Chapter 6 Gases. 2 Gases are highly compressible and occupy the full volume of their containers. When a gas is subjected to pressure, its volume decreases

Chapter 6

Gases

2

• Gases are highly compressible and occupy the full volume of their containers.

• When a gas is subjected to pressure, its volume decreases.

• Gases always form homogeneous mixtures with other gases.

• Gases only occupy about 0.1 % of the volume of their containers.

6.1 Properties of Gases

3

• Pressure is the force acting on an object per unit area:

• Gravity exerts a force on the earth’s atmosphere

Pressure

A

FP

4

Atmosphere Pressure and the Barometer• SI Units: 1 N (force) = 1 kg.m/s2; 1 Pa = 1 N/m2.• Atmospheric pressure is measured with a barometer.• If a tube is inserted into a container of mercury open to

the atmosphere, the mercury will rise 760 mm up the tube.

• Standard atmospheric pressure is the pressure required to support 760 mm of Hg in a column.

• Units: 1 atm = 760 mmHg = 760 torr = 1.01325 105 Pa = 101.325 kPa.

Pressure

5

Atmosphere Pressure and the Barometer

Pressure

P = g x h x d

6

Atmosphere Pressure and the Barometer

• The pressures of gases not open to the atmosphere are measured in manometers.

• A manometer consists of a bulb of gas attached to a U-tube containing Hg:– If Pgas < Patm then Pgas + Ph = Patm.

– If Pgas > Patm then Pgas = Patm + Ph.

Pressure

Example 1A

A barometer is filled with diethylene glycol (d=1.118g/cm3). The liquid height is found to be 9.25m. What is the barometric pressure expressed in millimeters of mercury?

Example 2A

Suppose the mercury level is 7.8mm higher in the arm open to the atmosphere than in the closed arm. What would the pressure of the gas be? Pressure atmosphere: 748.2mmHg

Example 3A

The 1.000 kg green cylinder has a diameter of 2.60 cm. What pressure, expressed in Torr, does this cylinder exert on the surface beneath it?

10/28

11

The Pressure-Volume Relationship: Boyle’s Law

• Weather balloons are used as a practical consequence to the relationship between pressure and volume of a gas.

• As the weather balloon ascends, the volume increases.• As the weather balloon gets further from the earth’s

surface, the atmospheric pressure decreases.• Boyle’s Law: the volume of a fixed quantity of gas is

inversely proportional to its pressure.• Boyle used a manometer to carry out the experiment.

6.2 The Gas Laws

12

13


• Mathematically:

• A plot of V versus P is not a straight line• A plot of V versus 1/P shows a straight line passing

through the origin.

The Gas Laws

PV

1constant constantPV

14


The Gas Laws

A 50.0 L cylinder contains nitrogen gas at a pressure of 21.5 atm. The contents of the cylinder are emptied into an evacuated tank of unknown volume with a new pressure of 1.55 atm. What is the volume of the tank?

Slide 16 of 41Copyright © 2011 Pearson

Canada Inc.

A 100 L vessel at 30 atm is attached to another vessel and gas is allowed to equilibrate between the two. The pressure was then found to be 10 atm. Use Boyle’s Law to determine the volume of the second container.

2. 100 L

3. 200 L

4. 300 L1. 50 L

5. 400 L

Slide 17 of 41Copyright © 2011 Pearson

Canada Inc.

A 100 L vessel at 30 atm is attached to another vessel and gas is allowed to equilibrate between the two. The pressure was then found to be 10 atm. Use Boyle’s Law to determine the volume of the second container.

2. 100 L

3. 200 L

4. 300 L1. 50 L

5. 400 L

18

The Temperature-Volume Relationship: Charles’s Law

• We know that hot air balloons expand when they are heated.

• Charles’s Law: the volume of a fixed quantity of gas at constant pressure increases as the temperature increases.

• Mathematically:

The Gas Laws

TV constant constantTV

19

20

The Temperature-Volume Relationship: Charles’s Law

• A plot of V versus T is a straight line.• When T is measured in C, the intercept on the

temperature axis is -273.15C. • We define absolute zero, 0 K = -273.15C.• Note the value of the constant reflects the assumptions:

amount of gas and pressure do not change.

The Gas Laws

A balloon is inflated to a volume of 2.5L inside a house kept at 24°C. The balloon is then taken outside where the temp is -25°C. What will the volume of the balloon be?

22

The Quantity-Volume Relationship: Avogadro’s Law

• Gay-Lussac’s Law of combining volumes: at a given temperature and pressure, the volumes of gases which react are ratios of small whole numbers.

The Gas Laws

23


• Avogadro’s Hypothesis: equal volumes of gas at the same temperature and pressure will contain the same number of molecules.

• Avogadro’s Law: the volume of gas at a given temperature and pressure is directly proportional to the number of moles of gas.

The Gas Laws

24


• Mathematically:

• We can show that 22.4 L of any gas at 0C contain 6.02 1023 gas molecules.

The Gas Laws

nV constant

25


The Gas Laws

27

• Consider the three gas laws.

• We can combine these into a general gas law:

6.3 & 6.4 The Ideal Gas Equation

), (constant 1

TnP

V

), (constant PnTV ),(constant TPnV

• Boyle’s Law:

• Charles’s Law:

• Avogadro’s Law:

PnT

V

28

• If R is the constant of proportionality (called the gas constant), then

• The ideal gas equation is:

• R = 0.08206 L·atm/mol·K = 8.314 J/mol·K

The Ideal Gas Equation

PnT

RV

nRTPV

29

30

• We define STP (standard temperature and pressure) = 0C, 273.15 K, 1 atm.

• Volume of 1 mol of gas at STP is:


L 41.22

atm 000.1K 15.273KL·atm/mol· 0.08206mol 1

PnRT

V

nRTPV

31

Relating the Ideal-Gas Equation and the Gas Laws

• If PV = nRT and n and T are constant, then PV = constant and we have Boyle’s law.

• Other laws can be generated similarly.• In general, if we have a gas under two sets of conditions,

then


22

22

11

11TnVP

TnVP

4A

What is the volume occupied by 20.2 g of NH3 at -25°C and 753 mmHg?

5A

How many moles of He are in a 5.00L storage tank filled with helium at 10.5 atm pressure at 30.0°C?

6A

• A 1.00 mL sample of N2 gas at 36.2°C and 2.14 atm is heated to 38.7°C and the pressure is changed to 1.02 atm. What volume does the gas occupy at this final temperature?

35

Gas Densities and Molar Mass• Density has units of mass over volume. • Rearranging the ideal-gas equation with M as molar mass

we get

Further Applications of the Ideal-Gas Equation

RTP

dV

nRTP

Vn

nRTPV

MM

36

Gas Densities and Molar Mass• The molar mass of a gas can be determined as follows:

Volumes of Gases in Chemical Reactions• The ideal-gas equation relates P, V, and T to number of

moles of gas.• The n can then be used in stoichiometric calculations.

Further Applications of the Ideal-Gas Equation

PdRTM

37

Example 1 • Calculate the density of NO2 gas at 0.970 atm

and 35°C.

d = PM

RT

d = (0.97 atm)(46.0658 g/mol)

(0.08206 L-atm/K-mol)(308K)

d = (44.683826 atm-g/mol)

(25.27448 L-atm/mol)

d = 1.7679424463 g/L

d = 1.8 g/L

38

Example 1 (10.37b)

• Calculate the molar mass of a gas if 2.50 g occupies 0.875 L at 685 torr and 35°C.

d = PM

RT

M = dRT

P

M = (2.50g/0.875L)(62.36L-Torr/K-mol)(308K)

685 torr

M = 54876.8 g-Torr/mol

685 Torr

M = 80.1121 g/mol

M = 80 g/mol

11/4

Example 9A

How many grams of NaN3 are needed to produce 20.0L of N2 gas at 30°C and 776mmHg?

2NaN3(s) 2Na(l) + 3N2(g)

Pathway: Find mole N2 then use balanced equation to get to grams of reactant.

6.5 Gases in Reactions

Use Stoichiometry to convert!• Law of combining volumes:

2NO(g) + O2(g) 2NO2(g)

2L NO(g) + 1L O2(g) 2L NO2(g)

Must be at the same T and P

10A

What volume of O2(g) is consumed per liter of NO(g) formed?

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

V

RTnP tottot

43

• Since gas molecules are so far apart, we can assume they behave independently.

• Dalton’s Law: in a gas mixture the total pressure is given by the sum of partial pressures of each component:

Each gas obeys the ideal gas equation:

6.6 Gas Mixtures and Partial Pressures

321total PPPP

44

N2 Ne H2

Volume: 1.0L 1.0 L 0.5 LPressure: 635 torr 212 torr 418 torr

•Consider the arrangement of bulbs shown in figure below. Each of the bulbs contains a gas at the pressure shown. What is the pressure of the system when all of the stopcocks are opened, assuming that the temperature remains constant?

45

635 torr + 212 torr + 418 torr = 1.0L 1.0 L 0.5 L

= 1056 torr

1056 torr / 2.5 = 422.4 torr

Example 11

What is the pressure exerted by a mixture of 1.0 g of H2 and 5.00 g He when the mixture is confined to a volume of 5.0L at 20°C?

47

• Combining the equations

Partial Pressures and Mole Fractions• Let ni be the number of moles of gas i exerting a partial

pressure Pi, then

where i is the mole fraction (ni/nt).

Gas Mixtures and Partial Pressures

VRT

nnnP 321total

totalPP ii

12A

A mixture of .197 mol CO2(g) and 0.00278 mole H2O(g) is held at 30°C and 2.5 atm. What are the partial pressures?

49

Collecting Gases over Water• It is common to synthesize gases and collect them by

displacing a volume of water.

• To calculate the amount of gas produced, we need to correct for the partial pressure of the water:


watergastotal PPP

50

Collecting Gases over Water


13A

2Al(s) + 6HCl(aq) 2AlCl3(aq) + 3H2(g)

If 35.5mL of H2(g) is collected over water at 26°C and a barometric pressure of 755mmHg, how many moles of HCl must have been consumed? (The vapor pressure of water 26°C is 25.2 mmHg)

52

• Theory developed to explain gas behavior.• Theory of moving molecules.• Assumptions:

– Gases consist of a large number of molecules in constant random motion.

– Volume of individual molecules negligible compared to volume of container.

– Intermolecular forces (forces between gas molecules) negligible.

6.7 / 6.8 Kinetic Molecular Theory

53

• Assumptions:– Energy can be transferred between molecules, but total

kinetic energy is constant at constant temperature.– Average kinetic energy of molecules is proportional to

temperature.

• Kinetic molecular theory gives us an understanding of pressure and temperature on the molecular level.

• Pressure of a gas results from the number of collisions per unit time on the walls of container.

Kinetic Molecular Theory

54


• Magnitude of pressure given by how often and how hard the molecules strike.

• Gas molecules have an average kinetic energy.

• Each molecule has a different energy.

55

• There is a spread of individual energies of gas molecules in any sample of gas.

• As the temperature increases, the average kinetic energy of the gas molecules increases.


56

57

• As kinetic energy increases, the velocity of the gas molecules increases.

• Root mean square speed, u, is the speed of a gas molecule having average kinetic energy.

• Average kinetic energy, , is related to root mean square speed:


221 mu

58

Application to Gas Laws• As volume increases at constant temperature, the average

kinetic of the gas remains constant. Therefore, u is constant. However, volume increases so the gas molecules have to travel further to hit the walls of the container. Therefore, pressure decreases.

• If temperature increases at constant volume, the average kinetic energy of the gas molecules increases. Therefore, there are more collisions with the container walls and the pressure increases.


59

Molecular Effusion and Diffusion• As kinetic energy increases, the velocity of the gas

molecules increases.• Average kinetic energy of a gas is related to its mass:

• Consider two gases at the same temperature: the lighter gas has a higher rms than the heavier gas.

• Mathematically:


221 mu

MRT

u3

14A

Which has the greater rms speed at 25°C, NH3(g) or HCl(g)? Calculate the Urms for the one with greater speed.

61

Molecular Effusion and Diffusion• The lower the molar mass, M, the higher the rms.


62


Graham’s Law of Effusion• As kinetic energy increases,

the velocity of the gas molecules increases.

• Effusion is the escape of a gas through a tiny hole (a balloon will deflate over time due to effusion).

• The rate of effusion can be quantified.

63

Graham’s Law of Effusion • Consider two gases with molar masses M1 and M2, the

relative rate of effusion is given by:

• Only those molecules that hit the small hole will escape through it.

• Therefore, the higher the rms the more likelihood of a gas molecule hitting the hole.


1

2

2

1MM

rr

64

Graham’s Law of Effusion • Consider two gases with molar masses M1 and M2, the

relative rate of effusion is given by:

• Only those molecules that hit the small hole will escape through it.

• Therefore, the higher the rms the more likelihood of a gas molecule hitting the hole.


1

2

2

1

2

1

2

13

3

MM

M

M RT

RT

uu

rr

65

Diffusion and Mean Free Path • Diffusion of a gas is the spread of the gas through space.• Diffusion is faster for light gas molecules.• Diffusion is significantly slower than rms speed (consider

someone opening a perfume bottle: it takes while to detect the odor but rms speed at 25C is about 1150 mi/hr).

• Diffusion is slowed by gas molecules colliding with each other.


66

Diffusion and Mean Free Path • Average distance of a gas molecule between collisions is

called mean free path.• At sea level, mean free path is about 6 10-6 cm.


15

If 2.2 x 10-4 mol N2(g) effuses through a tony hole

in 105 s, then how much H2(g) would effuse through the same oriffice in 105s?

16

A sample of Kr gas escapes through a tiny hold in 87.3s. The same amount of an unknown as escapes in 42.9 s under identical conditions. What is the MM of the unknown?

6.9 Real Gases

70

• From the ideal gas equation, we have

• For 1 mol of gas, PV/RT = 1 for all pressures.• In a real gas, PV/RT varies from 1 significantly.• The higher the pressure the more the deviation from ideal

behavior.

Real Gases: Deviations from Ideal Behavior

nRTPV

71

72

• From the ideal gas equation, we have

• For 1 mol of gas, PV/RT = 1 for all temperatures.• As temperature increases, the gases behave more ideally.• The assumptions in kinetic molecular theory show where

ideal gas behavior breaks down:– the molecules of a gas have finite volume;– molecules of a gas do attract each other.


nRTPV

73

74

• As the pressure on a gas increases, the molecules are forced closer together.

• As the molecules get closer together, the volume of the container gets smaller.

• The smaller the container, the more space the gas molecules begin to occupy.

• Therefore, the higher the pressure, the less the gas resembles an ideal gas.


75

• As the gas molecules get closer together, the smaller the intermolecular distance.


76

• The smaller the distance between gas molecules, the more likely attractive forces will develop between the molecules.

• Therefore, the less the gas resembles and ideal gas.• As temperature increases, the gas molecules move faster

and further apart.• Also, higher temperatures mean more energy available to

break intermolecular forces.


77

• Therefore, the higher the temperature, the more ideal the gas.


78

The van der Waals Equation• We add two terms to the ideal gas equation one to correct

for volume of molecules and the other to correct for intermolecular attractions

• The correction terms generate the van der Waals equation:

where a and b are empirical constants.


2

2

V

annbV

nRTP

17

Use the van der Waals equation to calculate the pressure exerted by 1.00 mol Cl2 gas confined to a volume of 2.00L at 273K. The value of a=6.49L2atm/mol-2 b=0.0562L/mol

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Chapter 6 Gases. 2 Gases are highly compressible and occupy the full volume of their containers. When a gas is subjected to pressure, its volume decreases