s tates of m atter. state of matter volumeshapedensity compressibility motion of molecules gas...

STATES OF MATTER

State of Matter

Volume Shape Density Compressibility Motion of Molecules

Gas

Liquid

Solid

Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.

GASESChapter 5

Elements that exist as gases at 250C and 1 atmosphere

PHYSICAL CHARACTERISTICS OF GASES

Assume the shape of their container Most compressible state of matter Mix evenly and completely (Diffusion) Very low density compared with solids

and liquids.Density usually measured in g/L of gas

Pressure = ForceArea

PRESSUREPressure (P): the force per unit area on a

surface.

What causes pressure?

collisions of the gas molecules with each other and with surfaces with which they come into contact.

depends on volume, temperature, and the number of molecules present.

CALCULATING FORCE

The SI Unit for force is the Newton (N)

Consider a person with a mass of 51 kg. At Earth’s surface, gravity has an acceleration of 9.8 m/s2. What is the value of force?

Force = mass x acceleration

Force = 51 kg × 9.8 m/s2 = 500 kg • m/s2 = 500 N

P = 1.5 N/cm2 P = 38.5 N/cm2 P = 77 N/cm2

The greater area the less pressure on the floor.

RELATIONSHIP BETWEEN PRESSURE, FORCE, AND AREA

BAROMETER

Barometer

device used to measure atmospheric pressure

Sea level 1 atm

4 miles 0.5 atm

10 miles 0.2 atm

UNITS OF PRESSURE

1 atm = 101.3 kPa = 760 mmHg = 760 torr

GAS VS. VAPOR

Gas:Substance found normally in the gaseous

state at normal temperature and pressure

Vapor:Gaseous form of any substance that is a

liquid or a solid at normal temperature and pressure

GAS LAWS

BOYLE’S LAW

As volume increases, pressure _________________

As volume decreases, pressure _________________

As pressure increases, volume ________________

As pressure decreases, volume _________________

PRESSURE AND VOLUME ARE: __________________ RELATED

As P (h) increases V decreases

Boyle’s Law

BOYLE’S LAW EQUATION

P 1/V

P x V = constant

P1 x V1 = P2 x V2

Constant temperatureConstant amount of gas

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?

P1 x V1 = P2 x V2

P1 = 726 mmHg

V1 = 946 mL

P2 = ?

V2 = 154 mL

P2 = P1 x V1

V2

726 mmHg x 946 mL154 mL

= = 4460 mmHg

CHARLES’ LAW

As temperature increases, volume ________________

As temperature decreases, volume _________________

As volume increases, temperature _________________

As volume decreases, temperature _________________

TEMPERATURE AND VOLUME ARE: __________________ RELATED

As T increases V increases

Variation of gas volume with temperatureat constant pressure.

Charles’ Law

CHARLES’ LAW EQUATION

V T

V / T = constant

T (K) = t (0C) + 273

Temperature must be in Kelvin

Constant pressureConstant amount of gas

V1 = V2

T1 T2

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 = 3.20 L

T1 = 398.15 K

V2 = 1.54 L

T2 = ?

T2 = V2 x T1

V1

1.54 L x 398.15 K3.20 L

= = 192 K

V1/T1 = V2/T2

AVOGADRO’S LAW

As amount increases, volume ________________

As amount decreases, volume _________________

As volume increases, amount _________________

As volume decreases, amount _________________

AMOUNT AND VOLUME ARE: __________________ RELATED

AVOGADRO’S LAW EQUATION

V number of moles (n)

V /n = constant

V1 = V2

n1 n2

Constant temperatureConstant pressure

If 22.3 mol of N2 gas has a volume of 15 L, how many moles of N2 gas will have a volume of 12 L?

V1 = 15 L

n1 = 22.3 mol

V2 = 12 L

n2 = ?

n2 = V2 x n1

V1

12 L x 22.3 mol15 L

= = 18 mol

V1/n1 = V2/n2

COMBINED GAS LAW EQUATION

Boyle’s law and Charles’s law can be combined into a single equation that can be used for situations in which temperature, pressure, and volume, all vary at the same time.

1 1 2 2

1 2

PV P V

T T

The temperature MUST be in Kelvin

A 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 10.0°C?

1 1 22

2 1

PV TV

P T 1 1 2 2

1 2

PV P V

T T

(1.08 atm)(50.0 L He)(283 K)

(0.855 atm)(298 K)60.0 L He1 1 2

22 1

PV TV

P T

V1 = 50.0 L

P1 = 1.08 atm

V2 = ?

P2 = .855 atm

T1 = 25oC T2 = 10oC

IDEAL GAS EQUATION

Charles’ law: V T(at constant n and P)

Avogadro’s law: V n(at constant P and T)

Boyle’s law: V (at constant n and T)1P

V nT

P

V = constant x = RnT

P

nT

PR is the gas constant

PV = nRT

0 0C and 1 atm

PV = nRT

R = PVnT

=(1 atm)(22.414L)

(1 mol)(273.15 K)

R = 0.0821 L • atm / (mol • K)

Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.

STP = STANDARD TEMPERATURE AND PRESSURE

NUMERICAL VALUES OF THE GAS CONSTANT “R”

ALWAYS MATCH UP YOUR UNITS!!!!

What is the volume (in liters) occupied by 49.8 g of HCl at STP?

PV = nRT

V = nRTP

T = 0 0C = 273.15 K

P = 1 atm

n = 49.8 g x 1 mol HCl36.45 g HCl

= 1.37 mol

V =1 atm

1.37 mol x 0.0821 x 273.15 KL•atmmol•K

V = 30.6 L

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

nRV

= PT

= constant

P1

T1

P2

T2

=

P1 = 1.20 atm

T1 = 291 K

P2 = ?

T2 = 358 K

P2 = P1 x T2

T1

= 1.20 atm x 358 K291 K

= 1.48 atm

Density (d) Calculations

d = mV =

PMRT

m is the mass of the gas in g

M is the molar mass of the gas

Molar Mass (M ) of a Gaseous Substance

dRTP

M = d is the density of the gas in g/L

V and T are

constant

P1 P2 Ptotal = P1 + P2

DALTON’S LAW OF PARTIAL PRESSURES

Daltons Law of Partial Pressure

•Gases mix homogeneously (form a solution) in any proportions

•Each gas in a mixture behaves as if it were the only gas present (assuming no chemical reactions). 

PT = PgasA + PgasB + PgasC + etc.

Examples:

1. An organic chemist was considering the pressures exerted by three gases (M, N, L) in a flask. The total pressure inside the flask was 456 mmHg. If gas M contributes 200 mmHg, and gas L contributes 10 mmHg, what is the pressure exerted by gas N.

Examples:

2. An organic chemist was considering the pressures exerted by three gases (M, N, L) in a flask. The total pressure inside the flask was 644 mmHg. If gas M contributes to 21% of the pressure, and gas N contributes 54% what are the pressures exerted by all three gases in mmHg.

MOLE FRACTIONS

Mole fraction of gas A = Moles of gas A_____ Total number of moles of gas

GAS AMOUNT IN MOLES

A 0.235B 1.025C 2.35D 0.78

Examples:1. Four gases are found in an atmospheric sample of gas. The data below indicates their respective amount. Determine the mole fraction of each.

GAS AMOUNT IN GRAMS

He 23.5

CO2 45.7

CH4 32.3

Steam

24.7

2. Four gases are found in an atmospheric sample of gas. The data below indicates their respective amount. Determine the mole fraction of each.

Pi = Xi PT

HOW DO PARTIAL PRESSURE AND MOLE FRACTION RELATE?

Where: Pi is the pressure i

Xi is the mole fraction of i

PT is the total pressureExamples: 1. Determine the partial pressure of oxygen, nitrogen, and argon using the following data. Total pressure of the system is 760 mmHg.

GAS AMOUNT IN MOLES

O2 20

N2 79

Ar 1

Example:

2. Determine the partial pressure of oxygen, nitrogen, and argon using the following data. Total pressure of the system is 760 mmHg.

GAS

AMOUNT IN GRAMS

O

2

45.6

N

2

32.2

Ar

100.76

WHAT IS VAPOR PRESSURE?

The pressure that exists above the surface of a liquid from particles escaping the surface of the liquid

VAPOR PRESSURE IS EFFECTED BY HOW VOLATILE THE LIQUID IS

BeforeEvaporation

At Equilibrium

COLLECTING GASES OVER WATER

•Dalton’s Law can be used to calculate the pressure of gas collected over water•Set-up for such a system is shown below

•The collection flask initially contains all water•When a reaction takes place in a reaction chamber•Gas travels through tubing attached to the reaction chamber•The gas displaces the water in the collection flask •When all the water is displaced the flask is stoppered and the gas and some water vapor is collected

To find the pressure attributed by the gas collected you need to subtract the pressure due to water vapor at a specific temperature from the total pressure.

Examples:1. A gas is collected by water displacement at 50oC and barometric pressure of 95 kPa. What is the pressure exerted by the dry gas?  2. Oxygen gas is collected by water displacement from the reaction of Na2O2 and water. The oxygen displaces 318 mL of water at 23oC and 1.000atm. What is the pressure of dry O2.

Gas Stoichiometry

What is the volume of CO2 produced at 370 C and 1.00 atm when 5.60 g of glucose are used up in the reaction:

C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)

g C6H12O6 mol C6H12O6 mol CO2 V CO2

5.60 g C6H12O6

1 mol C6H12O6

180 g C6H12O6

x6 mol CO2

1 mol C6H12O6

x = 0.187 mol CO2

V = nRT

P

0.187 mol x 0.0821 x 310.15 KL•atmmol•K

1.00 atm= = 4.76 L

Kinetic Molecular Theory of Gases

1. A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume.

2. Gas molecules are in constant motion in random directions. Collisions among molecules are perfectly elastic.

3. Gas molecules exert neither attractive nor repulsive forces on one another.

4. The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy

**Gases behave ideally at very high temperatures and low pressures**

Effect of intermolecular forces on the pressure exerted by a gas.

DIFFUSION AND EFFUSION

DIFFUSION: the gradual mixing of two or more gases due to their spontaneous, random motion (kinetic properties)

EFFUSION: process when the molecules of a gas confined in a container randomly pass through a tiny opening in the container

EFFUSION RATE VIDEO

GRAHAM’S LAW OF EFFUSION

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.

GRAHAM’S LAW- VISUAL PROBLEM

GRAHAM’S LAW PROBLEM

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

Given: identities of two gases, H2 and O2

Unknown: relative rates of effusion

Hydrogen = Compound AOxygen = Compound B

GRAHAM’S LAW PROBLEM

HINT: Always put the substance with the larger molar mass on top as compound B.

1. Calculate:

2. Rearrange the equation:rate of effusion of A = 3.98 rate of

effusion of B3. Write a sentence:

Hydrogen diffuses 3.98 times faster than Oxygen