Gases: Their Properties and Behavior Chapter 9. 2 Properties of Gases There are 5 important properties of gases: –Confined gases exerts pressure on the

Gases: Their Properties and BehaviorGases: Their Properties and BehaviorChapter 9

Chapter 9 2

Properties of Gases• There are 5 important properties of gases:

– Confined gases exerts pressure on the wall of a container uniformly

– Gases have low densities

– Gases can be compressed

– Gases can expand to fill their contained uniformly

– Gases mix completely with other gases in the same container

Chapter 9 3

Kinetic Molecular Theory of Gases• The Kinetic Molecular Theory of Gases is the model used

to explain the behavior of gases in nature.

• This theory presents physical properties of gases in terms of

the motion of individual molecules:

1. Average Kinetic Energy Kelvin Temperature

2. Gas molecules are points separated by a great distance

3. Particle volume is negligible compared to gas volume

4. Gas molecules are in rapid random motion

5. Gas collisions are perfectly elastic

6. Gas molecules experience no attraction or repulsion

Chapter 9 4

Properties that Describe a Gas

• These properties are all related to one another.• When one variable changes, it causes the other

three to react in a predictable manner.

Chapter 9 5

Gas Pressure (P)• Gas pressure (P) is the result of constantly moving

gas molecules striking the inside surface of their container.

Chapter 9 6

Atmospheric Pressure• Atmospheric pressure is the pressure exerted by the air

on the earth.

• Evangelista Torricelli invented the barometer in 1643 to measure atmospheric pressure.

• Atmospheric pressure is 760 mm of mercury or 1 atmosphere (atm) at sea level.

What happens to the atmospheric pressure as you go up in elevation?

Chapter 9 7

Measurement of Gas Pressure• Traditionally, the gas pressure inside of a container is

measured with a manometer

Chapter 9 8

Units of Pressure• Standard pressure is the atmospheric pressure at

sea level, 760 mm of mercury.

– Here is standard pressure expressed in other units:

Chapter 9 9

Gas Pressure Conversions

• The barometric pressure is 697.2 torr. What is the barometric pressure in atmospheres? In mm Hg? In Pascals (Pa)?

Chapter 9 10

Gas Law Problems

Chapter 9 11

Boyle’s Law (V and P)

• Mathematically, we write:

• For a before and after situation:

• Boyle’s Law states that the volume of a gas is inversely proportional to the pressure at constant temperature.

P 1 .V

P1V1 = P2V2

Chapter 9 12

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?

(1650 mm Hg )(1.50 L)7.00 L

= 354 mm Hg

P1V1 = P2V2rearranges to P1 V1

V2

= P2

Chapter 9 13

Charles’ Law (V and T)• In 1783, Jacques Charles discovered (while hot air ballooning) that

the volume of a gas is directly proportional to the temperature in Kelvin.



T V

V1

T1

V2

T2

=

Chapter 9 14

Charles’ Law Problem• A 275 L helium balloon is heated from 20C to 40C.

What is the final volume at constant P?

(275 L)(313 K)293 K

= 294 L

V1 T2

T1

= V2

V1

T1

V2

T2

= rearranges to

Chapter 9 15

Gay-Lussac’s Law (P and T)• In 1802, Joseph Gay-Lussac discovered that the pressure of a

gas is directly proportional to the temperature in Kelvin.



T P

P1

T1

P2

T2

=

Chapter 9 16

Gay-Lussac’s Law Problem• A steel container of nitrous oxide at 15.0 atm is cooled

from 25C to –40C. What is the final volume at constant V?

(15.0 atm)(298 K)233 K

= 11.7 atm

P1 T2

T1

= P2P1

T1

P2

T2

= rearranges to

Chapter 9 17

Avogadro’s Law (n and V)• In the previous laws, the amount of gas was always constant.

• However, the amount of a gas (n) is directly proportional to the volume of the gas, meaning that as the amount of gas increases, so does the volume.



n V

V1

n1

V2

n2

=

Chapter 9 18

Avogadro’s Law Problem• A steel container contains 2.6 mol of nitrous oxide with a volume

15.0 L. If the amount of nitrous oxide is increased to 8.4 mol, what is the final volume at constant T and P?

(15.0 L)(8.4 mol)2.6 mol

= 48.5 L

V1 n2

n1

= V2

V1

n1

V2

n2

= rearranges to

Chapter 9 19

Combined Gas Law• When we introduced Boyle’s, Charles’, 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 9 20

Combined Gas Law Problem

• Oxygen gas is normally sold in 49.0 L steel containers

at a pressure of 150.0 atm. What volume would the

gas occupy if the pressure was reduced to 1.02 atm

and the temperature raised from 20oC to 35oC?

Chapter 9 21

Molar Volume and STP• Standard temperature and pressure (STP) are defined as 0C and 1 atm.

• At standard temperature and pressure, one mole of any gas occupies 22.4 L.

• The volume occupied by one mole of gas (22.4 L) is called the molar volume.

1 mole Gas = 22.4 L

Chapter 9 22

Molar Volume Calculation – Volume to Moles

• A sample of methane, CH4, occupies 4.50 L at STP. How many moles of methane are present?

4.50 L CH4 × = 0.201 mol CH4

1 mol CH4

22.4 L CH4

Volume

MolarVolume

Moles

Chapter 9 23

Mole Unit Factors

• We now have three interpretations for the mole:

1 mol = 6.02 × 1023 particles

1 mol = molar mass

1 mol = 22.4 L (at STP for a gas)

• This gives us 3 unit factors to use to convert between moles, particles, mass, and volume.

Chapter 9 24

Mole Calculation - Grams to Volume • What is the mass of 3.36 L of ozone gas, O3, at STP?

= 7.20 g O33.36 L O3 × ×22.4 L O3

1 mol O3 48.00 g O3

1 mol O3

Grams

MolarMass

Moles

MolarVolume

Volume

Chapter 9 25

Mole Calculation – Molecules to Volume

• How many molecules of hydrogen gas, H2, occupy 0.500 L at STP?

0.500 L H2 ×1 mol H2

22.4 L H2

6.02×1023 molecules H2

1 mole H2

×

1.34 × 1022 molecules H2

Volume

MolarVolume

Moles

Avogadro’sNumber

Atoms

Chapter 9 26

Gas Density and Molar Mass• The density of a gas is much less than that of a liquid.

• We can calculate the density of any gas at STP easily.

• You can rearrange this equation to find the Molar mass of an unknown gas too!

= density, g/Lmolar mass in grams (MM)molar volume in liters (MV)

Chapter 9 27

Calculating Gas Density• What is the density of ammonia gas, NH3, at STP?

• 1.96 g of an unknown gas occupies 1.00 L at STP. What is the molar mass?

= 0.761 g/L17.04 g/mol22.4 L/mol

Chapter 9 28

The Ideal Gas Law• When working in the lab, you will not always be at STP.

• The four properties used in the measurement of a gas (Pressure, Volume, Temperature and moles) can be combined into a single gas law:

• Here, R is the ideal gas constant and has a value of:

• Note the units of R. When working problems with the Ideal Gas Law, your units of P, V, T and n must match those in the constant!

PV = nRT

0.0821 atmL/molK

Chapter 9 29

Ideal Gas Law Problem

• Sulfur hexafluoride (SF6) is a colorless,

odorless, very unreactive gas. Calculate the

pressure (in atm) exerted by 1.82 moles of

the gas in a steel vessel of volume 5.43 L at

69.5°C.

Chapter 9 30

• Density and Molar Mass Calculations:

• You can calculate the density or molar mass (M) of a gas.

• The density of a gas is usually very low under

atmospheric conditions.

TR

MP

V

Mnd

volume

mass

Ideal Gas Law and Molar Mass

Chapter 9 31

• What is the molar mass of a gas with a density of 1.342

g/L–1 at STP?

• What is the density of uranium hexafluoride, UF6, (MM =

352 g/mol) under conditions of STP?

• The density of a gaseous compound is 3.38 g/L–1 at 40°C

and 1.97 atm. What is its molar mass?

Ideal Gas Law and Molar Mass

Chapter 9 32

Gases in Chemical Reactions• Gases are involved as reactants

and/or products in numerous chemical reactions.

• Typically, the information given for a gas in a reaction is its Pressure (P), volume (V) (or amount of the gas (n)) and temperature (T).

• We use this information and the Ideal Gas Law to determine the moles of the gas (n) or the volume of the gas (V).

• Once we have this information, we can proceed with the problem as we would any other stoichiometry problem.

A (g) + X (s) → B (s) + Y (l)

Chapter 9 33

Reaction with a Gas• Hydrogen gas is formed when zinc metal reacts

with hydrochloric acid. How many liters of hydrogen gas at STP are produced when 15.8 g of zinc reacts?

Zn (s) + 2HCl (aq) → H2 (g) + ZnCl2 (aq)

Grams of Zn Moles of Zn Moles of Hh Liters of H2

MolarMass of Zn

MolarVolume

MoleRatio

Can use because at STP!

Chapter 9 34

Reaction with a Gas• Hydrogen gas is formed when zinc metal reacts with

hydrochloric acid. How many liters of hydrogen gas at a pressure of 755 atm and 35°C are produced when 15.8 g of zinc reacts?

Zn (s) + 2HCl (aq) → H2 (g) + ZnCl2 (aq)

Grams of Zn Moles of Zn Moles of Hh Liters of H2

MolarMass of Zn

Ideal GasLaw

MoleRatio

Use because not at STP!

Chapter 9 35

Dalton’s Law of Partial Pressures

• In a mixture of gases the total pressure, Ptot, is the sum

of the partial pressures of the gases:

• Dalton’s law allows us to work with mixtures of gases.

nV

RTP total

Ptot = P1 + P2 + P3 + etc.

Chapter 9 36

• For a two-component system, the moles of components A and B can be represented by the mole fractions (XA and XB).

• What is the mole fraction of each component in a mixture of 12.45 g of H2, 60.67 g of N2, and 2.38 g of NH3?


1 BABA

BB

BA

AA

XX

nn

nX

nn

nX

Chapter 9 37


• Mole fraction is related to the total pressure by:

• On a humid day in summer, the mole fraction of gaseous H2O (water vapor) in the air at 25°C can be as high as 0.0287. Assuming a total pressure of 0.977 atm, what is the partial pressure (in atm) of H2O in the air?

totii PXP

Chapter 9 38


• Exactly 2.0 moles of Ne and 3.0 moles of Ar were placed in a 40.0 L container at 25°C. What are the partial pressures of each gas and the total pressure?

• A sample of natural gas contains 6.25 moles of methane (CH4), 0.500 moles of ethane (C2H6), and 0.100 moles of propane (C3H8). If the total pressure of the gas is 1.50 atm, what are the partial pressures of the gases?

Chapter 9 39

Kinetic Molecular Theory of Gases• The Kinetic Molecular Theory of Gases is the

model used to explain the behavior of gases in nature.

• This theory presents physical properties of gases in

terms of the motion of individual molecules.

1. Average Kinetic Energy Kelvin Temperature

2. Gas molecules are points separated by a great distance

3. Particle volume is negligible compared to gas volume

4. Gas molecules are in rapid random motion

5. Gas collisions are perfectly elastic

6. Gas molecules experience no attraction or repulsion

Chapter 9 40

Kinetic Molecular Theory of Gases

Chapter 9 41

• Average Kinetic Energy (KE) is given by:

MM

RT

mN

RTu

N

RTmuKE

A

A

332

3

2

1

2

2


Chapter 9 42


• The Root–Mean–Square Speed (uRMS): is a measure of the average molecular speed of a particle of gas.

Taking square root of both sides gives the equation

MM

RTu

rms

3

MM

RTu

32

R = 8.314 J/mol K

Chapter 9 43

• Calculate the root–mean–square speeds (uRMS)

of helium atoms and nitrogen molecules in m/s

at 25°C.


Chapter 9 44

• Diffusion is the mixing

of different gases by

random molecular

motion and collision.

Graham’s Law: Diffusion and Effusion

• Effusion is when gas

molecules escape without

collision, through a tiny

hole into a vacuum.

Chapter 9 45

• Graham’s Law: The rate of effusion is proportional to its RMS speed (uRMS).

• For two gases at same temperature and pressure:

MM

RTrms

3 u

1

2

1

2

2

1

MM

MM

MM

MM

Rate

Rate


Rate

Chapter 9 46

• Under the same conditions, an unknown gas diffuses

0.644 times as fast as sulfur hexafluoride, SF6 (MM =

146 g/mol). What is the identity of the unknown gas

if it is also a hexafluoride?

• What are the relative rates of diffusion of the three

naturally occurring isotopes of neon: 20Ne, 21Ne, and 22Ne?


Chapter 9 47

• Deviations from Ideal behavior result from two key assumptions about ideal gases.

1. Molecules in gaseous state do not exert any force, either attractive or repulsive, on one another.

2. Volume of the molecules is negligibly small compared with that of the container.

• These assumptions breakdown at high pressures, low volumes and low temperatures.

Behavior of Real Gases

Chapter 9 48

• At STP, the volume occupied by a single molecule is very small relative to

its share of the total volume

– For example, a He atom (radius = 31 pm) has roughly the same space to move

about as a pea in a basketball

• Let’s say we increase the pressure of the system to 1000 atm, this will cause

a decrease in the volume the gas has to move about in

– Now our He atom is like a pea in a ping pong ball

• Therefore, at high pressures, the volume occupied by the gaseous molecules

is NOT negligible and must be considered.

• So the space the gas has to move around in is less than under Ideal

conditions!


VReal > VIdeal

Chapter 9 49

• At low volumes, particles are much closer together and

attractive forces become more important than at high volumes.

• This increase in intermolecular attractions pulls the molecules

away from the walls of the containers, meaning that they do not

hit the wall with as great a force, so the pressure is lower than

under ideal conditions.


PReal < PIdeal

Chapter 9 50

• A similar phenomenon is seen at low temperatures

(aka. The Flirting Effect)

– As molecules slow down, they have more time to interact

therefore increasing the effect of intermolecular forces.

• Again, this increase in intermolecular attractions pulls

the molecules away from the walls of the containers,

meaning that they do not hit the wall with as great a

force, so the pressure is lower than under ideal

conditions.


PReal < PIdeal

Chapter 9 51


Chapter 9 52

• Corrections for non-ideality require the van der Waals equation.

nRTbnVV

anP –

2

2

Correction for Intermolecular

Attractions

Correction for Molecular Volume


n = moles of gasa and b are constants given in the problem

PReal < PIdeal

VReal > VIdeal

Chapter 9 53

• Given that 3.50 moles of NH3 occupy 5.20 L at 47°C,

calculate the pressure of the gas (in atm) using: (a) the ideal gas equation

(b) the van der Waals equation. (a = 4.17, b = 0.0371)

• Calculate the pressure exerted by 4.37 moles of molecular

chlorine confined in a volume of 2.45 L at 38°C.

Compare the pressure with that calculated using the ideal

gas equation. (a = 6.49 and b = 0.0562)


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Gases: Their Properties and Behavior Chapter 9. 2 Properties of Gases There are 5 important properties of gases: –Confined gases exerts pressure on the