1. States of matter
solids
liquids
gases
fixed shape
shape of container
independent
Kinetic model of gases
very small volumes
in constant motion
“elastic” collisions

2. Pressure
= force
force =
collisions of particles
area
atmospheric pressure
14.7 lbs
sea level
ft2
vacuum =
absence of gas particles

3. atmospheric pressure
barometer
Torricelli
760 mm Hg =
1 atm
760 mm Hg =
760 torr
760 mm Hg =
x 105 Pa
could use water
10,300 mm H2O
34 feet

4. Boyle’s Law
Pressure
Volume PV
= constant

5. Boyle’s Law
Pressure
Volume PV
= constant
P1V1
= P2V2
A sample of Cl2 has a volume of 946 mL
at a pressure of 726 mm Hg.
If the volume is reduced to 154 mL,
calculate the pressure.
(726 mm Hg)
(946 mL) = P2
(154 mL)
P2 =
4460 mm Hg
4460 mm Hg 
760 mm Hg/atm =
5.97 atm

6. Charles’s Law
Volume
Temperature V
= constant T
Kelvin
V = 0
T = oC

7. Charles’s Law
Volume
Temperature V1 T1 V2 T2 = K
A container is filled with N2 in an ice bath (0o C)
Its volume is 58.7 mL.
It is then put in a bath of boiling isopropanol.
Its volume goes to 76.4 mL.
What is the b.p. of isopropanol?
58.7 mL
76.4 mL
T2 =
355.5 K = K
82.3o C T2

8. Avogadro’s Law
Volume
moles
Gay-Lussac
2 H2
+ O2 
2 H2O (g)
Law of combining volumes
200 mL H2
100 mL O2
200 mL H2O
Avogadro
equal volumes of gases
= equal numbers of molecules
1 mol gas
= x 1023 particles
= 22.4 L
at standard Temperature and Pressure
0.0oC (273.2 K)
1 atm (760 torr)

9. Ideal Gas Law
Boyle PV
= constant P V = n R T
Charles V R
= ideal gas constant
= constant T =
L atm
Avogadro V
= constant
mol K n
two types of problems
1. Predicting properties of a system
PV = nRT
no changes
2. Changing conditions
P1V1
= R =
P2V2
n1T1
n2T2

10. Calculate the pressure of
80.10 g. of CO2 gas
in a 5000 mL flask
at 70.0oC
1. Predicting properties of a system
PV = nRT
P =
V =
5000 mL
= 5.00 L
n =
80.10 g
÷ g/mol
= 1.82 mol
T =
70.2
= K
(P)
(5.00 L)
= (1.82 mol)
( L atm)
( K)
mol K
(P) =
10.25 atm

11. A small bubble (2.1 mL) rises from the bottom of a lake,
where the temperature is 8.00C
and the pressure is 6.4 atm, to the surface
where the temperature is 25.0oC.
What is the volume of the bubble at the surface?
initial
final
P1=
6.4 atm
P2=
1.0 atm
V1 =
2.1 x 10-3 L
V2 =
14 mL
T1 = K
T2 = K
2. Changing conditions
P1V1
n1T1
= R =
n2T2
P2V2