1 chapter 7 gases 7.1 properties of gases 7.2 gas pressure

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Chapter 7 Gases

7.1 Properties of Gases7.2 Gas Pressure

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Kinetic Theory of Gases

A gas consists of small particles that• move rapidly in straight lines • have essentially no attractive (or

repulsive) forces• are very far apart• have very small volumes compared

to the volumes of the containers they occupy

• have kinetic energies that increase with an increase in temperature

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Properties of Gases

• Gases are described in terms of four properties: pressure (P), volume (V), temperature (T), and amount (n).

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Gas pressure• is the force acting on a specific area

Pressure (P) = force area• has units of atm, mmHg, torr, lb/in.2 and

kilopascals(kPa).1 atm = 760 mmHg (exact)1 atm = 760 torr

1 atm = 14.7 lb/in.2

1 atm = 101.325 kPa

Gas Pressure

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Atmospheric Pressure

Atmospheric pressure

• is the pressure exerted by a column of air from the top of the atmosphere to the surface of the Earth

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Atmospheric Pressure (continued)

Atmospheric pressure• is about 1 atmosphere at

sea level • depends on the altitude

and the weather• is lower at high altitudes

where the density of air is less

• is higher on a rainy day than on a sunny day

Chapter 7 Gases

7.3Pressure and Volume

(Boyle’s Law)

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Boyle’s Law

Boyle’s law states that• the pressure of a gas

is inversely related to its volume when T and n are constant

• if the pressure (P) increases, then the volume (V) decreases

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In Boyle’s law• The product P x V is constant as long as T and n

do not change.P1V1 = 8.0 atm x 2.0 L = 16 atm L

P2V2 = 4.0 atm x 4.0 L = 16 atm L

P3V3 = 2.0 atm x 8.0 L = 16 atm L

• Boyle’s law can be stated as P1V1 = P2V2 (T, n constant)

PV Constant in Boyle’s Law

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Solving for a Gas Law Factor

The equation for Boyle’s law can be rearranged tosolve for any factor.

P1V1 = P2V2 Boyle’s Law

To solve for V2 , divide both sides by P2.P1V1 = P2V2

P2 P2

V1 x P1 = V2

P2

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Boyle’s Law and Breathing: Inhalation

During inhalation,• the lungs expand• the pressure in

the lungs decreases

• air flows towards the lower pressure in the lungs

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Boyle’s Law and Breathing: Exhalation

During exhalation,• lung volume

decreases• pressure within the

lungs increases• air flows from the

higher pressure in the lungs to the outside

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Guide to Calculations with Gas Laws

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Freon-12, CCl2F2, is used in refrigeration systems. What is the new volume (L) of an 8.0 L sample of Freon gas after its pressure is changed from 550 mmHg to 2200 mmHg at constant T?

STEP 1 Set up a data table:

Conditions 1 Conditions 2 Know Predict P1 = 550 mmHg P2 = 2200 mmHg P increasesV1 = 8.0 L V2 = ? V decreases

Calculation with Boyle’s Law

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STEP 2 Solve Boyle’s law for V2. When pressure increases, volume decreases.

P1V1 = P2V2

V2 = V1 x P1

P2 STEP 3 Set up problem

V2 = 8.0 L x 550 mmHg = 2.0 L 2200 mmHg

pressure ratio decreases volume

Calculation with Boyle’s Law (continued)

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Chapter 7 Gases

7.4Temperature and Volume

(Charles’s Law)

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Charles’s Law

In Charles’s law,• the Kelvin

temperature of a gas is directly related to the volume

• P and n are constant• when the

temperature of a gas increases, its volume increases

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• For two conditions, Charles’s law is writtenV1 = V2

(P and n constant)

T1 T2

• Rearranging Charles’s law to solve for V2 gives

T2 x V1 = V2 x T2

T1 T2

V2 = V1 x T2

T1

Charles’s Law: V and T

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A balloon has a volume of 785 mL at 21 °C. If thetemperature drops to 0 °C, what is the new volume ofthe balloon (P constant)?

STEP 1 Set up data table:

Conditions 1 Conditions 2 Know Predict

V1 = 785 mL V2 = ? V decreases

T1 = 21 °C T2 = 0 °C

= 294 K = 273 K T decreasesBe sure to use the Kelvin (K) temperature in gascalculations.

Calculations Using Charles’s Law

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Calculations Using Charles’s Law (continued)

STEP 2 Solve Charles’s law for V2: V1 = V2

T1 T2

V2 = V1 x T2

T1

Temperature factor decreases T

STEP 3 Set up calculation with data: V2 = 785 mL x 273 K = 729 mL

294 K

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Chapter 7 Gases

7.5Temperature and Pressure

(Gay-Lussac’s Law)

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Gay-Lussac’s Law: P and T

In Gay-Lussac’s law,• the pressure exerted by a

gas is directly related to the Kelvin temperature

• V and n are constant P1 = P2

T1 T2

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Chapter 7 Gases

7.6The Combined Gas Law

Summary of Gas Laws

The gas laws can be summarized as follows:

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• The combined gas law uses Boyle’s Law, Charles’s Law, and Gay-Lussac’s Law (n is constant).

P1 V1 = P2V2

T1 T2

Combined Gas Law

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A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29 °C. At what temperature (°C) will the helium have a volume of 90.0 mL and a pressure of 3.20 atm (n constant)?Step 1 Set up data table:Conditions 1 Conditions 2

P1 = 0.800 atm P2 = 3.20 atm

V1 = 0.180 L (180 mL) V2 = 90.0 mL

T1 = 29 °C + 273 = 302 K T2 = ?

Combined Gas Law Calculation

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STEP 2 Solve for T2 P1 V1 = P2 V2

T1 T2

T2 = T1 x P2 x V2

P1 V1

STEP 3 Substitute values to solve for unknown.

T2 = 302 K x 3.20 atm x 90.0 mL = 604 K 0.800 atm 180. mL

T2 = 604 K 273 = 331 °C

Combined Gas Law Calculation (continued)

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Chapter 7 Gases

7.7Volume and Moles(Avogadro’s Law)

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Avogadro's Law: Volume and Moles

Avogadro’s law states that

• the volume of a gas is directly related to the number of moles (n) of gas

• T and P are constant V1 = V2 n1 n2

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A gas has a pressure at 2.0 atm at 18 °C. What is the new pressure when the temperature is 62 °C? (V and n constant)STEP 1 Set up a data table:

Conditions 1 Conditions 2 Know Predict P1 = 2.0 atm P2 = ? P increases

T1 = 18 °C + 273 T2 = 62 °C + 273 T increases

= 291 K = 335 K

Calculation with Gay-Lussac’s Law

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Calculation with Gay-Lussac’s Law (continued)

STEP 2 Solve Gay-Lussac’s Law for P2: P1 = P2

T1 T2

P2 = P1 x T2

T1

STEP 3 Substitute values to solve for unknown: P2 = 2.0 atm x 335 K = 2.3 atm

291 KTemperature ratioincreases pressure

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The volumes of gases can be compared at STP(Standard Temperature and Pressure) when theyhave

• the same temperature

Standard temperature (T) = 0 °C or 273 K

• the same pressure

Standard pressure (P) = 1 atm (760 mmHg)

STP

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Molar VolumeThe molar volume of a gas • is measured at STP (standard temperature

and pressure)• is 22.4 L for 1 mole of any gas

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Molar Volume as a Conversion Factor

The molar volume at STP • has about the same volume as 3

basketballs• can be used to form 2 conversion

factors:

22.4 L and 1 mole 1 mole 22.4 L

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Guide to Using Molar Volume

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Using Molar Volume

What is the volume occupied by 2.75 moles of N2 gas at STP?

STEP 1 Given: 2.75 moles of N2

Need: Liters of N2

STEP 2 Write a plan:Use the molar volume to convert moles to liters.

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Using Molar Volume (continued)

STEP 3 Write equalities and conversion factors:1 mole of gas = 22.4 L

1 mole gas and 22.4 L 22.4 L 1 mole gas

STEP 4 Substitute data and solve:2.75 moles N2 x 22.4 L = 61.6 L of N2

1 mole N2

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Chapter 7 Gases

7.8The Ideal Gas Law

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• The relationship between the four properties (P, V, n, and T) of gases can be written equal to a constant R.

PV = RnT

• Rearranging this expression gives the expression called the ideal gas law.

PV = nRT

Ideal Gas Law

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The universal gas constant, R,• can be calculated using the molar volume at STP • when calculated at STP, uses a temperature of 273 K, a

pressure of 1.00 atm, a quantity of 1.00 mole of a gas, and a molar volume of 22.4 L.

P V R = PV = (1.00 atm)(22.4 L)

nT (1.00 mole)(273K) n T

= 0.0821 Latm mole K

Universal Gas Constant, R

Summary of Units for Ideal Gas Constants

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Guide to Using the Ideal Gas Law

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Chapter 7 Gases

7.9Partial Pressure (Dalton’s Law)

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The partial pressure of a gas• is the pressure of each gas in a mixture• is the pressure that gas would exert if it

were by itself in the container

Partial Pressure

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Dalton’s Law of Partial Pressures indicates that• pressure depends on the total number of gas

particles, not on the types of particles• the total pressure exerted by gases in a mixture

is the sum of the partial pressures of those gases

PT = P1 + P2 + P3 + .....

Dalton’s Law of Partial Pressures

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Dalton’s Law of Partial Pressures (continued)

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• For example, at STP, one mole of a pure gas in a volume of 22.4 L will exert the same pressure as one mole of a gas mixture in 22.4 L.

V = 22.4 L Gas mixtures

Total Pressure

0.5 mole O2

0.3 mole He0.2 mole Ar1.0 mole

1.0 mole N2

0.4 mole O2

0.6 mole He1.0 mole

1.0 atm 1.0 atm 1.0 atm

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Scuba Diving

• When a scuba diver is below the ocean surface, the increased pressure causes more N2(g) to dissolve in the blood.

• If a diver rises too fast, the dissolved N2 gas can form bubbles in the blood, a dangerous and painful condition called “the bends.”

• For deep descents, helium, which does not dissolve in the blood, is added to O2.

Guide to Solving for Partial Pressure

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Gases We Breathe

The air we breathe • is a gas mixture• contains mostly N2

and O2 and small amounts of other gases

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Blood Gases

• In the lungs, O2 enters the blood, while CO2 from the blood is released.

• In the tissues, O2 enters the cells, which releases CO2 into the blood.

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Blood Gases (continued)

In the body, • O2 flows into the tissues because the partial pressure

of O2 is higher in blood and lower in the tissues.

• CO2 flows out of the tissues because the partial pressure of CO2 is higher in the tissues and lower in the blood. Partial Pressures (mmHg) in Blood and Tissue

Gas Oxygenated Blood Deoxygenated Blood TissuesO2 100 mmHg 40 mmHg 30 mmHg or less

CO2 40 mmHg 46 mmHg 50 mmHg or greater

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Changes in Partial Pressures of Blood Gases During Breathing