1. Kinetic Molecular Theory
The Kinetic Molecular Theory explains the forces between molecules and the energy that they possess. This theory has 3 basic assumptions.

2. 1. Matter is composed of small particles (molecules)
1. Matter is composed of small particles (molecules). Volume is the measure of space in between the molecules and not the space the molecules contain themselves.

3. 2. The molecules are in constant motion.
   Solid - Molecules are held close to each other by their attractions of charge. They will bend and/or vibrate, but will stay in close proximity.

4.    Liquid - Molecules will flow or glide over one another, but stay toward the bottom of the container. Motion is a bit more random than that of a solid.

5. Gas - Molecules are in continual straight line motion
   Gas - Molecules are in continual straight line motion. The kinetic energy of the molecule is greater than the attractive force between them, thus they are much farther apart and move freely of each other.

6. 3. When the molecules collide with each other, or with the walls of a container, there is no loss of energy.

7. Properties of Gases Gases can be compressed into smaller volumes
↑ P causes ↓ V
so: ↑ P causes ↑density

8. Properties of Gases
Gases exert pressure on surroundings. So to confine a gas, pressure must be exerted on it. (Newton’s third law: for every action, there is an equal and opposite reaction)

9. Properties of Gases
Gases expand without limits, therefore a gas will completely fill the container it occupies

10. Gas Properties
Collisions between gas particles and between particles and the container walls are elastic collisions

11. Properties of Gases
Gas particles are in constant, rapid, random motion. Have kinetic energy

12. Gas Properties
Fluidity Gas particles glide easily past one another since the attractive forces are minimal.

13. Properties of Gases
Gases diffuse into one another. Gases mix completely, ie gases are miscible.
The amount of gas can be known if the pressure, temperature and volume are known.

14. TEMPERATURE AND KINETIC ENERGY
Different substances have the same average KE when they are at the same temperature.
The velocity of the particles depends on their atomic or molecular mass (formula weight). The greater their mass, the slower their velocity.

15. Pressure Pressure – force / unit area Units: Newtons / m2 (Pascals)
Atmospheres (atm)
mm Hg
Torr
1 atm = 14.7 psi = 1.01 x 105 Pa =
760 mm Hg = 760 Torr =101.3kPa

16. Pressure

18. GAS LAWS THREE VARIABLES THAT DETERMINE GAS PRESSURE:
Temperature
Amount (# of moles)
Volume

19. GAS LAWS THREE VARIABLES THAT DETERMINE GAS PRESSURE:
Temperature – the average Kinetic Energy of the gas. KE = ½ mv²
If the Temperature ↑ then the Pressure ↑

20. GAS LAWS THREE VARIABLES THAT DETERMINE GAS PRESSURE:
Amount (# of moles) – the number of particles of a gas in a closed container
If the Amount ↑ then the Pressure↑

21. GAS LAWS THREE VARIABLES THAT DETERMINE GAS PRESSURE:
Volume of the sealed, but flexible container of the gas
If the Volume ↑ then the Pressure ↓
If the Volume ↓ then the Pressure ↑

22. Gas Laws
Boyle’s Law
↑ P causes ↓ V
PV = constant
P1V1 = P2V2

23. The PEEP in the Vacuum

24. STP
Standard Temperature and Pressure (STP) is defined as 0oC and 1 atm of pressure.

26. A balloon initially occupies 12. 4 L at 1. 00 atm
A balloon initially occupies 12.4 L at 1.00 atm. What will be the volume at atm?
P1V1 = P2V2
P1=
1.00 atm
V1=
12.4 L
P2=
0.800 atm
V2= ?
1.00 atm * 12.4L = V2*0.800 atm
P1V1 = V2P2
V2=
15.5L

27. A 5. 00 L container is filled with gas at 1. 00 atm
A 5.00 L container is filled with gas at 1.00 atm. A piston then compresses the gas to L. What is the new pressure?
P2 = ?
P1 =
1.00 atm
V2 =
0.250L
V1 =
5.00 L
1.00 atm* 5.00 L = 0.250L *X
P1V1 = P2V2
P1V1 = V2 P2
X = 20.0atm

28. Example #3: 2. 50 L of a gas was at an unknown pressure
Example #3: 2.50 L of a gas was at an unknown pressure. However, at standard pressure, its volume was measured to be 8.00 L. What was the unknown pressure?
End E 4/7

29. As the temperature increases, the volume increases.
Charles Law
As the temperature increases, the volume increases.
(in K not oC)
↑ T causes ↑ V
V/T = constant
V1/T1 = V2/T2

32. Charles’s Law
If the amount of a gas and its pressure remain constant, then
As the temperature of a gas increases, the volume of the gas increases.
↑T =↑V
As the temperature of a gas decreases, the volume of the gas decreases
↓T =↓V

33. V1/T1 = V2/T2 (T2*V1)/T1 = V2 691 mL = V2
A 450 mL piston assembly containing gas initially
at 293 K is suddenly heated to 450 K.
What will be the new volume of the piston assembly?
V1=
450 mL
T1=
293 K
(T2*V1)/T1 = V2
V2= ?
T2=
450K
(450K *
450 mL)
/293 K
=V2
691 mL = V2

34. Example #2: 4. 40 L of a gas is collected at 50. 0°C
Example #2: 4.40 L of a gas is collected at 50.0°C. What will be its volume upon cooling to 25.0°C?

35. Example #3: 5.00 L of a gas is collected at 100 K and then allowed to expand to 20.0 L. What must the new temperature be in order to maintain the same pressure?

36. KELVINS
EVERY TEMPERATURE USED IN A CALCULATION MUST BE IN KELVINS, NOT DEGREES CELSIUS.

37. Increased Temperature = Increased Pressure
Gay –Lussac’s Law
The pressure of a fixed mass of gas at a constant volume varies directly with the Kelvin Temperature Or
Increased Temperature = Increased Pressure

38. Gay –Lussac’s Law
If the temperature of a container is increased, the pressure increases.
↑T =↑P
If the temperature of a container is decreased, the pressure decreases.
↓T =↓P

39. Don’t forget temperature is in Kelvin
Gay –Lussac’s Law
P P2
T T2
Don’t forget temperature is in Kelvin =

40. If the pressure of a chamber was 100 atm at 600 K, what will be the temperature of the chamber if the pressure was raised to 600 atm?
P1=
100 atm
P2=
600 atm
T1=
600K
T2=
? K
600 atm
100 atm
P1 = P2
T T2 =
? K
600K

41. 100 atm
600 atm =
600K
? K
? K *
100 atm =
600 atm *
600K

42. ? K = 600 atm * 600K 100 atm ? K = ? K = 3600K 100 atm 600 atm = 600K

43. Example #2: 5. 00 L of a gas is collected at 22. 0°C and 745. 0 mmHg
Example #2: 5.00 L of a gas is collected at 22.0°C and mmHg. When the temperature is changed to standard, what is the new pressure?

44. Avogadro's Law
Gives the relationship between volume and amount when pressure and temperature are held constant.
V1 / n1 = V2 /n2 or V1 = V2
n1 n2

45. Avogadro's Law
If the amount of gas in a container is increased, the volume increases.
↑n =↑V
If the amount of gas in a container is decreased, the volume decreases.
↓n =↓v

46. 5. 00 L of a gas is known to contain 0. 965 mol
5.00 L of a gas is known to contain mol. If the amount of gas is increased to 1.80 mol, what new volume will result (at an unchanged temperature and pressure)?

47. THE GAS LAWS Boyle’s P1V1 = P2V2 Charles’s V1 / T1 = V2 / T2
Gay-Lussac's P1 / T1 = P2 / T2
Avogadro's V1 / n1 = V2/ n2

48. Combined Gas Law Write Boyle's Law: P1V1 = P2V2
Multiply by Charles Law:
P1V12 / T1 = P2V22 / T2
Multiply by Gay-Lussac's Law:
P12V12 / T12 = P22V22 / T22
Take the square root to get the combined gas law:
P1V1 / T1 = P2V2 / T2 or
P1V1T2 = P2V2T1

49. 2. 00 L of a gas is collected at 25. 0°C and 745. 0 mmHg
2.00 L of a gas is collected at 25.0°C and mmHg. What is the volume at STP?
P1 = P2 =
V1= V2=
T1= T2=
760 mm Hg
745.0 mm Hg X
2.00 L
( )
(0+273)
P1V1 / T1 = P2V2 / T2
745.0mm Hg 2.00L / 298 K = 760 mm Hg *X / 273 K
V2 = 1.79 L

50. Ideal Gas Law To derive the Ideal Gas Law, we combine all the gas laws
Each of the laws has a constant, k. Each k is different.
The product of all the constants (k) is written as R. R depends on the units used. R

51. R =Universal Gas Constant (Write this down)
Ideal Gas Law
PV= nRT
P = Pressure
V = Volume (must be in Liters)
n = number of moles
T = Temperature(must be in Kelvin)
R =Universal Gas Constant (Write this down)
R = (L*atm)/(mol*K) or
R = (L*kPa)/(mol*K)
R = (L*mm Hg)/(mol*K)

52. How many moles of gases are contained in a can with a volume of 555 mL and a pressure of kPa at 20oC? (R = L*kPa)/(mol*K)
N = mole

53. Calculate the pressure exerted by 43 moles of nitrogen in a 65 L cylinder at 5oC?
(Since there are no pressure units given you can use any R value you want)
P= 15 atm
P= 1529 kPa
P= mmHg

54. If I have 4 moles of a gas at a pressure of 5
If I have 4 moles of a gas at a pressure of 5.6 atm and a volume of 12 liters, what is the temperature?
PV=nRT
4 mole P=
5.6 atm
n = R=
(L*atm)/(mol*K) V=
12 L T= x
5.6 atm *12 L = 4 mol * L* atm *X
mol K
67.2 atm L = x L atm/K
67.2atm L / L atm /K = x
204.6 K

55. A steel tank having a volume of 5. 82 L contains 8
A steel tank having a volume of 5.82 L contains 8.78 g of O2 under a pressure of 1.35 atm.  What is the temperature -- in degrees Celsius -- of the O2?
PV=nRT
R = (L*atm)/mol*K)
1.35 atm *5.82L = (8.78g/32g/mol)(0.0821L atm/mol K) T
7.874 atm L = T
0.0225atm L/K
T= K = = 76.5oC

56. A sample of an unknown weighing 2. 1025 grams is found to occupy 2
A sample of an unknown weighing grams is found to occupy L at 22.0°C and .9737atm How many moles of the gas are present?
(0.97 atm) (2.85 L) = (n) ( L atm / mol K) (295.0 K)
2.775 atm L = n *24.20 L atm/mol
2.775 atm L = n
L atm
mole
mole = n
Mole buck Question What is the molar mass of the unknown gas?