1 chapter 5 the gas laws. 2 5.1 pressure n force per unit area. n gas molecules fill container. n...

53
Chapter 5 Chapter 5 The Gas Laws The Gas Laws

Upload: wesley-moody

Post on 26-Dec-2015

215 views

Category:

Documents


0 download

TRANSCRIPT

Page 1: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

1

Chapter 5Chapter 5

The Gas LawsThe Gas Laws

Page 2: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

2

5.1 Pressure5.1 Pressure Force per unit area.Force per unit area. Gas molecules fill container.Gas molecules fill container. Molecules move around and hit Molecules move around and hit

sides.sides. Collisions are the force.Collisions are the force. Container has the area.Container has the area. Measured with a barometer.Measured with a barometer.

Page 3: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

3

BarometerBarometer The pressure of the The pressure of the

atmosphere at sea atmosphere at sea level will hold a level will hold a column of mercury column of mercury 760 mm Hg.760 mm Hg.

1 atm = 760 mm Hg1 atm = 760 mm Hg

1 atm Pressure

760 mm Hg

Vacuum

Page 4: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

4

Units of pressureUnits of pressure 1 atmosphere = 760 mm Hg1 atmosphere = 760 mm Hg 1 mm Hg = 1 torr1 mm Hg = 1 torr 1 atm = 101,325 Pascals = 101.325 kPa1 atm = 101,325 Pascals = 101.325 kPa Can make conversion factors from Can make conversion factors from

these.these. What is 724 mm Hg in kPa?What is 724 mm Hg in kPa? in torr?in torr? in atm?in atm?

Page 5: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

5

5.2 The Gas Laws5.2 The Gas Laws Boyle’s LawBoyle’s Law Pressure and volume are inversely related Pressure and volume are inversely related

at constant temperature.at constant temperature. PV= kPV= k As one goes up, the other goes down.As one goes up, the other goes down. PP11VV11 = P = P22 V V22 Holds Holds preciselyprecisely only at very low only at very low

pressures.pressures. A gas that A gas that strictly obeysstrictly obeys Boyle’s Law is Boyle’s Law is

called an called an ideal gasideal gas..

Page 6: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

6

(YDVD)(YDVD)

QuickTime™ and aSorenson Video decompressorare needed to see this picture.

Page 7: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

7

V

P (at constant T)

Page 8: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

8

ExamplesExamples 20.5 L of nitrogen at 25ºC and 742 20.5 L of nitrogen at 25ºC and 742

torr are compressed to 9.8 atm at torr are compressed to 9.8 atm at constant T. What is the new volume?constant T. What is the new volume?

30.6 mL of carbon dioxide at 740 torr 30.6 mL of carbon dioxide at 740 torr is expanded at constant temperature is expanded at constant temperature to 750 mL. What is the final pressure to 750 mL. What is the final pressure in kPa? in kPa?

Page 9: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

9

Charle’s LawCharle’s Law Volume of a gas varies directly with Volume of a gas varies directly with

the absolute temperature at constant the absolute temperature at constant pressure.pressure.

V = kT (if T is in Kelvin)V = kT (if T is in Kelvin)

VV1 1 = V = V22

T T11 = T = T22

Page 10: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

10

(YDVD)(YDVD)

QuickTime™ and aCinepak Codec by Radius decompressor

are needed to see this picture.

Page 11: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

11

V (

L)

T (ºC)

He

H2O

CH4

H2

-273.15ºC

Page 12: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

12

ExamplesExamples What would the final volume be if 247 What would the final volume be if 247

mL of gas at 22ºC is heated to 98ºC , mL of gas at 22ºC is heated to 98ºC , if the pressure is held constant?if the pressure is held constant?

At what temperature would 40.5 L of At what temperature would 40.5 L of gas at 23.4ºC have a volume of 81.0 gas at 23.4ºC have a volume of 81.0 L at constant pressure?L at constant pressure?

Page 13: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

13

Avogadro's LawAvogadro's Law Avagadro’sAvagadro’s At constant temperature and At constant temperature and

pressure, the volume of gas is directly pressure, the volume of gas is directly related to the number of moles.related to the number of moles.

V = k n (n is the number of moles)V = k n (n is the number of moles)

VV1 1 = V = V22

n n11 = n = n22

Page 14: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

14

Gay- Lussac LawGay- Lussac Law At constant volume, pressure and At constant volume, pressure and

absolute temperature are directly absolute temperature are directly related.related.

P = k TP = k T

PP1 1 = P = P22

T T11 = T = T22

Page 15: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

15

Combined Gas LawCombined Gas Law If the moles of gas remains constant, If the moles of gas remains constant,

use this formula and cancel out the use this formula and cancel out the other things that don’t change.other things that don’t change.

PP1 1 VV11 = P = P22 V V22

.. T T11 T T22

Page 16: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

16

ExamplesExamples A deodorant can has a volume of 175 A deodorant can has a volume of 175

mL and a pressure of 3.8 atm at 22ºC. mL and a pressure of 3.8 atm at 22ºC. What would the pressure be if the What would the pressure be if the can was heated to 100.ºC?can was heated to 100.ºC?

What volume of gas could the can What volume of gas could the can release at 22ºC and 743 torr?release at 22ºC and 743 torr?

Page 17: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

17

5.3 Ideal Gas Law5.3 Ideal Gas Law PV = nRTPV = nRT V = 22.41 L at 1 atm, 0ºC, n = 1 mole, V = 22.41 L at 1 atm, 0ºC, n = 1 mole,

what is R?what is R? R is the ideal gas constant.R is the ideal gas constant. R = 0.08206 L•atm/mol•K, (8.314 Kpa)R = 0.08206 L•atm/mol•K, (8.314 Kpa) The state of the gas now. The other The state of the gas now. The other

laws tell you about a gas when it laws tell you about a gas when it changes. changes.

(BDVD-ana)(BDVD-ana)

Page 18: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

18

Ideal Gas LawIdeal Gas Law An An equation of stateequation of state.. Independent of how you end up Independent of how you end up

where you are at. Does not depend where you are at. Does not depend on the path.on the path.

Given 3 you can determine the Given 3 you can determine the fourth.fourth.

An Empirical Equation - based on An Empirical Equation - based on experimental evidence.experimental evidence.

Page 19: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

19

Ideal Gas LawIdeal Gas Law A hypothetical substance - the ideal A hypothetical substance - the ideal

gas.gas. Gases only approach ideal behavior at Gases only approach ideal behavior at

low pressure (< 1 atm) and high low pressure (< 1 atm) and high temperature.temperature.

Low temperatures and high pressures Low temperatures and high pressures cause gases to deviate from ideal.cause gases to deviate from ideal.

Use the laws anyway, unless told to do Use the laws anyway, unless told to do otherwise. They give good estimates.otherwise. They give good estimates.

Page 20: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

20

ExamplesExamples A 47.3 L container containing 1.62 mol of A 47.3 L container containing 1.62 mol of

He is heated until the pressure reaches He is heated until the pressure reaches 1.85 atm. What is the temperature?1.85 atm. What is the temperature?

Kr gas in a 18.5 L cylinder exerts a Kr gas in a 18.5 L cylinder exerts a pressure of 8.61 atm at 24.8ºC What is pressure of 8.61 atm at 24.8ºC What is the mass of Kr?the mass of Kr?

A sample of gas has a volume of 4.18 L A sample of gas has a volume of 4.18 L at 29ºC and 732 torr. What would its at 29ºC and 732 torr. What would its volume be at 24.8ºC and 756 torr?volume be at 24.8ºC and 756 torr?

Page 21: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

21

Gas Density and Molar MassGas Density and Molar Mass D = m/VD = m/V Let Let MM stand for molar mass stand for molar mass MM = m/n = m/n n= PV/RTn= PV/RT MM = m = m

PV/RT PV/RT MM = mRT = m RT = DRT = mRT = m RT = DRT

PV PV V P V P P P

Page 22: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

22

Examples Examples What is the density of ammonia at What is the density of ammonia at

23ºC and 735 torr?23ºC and 735 torr? A compound has the empirical A compound has the empirical

formula CHCl. A 256 mL flask at formula CHCl. A 256 mL flask at 100.ºC and 750 torr contains 0.80 g of 100.ºC and 750 torr contains 0.80 g of the gaseous compound. What is the the gaseous compound. What is the molecular formula?molecular formula?

Page 23: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

23

5.4 Gas Stoichiometry5.4 Gas Stoichiometry Reactions happen in molesReactions happen in moles At Standard Temperature and At Standard Temperature and

Pressure (STP, 0ºC and 1 atm) 1 Pressure (STP, 0ºC and 1 atm) 1 mole of gas occupies 22.42 L.mole of gas occupies 22.42 L.

If not at STP, use the ideal gas law If not at STP, use the ideal gas law (PV=nRT) to calculate moles of (PV=nRT) to calculate moles of reactant or volume of product.reactant or volume of product.

Page 24: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

24

ExamplesExamples Mercury can be produced by the Mercury can be produced by the

following reactionfollowing reaction

What volume of oxygen gas can What volume of oxygen gas can be produced from 4.10 g of mercury be produced from 4.10 g of mercury (II) oxide at STP?(II) oxide at STP?

At 400.ºC and 740 torr?At 400.ºC and 740 torr?

→HgO Hg(l) + O (g) heat

2

Page 25: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

25

ExamplesExamples Using the following reactionUsing the following reaction

calculate the mass of sodium hydrogen calculate the mass of sodium hydrogen carbonate necessary to produce 2.87 L carbonate necessary to produce 2.87 L of carbon dioxide at 25ºC and 2.00 atm.of carbon dioxide at 25ºC and 2.00 atm.

If 27 L of gas are produced at 26ºC and If 27 L of gas are produced at 26ºC and 745 torr when 2.6 L of HCl are added, 745 torr when 2.6 L of HCl are added, what is the concentration of HCl?what is the concentration of HCl?

NaCl(aq) + CO (g) +H O(l)2 2

NaHCO3(s) + HCl(aq) →

Page 26: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

26

ExamplesExamples Consider the following reactionConsider the following reaction

What volume of NO at 1.0 atm and What volume of NO at 1.0 atm and 1000ºC can be produced from 10.0 L 1000ºC can be produced from 10.0 L of NHof NH33 and excess O and excess O22 at the same at the same

temperature and pressure?temperature and pressure? What volume of OWhat volume of O22 measured at STP measured at STP

will be consumed when 10.0 kg NHwill be consumed when 10.0 kg NH33 is is

reacted?reacted?

4NH (g) + 5 O 4 NO(g) + 6H O(g)3 22 ( )g →

Page 27: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

27

The Same reactionThe Same reaction

What mass of HWhat mass of H22O will be produced O will be produced

from 65.0 L of Ofrom 65.0 L of O22 and 75.0 L of NH and 75.0 L of NH33

both measured at STP? both measured at STP? What volume Of NO would be What volume Of NO would be

produced?produced? What mass of NO is produced from What mass of NO is produced from

500. L of NH3 at 250.0ºC and 3.00 atm?500. L of NH3 at 250.0ºC and 3.00 atm?

4NH (g) + 5 O 4 NO(g) + 6H O(g)3 22 ( )g →

Page 28: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

28

5.5 Dalton’s Law of Partial Pressures5.5 Dalton’s Law of Partial Pressures

The total pressure in a container is The total pressure in a container is the sum of the pressure each gas the sum of the pressure each gas would exert if it were alone in the would exert if it were alone in the container.container.

The total pressure is the sum of the The total pressure is the sum of the partial pressures.partial pressures.

PPTotalTotal = P = P11 + P + P22 + P + P33 + P + P44 + P + P55 ... ...

For each P = nRT/VFor each P = nRT/V

Page 29: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

29

Dalton's LawDalton's Law PPTotalTotal = n = n11RT + nRT + n22RT + nRT + n33RT +...RT +...

V V V V V V In the same container R, T and V are the In the same container R, T and V are the

same.same.

PPTotalTotal = (n = (n11+ n+ n22 + n + n33+...)RT+...)RT

V V

PPTotalTotal = (n = (nTotalTotal)RT)RT

V V

Page 30: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

30

The mole fractionThe mole fraction Ratio of moles of the substance to Ratio of moles of the substance to

the total moles.the total moles.

symbol is Greek letter chi symbol is Greek letter chi

= n= n11 = P= P1 1

n nTotal Total PPTotalTotal

Page 31: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

31

ExamplesExamples The partial pressure of nitrogen in air The partial pressure of nitrogen in air

is 592 torr. Air pressure is 752 torr, is 592 torr. Air pressure is 752 torr, what is the mole fraction of what is the mole fraction of nitrogen?nitrogen?

What is the partial pressure of What is the partial pressure of nitrogen if the container holding the nitrogen if the container holding the air is compressed to 5.25 atm?air is compressed to 5.25 atm?

Page 32: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

32

ExamplesExamples

3.50 L

O2

1.50 L

N2

2.70 atm When these valves are opened, what is When these valves are opened, what is

each partial pressure and the total each partial pressure and the total pressure?pressure?

4.00 L

CH4

4.58 atm 0.752 atm

Page 33: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

33

Vapor PressureVapor Pressure Water evaporates!Water evaporates! When that water evaporates, the When that water evaporates, the

vapor has a pressure.vapor has a pressure. Gases are often collected over water Gases are often collected over water

so the vapor pressure of water must so the vapor pressure of water must be subtracted from the total be subtracted from the total pressure.pressure.

It must be given.It must be given.

Page 34: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

34

ExampleExample NN22O can be produced by the O can be produced by the

following reactionfollowing reaction

what volume of N what volume of N22O collected O collected

over water at a total pressure of 94 over water at a total pressure of 94 kPa and 22ºC can be produced from kPa and 22ºC can be produced from 2.6 g of NH2.6 g of NH44NONO33? ( the vapor ? ( the vapor

pressure of water at 22ºC is 21 torr)pressure of water at 22ºC is 21 torr)€

NH4NO3(s) heat ⏐ → ⏐ ⏐ N2O (g) + 2H2O(l)

Page 35: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

35

ExampleExample NN22O can be produced by the O can be produced by the

following reactionfollowing reaction

what volume of N what volume of N22O collected O collected

over water at a total pressure of 94 over water at a total pressure of 94 kPa and 22ºC can be produced from kPa and 22ºC can be produced from 2.6 g of NH2.6 g of NH44NONO33? ( the vapor ? ( the vapor

pressure of water at 22ºC is 21 torr)pressure of water at 22ºC is 21 torr)€

NH4NO3(s) heat ⏐ → ⏐ ⏐ N2O (g) + 2H2O(l)

Ans: 0.877 L

Page 36: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

36

5.6 Kinetic Molecular Theory5.6 Kinetic Molecular Theory Theory tells why the things happen.Theory tells why the things happen. Explains why ideal gases behave the Explains why ideal gases behave the

way they do.way they do. Assumptions simplify the theory, but Assumptions simplify the theory, but

don’t work in real gases.don’t work in real gases. The particles are so small we can The particles are so small we can

ignore their volume.ignore their volume. The particles are in constant motion The particles are in constant motion

and their collisions cause pressure. and their collisions cause pressure.

Page 37: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

37

(YDVD) Mol motion(YDVD) Mol motion

QuickTime™ and aCinepak Codec by Radius decompressor

are needed to see this picture.

Page 38: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

38

Kinetic Molecular TheoryKinetic Molecular Theory The particles do not affect each other, The particles do not affect each other,

neither attracting or repelling.neither attracting or repelling. The average kinetic energy is proportional The average kinetic energy is proportional

to the Kelvin temperature.to the Kelvin temperature. Appendix 2 shows the derivation of the Appendix 2 shows the derivation of the

ideal gas law and the definition of ideal gas law and the definition of temperature.temperature.

We need the formula KE = 1/2 mvWe need the formula KE = 1/2 mv22

P.216 shows how PV/n=2/3(KE)P.216 shows how PV/n=2/3(KE)

Page 39: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

39

What it tells usWhat it tells us (KE)(KE)avgavg = 3/2 RT = 3/2 RT

This the meaning of temperature. (Higher This the meaning of temperature. (Higher TT means greater motion.) means greater motion.)

uu22 is the average particle velocity squared. is the average particle velocity squared. the root mean square velocity is the root mean square velocity is

u u 2 = 2 = uurmsrms Velocity is dependant on mass and Velocity is dependant on mass and

temperature. temperature. (BDVD-ana)(BDVD-ana)

Page 40: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

40

Combine these two equationsCombine these two equations (KE)(KE)avgavg = N = NAA(1/2 mu (1/2 mu 22 ) )

(KE)(KE)avgavg = 3/2 RT = 3/2 RT

Page 41: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

41

Combine these two equationsCombine these two equations (KE)(KE)avgavg = N = NAA(1/2 mu (1/2 mu 22 ) )

(KE)(KE)avgavg = 3/2 RT = 3/2 RT

Where M is the molar mass in Where M is the molar mass in kg/mole, and R has the units 8.3145 kg/mole, and R has the units 8.3145 J/Kmol.J/Kmol.

The velocity will be in m/sThe velocity will be in m/s

u = 3RT

Mrms

Page 42: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

42

Example Example Calculate the root mean square Calculate the root mean square

velocity of carbon dioxide at 25ºC.velocity of carbon dioxide at 25ºC. Calculate the root mean square Calculate the root mean square

velocity of hydrogen at 25ºC.velocity of hydrogen at 25ºC. Calculate the root mean square Calculate the root mean square

velocity of chlorine at 25ºC.velocity of chlorine at 25ºC.

Page 43: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

43

5.7 Effusion5.7 Effusion Passage of gas through a small hole, Passage of gas through a small hole,

into an evacuated chamber.into an evacuated chamber. The effusion rate measures how fast The effusion rate measures how fast

this happens.this happens. Graham’s Law - the rate of effusion Graham’s Law - the rate of effusion

is inversely proportional to the is inversely proportional to the square root of the mass of its square root of the mass of its particles.particles.

Rate of effusion for gas 1

Rate of effusion for gas 2=

MM

Page 44: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

44

EffusionEffusion (YDVD)(YDVD)

QuickTime™ and aAnimation decompressor

are needed to see this picture.

Page 45: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

45

DiffusionDiffusion The spreading of a gas through a The spreading of a gas through a

room.room. Slow considering molecules move at Slow considering molecules move at

100’s of meters per second.100’s of meters per second. Collisions with other molecules slow Collisions with other molecules slow

down diffusions.down diffusions. Best estimate is Graham’s Law.Best estimate is Graham’s Law.

Page 46: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

46

DiffusionDiffusion (YDVD)(YDVD)

QuickTime™ and aCinepak Codec by Radius decompressor

are needed to see this picture.

Page 47: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

47

ExamplesExamples Helium effuses through a porous cylinder Helium effuses through a porous cylinder

5.05 times faster than a compound . What is 5.05 times faster than a compound . What is it’s molar mass?it’s molar mass?

If 0.00251 mol of NHIf 0.00251 mol of NH33 effuse through a hole in effuse through a hole in

2.47 min, how much HCl would effuse in the 2.47 min, how much HCl would effuse in the same time?same time?

A sample of NA sample of N22 effuses through a hole in 38 effuses through a hole in 38

seconds. what must be the molecular weight seconds. what must be the molecular weight of gas that effuses in 55 seconds under of gas that effuses in 55 seconds under identical conditions? identical conditions?

Page 48: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

48

5.8 Real Gases5.8 Real Gases Real molecules do take up space and Real molecules do take up space and

they do interact with each other they do interact with each other (especially polar molecules).(especially polar molecules).

Need to add correction factors to the Need to add correction factors to the ideal gas law to account for these.ideal gas law to account for these.

Also, we must correct ideal gas Also, we must correct ideal gas behavior when at behavior when at high pressurehigh pressure (smaller volume) and (smaller volume) and low low temperaturetemperature (attractive forces (attractive forces become important).become important).

Page 49: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

49

Volume CorrectionVolume Correction The actual volume free to move in is less The actual volume free to move in is less

because of particle size.because of particle size. More molecules will have more effect.More molecules will have more effect. Corrected volume V’ = V - nbCorrected volume V’ = V - nb b is a constant that differs for each gas.b is a constant that differs for each gas.

P’ = nRTP’ = nRT

(V-nb) (V-nb)

Page 50: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

50

Pressure correctionPressure correction Because the molecules are attracted Because the molecules are attracted

to each other, the pressure on the to each other, the pressure on the container will be less than idealcontainer will be less than ideal

Depends on the number of Depends on the number of molecules per liter.molecules per liter.

Since two molecules interact, the Since two molecules interact, the effect must be squared.effect must be squared.

Pobserved = P’ - a

2

( )Vn

Page 51: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

51

AltogetherAltogether PPobsobs= nRT - a n = nRT - a n 22

V-nb VV-nb V

Called the Van der Wall’s equation if Called the Van der Wall’s equation if

rearrangedrearranged

Corrected Corrected Corrected Corrected Pressure Pressure Volume Volume

( )

( )P + an

V x V - nb nRTobs

⎛⎝⎜

⎞⎠⎟

⎣⎢⎢

⎦⎥⎥

=2

Page 52: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

52

Where does it come fromWhere does it come from a and b are determined by a and b are determined by

experiment.experiment. Different for each gas.Different for each gas. Bigger molecules have larger b.Bigger molecules have larger b. a depends on both size and polarity.a depends on both size and polarity. Once given, plug and chug.Once given, plug and chug.

Page 53: 1 Chapter 5 The Gas Laws. 2 5.1 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are

53

ExampleExample Calculate the pressure exerted by Calculate the pressure exerted by

0.5000 mol Cl0.5000 mol Cl22 in a 1.000 L container in a 1.000 L container

at 25.0ºCat 25.0ºC Using the ideal gas law.Using the ideal gas law. Van der Waal’s equationVan der Waal’s equation

– a = 6.49 atm La = 6.49 atm L22 /mol /mol22

– b = 0.0562 L/molb = 0.0562 L/mol