1. It’s a Gas, Gas, Gas!!!

2. Defining a Gas For a fixed number of particles Volume is determined by
Its temperature
Its pressure

3. What is Pressure? Pressure  Force exerted per unit area
Where does the force come from in a gas?
The particles colliding with the sides of the container
 collisions   pressure
and if you
 temperature   collisions

4. kilopascals and atmosphere
1 kilopascal (kPa) = 10 gms / 1 cm2
1 atmosphere = kPa

5. More Pressure Similarly you know that  collisions   pressure
                                             
Similarly you know that
 collisions   pressure
So if you
 volume   collisions

6. The Gas Laws

7. Charles’ Law  temperature   volume
for a fixed amount of gas at constant pressure, the temperature and volume are directly proportional:
 temperature   volume V T
- 273oC P1 P2
V V2
T T2 =

8. Guy Lussac’s Law  temperature   pressure
for a fixed amount of gas at a constant volume, the temperature and pressure are directly proportional:
 temperature   pressure
1 atm
2 atm P T
- 273oC V1 V2
P P2
T T2 =

9. Boyle’s Law  volume   pressure
for a fixed amount of gas at a constant temperature, the volume and pressure are inversely proportional:
 volume   pressure P V T1 T2
P1 V P2 V2 =

11. Ideal Gas Law – including moles
Suppose we include the number of moles in the Combined Gas Law, then we get:
Where n represents the number of moles in the gas.
P1 V P2 V2
n1 T n2 T2 =

12. Ideal Gas Law – R rated
Given this relationship, each side of the equation is a constant value or
Where n represents the number of moles in the gas and R represents the constant.
P V
n T = R

13. Ideal Gas Law – R value
Substituting in the values for a gas at STP, we get
Where n represents the number of moles in the gas and R represents the constant.
P V
n T
(101.3kPa) (22.4L) = R = =
8.31 kPa L
mol oK
(1 mol) (273oK)

14. Departure from Ideal Gas Law
Particles are not spherical
Intermolecular Forces are still present
Collisions are not completely elastic

15. Law of Partial Pressures
PT = P1 + P2 = …

16. Graham’s Law
RateA molar massB
RateB molar massA =

17. Graham’s Law – Proof! Rate1 molar mass2 Rate2 molar mass1 = KE = ½ mv2
If two gases are at same temperature
KE1 = KE2 or
½m1v12 = ½ m2v22
m1v12 = m2v22
v12 / v22 = m2 / m1
Rate molar mass2
Rate molar mass1 =