1. The GAS LAWS

2. Gas Properties Gases have mass Gases diffuse
Gases expand to fill containers
Gases exert pressure
Gases are compressible
Pressure & temperature are dependent

3. Gas Variables Units of volume (L) Units of amount (moles)
Volume (V)
Units of volume (L)
Amount (n)
Units of amount (moles)
Temperature (T)
Units of temperature (K)
Pressure (P)
Units of pressure (mmHg)
Units of pressure (KPa)
Units of pressure (atm)

4. A Little Review P1V1 = P2V2 V1 = V2 T1 T2 Boyle’s law
pressure & volume
as P then V
at constant T, n
P1V1 = P2V2
Charles’ law:
Temperature & volume
As T then V
At constant P, n
V1 = V2
T1 T2

5. A Little Review P1 = P2 T1 T2 Gay-Lussac’s law: Temperature & pressure
As P then T
At constant V, n
P1 = P2
T1 T2

6. A Problem
Blow up a balloon, measure its volume then take it to the top of Mt. Davis.
Explain what would happen to its volume.
What variables would you need to use to measure the change?
How do those variables change as you go up to the top of Mt. Davis?

7. PV=k1 V/T=k2 P/T=k3 Now What??
If we combine all of the relationships from the 3 laws covered thus far (Boyle’s, Charles’s, and Gay-Lussac’s) we can develop a mathematical equation that can solve for a situation where 3 variables change :
PV=k1
V/T=k2
P/T=k3

8. Combined gas law P1V1 T1 = P2V2 T2 P1V1T2 = P2V2T1
The law expresses the relationship between changes in volume, pressure, & temperature
# of moles is held constant
P1V1 T1 =
P2V2 T2
P1V1T2 = P2V2T1

9. Example problem - V1  - T1  - P2  - V2  - T2  1atm 2.0 atm ?
A gas with a volume of 4.0L at STP. What is its volume at 2.0atm and at 30°C?
- P1 
- V1 
- T1 
- P2 
- V2 
- T2 
1atm
2.0 atm ?
4.0 L
273K
30°C + 273
= 303K

10. 2.22L = V2 Example problem P1V1 T1 = P2V2 T2 (1 atm) (4.0L) (2 atm)
(273K)
(303K)
2.22L = V2

11. Gay-Lussac’s other Law
Gay-Lussac’s Law of combining volumes of gases
The law states that at constant temperature and pressure, the volumes of gaseous reactants and products can be expressed as ratios of small whole numbers.

12. Gay-Lussac’s other Law
In the early 1800s, French chemist Joseph Gay-Lussac studied gas volume relationships involving a chemical reaction between hydrogen and oxygen. He observed that 2 L of hydrogen can react with 1 L of oxygen to form 2 L of water vapor at constant temperature and pressure.
Hydrogen gas Oxygen gas  Water vapor
2 L (2 volumes) L (1 volume) L (2 volumes)
This reaction showed a simple and definite 2:1:2 relationship between the volumes of the reactants and the product.
Gay-Lussac also noticed simple and definite proportions by volume in other reactions of gases, such as in the reaction between hydrogen gas and nitrogen gas.
Hydrogen gas + Nitrogen gas  Ammonia gas (NH3)
(3 volumes) + (1 volume) = (2 volumes)

13. Gay-Lussac Contributions
This simple observation, combined with the insight of Avogadro, provided more understanding of how gases react and combine with each other. In fact it lead to the discovery of three vital chemistry concepts.
1) Diatomic Molecules
2) Gas Density and molecular mass comparison and determination
(Led to determination of Atomic mass)
3) Avogadro's Principle  The Mole

14. Avogadro’s Contribution
If the amount of gas in a
container is increased, the
volume increases.
container is decreased, the
volume decreases.
This can be proven simple by blowing up a balloon.

15. Amedeo Avogadro Amedeo Avogadro lived in Italy, 6.022 x1023  The Mole
And his work means an awful lot to chemistry.
Amedeo Avogadro may be dead – but,
We all should remember what he said.
6.022 x1023  The Mole

16. Chorus Equal Volumes of Gases, at the same temp and pressure,
Have the same number of molecules.
Amedeo…..Avogadro, that’s his hypothesis.

17. Amedeo Avogadro You got two balloons filled with different gases,
Equal volumes but they both have different masses,
If conditions are the same everywhere,
Tell me how the number of molecules compare.

18. Chorus Equal Volumes of Gases, at the same temp and pressure,
Have the same number of molecules.
Amedeo…..Avogadro, that’s his hypothesis.

19. Amedeo Avogadro Why is his hypothesis so great? You can use it to find
Mendeleev used it
Why is his hypothesis
so great?
You can use it to find
atomic weights.
Its based on the ratio of the masses,
Of equal volumes of those gases.
Atomic Mass

20. Chorus Equal Volumes of Gases, at the same temp and pressure,
Have the same number of molecules.
Amedeo…..Avogadro, that’s his hypothesis.

21. The GAS LAWS AT WORK

22. Amedeo Avogadro Avogadro helped shaped atomic theory
But when I think about it I
get sorta teary.
Cause no one believed in what he said,
Till four years after he was dead.

23. Amedeo Avogadro When you speak your mind and everyone ignores you,
No one’s standing up and rooting for you,
So what if they snicker and they laugh,
Someday they will want your autograph.

24. Chorus Equal Volumes of Gases, at the same temp and pressure,
Have the same number of molecules.
Amedeo…..Avogadro, that’s his hypothesis.

25. V = kn Avogadro’s Law So far we’ve compared all the variables except
the amount of a gas (n).
There is a lesser known law called Avogadro’s Law which relates V & n.
It turns out that they are directly related to each other.
As # of moles increases then V increases.
V = kn

26. Avogadro’s Law
Equal volumes of gases at the same T and P have the same number of molecules.
V = n (T/P) = kn
V and n are directly related.
twice as many molecules

27. Avogadro’s Law V1 n1 = V2 n2 Example :
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)?
5.00 L = V
0.965 mol mol
V2 = 9.33 L

28. If conditions are the same the volumes are the same!
IF the MOLE is BACK… STOICHIOMETRY is BACK !!
Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many liters of H2O are obtained from 3.5 liter of ammonia at STP?
If conditions are the same the volumes are the same!
4NH3 + 5O NO + 6H2O
3.5 L NH3
Y mol NH3
6 mol H2O
X L H2O
= L H2O 1
X L NH3
4 mol NH3
Y mol H2O

29. What is Standard molar Volume?
Experiments show that at STP, 1 mole of an ideal gas occupies L.

30. Complete worksheet problems 1- 6
How can we use 22.4L?
Example #1 : What is the volume of 10 moles of CO2 gas at STP?
Ex. #2 What is the density of H2O vapor at STP?
D = mass/volume
Molar mass H2O = 18.0 g/mol
Molar volume of H2O = 22.4 L/mol
D = 18.0g/mol
22.4 L/mol
10 mol CO2
22.4 L of CO2
= 224 L CO2 1
1 mole CO2
= g/L
Complete worksheet problems 1- 6

31. What would that formula look like? How would it be arranged?
Would it not be Grand if we could
have only one PERFECT law
So far we have held at least one variable constant for every law we have used.
What if we combined all the variables and created a constant to hold them together.
What would that formula look like?
How would it be arranged?

32. Temp is measured on the Kelvin Scale!
How about another song?
THE IDEAL GAS LAW
Temp is measured on the Kelvin Scale!
Pressure is P
Volume is V
n is number of mole
Temperature is T
P & V
PV = nRT

33. PV = nRT BOYLES + CHARLES + Gay-LUSSAC + AVOGADRO LAWS
THE IDEAL GAS LAW
BOYLES + CHARLES Gay-LUSSAC
+ AVOGADRO LAWS
COMBINED TOGETHER
PV = nRT

34. Ideal Gas Law PV = R nT PV = nRT
If we combine all of the laws together including Avogadro’s Law mentioned earlier we get: PV nT
= R
Where R is the universal gas constant
Normally
written as
PV = nRT

35. Ideal Gas Constant (R) R is a constant that connects all
of the four gas variables
R is dependent on the units of the variables for P, V, & T
Temp is always in Kelvin
Volume is in liters
Pressure is in either atm or mmHg or kPa

36. Ideal Gas Constant L•atm R=.0821 mol•K L•mmHg R=62.4 mol•K L•kPa
Because of the different pressure units there are 3 possibilities for our ideal gas constant
R=.0821
L•atm
mol•K
If pressure is given in atm
R=62.4
L•mmHg
mol•K
If pressure is given in mmHg
R=8.31
L•kPa
mol•K
If pressure is given in kPa

37. Using the Ideal Gas Law ? = .3635 mol What volume does 9.45g
of C2H2 occupy at STP?
.0821
L•atm
mol•K
P 
1atm
R  ?
V 
T 
273K
9.45g
26g
n 
= mol

38. PV = nRT V = 8.15L (1.0atm) (V) = (.0821 ) (.3635mol) (273K) (1.0atm)
( )
L•atm
mol•K
(.3635mol)
(273K)
(1.0atm)
(V)
(8.147L•atm) =
V = 8.15L

39. ? = 68.2 mol 303kPa 298K 3000g 44g P  R  V  T  n 
A camping stove propane tank holds 3000g of C3H8. How large a container would be needed to hold the same amount of propane as a gas at 25°C and a pressure of 303 kpa?
8.31
L•kPa
mol•K
P 
303kPa
R  ?
V 
T 
298K
3000g
44g
n 
= 68.2 mol

40. PV = nRT V = 557.7L (303kPa) (V) = (8.31 ) (68.2 mol) (298K) (303kPa)
( )
L•kPa
mol•K
(68.2 mol)
(298K)
(303kPa)
(V)
(168,970.4 L•kPa) =
V = 557.7L

41. The Ideal Gas Equations

42. Try These 5 problems = 1.96 g/L 44.0 g/mol 22.4 L/mol
1) Calculate the density of one mole of CO2 gas at STP. {D = m/V}
D = m/V Molar Mass = 44.0 g/mol ::
Molar volume = 22.4 L/mol
D =
44.0 g/mol
= 1.96 g/L
22.4 L/mol
2) What is the density of laughing gas(N2O4) released at 25°C and 1.02 atm?
D = PM/RT = (1.02 atm)(92g/mol)
(0.0821L*atm/mol*K)(298K)
= 3.83 g/L

43. Try These 5 problems
3) A g gas has a volume of 200 ml at STP. Is this gas
propane (C3H8), butane (C4H10) or something else?
Using the Ideal Gas Law PV = mRT/M  M = mRT/PV = (0.519 g)( L atm/mol K)(273K)
(1 Atm) (.200 L)
= 58.1 g/mol the molar mass of butane
4) A 1.25 g sample of a gaseous product of a chemical
reaction was found to have a volume of 350 ml at 20.0°C
and 750 mm Hg. What is the molar mass of this gas?
5) What is the volume of a gas whose molar mass is 90.2 g/mol.
1.39 g of the gas is contained at 755 mm Hg and 22 °C

44. Try These 5 problems
4) Using the Ideal Gas Law PV = mRT/M  M = mRT/PV
= (1.25 g)(62.4 L mm Hg/mol K)(293K)
(750 mm Hg) (.350 L) = 87.0 g/mol
5) Using the Ideal Gas Law PV = mRT/M  V = mRT/PM
= (1.39 g)(62.4 L mm Hg/mol K)(295K)
(755 mm Hg) (90.2 g/mol) =.376 L

45. PV = nRT (1.00 atm) (1.00 L) nH2O= (573K) (.0821L atm/mol K)
Ideal Gas Law & Stoichiometry
What volume of hydrogen gas must be burned to form 1.00 L of water vapor at 1.00 atm pressure and 300°C?
(hint: Solve for mole of H2O then convert)
PV = nRT
(1.00 atm)
(1.00 L)
nH2O=
(573K)
(.0821L atm/mol K)
nH2O= mols

46. Ideal Gas Law & Stoichiometry
2H2 + O2  2H2O
2 mol H2O
2 mol H2 =
1mol H2
22.4 L H2
mol
.476 L H2