General Chemistry

1. Chapter 5Chapter 5
The Gas LawsThe Gas Lawspppp

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 hitMolecules 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.

3. BarometerBarometer
The pressure of theThe pressure of the
atmosphere at seaatmosphere at sea
level will hold alevel will hold a
column of mercurycolumn of mercury
760 mm Hg.760 mm Hg.
1 atm = 760 mm Hg1 atm = 760 mm Hg
1 atm
Pressure
760
mm
Hg
Vacuum

4. ManometerManometer
Gas
h
Column ofColumn of
mercury tomercury to
measuremeasure
pressure.pressure.
h is how muchh is how much
lower thelower the
pressure ispressure is
than outside.than outside.

5. ManometerManometer
h is how muchh is how much
higher the gashigher the gas
pressure is thanpressure is than
the atmosphere.the atmosphere.
h
Gas

6. 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 these.Can make conversion factors from these.
What is 724 mm Hg in kPa? . . .What is 724 mm Hg in kPa? . . .
96.5 kPa96.5 kPa
In torr?In torr?
724 torr724 torr
In atm?In atm?
0.952 atm.0.952 atm.

7. 5.2 The Gas Laws of Boyle, Charles, and Avogadro5.2 The Gas Laws of Boyle, Charles, and Avogadro
Boyle’s LawBoyle’s Law
Pressure and volume are inverselyPressure and volume are inversely
related at constant temperature.related at constant temperature.
PV= kPV= k
As one goes up, the other goesAs one goes up, the other goes
down.down.
PP11VV11 = P= P22 VV22
GraphicallyGraphically

8. 8
1 atm
4 Liters
As the
pressure on
a gas
increases

9. 9
2 atm
2 Liters
As the
pressure on
a gas
increases
the volume
decreases
Pressure and
volume are
inversely
related

10. 10
A Boyle’s Law Relationship

11. 1
V
P (at constant T)

12. 2
V
1/P (at constant T)
Slope = k

13. 3
PV
P (at constant T)
CO2
O2
22.41Latm Ne

14. 4
ExampleExamplepppp
20.5 L20.5 L of nitrogen at 25.0ºC andof nitrogen at 25.0ºC and 742 torr742 torr areare
compressed tocompressed to 9.80 atm9.80 atm at constant T.at constant T.
What is theWhat is the new volumenew volume? Steps . . .? Steps . . .
PP11VV11 == PP22 VV22
(20.5 L)(742 torr)(20.5 L)(742 torr) == (9.80 atm x 760 torr/atm)(x)(9.80 atm x 760 torr/atm)(x)
x = 2.04 Lx = 2.04 L

15. 5
You try it!You try it!
30.6 mL of CO2 at 740 torr is expanded at
constant temperature to 750 mL. What is
the final pressure in kPa? Ans. . .
4.02 kPa. Steps . . .
PP11VV11 == PP22 VV22
((30.6 mL)(740 torr)30.6 mL)(740 torr) ==(750 mL) (x torr)(750 mL) (x torr)
x = 30.2 torrx = 30.2 torr
(30 torr)(101.325 kPa/ 760 torr) =(30 torr)(101.325 kPa/ 760 torr) = 4.02 kPa4.02 kPa

16. 6
Charles’ LawCharles’ Law
Volume of a gas varies directly withVolume of a gas varies directly with
the absolute temperature at constantthe absolute temperature at constant
pressure.pressure.
V = kT (if T is in Kelvin)V = kT (if T is in Kelvin)
VV11 = V= V22
TT11 = T= T22
GraphicallyGraphically

17. 17
Tr 51A Charles’ Law Relationship

18. 8
V(L)
T (ºC)
He
H2O
CH4
H2
-273.15ºC

19. 9
ExampleExamplepppp
What would the finalWhat would the final volumevolume be ifbe if 247 mL247 mL ofof
gas atgas at 22ºC22ºC is heated tois heated to 98ºC98ºC , if the, if the
pressure is held constantpressure is held constant? Steps.? Steps.
VV11/T/T11 == VV22/T/T22
(247 mL)/(22 + 273) K(247 mL)/(22 + 273) K == (x mL)/(273 + 98) K(x mL)/(273 + 98) K
x = 311 mlx = 311 ml

20. 0
Examples (do this)Examples (do this)
At what temperature would 40.5 L of gas atAt what temperature would 40.5 L of gas at
23.4ºC have a volume of 81.0 L at constant23.4ºC have a volume of 81.0 L at constant
pressure?pressure?
a) 173 ºC b) 280 ºC c) 320 ºC d) 593 ºCa) 173 ºC b) 280 ºC c) 320 ºC d) 593 ºC
320. ºC320. ºC (change K to ºC!)(change K to ºC!)

21. 1
Avogadro's LawAvogadro's Law
At constant temperature andAt constant temperature and
pressure, the volume of gas ispressure, the volume of gas is
directly related to the number ofdirectly related to the number of
moles.moles.
V = k n (n is the number of moles)V = k n (n is the number of moles)
VV11 = V= V22
nn11 = n= n22

22. 2
Figure 5.10 p. 195Figure 5.10 p. 195
Balloons HoldingBalloons Holding 1.0 L1.0 L
of Gas at 25º C and 1of Gas at 25º C and 1
atmatm. Each contains. Each contains
0.041 mol of gas, or 2.50.041 mol of gas, or 2.5
x 10x 102222
molecules,molecules, eveneven
thoughthough theirtheir massesmasses areare
different (different (equalequal
numbers, differentnumbers, different
massesmasses).).

23. 3
Gay- Lussac LawGay- Lussac Law
At constant volume, pressure andAt constant volume, pressure and
absolute temperature are directlyabsolute temperature are directly
related.related.
P = k TP = k T
PP11 = P= P22
TT11 = T= T22

24. 24
Tr 52A Gay-
Lussac’s Law
Relationship

25. 5
Combined Gas LawCombined Gas Lawpppp
If the moles of gasIf the moles of gas remain constantremain constant,,
use this formula and cancel out theuse this formula and cancel out the
other things that don’t change.other things that don’t change.
PP11 VV11 = P= P22 VV22
.. TT11 TT22

26. 6
ExamplesExamplespppp
A deodorant can has volume of 175 mL &A deodorant can has volume of 175 mL &
pressure of 3.8 atm at 22ºC. What pressurepressure of 3.8 atm at 22ºC. What pressure
if heated to 100.ºC?if heated to 100.ºC?
V is constant, so (PV is constant, so (P11VV11)/T)/T11 = (P= (P22VV22/T/T22))
(3.8)/(295) = (x)/373(3.8)/(295) = (x)/373
x = 4.8 atmx = 4.8 atm

27. 7
ExamplesExamples
A deodorant can has volume ofA deodorant can has volume of 175 mL175 mL &&
pressure ofpressure of 3.8 atm3.8 atm at 22ºC.at 22ºC. What volumeWhat volume
of gas could the can release atof gas could the can release at 22ºC22ºC andand
743743 torr?torr? Answer . . .Answer . . .
680 mL680 mL Steps. . .Steps. . .
((3.83.8)()(175175)/(273 + 22) = ()/(273 + 22) = (743/760743/760)(x)/()(x)/(273 + 22273 + 22))

28. 8
5.3 Ideal Gas Law5.3 Ideal Gas Lawpppp
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 KR = 0.08206 L atm/ mol K
R also = 62.396 L torr/k mol orR also = 62.396 L torr/k mol or
8.3145 J/k mol8.3145 J/k mol
Tells you about a gas isTells you about a gas is NOWNOW..
TheThe otherother laws tell you about a gaslaws tell you about a gas
when itwhen it changeschanges ((exam hint!exam hint!))..

29. 9
Ideal Gas LawIdeal Gas Law
AnAn equation of stateequation of state..
Independent of how you end upIndependent of how you end up
where you are at. Does not dependwhere you are at. Does not depend
on the path.on the path.
Given 3 measurements you canGiven 3 measurements you can
determine the fourth.determine the fourth.
An Empirical Equation - based onAn Empirical Equation - based on
experimentalexperimental evidence.evidence.

30. 0
Ideal Gas LawIdeal Gas Law
A hypothetical substance - the idealA hypothetical substance - the ideal
gasgas
Think of it as a limit.Think of it as a limit.
Gases only approach ideal behaviorGases only approach ideal behavior
at low pressure (< 1 atm) and highat low pressure (< 1 atm) and high
temperature.temperature.
Use the laws anyway, unless told toUse the laws anyway, unless told to
do otherwise.do otherwise.
They give good estimates.They give good estimates.

31. 1
ExamplesExamplespppp
AA 47.3 L47.3 L container withcontainer with 1.62 mol1.62 mol of Heof He
heated until pressure reachesheated until pressure reaches 1.85 atm1.85 atm..
What is the temperature? Steps . . .What is the temperature? Steps . . .
PVPV == nRnRTT
TT = (= (1.851.85)()(47.347.3)/(.)/(.08210821)()(1.621.62)) ==
658 K or658 K or 385 ºC385 ºC

32. 2
ExamplesExamplespppp
Kr gas in a 18.5 L cylinder exerts aKr gas in a 18.5 L cylinder exerts a
pressure of 8.61 atm at 24.8ºC. What ispressure of 8.61 atm at 24.8ºC. What is
the mass of Kr?the mass of Kr? Ans. . .Ans. . .
546 g546 g Steps . . .Steps . . .
n = PV/RT = 6.51 moln = PV/RT = 6.51 mol
Since MSince M = m/mol = m/n, m = M • n= m/mol = m/n, m = M • n
Get M from periodic tableGet M from periodic table
m = 83.8 • 6.51 = 546 gm = 83.8 • 6.51 = 546 g

33. 3
ExamplesExamples
A gas is 4.18 L at 29 ºC & 732 torr. WhatA gas is 4.18 L at 29 ºC & 732 torr. What
volume at 24.8ºC & 756 torr? Ans. . .volume at 24.8ºC & 756 torr? Ans. . .
Use combined law. Why? . . .Use combined law. Why? . . .
The conditions areThe conditions are changingchanging
x = 3.99 L (can convert torr to atm whenx = 3.99 L (can convert torr to atm when
calculating, but pressure units cancel out socalculating, but pressure units cancel out so
don’t need to for this problem).don’t need to for this problem).

34. 4
5.4 Gases and Stoichiometry5.4 Gases and Stoichiometry
Reactions happen in molesReactions happen in moles
At Standard Temperature andAt Standard Temperature and
Pressure (STP, 0ºC and 1 atm) 1Pressure (STP, 0ºC and 1 atm) 1
mole of gas occupies 22.41 L.mole of gas occupies 22.41 L.
If not at STP, use the ideal gas law toIf not at STP, use the ideal gas law to
calculate moles of reactant orcalculate moles of reactant or
volume of product.volume of product.

35. 5
Figure 5.11Figure 5.11
A Mole of Any Gas Occupies a Volume ofA Mole of Any Gas Occupies a Volume of
Approximately 22.4 L at STPApproximately 22.4 L at STP
How many particles are in each box?
6.022 x 1023
particles

36. 6
ExamplesExamplespppp
Can get mercury by the following:Can get mercury by the following:
What volume oxygen gas produced fromWhat volume oxygen gas produced from
4.10 g of mercury (II) oxide (M = 2174.10 g of mercury (II) oxide (M = 217
g/mol) at STP? Steps.g/mol) at STP? Steps.
Balance first! 2HgOBalance first! 2HgO →→ 2Hg + O2Hg + O22
Since at STP can then use stoich (spider)Since at STP can then use stoich (spider)
4.10 g HgO • 1 mol HgO/217g HgO • 1 mol O4.10 g HgO • 1 mol HgO/217g HgO • 1 mol O22/2 mol HgO • 22.4 L O/2 mol HgO • 22.4 L O22/1 mol O/1 mol O22
= 0.212 L or 212 mL= 0.212 L or 212 mL
(g)O+(l)HgHgO 2
heat
→

37. 7
ExamplesExamplespppp
Can get mercury by the following:Can get mercury by the following:
What volume oxygen gas can be producedWhat volume oxygen gas can be produced
from 4.10 g of mercury (II) oxide at 400.ºCfrom 4.10 g of mercury (II) oxide at 400.ºC
& 740. torr? (Hint: conditions are changed,& 740. torr? (Hint: conditions are changed,
so whatso what equation do you then use?) . . .equation do you then use?) . . .
1st, find molar mass of mercury(II) oxide.1st, find molar mass of mercury(II) oxide.
2nd, balance reactions & spider under STP2nd, balance reactions & spider under STP
3rd, then use the combined law to convert.3rd, then use the combined law to convert.
(1)(212)/273 = (740/760)(x)/(273 + 400)(1)(212)/273 = (740/760)(x)/(273 + 400)
0.537 L0.537 L
HgOheat
→Hg(l)+O2(g)

38. 8
ExamplesExamples
Using the following reactionUsing the following reaction
calculate the mass of sodium hydrogencalculate the mass of sodium hydrogen
carbonate (M = 84.01) that produces 2.87carbonate (M = 84.01) that produces 2.87
L of carbon dioxide at 25ºC & 2.00 atm.L of carbon dioxide at 25ºC & 2.00 atm.
a.a. Not at STP so use PV=nRT; ans. . .Not at STP so use PV=nRT; ans. . .
b.b. 19.7 g19.7 g
NaCl(aq) + CO (g) +H O(l)2 2
NaHCO (s) + HCl3 →

39. 9
ExamplesExamples
Using the following reactionUsing the following reaction
If 27 L of gas are produced at 26ºC and 745 torrIf 27 L of gas are produced at 26ºC and 745 torr
when 2.6 L of HCl are added what is thewhen 2.6 L of HCl are added what is the
concentration of HCl?concentration of HCl?
a.a. Use stoich to get moles HCl, then mole ratioUse stoich to get moles HCl, then mole ratio →→
moles COmoles CO22, then moles/L = concentration. Or,, then moles/L = concentration. Or,
use Chas law to correct to 24.2 L, then stoich touse Chas law to correct to 24.2 L, then stoich to
get 1.08 mol HCl = 1.08 mole COget 1.08 mol HCl = 1.08 mole CO22; so; so
concentration is 0.415concentration is 0.415 MM
NaCl(aq) + CO (g) +H O(l)2 2
NaHCO (s) + HCl3 →

40. 0
ExamplesExamples
Consider the following reactionConsider the following reaction
What volume NO at 1.0 atm & 1000ºCWhat volume NO at 1.0 atm & 1000ºC
produced from 10.0 L of NHproduced from 10.0 L of NH33 & xs O& xs O22 atat
samesame temperature & pressure?temperature & pressure?
T & P constant & cancel out so volume isT & P constant & cancel out so volume is
proportional to moles = 10 Lproportional to moles = 10 L
4NH3(g)+5O2(g)→4NO(g)+6H2O(g)

41. 1
You try it!You try it!
Consider the following reactionConsider the following reaction
What volume of OWhat volume of O22 at STP will beat STP will be
consumed when 10.0 kg NHconsumed when 10.0 kg NH33 is reacted?is reacted?
(Use “rounded” molar masses).(Use “rounded” molar masses). Ans. . .Ans. . .
watch Kg! Ans. Rounded towatch Kg! Ans. Rounded to 16 500 L16 500 L (from(from
calc ans of 16 470)calc ans of 16 470)
O(g)6H+NO(g)4)(O5+(g)4NH 223 →g

42. 2
The Same reaction with some twistsThe Same reaction with some twists pppp
What mass of HWhat mass of H22O produced from 65.0 L OO produced from 65.0 L O22
& 75.0 L of NH& 75.0 L of NH33 measured at STP? . . . Ans.measured at STP? . . . Ans.
As before, but must findAs before, but must find LR = O2 ans. 313 g313 g
What volume of NO would be produced?What volume of NO would be produced?
Above + mole ratio. Ans. =Above + mole ratio. Ans. = 52 L52 L
What mass of NO is produced from 500. L ofWhat mass of NO is produced from 500. L of
NHNH33 at 250.0ºC and 3.00 atm? Ans.at 250.0ºC and 3.00 atm? Ans.
non-STP so use combined gas law to get tonon-STP so use combined gas law to get to
STP (783 L), then “spider” Ans. =STP (783 L), then “spider” Ans. = 895 g895 g
4NH3(g)+5O2(g)→4NO(g)+6H2O(g)

43. 3
Gas Density and Molar MassGas Density and Molar Mass
D = m/VD = m/V
LetLet MM stand for molar massstand for molar mass
MM = m/n= m/n
n= PV/RTn= PV/RT
MM == mm
PV/RTPV/RT
MM = mRT = m RT = DRT= mRT = m RT = DRT
PVPV V PV P PP

44. 44
Finding M or D from ideal gas law
M = m
(PV/RT)
M = m RT
V P
Since m/V = density (D)
M = DRT = molar mass
P
Density varies directly with M and P
Density varies indirectly with T

45. 45
Quick Examplepp
What mass of ethene gas, C2H4, is in a
15.0 L tank at 4.40 atm & 305 K? (steps)
PV = mRT/M and M = 28 g/mol
m = PVM/RT
m = (4.40 atm)(15.0 L)(28 g/mol)
(0.0821 L atm/mol K)(305 K)
m = 73.8 g C2H4

46. 46
Another way to solve it (quick)pp
What mass of ethene gas, C2H4, is in a
15.0 L tank at 4.40 atm & 305 K? (steps)
PV = nRT, solve for n
n = PV/RT
n = (4.40 atm)(15.0 L)
(0.0821 L atm/mol K)(305 K)
n = 2.64 mol C2H4
Use spider to get g C2H4 (M = 28 g/mol)
Answer is 73.8 g C2H4

47. 47
Quick Example
A gas has a density of 3.36 g/L at 14 ºC
and 1.09 atm. What is its molar mass?
Use M = DRT/P to get an answer of . . .
72.6 g/mol. Calculation . . .
(3.36)(0.0821)(14 + 273)/(1.09) = 72.6 g/mol

48. 8
AP ExamplesAP Examples
Density of ammonia at 23ºC & 735 torr?Density of ammonia at 23ºC & 735 torr?
Ans.Ans.
D = MP/RT = 0.68 g/LD = MP/RT = 0.68 g/L

49. 9
AP ExamplesAP Examples
A compound has empirical formula CHCl. AA compound has empirical formula CHCl. A
256 mL flask at 100.ºC & 750 torr contains256 mL flask at 100.ºC & 750 torr contains
0.80 g of the0.80 g of the gaseousgaseous compound. What iscompound. What is
the molecular formula? Steps. . .the molecular formula? Steps. . .
What do you need to get the molecularWhat do you need to get the molecular
formula? . . .formula? . . .
The molecular mass, M.The molecular mass, M.
How do you get this?How do you get this?
M = g/mol. You are given the grams, so findM = g/mol. You are given the grams, so find
the number of moles using . . .the number of moles using . . .
PV = nRT and solve for n . . . Try it! . . .PV = nRT and solve for n . . . Try it! . . .

50. 0
AP Examples cont.AP Examples cont.
EF of CHCl. 0.80 g are in 256 mL flask atEF of CHCl. 0.80 g are in 256 mL flask at
100.ºC & 750 torr. What is the MF?100.ºC & 750 torr. What is the MF?
n = 0.0083 from n = PV/RTn = 0.0083 from n = PV/RT
What’s M? . . .What’s M? . . .
MM ≈≈ 97 from 0.80 g/0.0083 moles = g/mol97 from 0.80 g/0.0083 moles = g/mol
How do you find molecular formula from EFHow do you find molecular formula from EF
and MF?and MF?
Find how many Efs are in the MF (CHCl)Find how many Efs are in the MF (CHCl)
then multiply the subscripts of the EF bythen multiply the subscripts of the EF by
that number. Try it! . . .that number. Try it! . . .
97/4997/49 ≈≈ 2, so formula is C2, so formula is C22HH22ClCl22

51. 1
5.5 Dalton’s Law of Partial Pressures5.5 Dalton’s Law of Partial Pressures
The total pressure in a container isThe total pressure in a container is
the sum of the pressure each gasthe sum of the pressure each gas
would exert if it were alone in thewould exert if it were alone in the
container.container.
The total pressure is the sum of theThe total pressure is the sum of the
partial pressures.partial pressures.

52. 2
Figure 5.12 p. 206Figure 5.12 p. 206
Partial Pressure of EachPartial Pressure of Each
Gas in a MixtureGas in a Mixture

53. 53
We can find out the pressure in the fourth
container by adding up the pressure in the
first 3.
2 atm 1 atm 3 atm 6 atm

54. 4
Dalton’s LawDalton’s Law
PPTotalTotal = P= P11 + P+ P22 + P+ P33 + P+ P44 + P+ P55 ......
For each, P still = nRT/VFor each, P still = nRT/V

55. 5
Dalton's LawDalton's Law
PPTotalTotal = n= n11RT + nRT + n22RT + nRT + n33RT +...RT +...
VV VV VV
In the same container R, T and V areIn the same container R, T and V are
the same.the same.
PPTotalTotal = (n= (n11+ n+ n22 + n+ n33+...)RT+...)RT
VV
PPTotalTotal = (n= (nTotalTotal)RT)RT
VV

56. 56
Examplespp
In a balloon the total pressure is 1.3 atm. What
is the pressure of oxygen if the pressure of
nitrogen is 720 mm Hg? . . .
PTotal = P1 + P2 + P3 . . .
1.3 atm = Po2 + 720 mm Hg of N2
Convert to same units (we’ll use mm Hg)
1.3 atm(760 mm Hg/1 atm) = Po2 + 720 mm Hg
988 mm Hg = Po2 + 720 mm Hg so, . . .
Po2 = 988 mm Hg - 720 mm Hg = 268 mm Hg
Can also express as 0.35 atm.

57. 7
The mole fractionThe mole fraction
Ratio of moles of the substance toRatio of moles of the substance to
thethe totaltotal moles.moles.
symbol is Greek letter chisymbol is Greek letter chi χχ
χχ11 = n= n11 = P= P11
nnTotalTotal PPTotalTotal

58. 8
ExamplesExamplespppp
The partial pressure of nitrogen in air is 592The partial pressure of nitrogen in air is 592
torr. The partial pressure of oxygen in air istorr. The partial pressure of oxygen in air is
160 torr. What is the mole fraction of160 torr. What is the mole fraction of
oxygen? Ans. . .oxygen? Ans. . .
0.213 Why isn’t it 0.270? . . .0.213 Why isn’t it 0.270? . . .
Because you have to divide by theBecause you have to divide by the totaltotal
pressure, which is 752 torr.pressure, which is 752 torr.

59. 9
ExamplesExamples
The partial pressure of nitrogen in air is 592The partial pressure of nitrogen in air is 592
torr. Air pressure is 752 torr, what is thetorr. Air pressure is 752 torr, what is the
mole fraction of nitrogen? Steps. . .mole fraction of nitrogen? Steps. . .
ΧΧ = 592 torr/752 torr = 0.787= 592 torr/752 torr = 0.787
What is the partial pressure of nitrogen ifWhat is the partial pressure of nitrogen if
the container holding the air is compressedthe container holding the air is compressed
to 5.25 atm? . . .to 5.25 atm? . . .
0.787 x 5.25 atm = 4.13 atm0.787 x 5.25 atm = 4.13 atm
Note: can use mole fraction regardless ofNote: can use mole fraction regardless of
units (don’t have to convert).units (don’t have to convert).

60. 0
ExamplesExamplespppp
3.50 L
O2
1.50 L
N2
2.70 atm
When these valves are opened, what is each partialWhen these valves are opened, what is each partial
pressure and the total pressure? Hint.pressure and the total pressure? Hint.
Use mole fraction to get the new partial pressuresUse mole fraction to get the new partial pressures
(or Boyle’s law), then add to get total pressure. Ans.(or Boyle’s law), then add to get total pressure. Ans.
CHCH44 = 1.2 atm= 1.2 atm NN22 = 0.763= 0.763 OO22 = 0.294= 0.294 Sum = 2.127Sum = 2.127
4.00 L
CH4
4.58 atm 0.752 atm

61. 1
Vapor PressureVapor Pressure
Water evaporates!Water evaporates!
When that water evaporates, theWhen that water evaporates, the
vapor has a pressure.vapor has a pressure.
Gases are often collected over waterGases are often collected over water
so the vapor. pressure of water mustso the vapor. pressure of water must
be subtracted from the totalbe subtracted from the total
pressure.pressure.
It must be given.It must be given.

62. 2
Figure 5.13Figure 5.13
The Production of Oxygen by ThermalThe Production of Oxygen by Thermal
Decomposition of KCIODecomposition of KCIO33

63. 3
ExampleExamplepppp
NN22O can be produced by the followingO can be produced by the following
reactionreaction
what volume of Nwhat volume of N22O collected overO collected over
water at a total pressure of 94 kPa andwater at a total pressure of 94 kPa and
22ºC can be produced from 2.6 g of22ºC can be produced from 2.6 g of
NHNH44NONO33? (the vapor pressure of water? (the vapor pressure of water
at 22ºC is 21 torr). Steps . . .at 22ºC is 21 torr). Steps . . .
NH4NO3(s) heat
→N2O(g)+2H2O(l)

64. 4
ExampleExample
What vol. NWhat vol. N22O over water at total pressure of 94O over water at total pressure of 94
kPakPa & 22ºC can be produced from 2.6 g NH& 22ºC can be produced from 2.6 g NH44NONO33??
(Vp(Vpwaterwater at 22ºC is 21at 22ºC is 21 torrtorr))
Make sure reaction is balanced (it is)Make sure reaction is balanced (it is)
M of ammonium nitrate = 80. gM of ammonium nitrate = 80. g
Use spider to get moles of NUse spider to get moles of N220 = .0325, then use0 = .0325, then use
PV = nRT, but first have to:PV = nRT, but first have to:
Convert 21Convert 21 torrtorr to 2.8to 2.8 kPakPa (or(or vice versavice versa))
94.0 - 2.8 = 91.2 kPa N94.0 - 2.8 = 91.2 kPa N22OO
convert P to atm d/t R ( or use 8.31 for kPa)!!!convert P to atm d/t R ( or use 8.31 for kPa)!!!
Convert C to KConvert C to K
Ans. 6.6 LAns. 6.6 L
NH4NO3(s) heat
→N2O(g)+2H2O(l)

65. 5
5.6 Kinetic Molecular Theory5.6 Kinetic Molecular Theory
Theory tellsTheory tells whywhy the things happen.the things happen.
explains why ideal gases behave theexplains why ideal gases behave the
way they do.way they do.
AssumptionsAssumptions that simplify the theory,that simplify the theory,
but don’t work in real gases.but don’t work in real gases.
1 The particles areThe particles are so smallso small we canwe can
ignore their volumeignore their volume..
2 The particles are inThe particles are in constant motionconstant motion
and theirand their collisions cause pressurecollisions cause pressure..

66. 66
The Average speed of an oxygen
molecule is 1656 km/hr at 20ºC
The molecules don’t travel very
far without hitting each other so
they move in random directions.

67. 7
Kinetic Molecular TheoryKinetic Molecular Theory
3 The particles do not affect each other,The particles do not affect each other,
neither attracting or repellingneither attracting or repelling..
4 TheThe averageaverage kinetic energy iskinetic energy is
proportionalproportional to theto the KelvinKelvin temperature.temperature.
Appendix 2 shows the derivation of theAppendix 2 shows the derivation of the
ideal gas law and the definition ofideal gas law and the definition of
temperature.temperature.
We need the formula KE = 1/2 mvWe need the formula KE = 1/2 mv22

68. 8
What it tells usWhat it tells us
(KE)(KE)avgavg = 3/2 RT= 3/2 RT
This the meaning of temperature.This the meaning of temperature.
u is the particle velocity.u is the particle velocity.
u is the average particle velocity.u is the average particle velocity.
uu 22
is the average particle velocityis the average particle velocity
squared.squared.
the root mean square velocity isthe root mean square velocity is
√√ uu 2 =2 =
uurmsrms

69. 9
Combine these two equationsCombine these two equations
(KE)(KE)avgavg = n= nAA(1/2 mu(1/2 mu 22
)) (KE)(KE)avgavg ==
3/2 RT3/2 RT

70. 0
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 inWhere M is the molar mass in kgkg/mole/mole
((watch for test questionwatch for test question!!), and R has the), and R has the
units 8.3145units 8.3145 JJ/K mol./K mol.
The velocity will be inThe velocity will be in m/sm/s
u =
3RT
M
rms

71. 1
ExampleExamplepppp
Calculate the root mean square velocity ofCalculate the root mean square velocity of
carbon dioxide at 25ºC.carbon dioxide at 25ºC.
Must express R using joules & M = Kg/mol (not g,Must express R using joules & M = Kg/mol (not g,
d/t R)d/t R)
R = 8.3145 J/K•Mol, M =R = 8.3145 J/K•Mol, M = 0.00.04444 KgKg/mol ans. . ./mol ans. . .
411 m/sec411 m/sec
If time, do for hydrogen, for chlorine. Ans. . .If time, do for hydrogen, for chlorine. Ans. . .
Hydrogen is 1928 m/s; Chorine is 324 m/sHydrogen is 1928 m/s; Chorine is 324 m/s
Notice that the less massive hydrogen has aNotice that the less massive hydrogen has a
higher velocity.higher velocity.
urms =
3RT
M

72. 2
Range of velocitiesRange of velocities
The average distance a moleculeThe average distance a molecule
travels before colliding with anothertravels before colliding with another
is called the mean free path and isis called the mean free path and is
small (around 10small (around 10-7-7
m)m)

73. 3
Figure 5.18Figure 5.18
Path of OnePath of One
Particle in aParticle in a
Gas.Gas.
Any givenAny given
particle willparticle will
continuouslycontinuously
change its coursechange its course
as a result ofas a result of
collisions withcollisions with
other particles, asother particles, as
well as with thewell as with the
walls of itswalls of its
container.container.

74. 4
Range of velocitiesRange of velocities
Temperature is an average. There areTemperature is an average. There are
molecules ofmolecules of manymany speeds in thespeeds in the
average.average.
This is shown on a graph called aThis is shown on a graph called a
velocity distribution . . .velocity distribution . . .

75. 5
numberofparticles
Molecular Velocity
273 K

76. 6
numberofparticles
Molecular Velocity
273 K
1273 K

77. 7
numberofparticles
Molecular Velocity
273 K
1273 K
2273 K

78. 8
VelocityVelocity
Average increases as temperatureAverage increases as temperature
increases.increases.
Spread increases as temperatureSpread increases as temperature
increases.increases.

79. 9
5.7 Effusion & Diffusion5.7 Effusion & Diffusion
EffusionEffusion: Passage of gas through a: Passage of gas through a
small hole, into a vacuum.small hole, into a vacuum.
The effusion rate measures how fastThe effusion rate measures how fast
this happens.this happens.
Graham’s LawGraham’s Law: The rate of effusion is: The rate of effusion is
inversely proportional to the squareinversely proportional to the square
root of the mass of its particles.root of the mass of its particles.
Rateeffusionforgas1
Rateeffusionforgas2
=
M2
M1

80. 0
Figure 5.21 p. 219The Effusion of a Gas into an Evacuated ChamberFigure 5.21 p. 219The Effusion of a Gas into an Evacuated Chamber
Rate of effusion is inversely proportional to the
square root of the mass of the gas molecules.

81. 1
DerivingDeriving
The rate of effusion should beThe rate of effusion should be
proportional to uproportional to urmsrms
Effusion Rate 1 = uEffusion Rate 1 = urmsrms 11
Effusion Rate 2 = uEffusion Rate 2 = urmsrms 22

82. 2
Deriving Graham’s LawDeriving Graham’s Law
The rate of effusion should beThe rate of effusion should be
proportional to uproportional to urmsrms
Effusion Rate 1 = uEffusion Rate 1 = urmsrms 11
Effusion Rate 2 = uEffusion Rate 2 = urmsrms 22
effusionrate1
effusionrate2
=
urms 1
urms 2
=
3RT
M1
3RT
M2
=
M2
M1

83. 3
DiffusionDiffusion
The spreading of a gas through a room.The spreading of a gas through a room.
Slow, considering molecules move at 100’sSlow, considering molecules move at 100’s
of meters per second.of meters per second.
Collisions with other molecules slow downCollisions with other molecules slow down
diffusions.diffusions.
Best estimate of diffusion is Graham’s Law.Best estimate of diffusion is Graham’s Law.
Do not confuse rate with time.Do not confuse rate with time.
Time is in seconds (or min). Rate is in 1/sTime is in seconds (or min). Rate is in 1/s
(or 1/min), etc.(or 1/min), etc.
The following questions test yourThe following questions test your
understanding of this.understanding of this.

84. 4
Figure 5.23Figure 5.23
HCI(HCI(gg) and NH) and NH33((gg) Meet in a) Meet in a
TubeTube

85. 5
Whiteboard ExamplesWhiteboard Examplespppp
An element effuses through a porous cylinder 3.20An element effuses through a porous cylinder 3.20
timestimes slowerslower than helium. What is it’s molar mass?than helium. What is it’s molar mass?
Steps on whiteboard to get . . .Steps on whiteboard to get . . .
40.96 (Probably Argon = 39.948 g/mol)40.96 (Probably Argon = 39.948 g/mol)
0.00251 mol NH0.00251 mol NH33 effuses via a hole in 2.47 min, howeffuses via a hole in 2.47 min, how
much HCl would effuse in the same time? Your ans.?much HCl would effuse in the same time? Your ans.?
0.0017 mol HCl0.0017 mol HCl
Some NSome N22 effuses via a hole in 38 seconds. What iseffuses via a hole in 38 seconds. What is
molecular weight of gas that effuses in 55 secondsmolecular weight of gas that effuses in 55 seconds
under identical conditions? Your answer? . . . (hint:under identical conditions? Your answer? . . . (hint:
think if Nthink if N22 is faster or slower. If faster then the rateis faster or slower. If faster then the rate
must be > 1, if slower, then the rate is a fraction.must be > 1, if slower, then the rate is a fraction.
59 g/mol59 g/mol

86. 6
5.8 Real Gases5.8 Real Gases
Real moleculesReal molecules dodo take up space andtake up space and
theythey dodo interact with each otherinteract with each other
(especially polar molecules).(especially polar molecules).
Need to addNeed to add correction factorscorrection factors to theto the
ideal gas law to account for these.ideal gas law to account for these.

87. 7
VolumeVolume CorrectionCorrection
The actual volume free to move in is lessThe actual volume free to move in is less
because of particle size.because of particle size.
More molecules (More molecules (nn) will have more effect.) will have more effect.
Corrected volume V’ = V - nb. SubstituteCorrected volume V’ = V - nb. Substitute
this for V in PV = nRTthis for V in PV = nRT
bb is a constant that differs for each gas.is a constant that differs for each gas.
PP’’ == nRTnRT
(V-(V-nnbb))

88. 8
PressurePressure correctioncorrection
Because the molecules are attracted toBecause the molecules are attracted to
each other, the pressure on the containereach other, the pressure on the container
will bewill be lessless than idealthan ideal
Depends on the number of molecules perDepends on the number of molecules per
liter (liter (n ÷ Vn ÷ V). “a” is a unique constant.). “a” is a unique constant.
SinceSince twotwo molecules interact, the effectmolecules interact, the effect
must bemust be squaredsquared..

89. 9
Pressure correctionPressure correction
Because the molecules are attracted toBecause the molecules are attracted to
each other, the pressure on the containereach other, the pressure on the container
will be less than idealwill be less than ideal
Depends on the number of molecules perDepends on the number of molecules per
liter (liter (n ÷ Vn ÷ V). “a” is a unique constant). “a” is a unique constant
Since two molecules interact, the effectSince two molecules interact, the effect
must be squared.must be squared.
Pobserved = P’ - a
2
( )V
n

90. 0
AltogetherAltogether
PPobsobs= nRT - a n= nRT - a n 22
V-nb VV-nb V
Called the Van der Waal’s equation ifCalled the Van der Waal’s equation if
rearrangedrearranged
CorrectedCorrected CorrectedCorrected
PressurePressure VolumeVolume
“P”“P” xx “V”“V” == nRTnRT
( )
( )P + a
n
V
x V - nb nRTobs














=
2

91. 1
Where does it come fromWhere does it come from
a and b are determined bya 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.

92. 2
Figure 5.24 p. 222Figure 5.24 p. 222
Plots ofPlots of PV/nRTPV/nRT VersusVersus
PP for Several Gasesfor Several Gases
(200 K)(200 K)
Behavior is close toBehavior is close to
ideal ONLY at lowideal ONLY at low
pressures (less thanpressures (less than
1 atm)1 atm)

93. 3
Figure 5.25Figure 5.25
Plots ofPlots of
PV/nRTPV/nRT
VersusVersus PP forfor
Nitrogen GasNitrogen Gas
at Threeat Three
TemperaturesTemperatures
DeviationsDeviations
are smaller atare smaller at
higherhigher
temperatures.temperatures.

94. 4
RealReal vsvs Ideal Behavior (test question)Ideal Behavior (test question)
Gases act most ideally if at:Gases act most ideally if at:
Low pressureLow pressure
High temperature, and have . . .High temperature, and have . . .
Low mass and . . .Low mass and . . .
Low polarityLow polarity
This provides least deviation from theThis provides least deviation from the
expected collisions derived from theexpected collisions derived from the
kinetic molecular theory.kinetic molecular theory.
Be able to explainBe able to explain whywhy this is so.this is so.

95. 5
ExampleExample
Calculate the pressure exerted by 0.5000Calculate the pressure exerted by 0.5000
mol Clmol Cl22 in a 1.000 L container at 25.0ºCin a 1.000 L container at 25.0ºC
Using theUsing the idealideal gas law. . .gas law. . .
P = nRT/V =P = nRT/V = 12.2 atm12.2 atm
UsingUsing Van der Waal’sVan der Waal’s equation, whereequation, where
– a = 6.49 atm La = 6.49 atm L22
/mol/mol22
– b = 0.0562 L/mol . . .b = 0.0562 L/mol . . .
P = nRT/V - nb - a (n/v)P = nRT/V - nb - a (n/v)22
== 11.0 atm11.0 atm
Note the difference.Note the difference.

96. 6
5.9 Chemistry in the Atmosphere5.9 Chemistry in the Atmosphere
Acid RainAcid Rain
There will be AP questions on this. . .There will be AP questions on this. . .

97. 97
Sources of Acid Rain
l Power stations
l Oil refineries
l Coal with high S content
l Car and truck emissions
l Bacterial decomposition, and lightning
hitting N2 in the atmosphere.

98. 98
Acid Rain
l Unpolluted rain has a pH of 5.6
l Rain with a pH below 5.6 is “acid rain“
l CO2 in the air forms carbonic acid
CO2 + H2O H2CO3
l Adds to H+
of rain
H2CO3 H+
(aq) + HCO3-(aq)

99. 99
SO2 26 million tons in 1980
NO and NO2 22 million tons in 1980
Mt. St Helens (1980) 400,000 tons SO2
l Reactions with oxygen in air form SO3
2SO2 + O2 2 SO3
l Reactions with water in air form acids
SO3 + H2O H2SO4 sulfuric acid
NO + H2O HNO2 nitrous acid
HNO2 + H2O HNO3 nitric acid
Sources of Acid Rain

100. 100
Effects of Acid Rain
l Leaches Al from soil, which kills fish
l Fish kills in spring from runoff due to accumulation
of large amounts of acid in snow
l Dissolves waxy coatings that protect leaves from
bacteria
l Corrodes metals, textiles, paper and leather
l Dissolves minerals Mg, Ca, and K from the soil and
waxy coatings that protect leaves from bacteria.
l Read 5.9 in book for the rest of the story