GASES

SECTION 1 – GASES AND PRESSUREIn the chapter state of matter, you read about the kinetic-molecular theory,

which is based on the idea that particles of matter are always in motion. In this section, you will study the implications of the kinetic-molecular

theory of gases.You have learned that the temperature of a gas is related to the kinetic energy of the gas molecules. In this chapter, you will learn about other

properties of gases, including pressure, volume, an amount of gas present, and the relationship between these properties.

Pressure and Force■ If you blow air into a rubber balloon, the balloon will

increase in size■ The volume increase is caused by the collisions of

molecules of air with the inside walls of the balloon■ The collisions cause an outward push, or force,

against the inside walls

Pressure■ Pressure (P) ! the force per unit area on a surface

■ SI unit for force = Newton (N) ! force that will increase the speed of a one kilogram mass by one meter per second each second it is applied

■ At Earth’s surface, each kilogram of mass exerts 9.8 N of force, due to gravity

■ Gas molecules exert pressure on any surface with which they collide

■ The pressure exerted by a gas depends on volume, temperature, and the number of molecules present

Measuring Pressure■ Barometer ! a device used to measure

atmospheric pressure ■ Torricelli sealed a long glass tube at one

end and filled it with mercury■ Held open end with his thumb, he

inverted the tube into a dish of mercury without allowing any air to enter the tube

■ When he removed his thumb, the mercury column in the tube dropped to a height of about 760 mm above the surface of the mercury in the dish

■ He repeated the experiment with tubes of different diameters and lengths longer than 760 mm

■ In every case, the mercury dropped to a height of about 760 mm

Units of Pressure■ Many units used to measure

pressure■ mmHg ! millimeters of

mercury■ 1 mmHg = 1 torr ■ 1 atm ! atmosphere of

pressure = 760 mmHg■ SI unit ! Pascal ! the

pressure exerted by a force of one Newton (1N) acting on an area of one square meter

Standard Temperature and Pressure■ To compare volumes of gases, it is necessary to

know the temperature and pressure at which the volumes are measured

■ For purposes of comparison, scientists have agreed on standard conditions of exactly 1 atm pressure and 0°C

■ These conditions are called standard temperature and pressure and are commonly abbreviated STP

Dalton’s Law of Partial Pressures■ He found that in the absence of a chemical reaction, the pressure of a

gas mixture is the sum of the individual pressures of each gas alone■ The pressure of each gas in a mixture is called the partial pressure of

that gas ■ Dalton’s law of partial pressures ! the total pressure of a mixture of

gases is equal to the sum of the partial pressures of the component gases

■ The law is true regardless of the number of different gases that are present

■ Dalton’s law may be expressed as PT = P1 + P2 + P3 +…

Gases Collected by Water Displacement■ Gases produced in the laboratory are often

collected over water

■ The gas produced by the reaction displaces the water, which is more dense, in the collection bottle

■ You can apply Dalton’s law of partial pressures in calculating the pressures of gases collected in this way

■ A gas collected by water displacement is not pure but is always mixed with water vapor

■ That is because water molecules at the liquid surface evaporate and mix with the gas molecules

■ Water vapor, like other gases, exerts a pressure, known as water-vapor pressure

■ Suppose you wished to determine the total pressure of the gas and water vapor inside a collection bottle

■ You would raise the bottle until the water levels inside and outside the bottle were the same

■ At that point, the total pressure inside the bottle would be the same as the atmospheric pressure, Patm

■ According to Dalton’s law of partial pressures, the following is true

Patm = Pgas + PH2O

■ Oxygen gas from the decomposition of potassium chlorate, KClO3, was collected by water displacement. The barometric pressure and the temperature during the experiment were 731.0 torr and 20.0°C, respectively. What was the partial pressure of the oxygen collected?

■ Given: PT = Patm = 731.0 torr; ■ PH2O = 17.5 torr (vapor pressure of water at 20.0°C); ■ Patm = PO2 + PH2O

■ Unknown: PO2 in torr

■ Given: PT = Patm = 731.0 torr; ■ PH2O = 17.5 torr (vapor pressure of water at

20.0°C); ■ Patm = PO2 + PH2O

Unknown: PO2 in torr

PO2 = Patm − PH2O

PO2 = 731.0 torr − 17.5 torr = 713.5 torr

Some hydrogen gas is collected over water at 20.0°C.The levels of water inside and outside the gas-collection bottle are the same. The partial pressure of hydrogen is 742.5 torr. What is the barometric pressure at the time the gas is collected?

■ Answer 760.0 torr

Helium gas is collected over water at 25°C.What is the partial pressure of the helium, given that the barometric pressure is 750.0 mm Hg?

■ Answer 726.2 mm Hg

Practice Problems1. The average atmospheric pressure in Denver, Colorado, is

0.830 atm. Express this pressure (a) in mm Hg and (b) in kPa. 631 mm Hg 84.1 kPa2. Convert a pressure of 1.75 atm to kPa and to mm Hg. 177 kPa, 1330 mm Hg 3. Convert a pressure of 570. torr to atmospheres and to

kPa. 0.750 atm, 76.0 kPa

SECTION 2 – THE GAS LAWS

Scientists have been studying physical properties of gases for hundreds of years. In 1662, Robert Boyle discovered that gas pressure and volume are related mathematically. The observations of Boyle and others led to the development of the gas laws. The gas laws are simple mathematical relationships between the volume, temperature, pressure, and amount of a gas.

Boyle’s Law: Pressure-Volume■ Boyle’s law ! the volume of a fixed mass of gas varies

inversely with the pressure at constant temperature

■ Boyle’s law can be used to compare changing conditions for a gas

■ Using P1 and V1 to stand for initial conditions and P2 and V2 to stand for new conditions results in the following equations

P1V1 =k P2V2 = k

P1V1 = P2V2

Practice ProblemA sample of oxygen gas has a volume of 150. mL when its

pressure is 0.947 atm. What will the volume of the gas be at a pressure of 0.987 atm if the temperature remains constant?

■ Given: ■ V1 of O2 = 150. mL; ■ P1 of O2 = 0.947 atm; ■ P2 of O2 = 0.987 atm ■ Unknown: V2 of O2 in mL

Given: V1 of O2 = 150. mL; P1 of O2 = 0.947 atm; P2 of O2 = 0.987 atm

Unknown: V2 of O2 in mL

Rearrange the equation for Boyle’s law (P1V1 = P2V2) to obtain V2

Practice ProblemsA balloon filled with helium gas has a volume of 500 mL at a

pressure of 1 atm.The balloon is released and reaches an altitude of 6.5 km, where the pressure is 0.5 atm. Assuming that the temperature has remained the same, what volume does the gas occupy at this height?

■ Answer 1000 mL He

A gas has a pressure of 1.26 atm and occupies a volume of 7.40 L. If the gas is compressed to a volume of 2.93 L, what will its pressure be, assuming constant temperature?

■ Answer 3.18 atm

Divers know that the pressure exerted by the water increases about 100 kPa with every 10.2 m of depth. This means that at 10.2 m below the surface, the pressure is 201 kPa; at 20.4 m, the pressure is 301 kPa; and so forth. Given that the volume of a balloon is 3.5 L at STP and that the temperature of the water remains the same, what is the volume 51 m below the water’s surface?

■ Answer 0.59 L

Charles’s Law: Volume-Temperature■ Charles’s law ! the volume of a fixed mass of gas at

constant pressure varies directly with the Kelvin temperature

■The temperature −273.15°C is referred to as absolute zero and is given a value of zero in the Kelvin scale

K = 273 + °C

Practice ProblemsA 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?

Given: V1 of Ne = 752 mL; T1 of Ne = 25°C + 273 = 298 K; T2 of Ne = 50°C + 273 = 323 K – Note that Celsius temperatures have been converted to kelvins.This

is a very important step for working the problems in this chapter Unknown: V2 of Ne in mL

Given: V1 of Ne = 752 mL; T1 of Ne = 25°C + 273 = 298 K; T2 of Ne = 50°C + 273 = 323 K

Unknown: V2 of Ne in mL

A helium-filled balloon has a volume of 2.75 L at 20.°C.The volume of the balloon decreases to 2.46 L after it is placed outside on a cold day. What is the outside temperature in K? in °C?

■ Answer 262 K, or −11°C

A gas at 65°C occupies 4.22 L. At what Celsius temperature will the volume be 3.87 L, assuming the same pressure?

■Answer 37°C

Gay-Lussac’s Law: Pressure-Temperature

■ Gay-Lussac’s law: The pressure of a fixed mass of gas at constant volume varies directly with the Kelvin temperature

The gas in an aerosol can is at a pressure of 3.00 atm at 25°C. Directions on the can warn the user not to keep the can in a place where the temperature exceeds 52°C. What would the gas pressure in the can be at 52°C?

■ Given: ■ P1 of gas = 3.00 atm; ■ T1 of gas = 25°C + 273 = 298 K; ■ T2 of gas = 52°C + 273 = 325 K ■ Unknown: P2 of gas in atm

Given: P1 of gas = 3.00 atm; T1 of gas = 25°C + 273 = 298 K; T2 of gas = 52°C + 273 = 325 K

Unknown: P2 of gas in atm

Before a trip from New York to Boston, the pressure in an automobile tire is 1.8 atm at 20.°C. At the end of the trip, the pressure gauge reads 1.9 atm. What is the new Celsius temperature of the air inside the tire? (Assume tires with constant volume.)

■Answer 36°C

At 120.°C, the pressure of a sample of nitrogen is 1.07 atm. What will the pressure be at 205°C, assuming constant volume?

■Answer 1.30 atm

A sample of helium gas has a pressure of 1.20 atm at 22°C. At what Celsius temperature will the helium reach a pressure of 2.00 atm?

■Answer 219°C

The Combined Gas Law■ The combined gas law expresses the relationship between

pressure, volume, and temperature of a fixed amount of gas

Getting the other gas laws

If temperature is constant you get Boyle’s LawP1V1 = P2V2 If pressure is constant you get Charles’s Law

If volume is constant you get Gay-Lussac’s Law

Practice ProblemsA helium-filled balloon has a volume of 50.0 L at 25°C and 1.08

atm.What volume will it have at 0.855 atm and 10.°C?

Given: V1 of He = 50.0 L; T1 of He = 25°C + 273 = 298 K; T2 of He = 10°C + 273 = 283 K; P1 of He = 1.08 atm; P2 of He = 0.855 atm Unknown: V2 of He in L

Given: V1 of He = 50.0 L; T1 of He = 25°C + 273 = 298 K; T2 of He = 10°C + 273 = 283 K; P1 of He = 1.08 atm; P2 of He = 0.855 atm

Unknown: V2 of He in L

The volume of a gas is 27.5 mL at 22.0°C and 0.974 atm. What will the volume be at 15.0°C and 0.993 atm?

■Answer 26.3 mL

A 700. mL gas sample at STP is compressed to a volume of 200. mL, and the temperature is increased to 30.0°C. What is the new pressure of the gas in Pa?

■Answer 3.94 × 105 Pa

GAS VOLUMES AND THE IDEAL GAS LAW

In this section, you will study the relationships between the volumes of gases that react with each other. You’ll also learn about the relationship between

molar amount of gas in volume, and a single gas law that unifies all the basic gas laws into a single equation.

Measuring and Comparing the Volumes of Reacting Gases

■ Early 1800s, Gay-Lussac studied gas volume relationships with chemical reaction between H and O

■ Observed 2 L H can react with 1 L O to form 2 L of water vapor at constant temperature and pressure

Hydrogen gas + oxygen gas ! water vapor 2 L 1 L 2 L 2 volumes 1 volume 2 volumes

■ Reaction shows 2:1:2 relationship between volumes of reactants and product

■ Ratio applies to any proportions (mL, L, cm3)

■ Gay-Lussac also noticed ratios by volume between other reactions of gases

Hydrogen gas + chlorine gas ! hydrogen chloride gas 1 L 1 L 2 L

Law of Combining Volumes of Gases

■ 1808 Gay-Lussac summarized results in Gay-Lussac’s law of combining volumes of gases ! at constant temperature and pressure, the volumes of gaseous reactants and products can be expressed as ratios of small whole numbers

Avogadro’s Law

■ Important point of Dalton’s atomic theory: atoms are indivisible

■ Dalton also thought particles of gaseous elements exist in form of single atoms

■ Believed one atom of one element always combines with one atom of another element to form single particle of product

■ Gay-Lussac’s results presented problem for Dalton’s theory

■ Ex. Reactions like formation of water

Hydrogen gas + oxygen gas ! water vapor 2 L 1 L 2 L

■ Seems that oxygen involved would have to divide into two parts

■ 1811 Avogadro found way to explain Gay-Lussac’s simple ratios of combining volumes without violating Dalton’s idea of indivisible atoms

■ Rejected Dalton’s idea that reactant elements are always in monatomic form when they combine to form products

■ Reasoned these molecules could contain more than 1 atom

■Avogadro’s law ! equal volumes of gases at the same temperature and pressure contain equal numbers of molecules

■At the same temp and pressure, volume of any given gas varies directly with the number of molecules

1 mol CO2 atSTP = 22.4 L

1 mol O2 atSTP = 22.4 L

1 mol H2 atSTP = 22.4 L

■ Consider reaction of H and Cl to produce HCl

■ According to Avogadro’s law, equal volumes of H and Cl contain same number of molecules

■ b/c he rejected Dalton’s theory that elements are always monatomic, he concluded H and Cl components must each consist of 2 or more atoms joined together

■ Simplest assumption was that H and Cl molecules had 2 atoms each

■ Leads to following balanced equation:

H2(g) + Cl2(g) → 2HCl(g) 1 volume 1 volume 2 volumes 1 molecule 1 molecule 2 molecules

■ If the simplest formula for hydrogen chloride, HCl indicates molecule contains 1 H and 1 Cl

■ Then the simplest formulas for hydrogen and chlorine must be H2 and Cl2

■ Avogadro’s law also indicates that gas volume is directly proportional to the amount of gas, at given temp and pressure

V = kn

■ k = constant■ n = amount of gas in moles

Molar Volume of Gases

■ Remember 1 mol of substance contains 6.022 x 1023

■ According to Avogadro’s law, 1 mol of any gas occupies same volume as 1 mol of any other gas at same temperature and pressure, even though masses are different

■ Standard molar volume of gas ! volume occupied by 1 mol of gas at STP

■ = 22.4 L

■ Knowing volume of gas, you can use 1mol/22.4 L as conversion factor

■ Can find number of moles

■ Can find mass

■ Can also use molar volume to find volume if you have number of moles or mass

Sample Problem

■ A chemical reaction produces 0.0680 mol of oxygen gas. What volume in liters is occupied by this gas sample at STP?

1. Analyze

■ Given: molar mass of O2 = 0.0680 mol

■ Unknown: volume of O2 in liters at STP

2. Plan

■ moles of O2 → liters of O2 at STP

■ The standard molar volume can be used to find the volume of a known molar amount of a gas at STP

3. Compute

Practice Problems■ At STP, what is the volume of 7.08 mol of nitrogen

gas? ■ 159 L N2

■ A sample of hydrogen gas occupies 14.1 L at STP. How many moles of the gas are present?

■ 0.629 mol H2

■ At STP, a sample of neon gas occupies 550. cm3. How many moles of neon gas does this represent?

■ 0.0246 mol Ne

Sample Problem

■ A chemical reaction produced 98.0 mL of sulfur dioxide gas, SO2, at STP. What was the mass (in grams) of the gas produced?

1. Analyze

■ Given: volume of SO2 at STP = 98.0 mL■ Unknown: mass of SO2 in grams

2. Plan

■ liters of SO2 at STP→moles of SO2→grams of SO2

3. Compute

■ = 0.280 g SO2

Practice Problems■ What is the mass of 1.33 × 104 mL of oxygen gas at

STP? ■ 19.0 g O2

■ What is the volume of 77.0 g of nitrogen dioxide gas at STP?

■ 37.5 L NO2

■ At STP, 3 L of chlorine is produced during a chemical reaction. What is the mass of this gas?

■ 9 g Cl2

■ A gas sample can be characterized by 4 quantities

1. Pressure2. Volume3. Temperature4. Number of moles

■ Number of moles present always affects at least one of the other 3 quantities

■ Collision rate per unit area of container wall depends on number of moles

■ Increase moles, increase collision rate, increase pressure

■ Pressure, volume, temperature, moles are all interrelated

■ A mathematical relationship exists to describe behavior of gas for any combination of these conditions

■ Ideal gas law ! mathematical relationship among pressure, volume, temperature, and number of moles of a gas

Derivation of Ideal Gas Law

■ Derived by combining the other gas laws■ Boyle’s law: at constant temp, the volume of a given

mass of gas is inversely proportional to the pressure.

■ Charles’s law: At constant pressure, volume of given mass of gas is directly proportional to Kelvin temperature

V α T

■ Avogadro’s law: at constant temp and pressure, volume of given mass of gas is directly proportional to the number of moles

V α n

■ Volume is proportional to pressure, temp and moles in each equation

■ Combine the 3:

Can change proportion to equality by adding constant, this time R

■ This equation says the volume of a gas varies directly with the number of moles of gas and its Kelvin temperature

■ Volume also varies inversely with pressure

■ Ideal gas law combines Boyle’s, Charles’s, Gay-Lussac’s, and Avogadro’s laws

■ Ex: PV = nRT ! n and T are constant, nRT is constant b/c R is also constant

■ This makes PV = constant which is Boyle’s law

The Ideal Gas Constant - R■ Value depends on units for volume, pressure, temp

Unit of R Value of R

Unit of P Unit of V Unit of T Unit of n

62.4 mmHg L K Mol

0.0821 Atm L K Mol

8.314 Pa m3 K Mol

8.314 kPa L K Mol

Sample Problem

■ What is the pressure in atmospheres exerted by a 0.500 mol sample of nitrogen gas in a 10.0 L container at 298 K?

1. Analyze

■ Given: ■ V of N2 = 10.0 L ■ n of N2 = 0.500 mol ■ T of N2 = 298 K

■ Unknown: ■ P of N2 in atm

2. Plan

■ n,V,T → P■ The gas sample undergoes no change in conditions■ Therefore, the ideal gas law can be rearranged and

used to find the pressure as follows

3. Compute

= 1.22 atm

Practice Problems

■ What pressure, in atmospheres, is exerted by 0.325 mol of hydrogen gas in a 4.08 L container at 35°C?

■ 2.01 atm

■ A gas sample occupies 8.77 L at 20°C.What is the pressure, in atmospheres, given that there are 1.45 mol of gas in the sample?

■ 3.98 atm

■ What is the volume, in liters, of 0.250 mol of oxygen gas at 20.0°C and 0.974 atm pressure?

■ 6.17 L O2

■ A sample that contains 4.38 mol of a gas at 250 K has a pressure of 0.857 atm. What is the volume?

■ 105 L■ How many liters are occupied by 0.909 mol of

nitrogen at 125°C and 0.901 atm pressure? ■ 33.0 L N2

■ What mass of chlorine gas, Cl2, in grams, is contained in a 10.0 L tank at 27°C and 3.50 atm of pressure?

■ 101 g Cl2■ How many grams of carbon dioxide gas are there

in a 45.1 L container at 34°C and 1.04 atm? ■ 81.9 g CO2

■ What is the mass, in grams, of oxygen gas in a 12.5 L container at 45°C and 7.22 atm?

■ 111 g O2

■ A sample of carbon dioxide with a mass of 0.30 g was placed in a 250 mL container at 400. K. What is the pressure exerted by the gas?

■ 0.90 atm

SECTION 4Effusion and

Diffusion

Graham’s Law of Effusion

■ Rates of effusion and diffusion depend on relative velocities of gas molecules

■ Velocity varies inversely with mass (lighter molecules move faster)

■ Average kinetic energy ½ mv2■ For two gases, A and B, at same temp:

½ MAvA2 = ½ MBvB2

■ MA and MB = molar masses of A and B■ Multiply by 2

MAvA2 = MBvB2

MAvA2 = MBvB

2

■ Suppose you wanted to compare the velocities of the two gases

■ You would first rearrange the equation above to give the velocities as a ratio

■ Take square root of each side

This equation shows that velocities of two gases are inversely proportional to the square roots of their molar masses

■ b/c rates of effusion are directly proportional to molecular velocities, can write

Graham’s Law of Effusion

■ The rates of effusion of gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses

Application of Graham’s Law■ Graham’s experiments dealt with densities of gases■ Density varies directly with molar mass■ So…square roots of molar masses from equation can be

replaced with square roots of densities

Sample Problem

■ Compare the rates of effusion of hydrogen and oxygen at the same temperature and pressure.

1. Analyze

■ Given: ■ identities of two gases, H2 and O2

■ Unknown: ■ relative rates of effusion

2. Plan

■ molar mass ratio → ratio of rates of effusion

■ The ratio of the rates of effusion of two gases at the same temperature and pressure can be found from Graham’s law

3. Compute

■ Hydrogen effuses 3.98 times faster than oxygen.

Practice Problems■ A sample of hydrogen effuses through a porous

container about 9 times faster than an unknown gas. Estimate the molar mass of the unknown gas.

■ 160 g/mol■ Compare the rate of effusion of carbon dioxide

with that of hydrogen chloride at the same temperature and pressure.

■ CO2 will effuse about 0.9 times as fast as HCl

■ If a molecule of neon gas travels at an average of 400 m/s at a given temperature, estimate the average speed of a molecule of butane gas, C4H10, at the same temperature.

■ about 235 m/s