1) When a hot air balloon is heated, some of the air escapes from the top, lowering the density inside and making the balloon buoyant. 2) The atmosphere protects the planet and provides chemicals necessary for life, including oxygen for metabolism, nitrogen to dilute oxygen, and carbon dioxide and water vapor that trap heat. 3) Gases have indefinite volume and shape, low densities, and high kinetic energies according to their temperature as described by the kinetic molecular theory.

1. The gases
The Gaseous State of Matter
The air in a hot air balloon expands
When it is heated. Some of the air
escapes from the top of the balloon,
lowering the air density inside the
balloon, making the balloon buoyant.

2. The atmosphere
The atmosphere protects the planet and
provides chemicals necessary for life.

3. The Atmosphere
Each of the gases, N2, O2, CO2 and H2O among others
in the atmosphere, serve a purpose
O2 supports metabolism
N2 dilutes the O2 so explosive combustion does not
take place and is also part of proteins
CO2 and H2O trap heat

4. General Properties
Gases
• Have an indefinite volume
Expand to fill a container
• Have an indefinite shape
Take the shape of a container
• Have low densities
d air = 1.2 g / L at 25°C
d H2O = 1.0 g / mL
• Have high kinetic energies

5. Kinetic Molecular Theory (KMT)
Assumptions of the KMT and ideal gases include:
1. Gases consist of tiny particles
2. The distance between particles is large compared
with the size of the particles.
3. Gas particles have no attraction for each other
4. Gas particles move in straight lines in all directions,
colliding frequently with each other and with the
walls of the container.

6. Kinetic Molecular Theory
Assumptions of the KMT (continued):
5. Collisions are perfectly elastic (no energy is lost in
the collision).
6. The average kinetic energy for particles is the same
for all gases at the same temperature.
1 2
KE = mv where m is mass and v is velocity
2
7. The average kinetic energy is directly proportional
to the Kelvin temperature.

7. Diffusion

8. Effusion
Gas molecules pass through a very small opening from
a container at higher pressure of one at lower
pressure.
Graham’s law of effusion:
rate of effusion of gas A
density B
molar mass B
=
=
rate of effusion of gas B
density A
molar mass A

9. Graham’s law of effusion
Rates of effusion of gases are inversely proportional
to the square roots of their molar masses
© 2013 John Wiley & Sons, Inc. All rights reserved.

10. Measurement of Pressure
Pressure =
Force
Area
Pressure depends on the
• Number of gas molecules
• Temperature of the gas
• Volume the gas occupies

11. Atmospheric Pressure
Atmospheric pressure is due to the mass of the
atmospheric gases pressing down on the earth’s
surface.

12. Barometer

13. Pressure Conversions
Convert 675 mm Hg to atm. Note: 760 mm Hg = 1 atm
1 atm
675 mm Hg ×
= 0.888 atm
760 mm Hg
Convert 675 mm Hg to torr. Note: 760 mm Hg = 760
torr.
760 torr
675 mm Hg ×
= 675 torr
760 mm Hg

14. Dependence of Pressure on
Number of Molecules
P is proportional to n
(number of molecules)
at Tc (constant T) and
Vc (constant V).
The increased pressure
is due to more frequent
collisions with walls of
the container as well
increased force of each
collision.

15. Dependence of Pressure on
Temperature
P is proportional to T at nc
(constant number of moles)
and Vc.
The increased pressure is
due to
• more frequent collisions
• higher energy collisions

16. Boyle’s Law
1
At Tc and nc : V α
P
or PV1 = PV2
1
2
What happens to V if you double P?
• V decreases by half!
What happens to P if you double V?
• P decreases by half!

17. Boyle’s Law
PV1 = PV2
1
2
A sample of argon gas occupies 500.0 mL at 920. torr.
Calculate the pressure of the gas if the volume is
increased to 937 mL at constant temperature.
Knowns V1 = 500 mL P1 = 920. torr V2 = 937 mL
Set-Up
PV1
P2 = 1
V2
Calculate
920. torr × 500. mL
P2 =
= 491 torr
937 mL

18. Boyle’s Law
Another approach to the same problem:
Since volume increased from 500. mL to 937 ml, the
pressure of 920. torr must decrease.
Multiply the pressure by a volume ratio that decreases
the pressure:
 500. mL
P2 = 920. torr 
 937 mL

÷= 491 torr


19. Charles’ Law
• The volume of an ideal gas at
absolute zero (-273°C) is zero.
• Real gases condense at their
boiling point so it is not
possible to have a gas with zero
volume.
• The gas laws are based on
Kelvin temperature.
• All gas law problems must be
worked in Kelvin!
At Pc and nc :
V α T or
V1
V2
=
T1
T2

20. Charles’ Law
V1
V2
=
T1
T2
A 2.0 L He balloon at 25°C is taken outside on a cold
winter day at -15°C. What is the volume of the
balloon if the pressure remains constant?
Knowns V1 = 2.0 L T1 = 25°C= 298 K T2 = -15°C = 258 K
Set-Up
Calculate
V1T2
V1
V2
rearranged gives V2 =
=
T1
T1
T2
(2.0 L)(258 K)
V2 =
= 1.7 L
298 K

21. Charles’ Law
Another approach to the same problem:
Since T decreased from 25°C to -15°C, the volume of
the 2.0L balloon must decrease.
Multiply the volume by a Kelvin temperature ratio
that decreases the volume:
 258K
P2 = 2.0L 
 298 K

÷= 1.7L


22. Gay-Lussac’s Law
At Vc and nc : P α T
or
P
P2
1
=
T1
T2

23. Combined Gas Laws
Used for calculating the results
of changes in gas conditions.
• Boyle’s Law where Tc
• Charles’ Law where
Pc
PV1
1
T1
=
P2V2
T2
PV1 = PV2
1
2
V1
V2
=
T1
T2
P
P2
1
=
T1
T2
• Gay Lussacs’ Law where
Vc
P1 and P2 , V1 and V2 can be any units as long as they
are the same. T1 and T2 must be in Kelvin.

24. Combined Gas Law
PV1
1
T1
=
PV2
2
T2
If a sample of air occupies 500. mL at STP, what is the
volume at 85°C and 560 torr?
STP: Standard Temperature 273K or 0°C
Standard Pressure 1 atm or 760 torr
Knowns
Set-Up
Calculate
V1 = 500. mL T1 =273K
P1= 760 torr
T2 = 85°C = 358K
P2= 560 torr
PV1T2
V2 = 1
T1 P2
(760 torr)(500. mL)(358K)
V2 =
= 890. ml
(273K)(560 torr)

25. Combined Gas Law
PV1
1
T1
=
PV2
2
T2
A sample of oxygen gas occupies 500.0 mL at 722 torr
and –25°C. Calculate the temperature in °C if the gas
has a volume of 2.53 L at 491 mmHg.
Knowns V1 = 500. mL T1 = -25°C = 248K P1= 722 torr
V2 = 2.53 L = 2530 mL
P2= 560 torr
T1 PV2
T2 = 2
Set-Up
PV1
1
Calculate
( 491 torr ) ( 2530 ml ) ( 248K ) =853K =
T2 =
( 722 torr ) ( 500.0 ml )
580°C

26. Dalton’s Law of Partial Pressures
The total pressure of a mixture of gases is the sum of
the partial pressures exerted by each of the gases in
the mixture.
PTotal = PA + PB + PC + ….
Atmospheric pressure is the result of the combined
pressure of the nitrogen and oxygen and other trace
gases in air.
PAir = PN2 + PO2 + PAr + PCO2 + PH 2O + ….

27. Collecting Gas Over Water
• Gases collected over water contain both the gas and
water vapor.
• The vapor pressure of water is
constant at a given temperature
• Pressure in the bottle is
equalized so that the Pinside = Patm
Patm = Pgas + PH 2 O

28. Avogadro’s Law
Equal volumes of different gases at the same T and P
contain the same number of molecules.
The ratio is
the same:
1 volume
1 molecule
1 mol
1 volume
1 molecule
1 mol
2 volumes
2 molecules
2 mol

29. Mole-Mass-Volume Relationships
Molar Volume: One mole of any gas occupies 22.4 L
at STP.
Determine the molar mass of a gas, if 3.94 g of the gas
occupied a volume of 3.52 L at STP.
Knowns m = 3.94 g V = 3.52 L T = 273 K P = 1 atm
Set-Up
Calculate
22.4 L
1 mol = 22.4 L so the conversion factor is
1mol
 3.94 g   22.4 L 

÷
÷ = 58.1g/mol
 1.52 L   1 mol 

30. Density of Gases
mass
g
d=
=
volume
L
Calculate the density of nitrogen gas at STP.
dSTP
 1 mol 
= molar mass 
÷
 22.4 L 
 28.02 g  1 mol 
dSTP = 
= 1.25g/L
÷
÷particular temperature,
Note that densities are mol  22.4 La
 1 always cited for 
since gas densities decrease as temperature increases.

31. Ideal Gas Law
L×
atm
PV = nRT where R = 0.0821
mol ×K
Calculate the volume of 1 mole of any gas at STP.
Knowns
Set-Up
n = 1 mole T = 273K P = 1 atm
nRT
V =
P
Calculate
Molar volume!
L ×atm
)(273 K)
mol ×K
= 22.4 L
(1 atm)
(1 mol)(0.0821
V =

32. Ideal Gas Law
L×
atm
PV = nRT where R = 0.0821
mol ×K
How many moles of Ar are contained in 1.3L at 24°C
and 745 mm Hg?
Knowns V = 1.3 L T = 24°C = 297 K
P = 745 mm Hg = 0.980 atm
Set-Up
Calculate
PV
n =
RT
(0.980 atm)(1.3 L)
n=
=0.052 mol
L ×atm
(0.0821
)(297 K)
mol ×K

33. Ideal Gas Law
Calculate the molar mass (M) of an unknown gas, if 4.12
g occupy a volume of 943mL at 23°C and 751 torr.
Knowns m =4.12 g
V = 943 mL = 0.943 L
T = 23°C = 296 K
P = 751 torr = 0.988 atm
Set-Up
g
n=
M
so
g
PV = RT
M
gRT
M=
PV
L ×atm
(4.12 g)(0.0821
)(296 K)
mol ×K
Calculate M =
=107 g/mol
(0.988 atm)(0.943 L)

34. Gas Stoichiometry
• Convert between moles and volume using the Molar
Volume if the conditions are at STP : 1 mol = 22.4 L.
• Use the Ideal Gas Law if the conditions are not at
STP.

35. Gas Stoichiometry
Calculate the number of moles of phosphorus needed to
react with 4.0L of hydrogen gas at 273 K and 1 atm.
P4(s) + 6H2(g)  4PH3(g)
Knowns
V = 4.0 L T = 273 K
Solution Map
Calculate
P = 1 atm
L H2  mol H2  mol P4
 1 mol H2   1 mol P4  = 0.030 mol P4
4.0 L H2 
÷ 
÷
22.4L   6 mol H2 


36. Gas Stoichiometry
What volume of oxygen at 760 torr and 25°C are needed to react
completely with 3.2 g C2H6?
2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(l)
Knowns
m = 3.2 g C2H6
Solution Map
Calculate
T = 25°C = 298K P = 1 atm
m C2H6  mol C2H6  mol O2  volume O2
 1 mol C 2 H 6   7 mol O2 
3.2g C2 H 6 
÷
÷= 0.37mol O2
 30.08g C2 H 6   2 mol C2 H 6 
L ×atm
(0.37 mol)(0.0821
)(298 K)
mol ×K
V =
= 9.1 L
(1 atm)

37. Volume-Volume Calculations
Calculate the volume of nitrogen needed to react with
9.0L of hydrogen gas at 450K and 5.00 atm.
N2(g) + 3H2(g)  2NH3(g)
Knowns
Solution Map
Calculate
V = 9.0 L T = 450K P = 5.00 atm
Assume T and P for both gases are the same.
Use volume ratio instead of mole ratio!
L H2  L N2
9.0 L H2
 1 L N2 

÷
3 L H2 

= 3.0 L N2

38. Real Gases
Most real gases behave like ideal gases under ordinary
temperature and pressure conditions.
Conditions where real gases don’t behave ideally:
• At high P because the distance between particles is
too small and the molecules are too crowded
together.
• At low T because gas molecules begin to attract each
other.
High P and low T are used to condense gases.

39. The law of effusion and diffusion
The postulate of the kinetic theory states that the
temperature of a system is proportional to the average
kinetic energy of its particles and nothing else.
In other words, two gases at the same temperature, such
as and must have the same average kinetic energy,
hence
multiplying by 2 gives:

40. Law of Effusion and diffusion
Then rearranging gives:
Taking out the square root both sides we get