gases

1

GasesGases

2

Characteristics of GasesCharacteristics of Gases

There are three phases for all substances: There are three phases for all substances: solid, liquid and gases.solid, liquid and gases.

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

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

Gases always form homogeneous mixtures Gases always form homogeneous mixtures with other gases.with other gases.

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

3

PressurePressure

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

Units: psiUnits: psi pounds per square pounds per square inchinch

AF

P

4

PressurePressure

SI Units: 1 N = 1 kg.m/sSI Units: 1 N = 1 kg.m/s22; 1 Pa = 1 ; 1 Pa = 1 N/mN/m22..

Atmospheric pressure is measured Atmospheric pressure is measured with a barometer.with a barometer.

If a tube is inserted into a container If a tube is inserted into a container of mercury open to the atmosphere, of mercury open to the atmosphere, the mercury will rise 760 mm up the the mercury will rise 760 mm up the tube.tube.

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

5

PressurePressure

force per unit areaforce per unit area standard pressurestandard pressure

– 76 cm Hg76 cm Hg– 760 mm Hg760 mm Hg– 760 torr760 torr– 1 atmosphere1 atmosphere– 101.3 kPa101.3 kPa

Units: 1 atm = 760 Units: 1 atm = 760 mmHg = 760 torr = mmHg = 760 torr = 1.01325 1.01325 10 1055 Pa = Pa = 101.325 kPa.101.325 kPa.

density = 13.6 g/mL

6

The Gas LawsThe Gas Laws

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

As the weather balloon goes up, the volume As the weather balloon goes up, the volume increases.increases.

As the weather balloon gets further from As the weather balloon gets further from the earth’s surface, the atmospheric the earth’s surface, the atmospheric pressure decreases.pressure decreases.

Boyle’s Law: the volume of a fixed quantity Boyle’s Law: the volume of a fixed quantity of gas is inversely proportional to its of gas is inversely proportional to its pressure.pressure.

7

Boyle’s LawsBoyle’s Laws

Mathematically:Mathematically:

A plot of V versus P is a hyperbola.A plot of V versus P is a hyperbola. Similarly, a plot of V versus 1/P must be Similarly, a plot of V versus 1/P must be

a straight line passing through the a straight line passing through the origin.origin.

PV

1constant constant PV

8

Boyle’s LawsBoyle’s Laws

9

Boyle’s LawBoyle’s Law

V1/P V= k (1/P) or PV = k P1V1 = k1 P2V2 = k2

kk1 1 = k= k2 2 (for same sample of gas @ (for same sample of gas @ same T)same T)

PP11VV11 = P = P22VV22 Boyle’s Law (math form) Boyle’s Law (math form) think in terms of balloons!!!think in terms of balloons!!!

10

Boyle’s LawBoyle’s Law Example At 25Example At 2500C a sample of He has a C a sample of He has a

volume of 400 mL under a pressure of volume of 400 mL under a pressure of 760 torr. What volume would it occupy 760 torr. What volume would it occupy under a pressure of 2.00 atm at the under a pressure of 2.00 atm at the same T?same T?

11



2

112

2211

P

V PV

or V PV P

12



P V P V or

VP V

P

760 torr mL

1520 torr

mL

1 1 2 2

21 1

2

400

200

13


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

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


TV constant constant TV

14

Charles’ LawCharles’ Law

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

on the temperature axis is -273.15on the temperature axis is -273.15C. C. We define absolute zero, 0 K = -We define absolute zero, 0 K = -

273.15273.15C.C. Note the value of the constant reflects Note the value of the constant reflects

the assumptions: amount of gas and the assumptions: amount of gas and pressure.pressure.

15


16


volume of a gas is directly proportional volume of a gas is directly proportional to the absolute temperature at constant to the absolute temperature at constant pressurepressureK = C + 273o

17


volume of a gas is directly volume of a gas is directly proportional to the absolute proportional to the absolute temperature at constant pressuretemperature at constant pressure

K = C + 273

mathematical forms of Charles's Law

V T or V = kT or V

Tk

o

18


volume of a gas is directly volume of a gas is directly proportional to the absolute proportional to the absolute temperature at constant pressuretemperature at constant pressure

form usefulmost in the T

V

T

V

so equal are sk' ehowever thk T

V andk

T

V

kT

Vor kT = Vor TV

Law sCharles' of forms almathematic

273 + C= K

2

2

1

1

2

2

1

1

o

19


Example A sample of hydrogen, Example A sample of hydrogen, H2, occupies 100 mL at 25H2, occupies 100 mL at 2500C and C and 1.00 atm. What volume would it 1.00 atm. What volume would it occupy at 50occupy at 5000C under the same C under the same pressure?pressure?

T1 = 25 + 273 = 298 T1 = 25 + 273 = 298

T2 = 50 + 273 = 323T2 = 50 + 273 = 323

20


Example A sample of hydrogen, Example A sample of hydrogen, H2, occupies 100 mL at 25H2, occupies 100 mL at 2500C and C and 1.00 atm. What volume would it 1.00 atm. What volume would it occupy at 50occupy at 5000C under the same C under the same pressure?pressure?V

T

V

T so V =

V T

T

V =100 mL 323 K

298 K mL

1

1

2

22

1 2

1

2

108

21

Standard Temperature & Standard Temperature & PressurePressure

STPSTP– P = 1.00000 atm or 101.3 kPaP = 1.00000 atm or 101.3 kPa– T = 273 K or 0T = 273 K or 000CC

22


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

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

23

Avogadro’s LawAvogadro’s Law


V = constant V = constant nn.. We can show that 22.4 L of any gas at We can show that 22.4 L of any gas at

00C contain 6.02 C contain 6.02 10 102323 gas molecules. gas molecules.

24

The Ideal Gas EquationThe Ideal Gas Equation Summarizing the Gas LawsSummarizing the Gas Laws

Boyle: (constant n, T)Boyle: (constant n, T)

Charles: Charles: V V T T (constant n, P) (constant n, P)

Avogadro:Avogadro: V V n n (constant P, T). (constant P, T). Combined:Combined:

Ideal gas equation:Ideal gas equation:

PnT

V

PnT

RV

25

The Ideal Gas EquationThe Ideal Gas Equation

Ideal gas equation: Ideal gas equation:

PV = nRTPV = nRT.. R = gas constant = 0.08206 L•atm/mol-R = gas constant = 0.08206 L•atm/mol-

K.K. We define STP (standard temperature We define STP (standard temperature

and pressure) = 0and pressure) = 0C, 273.15 K, 1 atm.C, 273.15 K, 1 atm. Volume of 1 mol of gas at STP is 22.4 L.Volume of 1 mol of gas at STP is 22.4 L.

26

Ideal Gas LawIdeal Gas Law

R has several values depending upon the R has several values depending upon the choice of unitschoice of unitsR = 8.314 J/mol KR = 8.314 kg m2/s2 K molR = 8.314 dm3 kPa/K molR = 1.987 cal/K mol

R = PV

nT

1.00 atm L

1.00 mol K

L atm

mol K

22 4

273

0 0821

.

.

27

Ideal Gas LawIdeal Gas Law Example: What volume would 50.0 g of Example: What volume would 50.0 g of

ethane, Cethane, C22HH66, occupy at 140, occupy at 140ooC under a C under a pressure of 1820 torr?pressure of 1820 torr?

28

Example: What volume would 50.0 g of Example: What volume would 50.0 g of ethane, Cethane, C22HH66, occupy at 140, occupy at 140ooC under a C under a pressure of 1820 torr?pressure of 1820 torr?

T = 140 + 273 = 413 KP = 1820 torr (1 atm/760 torr) = 2.39 atm50 g (1 mol/30 g) = 1.67 mol


29

V = n R T

P


30

V = n R T

P

mol L atmmol K

K

torr1 atm

760 torr

L

167 0 0821 413

1820

236

. .

.


31

Example: Calculate the number of moles Example: Calculate the number of moles in, and the mass of, an 8.96 L sample of in, and the mass of, an 8.96 L sample of methane, CHmethane, CH44, measured at standard , measured at standard conditions.conditions.


32

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

.

? .


33

Example: Calculate the pressure exerted Example: Calculate the pressure exerted by 50.0 g of ethane, Cby 50.0 g of ethane, C22HH66, in a 25.0 L , in a 25.0 L container at 25container at 25ooC.C.


34

atm 63.1

L 25.0

K 298K mol

atm L0821.0mol 67.1

V

T Rn = P

K 298 = T & mol 1.67 =n :Example From


35

If If PVPV = = nRTnRT and and nn and and T T are constant, are constant, then then PV PV = constant and we have = constant and we have Boyle’s law.Boyle’s law.

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

sets of conditions, thensets of conditions, then

The Ideal Gas LawThe Ideal Gas Law

22

22

11

11TnVP

TnVP

36

Ideal Gas LawIdeal Gas Law Example: A sample of nitrogen gas, NExample: A sample of nitrogen gas, N22, ,

occupies 750 mL at 75occupies 750 mL at 7500C under a C under a pressure of 810 torr. What volume pressure of 810 torr. What volume would it occupy at standard conditions?would it occupy at standard conditions?V = 750 mL V = ?

T = 348 K T = 273 K

P = 810 torr P = 760 torr

V =P V T

P T

1 2

1 2

1 2

21 1 2

2 1

37

Example: A sample of nitrogen gas, NExample: A sample of nitrogen gas, N22, , occupies 750 mL at 75occupies 750 mL at 7500C under a C under a pressure of 810 torr. What volume pressure of 810 torr. What volume would it occupy at standard conditions?would it occupy at standard conditions?

V = 750 mL V = ?

T = 348 K T = 273 K

P = 810 torr P = 760 torr

V =P V T

P T

torr mL K

760 torr K

mL

1 2

1 2

1 2

21 1 2

2 1

810 750 273

348

627

38

Ideal Gas LawsIdeal Gas Laws

A sample of methane, CHA sample of methane, CH44, occupies 260 , occupies 260 mL at 32mL at 32ooC under a pressure of 0.500 C under a pressure of 0.500 atm. At what temperature would it atm. At what temperature would it occupy 500 mL under a pressure of 1200 occupy 500 mL under a pressure of 1200 torr?torr?

39

Ideal Gas LawsIdeal Gas Laws

V = 260 mL V = 500 mL

P = 0.500 atm = 380 torr P = 1200 torr

T = 305 K T = ?

T =T P V

P V

K 1200 torr mL

380 torr mL

= 1852 K or 1579 C

1 2

1 2

1 2

21 2 2

1 1

o

305 500

260

40

Density has units of mass over volume. Density has units of mass over volume. Rearranging the ideal-gas equation with Rearranging the ideal-gas equation with

MM as molar mass we get as molar mass we get

Molar Mass and DensityMolar Mass and Density

RTP

dV

nRTP

Vn

MM

41

The molar mass of a gas can be The molar mass of a gas can be determined as follows:determined as follows:

Molar Mass and DensityMolar Mass and Density

PdRTM

Volumes of Gases in Chemical Volumes of Gases in Chemical ReactionsReactions

The ideal-gas equation relates The ideal-gas equation relates PP, , VV, and , and TT to number of moles of gas. to number of moles of gas.

The The nn can then be used in can then be used in stoichiometric calculations.stoichiometric calculations.

42

Molar Mass and DensityMolar Mass and Density Example: One mole of a gas occupies Example: One mole of a gas occupies

36.5 L and its density is 1.36 g/L at a 36.5 L and its density is 1.36 g/L at a given T & P.given T & P.(a)(a) What is its molar mass?What is its molar mass?

(b)(b) What is its density at STP?What is its density at STP?

43



(b)(b) What is its density at STP?What is its density at STP?? . ..

gmol

Lmol

gL

g / mol 36 5 13649 6

44



(b)(b) What is its density at STP?What is its density at STP?? . ..

.

..

gmol

Lmol

gL

g / mol

? g

L

g

mol

mol

L g / L

STP

36 5 13649 6

49 6 1

22 42 21

45

Determination of Molar Mass & Determination of Molar Mass & Formulas of Gaseous CompoundsFormulas of Gaseous Compounds

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

100 g of compound contains:100 g of compound contains:

80 g of C & 20 g of H80 g of C & 20 g of H

46

H mol 8.19H g 1.01

H mol 1H g 20.0 = atoms H mol ?

C mol 67.6C g 12.0

C mol 1C g 80.0 = atoms C mol ?


47

mol 0244.0K 273

K molatm L

0821.0

L 546.0atm 00.1

RT

PV=n

15 = mass with CH 36.67

19.8is formulasimplest

H mol 8.19H g 1.01


C mol 67.6C g 12.0


3


48

molar mass = 0.732 g

0.0244 mol

g

mol

CH C H3 2 6

30 0

30 0

1502 2

.

.

.


49

Example: A 1.74 g sample of a compound Example: A 1.74 g sample of a compound that contains only carbon & hydrogen that contains only carbon & hydrogen contains 1.44 g of C and 0.300 g of H. At contains 1.44 g of C and 0.300 g of H. At STP 101 mL of the gas has a mass of STP 101 mL of the gas has a mass of 0.262 gram. What is its molecular (true) 0.262 gram. What is its molecular (true) formula?formula?


50

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


51

? g

mol

g

0.00451 mol g / mol

58.1

29C H C H2 5 4 10

0 262581

2 2

..


52

Example: A 250 mL flask contains 0.423 Example: A 250 mL flask contains 0.423 g of vapor at 100g of vapor at 10000C and 746 torr. What C and 746 torr. What is molar mass of compound?is molar mass of compound?

n =

PVRT

atm L

L atmmol K

K mol

746760

0 250

0 0821 3730 00801

.

..


53

Example: A 250 mL flask contains 0.423 Example: A 250 mL flask contains 0.423 g of vapor at 100g of vapor at 10000C and 746 torr. What C and 746 torr. What is molar mass of compound?is molar mass of compound?

n =

PV

RT

atm L

L atm

mol K K

mol

? g

mol

g

0.00801 mol g / mol

746

7600 250

0 0821 3730 00801

0 42352 8

.

..

..


54

Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures

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

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

PPtt = = PP11 + + PP22 + + PP33 + … + …

55

Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures

Each gas obeys the ideal gas equation:Each gas obeys the ideal gas equation:

Combining equations:Combining equations:

VRT

nP ii

VRT

nnnPt 321

56

Mole FractionsMole Fractions

Let Let nnii be the number of moles of gas be the number of moles of gas ii exerting a partial pressure exerting a partial pressure PPii, then, then

Pi = iPt

where where ii is the mole fraction ( is the mole fraction (nnii//nntt).).

57

It is common to synthesize gases and It is common to synthesize gases and collect them by displacing a volume of collect them by displacing a volume of water.water.

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

Ptotal = Pgas + Pwater

Collecting Gases over waterCollecting Gases over water

58

Collecting Gases over waterCollecting Gases over water

59

Dalton’s Law of Partial PressuresDalton’s Law of Partial Pressures

Example: If 100 mL of hydrogen, Example: If 100 mL of hydrogen, measured at 25measured at 2500C & 3.00 atm pressure, C & 3.00 atm pressure, and 100 mL of oxygen, measured at 25and 100 mL of oxygen, measured at 2500C C & 2.00 atm pressure, were forced into & 2.00 atm pressure, were forced into one of the containers at 25one of the containers at 2500C, what would C, what would be the pressure of the mixture of gases?be the pressure of the mixture of gases?

60

Example: If 100 mL of hydrogen, Example: If 100 mL of hydrogen, measured at 25measured at 2500C & 3.00 atm pressure, C & 3.00 atm pressure, and 100 mL of oxygen, measured at 25and 100 mL of oxygen, measured at 2500C C & 2.00 atm pressure, were forced into & 2.00 atm pressure, were forced into one of the containers at 25one of the containers at 2500C, what would C, what would be the pressure of the mixture of gases?be the pressure of the mixture of gases?

P P P

3.00 atm + 2.00 atm

= 5.00 atm

Total H O2 2

Dalton’s Law of Partial PressuresDalton’s Law of Partial Pressures

61

Vapor PressureVapor Pressure

Pressure exerted by gas over the liquid Pressure exerted by gas over the liquid at equilibrium.at equilibrium.

62

Example: A sample of hydrogen was Example: A sample of hydrogen was collected by displacement of water at collected by displacement of water at 252500C. The atmospheric pressure was 748 C. The atmospheric pressure was 748 torr. What pressure would the dry torr. What pressure would the dry hydrogen exert in the same container?hydrogen exert in the same container?


63

Example: A sample of hydrogen was Example: A sample of hydrogen was collected by displacement of water at collected by displacement of water at 252500C. The atmospheric pressure was 748 C. The atmospheric pressure was 748 torr. What pressure would the dry torr. What pressure would the dry hydrogen exert in the same container?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


64

Example: A sample of oxygen was Example: A sample of oxygen was collected by displacement of water. The collected by displacement of water. The oxygen occupied 742 mL at 27oxygen occupied 742 mL at 27ooC. The C. The barometric pressure was 753 torr. What barometric pressure was 753 torr. What volume would the dry oxygen occupy at volume would the dry oxygen occupy at STP?STP?


65

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


66

Kinetic-Molecular TheoryKinetic-Molecular Theory

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

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

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

– Intermolecular forces (forces between gas molecules) negligible.Intermolecular forces (forces between gas molecules) negligible.– Energy can be transferred between molecules, but total kinetic Energy can be transferred between molecules, but total kinetic

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

temperature.temperature.

67

Kinetic Molecular TheoryKinetic Molecular Theory

Postulate 1Postulate 1– gases are discrete molecules that are far gases are discrete molecules that are far

apartapart– have few intermolecular attractionshave few intermolecular attractions– molecular volume small compared to gas’s molecular volume small compared to gas’s

volumevolume Proof - Gases are easily compressible.Proof - Gases are easily compressible.

68

Postulate 2Postulate 2– molecules are in constant straight line molecules are in constant straight line

motionmotion– varying velocitiesvarying velocities– molecules have elastic collisionsmolecules have elastic collisions

Proof - A sealed, confined gas exhibits no Proof - A sealed, confined gas exhibits no pressure drop over time.pressure drop over time.


69

Postulate 3Postulate 3– kinetic energy is proportional to absolute Tkinetic energy is proportional to absolute T– average kinetic energies of molecules of average kinetic energies of molecules of

different gases are equal at a given Tdifferent gases are equal at a given T Proof - Motion increases as temperature Proof - Motion increases as temperature

increases.increases.


70

Boyle’s & Dalton’s LawBoyle’s & Dalton’s Law– PP 1/V as V increases collisions decrease so 1/V as V increases collisions decrease so

P dropsP drops

– PPtotaltotal = P = PAA + P + PBB + P + PCC + .....because few + .....because few intermolecular attractionsintermolecular attractions

Charles’ LawCharles’ Law– V V T increased T raises molecular velocities, T increased T raises molecular velocities,

V increases to keep P constantV increases to keep P constant


71


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

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

72


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

Gas molecules have an average kinetic Gas molecules have an average kinetic energy.energy.

Each molecule has a different energy.Each molecule has a different energy. There is a spread of individual energies of There is a spread of individual energies of

gas molecules in any sample of gas.gas molecules in any sample of gas. As the temperature increases, the average As the temperature increases, the average

kinetic energy of the gas molecules kinetic energy of the gas molecules increases.increases.

73


74


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

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

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

= ½m= ½muu22

75


As volume increases at constant temperature, As volume increases at constant temperature, the average kinetic energy of the gas the average kinetic energy of the gas remains constant. Therefore, remains constant. Therefore, uu is constant. is constant. However, volume increases so the gas However, volume increases so the gas molecules have to travel further to hit the molecules have to travel further to hit the walls of the container. Therefore, pressure walls of the container. Therefore, pressure decreases.decreases.

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

76

Effusion & DiffusionEffusion & Diffusion

diffusion - intermingling of gasesdiffusion - intermingling of gases effusion - gases pass through porous effusion - gases pass through porous

containerscontainers rates are inversely proportional to square rates are inversely proportional to square

roots of molecular weights or densitiesroots of molecular weights or densities

77

effusion - gases pass through porous containerseffusion - gases pass through porous containers– Effusion is the escape of a gas through a tiny hole (a Effusion is the escape of a gas through a tiny hole (a

balloon will deflate over time due to effusion).balloon will deflate over time due to effusion). rates are inversely proportional to square roots rates are inversely proportional to square roots

of molecular weights or densitiesof molecular weights or densities

R

R

M

M or

R

R

D

D

1

2

2

1

1

2

2

1


78

Diffusion of a gas is the spread of the gas Diffusion of a gas is the spread of the gas through space.through space.

Diffusion is faster for light gas molecules.Diffusion is faster for light gas molecules. Diffusion is significantly slower than rms Diffusion is significantly slower than rms

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

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

Average distance of a gas molecule between Average distance of a gas molecule between collisions is called mean free path.collisions is called mean free path.


79

At sea level, mean free path is about 6 At sea level, mean free path is about 6 1010-6-6 cm. cm.


80

Calculate the ratio of the rate of effusion of He to Calculate the ratio of the rate of effusion of He to that of sulfur dioxide, SOthat of sulfur dioxide, SO22, at the same T & P., at the same T & P.


81

Calculate the ratio of the rate of effusion of He to Calculate the ratio of the rate of effusion of He to that of sulfur dioxide, SOthat of sulfur dioxide, SO22, at the same T & P., at the same T & P.

R

R

M

M

g / mol

4.0 g / mol

R R

He

SO

SO

He

He SO

2

2

2

641

16 4 4

.


82

Example: A sample of hydrogen, HExample: A sample of hydrogen, H22, was , was found to effuse through a pinhole 5.2 found to effuse through a pinhole 5.2 times as rapidly as the same volume of times as rapidly as the same volume of unkown gas ( at the same T & P). What unkown gas ( at the same T & P). What is the molecular weight of the unknown is the molecular weight of the unknown gas?gas?


83

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


84

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

Average kinetic energy of a gas is related to Average kinetic energy of a gas is related to its mass:its mass:

= ½mu2

Average SpeedAverage Speed

85

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


The lower the molar mass, The lower the molar mass, MM, the higher the , the higher the rms for that gas at a constant temperature.rms for that gas at a constant temperature.

MRT

u3


86


87

Real GasesReal Gases

real gases behave ideally at ordinary T’s real gases behave ideally at ordinary T’s & P’s& P’s

at low T and high P - large deviationsat low T and high P - large deviations– molecules are on top of one anothermolecules are on top of one another– molecular interactions become importantmolecular interactions become important

88

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

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

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

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

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


89


90


91

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

Therefore, the less the gas resembles and Therefore, the less the gas resembles and ideal gas.ideal gas.

As temperature increases, the gas molecules As temperature increases, the gas molecules move faster and further apart.move faster and further apart.

Also, higher temperatures mean more energy Also, higher temperatures mean more energy available to break intermolecular forces.available to break intermolecular forces.

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


92


93

van der Waals’s equationvan der Waals’s equation

a & b are van der Waal’s constantsa & b are van der Waal’s constants

different values for each gasdifferent values for each gas

P + n a

VV nb nRT

2

2


94

van der Waals’s equationvan der Waals’s equation– a accounts for intermolecular attractionsa accounts for intermolecular attractions– b accounts for volume of gas moleculesb accounts for volume of gas molecules– large V’s a & b reduce to small quantitieslarge V’s a & b reduce to small quantities

van der Waal’s equation reduces to van der Waal’s equation reduces to ideal gas law at high temperatures and ideal gas law at high temperatures and low pressureslow pressures


95

attractive forces in nonpolar gases attractive forces in nonpolar gases – London ForcesLondon Forces

• noble gases- instantaneous induced dipole noble gases- instantaneous induced dipole momentsmoments

attractive forces in polar gases attractive forces in polar gases – dipole-dipole or H-bondsdipole-dipole or H-bonds

• water molecules can H-bondwater molecules can H-bond• why ice floats why ice floats


96


97

Example: Calculate the pressure exerted Example: Calculate the pressure exerted by 84.0 g of ammonia, NHby 84.0 g of ammonia, NH33, in a 5.00 L at , in a 5.00 L at 20020000C using the ideal gas law.C using the ideal gas law.


98

n = 84.0 g NH mol

17.0 g mol

P = nRTV

molL atmmol K

K

5.00 L atm

3

14 94

4 94 0 0821 473

38 4

.

. .

.


99

Example: Solve previous example using Example: Solve previous example using the van der Waal’s equation.the van der Waal’s equation.


100

Example: Solve previous example using Example: Solve previous example using the van der Waal’s equation.the van der Waal’s equation.

n = 4.94 mol a = 4.17 L atm

mol b = 0.0371L

mol

P + n a

VV - nb nRT

P =nRT

V - nbn a

V

2

2

2

2

2

2


101

P

P

P atm

a 7.6% difference

4 94 0 0821 473

5 00 4 94 0 0371

4 94 417

5 00

19184 817

4 07 39 8 41

35 7

2

2

. .

. ( . )( . )

. .

.

..

. ( . . )

.


102

We can’t forget about moles!We can’t forget about moles!

Example: What volume of oxygen Example: What volume of oxygen measured at STP, can be produced by measured at STP, can be produced by the thermal decomposition of 120 g of the thermal decomposition of 120 g of KClOKClO33??

103

Mass-Volume Relationships in Mass-Volume Relationships in Reactions Involving GasesReactions Involving Gases

2 mol 2 mol 2 mol 3 mol2 mol 3 mol

2(122.6g) 2(122.6g) 2(74.6g) 3(32g)2(74.6g) 3(32g) 3 moles of oxygen 3 moles of oxygen

equal to 3(22.4L) equal to 3(22.4L) oror

67.2 L at STP67.2 L at STP

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

22 & (

104

? L O 120 g KClO mol KClO

122.6 g KClO

mol O

2 mol KClO

L O

mol O L O

STP 2 33

3

2

3

STP 2

2STP 2

1 3

22 4

132 9

..

Mass-Volume Relationships in Mass-Volume Relationships in Reactions Involving GasesReactions Involving Gases

105

Synthesis QuestionSynthesis Question

The lethal dose for hydrogen sulfide is 6 ppm. The lethal dose for hydrogen sulfide is 6 ppm. In other words, if in 1 million molecules of air In other words, if in 1 million molecules of air there are six hydrogen sulfide molecules then there are six hydrogen sulfide molecules then that air would be deadly to breathe. How many that air would be deadly to breathe. How many hydrogen sulfide molecules would be required hydrogen sulfide molecules would be required to reach the lethal dose in a room that is 77 to reach the lethal dose in a room that is 77 feet long, 62 feet wide and 50 feet tall at 1.0 feet long, 62 feet wide and 50 feet tall at 1.0 atm and 25.0atm and 25.0oo C? C?

106

363

39

39

cm 106.76cm 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


107

S Hof molecules 1066.1air of molecules 10

S Hof molecule 1air of molecules 1066.1 dose Lethal

air of molecules 1066.1mol

molecules 106.022mol 276,000

air of mol 276,000 K2980.0821

L106.76atm 1

RT

PV n

K298 25 273 T

L106.76cm 1000

L1cm 106.76

223

6229

2923

Kmolatm L

6

63

39


108

Group QuestionGroup Question

Tires on a car are typically filled to a Tires on a car are typically filled to a pressure of 35 psi at 300K. A tire is 16 pressure of 35 psi at 300K. A tire is 16 inches in radius and 8.0 inches in inches in radius and 8.0 inches in thickness. The wheel that the tire is thickness. The wheel that the tire is mounted on is 6.0 inches in radius. What mounted on is 6.0 inches in radius. What is the mass of the air in a tire?is the mass of the air in a tire?

gases

Documents