1 CHAPTER 10 GASES AND KINETIC-MOLECULAR THEORY

1

CHAPTER 10

GASES AND KINETIC-MOLECULAR THEORY

2

CHAPTER GOALS

1. Comparison of Solids, Liquids, and Gases2. Composition of the Atmosphere and Some

Common Properties of Gases3. Pressure4. Boyle’s Law: The Volume-Pressure Relationship5. Charles’ Law: The Volume-Temperature

Relationship; The Absolute Temperature Scale6. Standard Temperature and Pressure7. The Combined Gas Law Equation8. Avogadro’s Law and the Standard Molar

Volume

3

CHAPTER GOALS

9. Summary of Gas Laws: The Ideal Gas Equation

10. Determination of Molecular Weights and Molecular Formulas of Gaseous Substances

11. Dalton’s Law of Partial Pressures12. Mass-Volume Relationships in Reactions

Involving Gases 13. The Kinetic-Molecular Theory14. Diffusion and Effusion of Gases15. Real Gases: Deviations from Ideality

4

Comparison of Solids, Liquids, and Gases

The density of gases is much less than that of solids or liquids.

Densities (g/mL)

Solid Liquid Gas

H2O 0.917 0.998 0.000588

CCl4 1.70 1.59 0.00503

Gas molecules must be very far apart compared to liquids and solids.

5

Composition of the Atmosphere and Some Common Properties of Gases

Gas % by Volume

N2 78.09

O2 20.94

Ar 0.93

CO2 0.03

He, Ne, Kr, Xe 0.002

CH4 0.00015

H2 0.00005

Composition of Dry Air

6

Pressure

Pressure is force per unit area. lb/in2

N/m2

Gas pressure as most people think of it.

7

Pressure

Atmospheric pressure is measured using a barometer.

Definitions of standard pressure 76 cm Hg 760 mm Hg 760 torr 1 atmosphere 101.3 kPa

Hg density = 13.6 g/mL

8

Boyle’s Law: The Volume-Pressure Relationship

V 1/P or V= k (1/P) or PV = k P1V1 = k1 for one sample of a gas. P2V2 = k2 for a second sample of a gas. k1 = k2 for the same sample of a gas

at the same T. Thus we can write Boyle’s Law

mathematically as P1V1 = P2V2

9


Example 12-1: At 25oC a sample of He has a volume of 4.00 x 102 mL under a pressure of 7.60 x 102 torr. What volume would it occupy under a pressure of 2.00 atm at the same T?

torr1520

mL 400 torr760

P

V PV

V PV P

2

112

2211

2211 V PV P

mL 1000.2

torr1520

mL 400 torr760

P

V PV

V PV P

2

2

112

2211

10


Notice that in Boyle’s law we can use any pressure or volume units as long as we consistently use the same units for both P1 and P2 or V1 and V2.

Use your intuition to help you decide if the volume will go up or down as the pressure is changed and vice versa.

11

Charles’ Law: The Volume-Temperature Relationship;

The Absolute Temperature Scale

0

5

10

15

20

25

30

35

0 50 100 150 200 250 300 350 400

Volume (L) vs.

Temperature (K)

Gases liquefy before reaching 0K

absolute zero = -273.15 0C

12



Charles’s law states that the volume of a gas is directly proportional to the absolute temperature at constant pressure. Gas laws must use the Kelvin scale to

be correct. Relationship between Kelvin and

centigrade.

K = C + 273o

13



Mathematical form of Charles’ law.

kT

Vor kT = Vor TV

form usefulmost in the T

V

T

V

so equal are sk' ehowever thk T

V andk

T

V

kT

Vor kT = Vor TV

2

2

1

1

2

2

1

1

14



Example 12-2: A sample of hydrogen, H2, occupies 1.00 x 102 mL at 25.0oC and 1.00 atm. What volume would it occupy at 50.0oC under the same pressure?

T1 = 25 + 273 = 298

T2 = 50 + 273 = 323

mL 108K 298

K 323mL 101.00=V

T

TV=V

T

V

T

V

2

2

1

212

2

2

1

1

15

Standard Temperature and Pressure

Standard temperature and pressure is given the symbol STP. It is a reference point for some gas

calculations. Standard P 1.00000 atm or 101.3 kPa Standard T 273.15 K or 0.00oC

16

The Combined Gas Law Equation

Boyle’s and Charles’ Laws combined into one statement is called the combined gas law equation. Useful when the V, T, and P of a gas are

changing.

2

2

1

12211 T

V

T

V VPVP

Law Charles' Law sBoyle'

2

22

1

11

2

2

1

12211

T

V P

T

V P k

T

V P

:is law gas combined The :gas of samplegiven aFor

T

V

T

V VPVP

Law Charles' Law sBoyle'

17


Example 12-3: A sample of nitrogen gas, N2, occupies 7.50 x 102 mL at 75.00C under a pressure of 8.10 x 102 torr. What volume would it occupy at STP?

12

2112

21

21

21

T P

T V P=Vfor Solve

torr760=P torr 810=P

K 273=TK 348=T

?=V mL 750=V

mL 627

K 348 torr760

K 273mL 750 torr810

T P

T V P=Vfor Solve

torr760=P torr 810=P

K 273=TK 348=T

?=V mL 750=V

12

2112

21

21

21

18

The Combined Gas Law Equation Example 12-4 : A sample of methane,

CH4, occupies 2.60 x 102 mL at 32oC under a pressure of 0.500 atm. At what temperature would it occupy 5.00 x 102 mL under a pressure of 1.20 x 103 torr?

You do it!You do it!

19


C1580 K 1852=

mL 260 torr380

mL 500 torr1200K 305

V P

V P T= T

? = T K 305 = T

torr 380 =

torr1200 = P atm 0.500 = P

mL 500 = V mL 260 = V

o

11

2212

21

21

21

20

Avogadro’s Law and theStandard Molar Volume

21


Avogadro’s Law states that at the same temperature and pressure, equal volumes of two gases contain the same number of molecules (or moles) of gas.

If we set the temperature and pressure for any gas to be STP, then one mole of that gas has a volume called the standard molar volumestandard molar volume.

The standard molar volume is 22.4 L at STP. This is another way to measure moles. For gases, the volume is proportional to the number

of moles. 11.2 L of a gas at STP = 0.500 mole

44.8 L = ? moles

22


Example 12-5: One mole of a gas occupies 36.5 L and its density is 1.36 g/L at a given temperature and pressure. (a) What is its molar mass? (b) What is its density at STP?

? . ..

gmol

Lmol

gL

g / mol 36 5 13649 6

g/L 21.2L 4.22

mol 1

mol

g 6.49

L

g ?

STP

23

Summary of Gas Laws:The Ideal Gas Law

Boyle’s Law - V 1/P (at constant T & n) Charles’ Law – V T (at constant P & n) Avogadro’s Law – V n (at constant T & P) Combine these three laws into one statement

V nT/P Convert the proportionality into an equality.

V = nRT/P This provides the Ideal Gas Law.

PV = nRT R is a proportionality constant called the universal gas

constant.

24


We must determine the value of R. Recognize that for one mole of a gas at 1.00

atm, and 273 K (STP), the volume is 22.4 L. Use these values in the ideal gas law.

R = PV

nT

1.00 atm L

1.00 mol K

L atm

mol K

22 4

273

0 0821

.

.

25


R has other values if the units are changed.

R = 8.314 J/mol K Use this value in thermodynamics.

R = 8.314 kg m2/s2 K mol Use this later in this chapter for gas velocities.

R = 8.314 dm3 kPa/K mol This is R in all metric units.

R = 1.987 cal/K mol This the value of R in calories rather than J.

26


Example 12-6: What volume would 50.0 g of ethane, C2H6, occupy at 1.40 x 102 oC under a pressure of 1.82 x 103 torr?

To use the ideal gas law correctly, it is very important that all of your values be in the correct units!

1. T = 140 + 273 = 413 K2. P = 1820 torr (1 atm/760 torr) = 2.39 atm3. 50 g (1 mol/30 g) = 1.67 mol

27


atm 39.2

K 413K mol

atm L 0821.0mol 67.1

P

T Rn = V

P

T Rn = V

L 6.23atm 39.2

K 413K mol

atm L 0821.0mol 67.1

P

T Rn = V

28


Example 12-7: Calculate the number of moles in, and the mass of, an 8.96 L sample of methane, CH4, measured at standard conditions.


29


n =

PV

RT

1.00 atm L

0.0821L atm

mol K K

0.400 mol CH

g CH mol16.0 g

mol 6.40 g

4

4

8 96

273

0 400

.

? .

30


Example 12-8: Calculate the pressure exerted by 50.0 g of ethane, C2H6, in a 25.0 L container at 25.0oC.


atm 63.1PL 25.0

K 298K mol

atm L0.0821mol 1.67

P

V

T Rn = P

K 298 = T and mol 1.67 =n

31

Determination of Molecular Weights and Molecular Formulas of Gaseous Substances

Example 12-9: A compound that contains only carbon and hydrogen is 80.0% carbon and 20.0% hydrogen by mass. At STP, 546 mL of the gas has a mass of 0.732 g . What is the molecular (true) formula for the compound?

100 g of compound contains 80 g of C and 20 g of H.

32


C mol 67.6C g 12.0

C mol 1C g 80.0 = atoms C mol ?

15 = mass with CH is formula empirical the36.67

19.8

ratio.number holesmallest w theDetermine

H mol 8.19H g 1.01

H mol 1H g 20.0 = atoms H mol ?

C mol 67.6C g 12.0


3

H mol 8.19H g 1.01


C mol 67.6C g 12.0


33


6223 HCCH

doubled. formula empirical the

is formulamolecular theThus

20.15

0.30

mass empirical

mass actualmol

g0.30

mol 0.0244

g 0.732 =n

moles. ofnumber by the divided mass

theis massmolar theRemember,

34


Example 12-10: A 1.74 g sample of a compound that contains only carbon and hydrogen contains 1.44 g of carbon and 0.300 g of hydrogen. At STP 101 mL of the gas has a mass of 0.262 gram. What is its molecular formula?


35


mol 00451.0K 273

K molatm L

0.0821

L 0.101atm 00.1

RT

PV=n

29 = mass with HC5.20.120

0.297

H mol 297.0H g 1.01


C mol 120.0C g 0.12


52

36


104252 HCHC229

58.1

g/mol 1.58mol 0.00451

g 262.0

mol

g ?

37

Dalton’s Law of Partial Pressures

Dalton’s law states that the pressure exerted by a mixture of gases is the sum of the partial pressures of the individual gases.

Ptotal = PA + PB + PC + .....

38


Example 12-11: If 1.00 x 102 mL of hydrogen, measured at 25.0 oC and 3.00 atm pressure, and 1.00 x 102 mL of oxygen, measured at 25.0 oC and 2.00 atm pressure, were forced into one of the containers at 25.0 oC, what would be the pressure of the mixture of gases?

P P P

3.00 atm + 2.00 atm

= 5.00 atm

Total H O2 2

39


Vapor Pressure is the pressure exerted by a substance’s vapor over the substance’s liquid at equilibrium.

40


Example 12-12: A sample of hydrogen was collected by displacement of water at 25.0 oC. The atmospheric pressure was 748 torr. What pressure would the dry hydrogen exert in the same container?

P P P P P P

P torr = 724 torr

P from table

Total H H O H Total H O

H

H O

2 2 2 2

2

2

748 24

41


Example 12-13: A sample of oxygen was collected by displacement of water. The oxygen occupied 742 mL at 27.0 oC. The barometric pressure was 753 torr. What volume would the dry oxygen occupy at STP?


42


V 742 mL V ?

T K T K

P = 726 torr P torr

V mL273 K300 K

726 torr torr

645 mL @ STP

1 2

1 2

1 2

2

300 273

753 27 760

742760

43

Mass-Volume Relationships in Reactions Involving Gases

44


2 mol KClO3 yields 2 mol KCl and 3 mol O2

2(122.6g) yields 2 (74.6g) and 3 (32.0g)Those 3 moles of O2 can also be thought of as:

3(22.4L) or67.2 L at STP

g)(2(s)&MnO

(s)3 O 3 + KCl 2KClO 2 2

•In this section we are looking at reaction stoichiometry, like in Chapter 3, just including gases in the calculations.

45


Example 12-14: What volume of oxygen measured at STP, can be produced by the thermal decomposition of 120.0 g of KClO3?


46


2STP2STP

2

2STP

3

2

3

332STP

O L 9.32O L ?

O mol 1

O L 4.22

KClO mol 2

O mol 3

KClO g 122.6

KClO mol 1 KClO g 120.0O L ?

47

The Kinetic-Molecular Theory The basic assumptions of kinetic-

molecular theory are: Postulate 1

Gases consist of discrete molecules that are relatively far apart.

Gases have few intermolecular attractions. The volume of individual molecules is very

small compared to the gas’s volume. Proof - Gases are easily compressible.

48

The Kinetic-Molecular Theory

Postulate 2 Gas molecules are in constant, random,

straight line motion with varying velocities.

Proof - Brownian motion displays molecular motion.

49


Postulate 3 Gas molecules have elastic collisions

with themselves and the container. Total energy is conserved during a

collision. Proof - A sealed, confined gas

exhibits no pressure drop over time.

50

The Kinetic-Molecular Theory Postulate 4

The kinetic energy of the molecules is proportional to the absolute temperature.

The average kinetic energies of molecules of different gases are equal at a given temperature.

Proof - Brownian motion increases as temperature increases.

51

The Kinetic-Molecular Theory• The kinetic energy of the molecules is

proportional to the absolute temperature. The kinetic energy of the molecules is proportional to the absolute temperature.• Displayed in a Maxwellian distribution.

52


The gas laws that we have looked at earlier in this chapter are proofs that kinetic-molecular theory is the basis of gaseous behavior.

Boyle’s Law P 1/V As the V increases the molecular collisions with container

walls decrease and the P decreases. Dalton’s Law

Ptotal = PA + PB + PC + ..... Because gases have few intermolecular attractions, their

pressures are independent of other gases in the container.

Charles’ Law V T An increase in temperature raises the molecular

velocities, thus the V increases to keep the P constant.

53


54


uRT

Mrmsm

3

The root-mean square velocity of gases is a very close approximation to the average gas velocity.

Calculating the root-mean square velocity is simple:

To calculate this correctly: The value of R = 8.314 kg m2/s2 K mol And M must be in kg/mol.

55


Example 12-17: What is the root mean square velocity of N2 molecules at room T, 25.0oC?

kg/mol 028.0

K 298molK sec

m kg8.3143

u2

2

rms

u

3 8.314kg m

sec K mol K

kg / mol

m / s = 1159 mi / hr

rms

2

2

298

0 028

515

.

56


Example 12-18: What is the root mean square velocity of He atoms at room T, 25.0oC?


57


u

3 8.314kg m

sec K mol K

kg / mol

m / s = 3067 mi / hr

rms

2

2

298

0 004

1363

.

Can you think of a physical situation that proves He molecules have a velocity that is so much greater than N2 molecules?

What happens to your voice when you breathe He?

58

Diffusion and Effusion of Gases Diffusion is the intermingling of

gases. Effusion is the escape of gases

through tiny holes.

59

Diffusion and Effusion of Gases

This is a demonstration of diffusion.

60


The rate of effusion is inversely proportional to the square roots of the molecular weights or densities.

1

2

2

1

1

2

2

1

D

D

R

R

or

M

M

R

R

61


Example 12-15: Calculate the ratio of the rate of effusion of He to that of sulfur dioxide, SO2, at the same temperature and pressure.

R

R

M

M

g / mol4.0 g / mol

R R

He

SO

SO

He

He SO

2

2

2

641

16 4 4

.

62


Example 12-16: A sample of hydrogen, H2, was found to effuse through a pinhole 5.2 times as rapidly as the same volume of unknown gas (at the same temperature and pressure). What is the molecular weight of the unknown gas?


g/mol 54 = g/mol) 0.2(27M

g/mol 0.2

M27

g/mol 0.2

M2.5

M

M

R

R

unk

unk

unk

H

unk

unk

H

2

2

63

Real Gases: Deviations from Ideality

Real gases behave ideally at ordinary temperatures and pressures.

At low temperatures and high pressures real gases do not behave ideally.

The reasons for the deviations from ideality are:

1. The molecules are very close to one another, thus their volume is important.

2. The molecular interactions also become important.

64

Real Gases:Deviations from Ideality van der Waals’ equation accounts for the

behavior of real gases at low temperatures and high pressures.

P + n a

VV nb nRT

2

2

• The van der Waals constants a and b take into account two things:1. a accounts for intermolecular attraction2. b accounts for volume of gas molecules

• At large volumes a and b are relatively small and van der Waal’s equation reduces to ideal gas law at high temperatures and low pressures.

65

Real Gases:Deviations from Ideality

What are the intermolecular forces in gases that cause them to deviate from ideality?

1. For nonpolar gases the attractive forces are London Forces

2. For polar gases the attractive forces are dipole-dipole attractions or hydrogen bonds.

66


Example 12-19: Calculate the pressure exerted by 84.0 g of ammonia, NH3, in a 5.00 L container at 200. oC using the ideal gas law.


atm 4.38PL 5.00

K 473K mol

atm L0821.0mol 94.4

V

nRT = P

mol 94.4g 17.0

mol 1NH g 84.0 =n 3

67


Example 12-20: Solve Example 12-19 using the van der Waal’s equation.

2

2

2

2

2

2

V

an

nb-V

nRT=P

nRTnb-VV

an + P

molL0.0371=b

mol

atm L 4.17 = a mol 4.94 =n

68


ideal from difference 7.6% a is which atm 7.35P

)atm 1.4atm 8.39(atm 07.4L 817.4

atm L 8.191P

L 00.5

17.4mol 94.4

)0371.0)(mol 94.4(L 00.5

K4730821.0 mol 94.4P 2

molatm L2

molL

K molatm L

2

2

69

Synthesis Question

The lethal dose for hydrogen sulfide is 6.0 ppm. In other words, if in 1 million molecules of air there are six hydrogen sulfide molecules then that air would be deadly to breathe. How many hydrogen sulfide molecules would be required to reach the lethal dose in a room that is 77 feet long, 62 feet wide and 50. feet tall at 1.0 atm and 25.0 oC?

70

Synthesis Question

L106.76cm 1000

L1cm 106.76

cm 106.76cm 1524cm 1890cm 2347V

cm 1524in

cm 54.2

ft

in 12ft 05

cm 1890in

cm 54.2

ft

in 12ft 26

cm 2347in

cm 54.2

ft

in 12ft 77

63

39

39

71

Synthesis Question

SH of molecules 1096.9air of molecules 10

SH of molecule 6.0air of molecules 1066.1 dose Lethal

air of molecules 1066.1mol

molecules 106.022mol 276,000

air of mol 276,000K 2980.0821

L 106.76atm 1

RT

PVn

K 298 25 273 T

L 106.76cm 1000

L 1cm 106.76

223

6229

2923

K molatm L

6

63

39

72

Group Question

Tires on a car are typically filled to a pressure of 35 psi at 3.00 x 102 K. A tire is 16 inches in radius and 8.0 inches in thickness. The wheel that the tire is mounted on is 6.0 inches in radius. What is the mass of the air in the tire?

73

End of Chapter 12

Gases are the simplest state of matter.

Liquids and solids are more complex. They are the subject of

Chapter 13.

Documents

1 CHAPTER 10 GASES AND KINETIC-MOLECULAR THEORY