1. 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

2. Force Pressure = Area Units of Pressure 1 pascal (Pa) = 1 N/m2
Barometer
Pressure =
Units of Pressure
1 pascal (Pa) = 1 N/m2
1 atm = 760 mmHg = 760 torr
1 atm = 101,325 Pa

3. 10 miles
0.2 atm
4 miles
0.5 atm
Sea level
1 atm

5. Boyle’s Law P a 1/V Constant temperature Constant amount of gas
P x V = constant
Boyle's Law ( scuba style! ) ACSBR physics mini project – YouTube
P1 x V1 = P2 x V2

6. P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL V2 = 154 mL P1 x V1
A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?
P1 x V1 = P2 x V2
P1 = 726 mmHg
P2 = ?
V1 = 946 mL
V2 = 154 mL
P1 x V1 V2
726 mmHg x 946 mL
154 mL =
P2 =
= 4460 mmHg

7. As T increases
V increases

8. Variation of gas volume with temperature at constant pressure.
Charles’ & Gay-Lussac’s Law
V a T
Temperature must be
in Kelvin
V = constant x T
V1/T1 = V2/T2
T (K) = t (0C)

9. V1/T1 = V2/T2 V1 = 3.20 L V2 = 1.54 L T1 = 398.15 K T2 = ? V2 x T1 V1
A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant?
V1/T1 = V2/T2
V1 = 3.20 L
V2 = 1.54 L
T1 = K
T2 = ?
V2 x T1 V1
1.54 L x K
3.20 L =
T2 =
= 192 K
charles law demonstration - Google Videos
Giant Koosh Ball in Liquid Nitrogen! - YouTube

10. 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

11. 4NH3 + 5O2 4NO + 6H2O 1 mole NH3 1 mole NO At constant T and P
Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure?
4NH3 + 5O NO + 6H2O
1 mole NH mole NO
At constant T and P
1 volume NH volume NO

12. Ideal Gas Law 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)
V a nT P
V = constant x = R nT P
R is the gas constant
PV = nRT

13. PV = nRT PV (1 atm)(22.414L) R = = nT (1 mol)(273.15 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.
PV = nRT
R = PV nT =
(1 atm)(22.414L)
(1 mol)( K)
R = L • atm / (mol • K)

14. PV = nRT nRT V = P 1.37 mol x 0.0821 x 273.15 K V = 1 atm V = 30.6 L
What is the volume (in liters) occupied by 49.8 g of HCl at STP?
T = 0 0C = K
P = 1 atm
PV = nRT
n = 49.8 g x
1 mol HCl
36.45 g HCl
= 1.37 mol
V =
nRT P
V =
1 atm
1.37 mol x x K
L•atm
mol•K
V = 30.6 L

15. PV = nRT n, V and R are constant nR V = P T = constant P1 T1 P2 T2 =
Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)?
PV = nRT
n, V and R are constant nR V = P T
= constant
P1 = 1.20 atm
T1 = 291 K
P2 = ?
T2 = 358 K P1 T1 P2 T2 =
P2 = P1 x T2 T1
= 1.20 atm x
358 K
291 K
= 1.48 atm
Egg in a Bottle.wmv - YouTube

16. 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

17. Deviations from Ideal Behavior of Gases
Deviation from ideal behavior is large at high pressure and low temperature
At lower pressures and high temperatures, the deviation from ideal behavior is typically small, and the ideal gas law can be used to predict behavior with little error.

18. Deviation from ideal behavior as a function of temperature
As temperature is decreased below a critical value, the deviation from ideal gas behavior becomes severe, because the gas CONDENSES to become a LIQUID.

19. J. D. van der Waals corrected the ideal gas equation in a simple, but useful, way
(a) In an ideal gas, molecules would travel in straight lines. (b) In a real gas, the paths would curve due to the attractions between molecules.

20. Mixtures of Gases Dalton's law of partial pressure states:
the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases.

21. Dalton's Law of Partial Pressure PT = P1 + P2 + P3 + …….

22. Partial Pressure in terms of mole fractionOR
XA Ptotal = PA
(XA = mole fraction of A and PA = partial pressure of gas A)

23. Mole fraction What is the mole Fraction of Gas A in mixture I?
What is the mole fraction of Gas B in mixture II?

24. XA = 3 = 0.2 3+4+5 PA = 2.5 atm PT = PA/ XA = 2.5/0.2 = 12.5 atm
Example: If there are 3 moles of gas A, 4 moles of gas B and 5 moles of gas C in a mixture of gases and the pressure of A is found to be 2.5 atm, what is the total pressure of the sample of gases?
XA = = 0.2
3+4+5
PA = 2.5 atm
PT = PA/ XA = 2.5/0.2 = 12.5 atm

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

26. Graham’s Law of Effusion
Graham’s law states that the rates of effusion of two gases are inversely proportional to the square roots of their molar masses at the same temperature and pressure:
Graham’s Law of Effusion
diffusion of a gas - YouTube

27. Graham’s Law of Effusion
The velocity of effusion is also inversely proportional to the molar masses:

28. Graham’s Law of Effusion
But the time required for effusion to take is directly proportional to the molar masses:
The density of the gas is also directly proportional to the molar masses:

29. Graham’s Law of Effusion
Compare the rate of effusion for hydrogen and oxygen gases.