1. Gases

2. Gases I. comparison of solids, liquids and Gasses
A. fluids- things that flow - gases and liquids
B. condensed states - liquids and solids
C. Vapor - refers to a gas that is formed by
evaporation of a liquid or the sublimation
of a solid (sublimation - going directly
from a solid to a gas)

3. II some common properties of gases
A. Gases can be compressed into smaller volumes;
that is, their densities can be increased by
applying increased pressure.
B. Gases exert pressure on their surroundings;
in turn, pressure must be exerted to confine
gases.
C. Gases expand without limits, and so gas
samples completely and uniformly occupy
the volume of any container.

4. D. Gases diffuse into each other, and so samples of
gas placed in the same container mix completely.
Conversely,different gases in a mixture do not
separate on standing.
1. diffusion - going from a high concentration
to a low concentration

5. E. expand to fill container
F. take shape of container
G. some are invisible

6. H. some are combustible
I. four properties determine the physical
behavior of a gas
1. amount of gas
2. volume of the gas
3. temperature of the gas
4. pressure of the gas

7. Elements that exist as gases at 250C and 1 atmosphere
5.1

8. 5.1

9. 1. Pressure = force = Newtons = pascal Area m2
J. exert Pressure
1. Pressure = force = Newtons = pascal
Area m2
a. Named after Blaise Pascal ( ) -
father of modern Hydraulics
b. units Newtons/m2 = 1 Pascal (Pa)
c. 1 Pascal is rather small - more comely
used unit is the kilopascal
2. liquid pressure = g x h x d

10. (force = mass x acceleration)
Area
Barometer
Pressure =
(force = mass x acceleration)
Units of Pressure
1 pascal (Pa) = 1 N/m2
1 atm = 760 mm Hg = 760 torr
= 101,325 Pa = 14.7 psi = in. Hg
5.2

11. 10 miles 0.2 atm 4 miles 0.5 atm Sea level 1 atm
5.2

12. 4. pressure of gases is due to the billion
collision/sec of the gas molecules with
the walls of the container
5. Barometer - device used to measure gas pressure
a. invented by Evangelista Torricelli

13. Boyle’s Law
P α 1/V
This means Pressure and Volume are INVERSELY PROPORTIONAL if moles and temperature are constant (do not change). For example, P goes up as V goes down.
P1V1 = P2 V2
Robert Boyle ( ). Son of Earl of Cork, Ireland.

15. Gas Law Problems
have 5 steps

16. e.g. A gas under a pressure of 750. mm of Hg
occupies 25 liters. What will its volume be
under a pressure of 800. mm of Hg?
Temperature remains constant. P1
= 750. mm of Hg
= 25 L
Step 1 V1 V2
= X P2
= 800. mm or Hg
P1V1 = P2V2 - base equation
Step 2

17. P1V1 = V2 - working equation P2
Divide by P2 on both sides
P1V1 = P2V2 P2 P2
Make sure your unkown is on the top
P1V1 = V working equation P2
Step 3
(750. mm of Hg)( 25 L) =
(800. mm of Hg)
23 L
Step 5
Step 4

18. Charles’s Law If n and P are constant, then V α T
V and T are directly proportional.
V V2 =
T T2
If one temperature goes up, the volume goes up!
Jacques Charles ( ). Isolated boron and studied gases. Balloonist.

19. There are 3 temperature scales
remember - temperature is a measure of molecular
motions
Fahrenheit
Celsius
Can you have F and C temperatures below 0?
If temperature is a measure of molecular movement -
How can you have negative temperatures?

20. We had to come up with a new temperature scale
Kelvin
K = C + 273
What is 22 C in Kelvin?
K = = 295 K

21. Absolute zero Really Really Cold 0 K = no molecular motion
0 K = -273 C = -460 F
Really Really Cold

23. A gas occupies a volume of 2.5 liters at a temperature
Example of Charles Law
A gas occupies a volume of 2.5 liters at a temperature
of 22 C. What temperature is required for the gas to
occupy 3.0 liters V1
= 2.5 l
= 22 C = 295 K T1 V2
= 3.0 liters T2
= x
Step 1

24. V1 = V2
T1 T2
Step 2
First you must move your unknown to the top
Multiply both sides by T2
V1 = V2
T1 T2 T2 T2 T1
T2 V1 = V2 T1 T1 V1 V1

25. T2 = V2T1 V1 Step 3 = (3.0 liters)(295 K) 2.5 L Step 4 354 K = 350 K

26. Gay-Lussac’s Law If n and V are constant, then P α T
P and T are directly proportional.
P P2 =
T T2
If one temperature goes up, the pressure goes up!
Joseph Louis Gay-Lussac ( )

27. Combined Gas Law
The good news is that you don’t have to remember all three gas laws! Since they are all related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION!
P1 V P2 V2 =
T T2

28. By doing step 1, you will be able to
select the right equations.
If any on the variables are held
constant, disregard them.

29. e.g. A gas occupies 10.0 L at 700. mm of Hg and
a temperature of 30.0 C. What volume will the gas
Occupy at 800. mm of Hg and 20.0 C
V1=10.0 L
P1= 700. mm of Hg
T1 = = K
Step 1
V2 = x
P2 = 800. mm of Hg
T2= C = K

30. Step 2
P1V1 = P2V2
T T2
T2P1V1 = V2
Step 3
P2T1
(293.0 K)(700. mm of Hg)(10.0 L)
(800. mm of Hg)(303.0 K)
Step 4
8.46 L = V2
Step 5

31. What temperature is required to change 250 ml of a gas
at 50.0 C and mm of Hg to 150 ml at mm of Hg
V1= 250
P1= mm of Hg
T1 = = K
V2 = 150 ml
P2 = mm of Hg
T2=x

32. Now just put your numbers in
P1V1 = P2V2
T T2 T2 T2
P1V1 = P2V2
T T2 T1
T2P1V1 T1
= P2V2 T1
P1V1
P1V1
T2 = P2V2T1
P1V1
Now just put your numbers in

33. on handout - due question 7-15
7-10 are Charles law
11-15 are combined Gas Laws
Stop here

34. Standard Pressure = 1 atm (or an equivalent)
Standard Temperature = 0 deg C (273 K)
STP allows us to compare amounts of gases between different pressures and temperatures

35. Avogadro’s Law V a number of moles (n) V = constant x n V1/n1 = V2/n2
Constant temperature
Constant pressure
V = constant x n
V1/n1 = V2/n2
5.3

36. Ideal Gas Equation 1 Boyle’s law: V a (at constant n and T) P
Charles’ law: V a T (at constant n and P)
Avogadro’s law: V a n (at constant P and T)
n = number of moles
V a nT P
V = constant x = R nT P
R is the gas constant
PV = nRT
5.4

37. R = 0.082057 L • atm / (mol • K) or .0821 L Atm mole K
The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).
Experiments show that at STP, 1 mole of an ideal gas occupies L.
22.4 L is standard volume
PV = nRT
R = PV nT =
(1 atm)(22.414L)
(1 mol)( K)
R = L • atm / (mol • K) or L Atm
mole K
5.4

38. When you use the ideal gas equation
Pressure must be in atmospheres
1 atm = 760 mm of Hg
Temperature must be in Kelvin
volume in liters
R is the ideal gas constant = L Atm
mole K

39. How many liters will 12.0 g of O2 occupy at
780.0 mm of Hg and 30.0 C
P =
(780.0 mm of Hg) (1atm ) = Atm
760 mm of Hg
V = X
(12.0 g O2)( 1mole ) = .375 moles
32.0 g O2
n =
L atm/mole K
R =
T =
K = C = K

40. PV = nRT
V = nRT/P
L atm
mole K K V=
.375 moles
1.026 Atm
V = L

41. 65 g of a CH4 occupies 30.0 L at a temperature of
10.0 C. What pressure must be exerted on this gas?
P = X
V= 30.0 L
n = (65 g CH4)( 1 mole CH4) = moles CH4
(16 g CH4)
R = L Atm
Mole K
T = 10.0C = K

42. PV = nRT
P = nRT V
P = (4.1 mole CH4)(.0821 L Atm ) ( K)
(30.0 L) ( mole K )
P = 3.1 Atm

43. What is the molecular weight of a 2.0 g sample of a
gas that occupies 5.0 L at a pressure of Atm
at 22 C.
P= Atm
V= 5.0 L
n = x
R = L Atm/mole K
T = 22 C ==295 K
PV= nRT

44. remember the units for M.W. is g/mole
PV= nRT
PV = n RT
(0.303 Atm)(5.0 L)(mole K)
(295 K) ( L Atm)
n= moles
remember the units for M.W. is g/mole
M.W. = g
.0625 moles
= 32 g/mole

45. e.G How many liters will .639 mole of Cl2 occupy
under standard conditions

46. x PV = nRT P = V = n = R = T = 1 Atm V = nRT P .639 moles Cl2
.0821 L Atm
Mole K
273 K
V =
(.639 moles Cl2)
(.0821 L Atm)
(273 K)
14.3 L =
(1 Atm)
Mole K

47. You need your books today
On page 358 – do questions 16-20

48. C6H12O6 (s) + 6O2 (g) ===> 6CO2 (g) + 6H2O (l)
Gas Stoichiometry
Mass - Volume problems
What is the volume of CO2 produced when 5.60 g of glucose are used up in the reaction at STP:
C6H12O6 (s) + 6O2 (g) ===> 6CO2 (g) + 6H2O (l)
g C6H12O mol C6H12O mol CO V CO2
5.5

49. C6H12O6 (s) + 6O2 (g) ===> 6CO2 (g) + 6H2O (l)
1 mole
6 moles
C6H12O6 (s) + 6O2 (g) ===> 6CO2 (g) + 6H2O (l)
x L
5.60 g
22.4 L/mol
180. g/mole
molar volume ) ( ) ( ) (
1 mole C6H12O6
6 moles CO2
(5.60 gC6H12O6)
22.4 L CO2
180. g C6H12O6
1 mole C6H12O6
1 mole CO2
4.18 L

50. When a problem gives you grams - change it to moles
How many liters of Cl2 is required to react
with 50.0 g of K to produce KCl at STP?
K Cl > KCl
2 mole
1 mole 2 2
50.0 g
X L
39.1 g/mole
22.4 L/mole
When a problem gives you grams - change it to moles
( 1 mole K )
(1 mole Cl2 )
(50.0 g K)
(22.4 L)
(39.1g K)
( 2 moles K )
( 1 mole Cl2 )
= L

51. Volume - mass problems

52. During a collision, automobile air bags are inflated by N2
gas formed by the explosive decomposition
of sodium azide at STP
What mass of sodium azide would be needed to inflate
31.0 L 2
NaN3 ====> Na N2 2 3
x g
31.0 L
65.0 g/mole
22.4 L/mole
(2 moles NaN3)
(65.0 g NaN3)
(1 mole N2)
(31.0 L N2)
( 1 mole NaN3)
( 22.4 L N2)
( 3 moles N2)

53. Try these two
How many liters of Hydrogen are produced from
110.0 g of water at STP?
H2O => H O2
How many grams of Na are required to produce L
of H2 when it reacts with water at STP
Na H2O => NaOH H2

54. dRT P M = d = m V = = 2.21 1 atm x 0.0821 x 300.15 K M = M =
A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.00C. What is the molar mass of the gas?
dRT P
M =
d = m V
4.65 g
2.10 L =
= 2.21 g L
2.21 g L
1 atm
x x K
L•atm
mol•K
M =
M =
54.6 g/mol
5.3

55. Gas Stoichiometry continued
Volume - Volume problems
As the name suggest - you start with a volume and
end with a volume
What volume of O2 is required to completely burn
70.0 L of CH4?
CH O2 ====> CO H2O 2 2
As always - make sure you balance all equations

56. You can see that a balanced chemical equation can
CH O2 ====> CO H2O
(70.0 L of CH4)
( 2 liters of O2 )
= L of O2
( 1 liter of CH4 )
You can see that a balanced chemical equation can
be used for a number of things

57. 1 How many liters of O2 are produced when 50.0 L of H2O2 decomposes
H2O2 ===> H2O O2
2. How many liters of O2 are required to burn L of propane?
C3H O2 ==> CO H2O

58. d is the density of the gas in g/L
Density (d) Calculations
m is the mass of the gas in g m V = PM RT
d =
M is the molar mass of the gas
Molar Mass (M ) of a Gaseous Substance
dRT P
M =
d is the density of the gas in g/L
5.4

59. Dalton’s Law of Partial Pressures
V and T are constant P1 P2
Ptotal = P1 + P2
5.6

60. PA = nART V PB = nBRT V PT = PA + PB XA = nA nA + nB XB = nB nA + nB
Consider a case in which two gases, A and B, are in a container of volume V.
PA =
nART V
nA is the number of moles of A
PB =
nBRT V
nB is the number of moles of B
mole fraction (Xi) = ni nT
PT = PA + PB
XA = nA
nA + nB
XB = nB
nA + nB
Pi = Xi PT
PB = XB PT
PA = XA PT
5.6

61. 5.6

62. Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties.
NH4Cl
NH3
17 g/mol
HCl
36 g/mol
5.7

63. GAS DIFFUSION AND EFFUSION
diffusion is the gradual mixing of molecules of different gases.
effusion is the movement of molecules through a small hole into an empty container.

64. GAS DIFFUSION AND EFFUSION
Graham’s law governs effusion and diffusion of gas molecules. KE=1/2 mv2
Rate of effusion is inversely proportional to its molar mass.
Thomas Graham, Professor in Glasgow and London.

65. GAS DIFFUSION AND EFFUSION
Molecules effuse thru holes in a rubber balloon, for example, at a rate (= moles/time) that is
proportional to T
inversely proportional to M.
Therefore, He effuses more rapidly than O2 at same T. He

66. Gas Diffusion relation of mass to rate of diffusion
HCl and NH3 diffuse from opposite ends of tube.
Gases meet to form NH4Cl
HCl heavier than NH3
Therefore, NH4Cl forms closer to HCl end of tube.

67. (8.24 mols CH4 + 0.421mols C2H6 + 0.116 mols C3H8.) Xpropane =
A sample of natural gas contains 8.24 moles of CH4, moles of C2H6, and moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)?
Pi = Xi PT
PT = 1.37 atm
mols C3H8.
(8.24 mols CH mols C2H mols C3H8.)
Xpropane =
Xpropane =
0.0132
Ppropane = x 1.37 atm
= atm
5.6

68. PT = PO + PH O 2KClO3 (s) 2KCl (s) + 3O2 (g) 5.6
Bottle full of oxygen gas and water vapor
2KClO3 (s) KCl (s) + 3O2 (g)
PT = PO + PH O 2
5.6

69. Kinetic theory of gases and …
Compressibility of Gases
Boyle’s Law
P a collision rate with wall
Collision rate a number density
Number density a 1/V
P a 1/V
Charles’ Law
Collision rate a average kinetic energy of gas molecules
Average kinetic energy a T
P a T
5.7

70. Kinetic Molecular Theory of Gases
A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume.
Gas molecules are in constant motion in random directions. Collisions among molecules are perfectly elastic.
Gas molecules exert neither attractive nor repulsive forces on one another.
The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy
5.7

71. Kinetic theory of gases and …
Avogadro’s Law
P a collision rate with wall
Collision rate a number density
Number density a n
P a n
Dalton’s Law of Partial Pressures
Molecules do not attract or repel one another
P exerted by one type of molecule is unaffected by the presence of another gas
Ptotal = SPi
5.7

72. What volume of O2 is required to react with 3.00 g of H2
at 22 C and 1.20 atm 2
H2 + O2 ===> H2O 2
( 1 mole H2)
( 1 mole of O2 )
(3.00g H2)
( 2.02 g H2)
( 2 moles H2 )
= .743 moles O2

73. P =
V =
n =
R =
T =
2.0 Atm
PV = nRT
250.0 L
PV = n RT x
.0821 L Atm
Mole K
20.0 C = K
(2.0 Atm)
(250.0 L)
Mole K
21 moles =
0821 L Atm
293.0 K

74. 1. How many grams of Na is required to react with water to
produce L of H2 at 20.0 C and 2.0 atm?
NaOH is the other product.
Na H2O > NaOH H2 2 2 2
Because you are looking for g and start with a volume,
it is a volume - mass problem
Start with the Ideal gas equation

75. (21 moles of H2)
( 2 moles of Na )
( 23 g Na)
(1 mole H2)
(1 mole of Na)
= g Na