1. Unit 8
Gas Laws!

2. Kinetic Molecular Theory
Particles in an ideal gas…
have no volume.
have elastic collisions.
are in constant, random, straight-line motion.
don’t attract or repel each other.
have an avg. KE directly related to Kelvin temperature.

3. Real Gases Particles in a REAL gas… Gas behavior is most ideal…
have their own volume
attract each other
Gas behavior is most ideal…
at low pressures
at high temperatures
in nonpolar atoms/molecules

4. The Nature of Gases Gases expand to fill their containers
Gases are fluids – they flow
Gases have low density
1/1000 the density of the equivalent liquid or solid
Gases are compressible
Gases effuse and diffuse

5. Diffusion Diffusion: describes the mixing of gases.
The rate of diffusion is the rate of the gases mixing.

6. Effusion
Effusion: describes the passage of gas into an evacuated chamber.

7. Kinetic Molecular Theory
Kinetic Molecular Theory -model for ideal gases Ideal gases are imaginary gases that perfectly fit all of the assumptions of the kinetic molecular theory.
Gases consist of tiny particles that are far apart
relative to their size.
Collisions between gas particles, between other
particles, and the walls of the container are
elastic collisions

8. Kinetic Molecular Theory (cont. )
Kinetic Molecular Theory (cont.) -model for ideal gases Ideal gases are imaginary gases that perfectly fit all of the assumptions of the kinetic molecular theory.
No kinetic energy is lost in elastic
collisions
There are no forces of attraction between gas
particles
The average kinetic energy of gas particles
depends on temperature, not on the identity
of the particle.

9. The Meaning of Temperature
Kelvin temperature is an index of the random motions of gas particles (higher T means greater motion.)
K = ºC + 273

10. Pressure
Is caused by the collisions of molecules with the walls of a container
is equal to force/unit area
SI units = Newton/meter2 = 1 Pascal (Pa)
1 standard atmosphere = 101,325 Pa
1 standard atmosphere = 1 atm = 760 mm Hg = 760 torr

11. Measuring Pressure The first device for measuring atmospheric
pressure was developed by Evangelista Torricelli
during the 17th century.
The device was called a “barometer”
Baro = weight
Meter = measure

12. An Early Barometer The normal pressure due to
the atmosphere at sea level
can support a column of
mercury that is 760 mm high.

13. Practice Problem
The average atmospheric pressure in Denver, Colorado is atm. Express this pressure in (a) millimeters of Hg and (b) kPa.

14. Practice Problem
The average atmospheric pressure in Denver, Colorado is atm. Express this pressure in (a) millimeters of Hg and (b) kPa.
760 mm Hg = 1 atm
kPa = 1 atm
760 torr = 1 am
631 mm Hg
84.1 kPa

15. More Practice Problem 0.985 atm to mm Hg B) 642 Torr to kPa
849 Torr to atm
D) 4.00 atm to Torr

16. More Practice Problem 0.985 atm to mm Hg 749 mm Hg
B) Torr to kPa kPa
849 Torr to atm atm
D) atm to Torr Torr

17. Standard Temperature and Pressure “STP”
P = 1 atmosphere, 760 torr
T = 0°C, 273 Kelvins
The molar volume of an ideal gas is liters at STP

18. Standard Temperature & Pressure
STP
Standard Temperature & Pressure
0°C K
1 atm kPa
-OR-

19. Converting Celsius to Kelvin
Gas law problems involving temperature require
that the temperature be in KELVINS!
Kelvins = C + 273
°C = Kelvins - 273
Absolute zero-the temperature where all molecular motion ceases – O K

20. Convert the following into Kelvin:
CONVERT THE FOLLOWING INTO CELSIUS:
100K
200 K
273 K
350 K
299 K
4 K
Convert the following into Kelvin:
0o C -50o C 90o C -20o C 22 o C 100 o C

21. Convert the following into Kelvin:
CONVERT THE FOLLOWING INTO CELSIUS:
100K
200 K
273 K
350 K
299 K
4 K
Convert the following into Kelvin:
-173 C
-73 C
0 C
77 C
26 C
-269 C
273 K
223 K
363 K
253 K
295 K
373 K
0o C -50o C 90o C -20o C 22 o C 100 o C

22. THE GAS LAWS
We use FOUR variables to describe a sample of gas in calculations:
1) amount in moles (n)
2) volume (V)
3) temperature (T)
4) pressure (P)

23. THE GAS LAWS
These four variables (moles, V, T, and P) are related through some simple gas laws that show how one of the variables changes as a second variables changes, and the other two remain constant

24. The Combined Gas Law The combined gas law expresses the relationship
between pressure, volume and temperature of a
fixed amount of gas.

25. Practice Problem
A helium-filled balloon has a volume of 50.0 L at 25 C and 1.08 atm. What volume will it have a atm and 10.0 C? (remember T must be in Kelvin)

26. Practice Problem
A helium-filled balloon has a volume of 50.0 L at 25 C and 1.08 atm. What volume will it have a atm and 10.0 C?
P1V1/T1 = P2V2/T2
P1V1T2/P2T1 = V2
V2 = 60.0 L He

27. COMBINED GAS LAW P T V
A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP.
COMBINED GAS LAW
GIVEN:
V1 = 7.84 cm3
P1 = 71.8 kPa
T1 = 25°C = 298 K
V2 = ?
P2 = kPa
T2 = 273 K
P T V
WORK:
P1V1T2 = P2V2T1
(71.8 kPa)(7.84 cm3)(273 K)
=( kPa) V2 (298 K)
V2 = 5.09 cm3
C. Johannesson

28. Boyle’s Law Pressure ´ Volume = Constant P1V1 = P2V2 (T = constant)
Pressure is inversely proportional (as one goes up,
other goes down) to volume when temperature is held constant.
Pressure ´ Volume = Constant
P1V1 = P2V2 (T = constant)

29. Boyle’s Law P1V1 = P2V2 Inverse Relationship T is constant
If one goes up, the other goes down
If one goes down, the other goes up

30. A Graph of Boyle’s Law

31. Practice Problem Pressure ´ Volume = Constant
A sample of oxygen gas has a volume of mL when its pressure is atm. What will the volume of the gas be at a pressure of atm if the temperature remains constant?
Pressure ´ Volume = Constant
P1V1 = P2V2 (T = constant)
(P1V1)/P2 = V2
V2 = 144 mL O2

32. A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa.
BOYLE’S LAW
GIVEN:
V1 = 100. mL
P1 = 150. kPa
V2 = ?
P2 = 200. kPa P V
WORK:
P1V1T2 = P2V2T1
(150.kPa)(100.mL)=(200.kPa)V2
V2 = 75.0 mL

33. Charles’ Law
The volume of a gas is directly proportional to temperature.
(P = constant)

34. Charles’ Law V1/T1 = V2/T2 Direct Relationship P is constant
Temp in Kelvin!!! (°C + 273)
As one goes up, so does the other
As one goes down, so does the other
higher temp = more kinetic energy, more collisions, so volume increases

35. Practice Problem V1T2/T1 = V2 V2 = 815 mL Ne
A sample of neon gas occupies a volume of 752 mL at 25 C. What volume will the gas occupy at 50 C if the pressure remains constant?
V1T2/T1 = V2
V2 = 815 mL Ne

36. A gas occupies 473 cm3 at 36°C. Find its volume at 94°C.
CHARLES’ LAW
GIVEN:
V1 = 473 cm3
T1 = 36°C = 309K
V2 = ?
T2 = 94°C = 367K T V
WORK:
P1V1T2 = P2V2T1
(473 cm3)(367 K)=V2(309 K)
V2 = 562 cm3
C. Johannesson

37. Some Vids:
wbURBw&NR=1
Explain on your own sheet of paper, using your newly gained knowledge of pressure, volume, and temperature what is occurring in these videos.

38. Gay Lussac’s Law The pressure and temperature of a gas are
directly related, provided that the volume
remains constant.

39. Gay-Lussac’s Law P T P1/T1 = P2/T2 Direct Relationship
Temp in Kelvin!!!
Volume is constant
Sample: The temperature of a gas goes from 30 °C to 50 °C. The starting pressure is 760 mm Hg, what is the ending pressure? P T

40. Practice Problem
The gas in a container is at a pressure of 3.00 atm at 25 C. Directions on the container warn the user not to keep it in a place where the temperature exceeds 52 C. What would the gas pressure in the container be at 52 C?

41. Practice Problem P1/T1 = P2/T2 P1T2/T1 = P2 P2 =3.27 atm
The gas in a container is at a pressure of 3.00 atm at 25 C. Directions on the container warn the user not to keep it in a place where the temperature exceeds 52 C. What would the gas pressure in the container be at 52 C?
P1/T1 = P2/T2
P1T2/T1 = P2
P2 =3.27 atm

42. A gas’ pressure is 765 torr at 23°C
A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr?
GAY-LUSSAC’S LAW
GIVEN:
P1 = 765 torr
T1 = 23°C = 296K
P2 = 560. torr
T2 = ? P T
WORK:
P1V1T2 = P2V2T1
(765 torr)T2 = (560. torr)(309K)
T2 = 226 K = -47°C
C. Johannesson

43. Ideal Gas Law PV = nRT P = pressure in atm V = volume in liters
n = moles
R = proportionality constant
= L atm/ mol·K
T = temperature in Kelvins

44. UNIVERSAL GAS CONSTANT
A. Ideal Gas Law
PV=nRT
UNIVERSAL GAS CONSTANT
R= Latm/molK
R=8.315 dm3kPa/molK
You don’t need to memorize these values!

45. Sample Problems…
You have 4 moles of helium gas at 1.5 atm and 25 L, what is the temperature?
How many moles of argon gas are present in 500 mL at 3 atm and 30 ºC?

46. Calculate the pressure in atmospheres of 0
Calculate the pressure in atmospheres of mol of He at 16°C & occupying 3.25 L.
GIVEN:
P = ? atm
n = mol
T = 16°C = 289 K
V = 3.25 L
R = Latm/molK
WORK:
PV = nRT
P(3.25)=(0.412)(0.0821)(289)
L mol Latm/molK K
P = 3.01 atm

47. V = ? n = 85 g T = 25°C = 298 K P = 104.5 kPa R = 8.315 dm3kPa/molK
Find the volume of 85 g of O2 at 25°C and kPa.
GIVEN:
V = ?
n = 85 g
T = 25°C = 298 K
P = kPa
R = dm3kPa/molK
WORK:
85 g 1 mol = 2.7 mol
32.00 g
= 2.7 mol
PV = nRT
(104.5)V=(2.7) (8.315) (298)
kPa mol dm3kPa/molK K
V = 64 dm3
C. Johannesson

48. Dalton’s Law of Partial Pressures
For a mixture of gases in a container,
PTotal = P1 + P2 + P
This is particularly useful in calculating the pressure of gases collected over water.

49. Temp (oC)
Vapor Pressure (mmHg)
-10
2.2 40
55.3
4.6 60
149.4 5
6.5 80
355.1 10
9.2 95
634 11
9.8 96
658 12
10.5 97
682 13
11.2 98
707 14
11.9 99
733 15
12.8
100
760 20
17.5
101
788 25
23.8
110
1074.6 30
31.8
120
1489 37
47.1
200
11659

50. Practice Problem
If I place 3 moles of N2 and 4 moles of O2 in a 35 mL container at a temperature of 25°C, what will the pressure of the resulting mixture of gases be?

51. Practice Problem Pressure of O2 = 2.796 atm
Use: PV = nRT to find pressure of each gas then add the two pressures together.
Pressure of N2 = atm
Pressure of O2 = atm
Total pressure of container = atm

52. Gas Stoichiometry
For gaseous reactants or products, the coefficients in chemical equations not only indicate molar amounts and mole ratios, but also volume ratios.
2 CO(g) + O2 (g)  2 CO2 (g)
2 volumes CO
1 volume O2

53. Gas Stoichiometry Moles  Liters of a Gas: STP - use 22.4 L/mol
Non-STP - use ideal gas law
Non-STP
Given liters of gas?
start with ideal gas law
Looking for liters of gas?
start with stoichiometry conv.

54. Practice Problem
Propane, C3H8 is a gas that is sometimes used as a fuel for cooking and heating.
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
What will be the volume, in L, of oxygen required for the complete combustion of L of propane?
What will be the volume of CO2 produced in the reaction? Assume all volume measurements are made at the same temperature and pressure.

55. Practice Problem
Propane, C3H8 is a gas that is sometimes used as a fuel for cooking and heating.
C2H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
What will be the volume, in L, of oxygen required for the complete combustion of L of propane?
1.75 L O2
b. What will be the volume of CO2 produced in the reaction? Assume all volume measurements are made at the same temperature and pressure.
1.05 L CO2

56. Practice Problem 2H2(g) + O2(g)  2H2O(l)
How many liters of water can be made from 55 grams of oxygen gas and an excess of hydrogen at STP?

57. Practice Problem 2H2(g) + O2(g)  2H2O(l)
How many liters of water can be made from 55 grams of oxygen gas and an excess of hydrogen at STP? = 77 L H2O

58. Standard Molar Volume
Equal volumes of all gases at the same temperature and pressure contain the same number of molecules.
- Amedeo Avogadro

59. Standard Molar Volume

60. Gas Stoichiometry
1 mole = 22.4 liters
At STP

61. Practice Problem What volume does 0.0685 mol of gas occupy at STP?
What quantity of gas, in moles, is contained in L at STP?

62. Practice Problem What volume does 0.0685 mol of gas occupy at STP?
2. What quantity of gas, in moles, is contained in 2.21 L at STP?
mol

63. Example 1 CaCO3  CaO + CO2 5.25 g ? L non-STP 5.25 g CaCO3 1 mol
What volume of CO2 forms from g of CaCO3 at 103 kPa & 25ºC?
CaCO3  CaO CO2
5.25 g
? L non-STP
Looking for liters: Start with stoich and calculate moles of CO2.
5.25 g
CaCO3
1 mol
CaCO3
100.09g
1 mol
CO2
CaCO3
= 1.26 mol CO2
Plug this into the Ideal Gas Law to find liters.

64. Example 2 V = 1.26 dm3 CO2 P = 103 kPa V = ? n = 1.26 mol
What volume of CO2 forms from g of CaCO3 at 103 kPa & 25ºC?
GIVEN:
P = 103 kPa
V = ?
n = 1.26 mol
T = 25°C = 298 K
R = dm3kPa/molK
WORK:
PV = nRT
(103 kPa)V =(1mol)(8.315dm3kPa/molK)(298K)
V = 1.26 dm3 CO2
C. Johannesson

65. Given liters: Start with Ideal Gas Law and calculate moles of O2.
Example 3
How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C?
4 Al O2  2 Al2O3
15.0 L
non-STP
? g
GIVEN:
P = 97.3 kPa
V = 15.0 L
n = ?
T = 21°C = 294 K
R = dm3kPa/molK
WORK:
PV = nRT
(97.3 kPa) (15.0 L) = n (8.315dm3kPa/molK) (294K)
n = mol O2
Given liters: Start with Ideal Gas Law and calculate moles of O2.
NEXT 
C. Johannesson

66. Use stoich to convert moles of O2 to grams Al2O3.
Example 4
How many grams of Al2O3 are formed from L of O2 at 97.3 kPa & 21°C?
4 Al O2  2 Al2O3
15.0L
non-STP
? g
Use stoich to convert moles of O2 to grams Al2O3.
0.597
mol O2
2 mol Al2O3
3 mol O2
g Al2O3
1 mol
Al2O3
= 40.6 g Al2O3