that occurs once the gas sample has cooled? The pressure of the gas increases. The volume of the gas increases. The pressure of the gas decreases. The volume of the gas decreases. Both volume and pressure change. d) is correct. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! The gas sample is then cooled to a temperature of 15°C. Solve for the new condition. (Hint: A moveable piston keeps the pressure constant overall, so what condition will change?) 5.43 L The new volume is 5.43 L (use Charles’s law to solve for the new volume). (6.00 L) / (45+273) = (V2) / (15+273) Copyright © Cengage Learning. All rights reserved

Diffusion – the mixing of gases. Effusion – describes the passage of a gas through a tiny orifice into an evacuated chamber. Rate of effusion measures the speed at which the gas is transferred into the chamber. Copyright © Cengage Learning. All rights reserved

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Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. M2 M1  r1 r2 = NH4Cl NH3 17 g/mol HCl 36 g/mol

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An ideal gas is a hypothetical concept. No gas exactly follows the ideal gas law. We must correct for non-ideal gas behavior when: Pressure of the gas is high. Temperature is low. Under these conditions: Concentration of gas particles is high. Attractive forces become important. Copyright © Cengage Learning. All rights reserved

Gases behave most ideally at low pressure and high temperatures.
Ideal Gases Vs. Real Gases In reality there is no such thing as an ideal gas. Instead this is a useful model to explain gas behavior. Non-polar gases behave more ideally than polar gases because attractive forces are present in polar gases.

Deviations from Ideal Behavior
1 mole of ideal gas Repulsive Forces PV = nRT n = PV RT = 1.0 Attractive Forces

Plots of PV/nRT Versus P for Nitrogen Gas at Three Temperatures

For a real gas, the actual observed pressure is lower than the pressure expected for an ideal gas due to the intermolecular attractions that occur in real gases. Copyright © Cengage Learning. All rights reserved

Values of the van der Waals Constants for Some Gases
The value of a reflects how much of a correction must be made to adjust the observed pressure up to the expected ideal pressure. A low value for a reflects weak intermolecular forces among the gas molecules. Copyright © Cengage Learning. All rights reserved

A Problem to Consider The following values for SO2
If sulfur dioxide were an “ideal” gas, the pressure at 0 oC exerted by mol occupying L would be atm. Use the van der Waals equation to estimate the “real” pressure. The following values for SO2 a = L2.atm/mol2 b = L/mol 29

A Problem to Consider First, let’s rearrange the van der Waals equation to solve for pressure. R= L. atm/mol. K T = K V = L a = L2.atm/mol2 b = L/mol 29

A Problem to Consider The “real” pressure exerted by 1.00 mol of SO2 at STP is slightly less than the “ideal” pressure. 29

Air Pollution Two main sources: Transportation Production of electricity Combustion of petroleum produces CO, CO2, NO, and NO2, along with unburned molecules from petroleum. Copyright © Cengage Learning. All rights reserved

Nitrogen Oxides (Due to Cars and Trucks)
At high temperatures, N2 and O2 react to form NO, which oxidizes to NO2. The NO2 breaks up into nitric oxide and free oxygen atoms. Oxygen atoms combine with O2 to form ozone (O3). Copyright © Cengage Learning. All rights reserved

Concentration for Some Smog Components vs. Time of Day

Sulfur Oxides (Due to Burning Coal for Electricity)
Sulfur produces SO2 when burned. SO2 oxidizes into SO3, which combines with water droplets in the air to form sulfuric acid. Copyright © Cengage Learning. All rights reserved

Sulfur Oxides (Due to Burning Coal for Electricity)
Sulfuric acid is very corrosive and produces acid rain. Use of a scrubber removes SO2 from the exhaust gas when burning coal. Copyright © Cengage Learning. All rights reserved

A Schematic Diagram of a Scrubber

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