1. GAS LAWS
Add a picture or 2.

2. Elements that exist as gases at 250C and 1 atmosphere

4. Pressure Remember: Pressure = Force / area or mass x gravity / area
Force = mass x gravity
Get rid of per…use force/area.
Explain force: mass x gravity so pressure = mass x gravity/area
Remember to explain what the pictures mean wrt pressure…that molecules exert an outward pressure on their container.

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

6. Measuring Pressure Devices used: Barometer-Invented by Torricelli
Pressure Gauge
Tire Gauge
Sphygmomanometer
What was invented by Torricelli?
Explain what each does and how it uses pressure (actually this might be a good unit project)

7. Units of Pressure
Standard atmospheric pressure (1 atm) = mm Hg or 760 torr (1 mm Hg = 1 torr)
In the United States we use inches in atmospheric pressure.
Pascal- used in science as a unit of pressure (Pa).
Kilopascals are used as well (KPa).
Good…remember periods…kids notice grammatical errors or omissions

8. Pressure Column height measures Pressure of atmosphere
1 standard atmosphere (atm) *
= 760 mm Hg (or torr) *
= inches Hg *
= 14.7 pounds/in2 (psi) *HD only
= kPa (SI unit is PASCAL) * HD only
= about 34 feet of water!
* Memorize these!

9. Practice How many torrs are in 4.5 atm?
The barometer reads 1,760 torr, how many atm?
I have a reading of 3700 torr, convert to atm.
You have 3 atm, how many torr?
good

10. Boyle’s Law Shows the relationship between Pressure and Volume.
The Law assumes that the temperature does not change.
Proposed by Robert Boyle.
What assumes? Boyle’s Law.
What does the lower right picture have to do with the law? You will need to explain it.
Some of the kids are going to struggle with volume…they know what it is but may have problems calculating it.

11. Boyle’s Law P1V1 = P2V2 P1= Pressure 1 V1= Volume 1 P2= Pressure 2
This formula works with any unit of Pressure (torr, Pa)
Any unit of volume works (mL, L).
Make sure you have the same units on both sides! If not, change one to match the other.
If one side has mL and the other has L, convert one to match the other.
Make sure it is written somewhere on the slide what P1, P2, V1, and V2 mean. The kids need to be able to go back to the lecture during homework and be reminded of how to do things.
Add the possible units of volume….L, ml, etc is fine.
Explain your 3rd statement

12. Boyle’s Law
A gas measured a volume of 100 mL under pressure of 740 torr. What would be the volume under a pressure of 780 torr with constant temperature?
Step 1: Figure out what you have.
P1 = 740 torr
V1= 100 mL
P2= 780 torr
V2= X
good

13. Boyle’s Law Step 2: Plug the numbers into the formula P1V1 = P2V2
(740 torr)(100 mL) = (780 torr)(X)
Step 3: Solve for x
74000 = 780X
X= mL
Get rid of the bullets on the formula and solving for x work…keep on step 2 and 3

14. Boyle’s Law So the new volume in this case is 94.87 mL
Since we increased the pressure, the volume is decreased.
The formula is a proportion. If something increases, something else will decrease!
Explain directly and indirectly proportional.
Include a picture with this so they can see what you are talking about…eg balloons changing volume under pressure. They are going to get very confused with this cuz it is kind of abstract.

15. Practice
P1V1 = P2V2
A sample of gas is confined to a 100 mL flask under pressure of 740 torr. If the same gas were transferred to a 50 mL flask, what’s the new pressure?
P2 = P1V1/V2
P1 =
V1 =
P2 =
V2 =
740 torr
100 ml X
50 ml
P2 = (740)(100)/50
good
P2 = 74000/50
P2 = 1480 torr

16. Practice
P1V1 = P2V2
2. You are given a gas that you measure under a pressure of 720 Pa. When the pressure is changed to 760 Pa, the volume became 580 mL. What is the first volume?
V1 = P2V2/P1
P1 =
V1 =
P2 =
V2 =
720 Pa X
760 Pa
580 ml
V1 = (760)(580)/720
good
V1 = /720
V1 = ml

17. P1V1 = P2V2
Practice
A pressure on 134 mL of air is changed to 1200 torr at a constant temperature, if the new volume is 45 mL what is the original pressure?
P1 = P2V2/V1
P1 =
V1 =
P2 =
V2 = X
134 ml
1200 torr
45 ml
P1 = (1200)(45)/134
good
P1 = 54000/134
P1 = torr

18. P1V1 = P2V2
Practice
4. An amount of Oxygen occupies 2 L when under pressure of 680 torr. If the volume is increased to 3 L what is the new pressure?
P2 = P1V1/V2
P1 =
V1 =
P2 =
V2 =
680 torr
2 L X 3L
P2 = (740)(100)/50
good
P2 = 74000/50
P2 = 1480 torr

19. Reminders Check your units
Temperature must remain constant in Boyle’s Law.
P= Pressure (torr, Pa, KPa).
V= Volume (mL, L).
Remind them here of what p and v are.
Add pictures to show volume and pressure.
Add periods.
“If you don’t exhale on your way up Cookie, your lungs will explode!
Robert De Niro as Chief Saturday

21. When the temperature increases, the volume increases.
Good picture…might want to add some explanation so they can look back and remember what the slide was about.

22. As T increases
V increases

23. Charles Law
Shows a relationship between Volume and Temperature. One thing increase, so does the other. Both behave the same way.
Pressure remains constant.
V= Volume
T= Temperature
Good…make sure to explain the proportion happening here

24. Few things first… ALL TEMPERATURES MUST BE CONVERTED TO KELVIN!
The Law only works with Kelvin
If you have Celsius, make sure you convert it to Kelvin.
Remember Kelvin = Celsius + 273

25. Temperature Conversion Practice=
358K
Convert 85°C to K. ____________________
Convert 376K to °C. ____________________
Convert 154K to °C. ____________________
Convert -65°C to K. ____________________
Convert 0°C to K. _____________________
Convert 0K to °C. _____________________ =
103 °C =
-119 °C
208K =
273K = =
-273 °C

26. Few Practice Problems
V1/T1 = V2/T2
You heat 100 mL of a gas at 25 C to 80 C. What is the new volume of the gas?
V2 = V1T2/T1
V2 = (100)(353)/298
V1 =
T1 =
V2 =
T2 =
100 ml
25ºC X
80ºC
Add some pictures about charles law
+ 273 =
298K
V2 = 35300/298
V2 = ml
+ 273 =
353K

27. Few Practice Problems
V1/T1 = V2/T2
You have 200 mL of a gas at 55 C and you freeze it to 0 C. What is the new volume?
V2 = V1T2/T1
V2 = (200)(273)/328
V1 =
T1 =
V2 =
T2 =
200 ml
55ºC X
0ºC
Add some pictures about charles law
+ 273 =
328K
V2 = 54600/328
V2 = ml
+ 273 =
273K

28. Few Practice Problems
V1/T1 = V2/T2
A 150 mL sample of gas is at 125 K. After cooling the new volume is 80 mL. What is the new temperature?
T2 = V2T1/V2
T2 = (80)(125)/150
V1 =
T1 =
V2 =
T2 =
150 ml
125K
80 ml X
Add some pictures about charles law
T2 = 10000/150
T2 = 66.67K

29. Gay-Lussac’s Law P1/T1 = P2/T2
Shows relationship between pressure (P) and temperature (T). If Temperature increases, so does the pressure.
Based on Charles’ Law.
Volume remains constant.
Like Charles’ Law, all temperature units must be in KELVIN.
Used in pressure cookers and autoclaves.

30. Practice
P1/T1 = P2/T2
A soda bottle with a temperature of 25ºC and 3 atm of pressure was put into a freezer with a temperature of –1°C. What is the pressure on the bottle inside the freezer?
P2 = P1T2/T1
P1 =
T1 =
P2 =
T2 =
3 atm
25ºC X
-1ºC
P2 = (3)(272)/298
+ 273 =
298K
P2 = 816/298
P2 = 2.74 atm
+ 273 =
272K

31. Combined Gas Law
Combines Boyle’s Law, Gay-Lussac’s Law, and Charles’ Law.
Shows the relationship of Pressure, Volume and Temperature.
Have you explained gay-lussacs law? How can they get the combined law without it? Add pictures

32. Combined Gas Law Derived from Boyle’s, Gay-Lussac, and Charles Law.
Used when both temperature and pressure changes.
Can be used to find a constant.
If two things increase, one thing will decrease.
Math was used to derive the law.
Remember to explain the proportions…as one goes up, what goes down/up?

33. Combined Gas Law (P, V, T)
Combined Gas Law: nR = PV T

34. 1. A gas has a volume of 800. 0 mL at -23. 00 °C and 300. 0 torr
1. A gas has a volume of mL at °C and torr. What would the volume of the gas be at °C and torr of pressure?
P1V1 = P2V2
T T2
P1 =
V1 =
T1 =
P2 =
V2 =
T2 =
300 torr
800 ml
-23ºC
600 torr X
227.0ºC
V2 = P1V1T2
T1 P2
+ 273 =
250K
V2 = (300)(800)(500)
(250)(600)
V2 = 800 ml
+ 273 =
500K

35. Remember this…
When you are working with Charles’ Law and Combined Gas Law, make sure you have the proper units!
The laws help us predict the behavior of gases under different circumstances.
Go back and explain why all these laws are needed…the kids need reasoning behind what they are learning…helps it stick and helps them want to learn more (sometimes)

36. How is this possible?
KC-135 Jet
Rail Car 

37. Avogadro’s Law V1/n1 = V2/n2
Allows us to calculate the number of moles (n) of a gas with in a volume (V).
Pressure and Temperature remains the same.
If volume increases, so does the number of moles.

39. Practice
V1/n1 = V2/n2
In 150 mL sample, there are 2.5 grams of Cl2 gas, if I double the volume to 300 mL how much Cl2 gas in moles, is in the new sample?
n2 = V2 n1/ V1
V1 =
n1 =
V2 =
n2 =
150 ml
2.5g
300 ml x
n2 = (300)(0.147)/150
n2 = 44.1/150
n2 = m
÷ 17g/m =
0.147 m

40. Ideal gas law Formula: PV= nRT P= Pressure in atmospheres
V= Volume in Liters
n= number of moles of a gas
R= constant, L • atm / mol • K
T= Temperature in Kelvin

41. Ideal Gas Law
An ideal gas is a gas that obeys Boyle’s and Charles’ Law.
Based on the Kinetic Theory.
Shows the relationship between Pressure, Volume, number of moles and Temperature.
Make sure everything has the proper unit.
Can be used to figure out density of gas or determine the behavior of a tire.

42. Practice
PV = nRT
How many moles of gas are contained in mL at 21.0 °C and mm Hg pressure?
÷ 760 mmHg/atm = atm
P =
V =
n =
R = Latm/molK
T =
750.0 mm Hg
890.0 ml X
21.0ºC
÷ 1000 ml/L = 0.89 L
n = PV/RT
n = (0.987)(0.89)/(0.0821)(294)
n = 0.878/24.1
n = m
+ 273 = 294 K

43. And now, we pause for this commercial message from STP
OK, so it’s really not THIS kind of STP…
STP in chemistry stands for Standard Temperature and Pressure
Standard Pressure = 1 atm (or an equivalent)
Standard Temperature = 0 deg C (273 K)
STP allows us to compare amounts of gases between different pressures and temperatures

44. 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)

45. 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

46. GAS DENSITY
Low
density
22.4 L of ANY gas AT STP = 1 mole
High
density

47. 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

48. Gas Stoichiometry
What is the volume of CO2 produced at 370 C and 1.00 atm when 5.60 g of glucose are used up in the reaction:
C6H12O6 (s) + 6O2 (g) CO2 (g) + 6H2O (l)
g C6H12O mol C6H12O mol CO V CO2
1 mol C6H12O6
180 g C6H12O6 x
6 mol CO2
1 mol C6H12O6 x
5.60 g C6H12O6
= mol CO2
0.187 mol x x K
L•atm
mol•K
1.00 atm =
nRT P
V =
= 4.76 L

49. Gases and Stoichiometry
2 H2O2 (l) ---> 2 H2O (g) + O2 (g)
Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the volume of O2 at STP?
Bombardier beetle uses decomposition of hydrogen peroxide to defend itself.

50. Deviations from Ideal Gas Law
Real molecules have volume.
The ideal gas consumes the entire amount of available volume. It does not account for the volume of the molecules themselves.
There are intermolecular forces.
An ideal gas assumes there are no attractions between molecules. Attractions slow down the molecules and reduce the amount of collisions.
Otherwise a gas could not condense to become a liquid.

51. Gases in the Air The % of gases in air Partial pressure (STP)
78.08% N mm Hg
20.95% O mm Hg
0.94% Ar mm Hg
0.03% CO mm Hg
PAIR = PN + PO + PAr + PCO = 760 mm Hg
Total Pressure mm Hg 2 2 2

52. Dalton’s Law of Partial Pressures
V and T are constant P1 P2
Ptotal = P1 + P2

53. Dalton’s Law
John Dalton

54. Consider a case in which two gases, A and B, are in a container of volume V.
PA =
nART V
nA is the number of moles of A
PB =
nBRT V
nB is the number of moles of B
XA = nA
nA + nB
XB = nB
nA + nB
PT = PA + PB
PA = XA PT
PB = XB PT
Pi = Xi PT

55. A sample of natural gas contains 8. 24 moles of CH4, 0
A sample of natural gas contains 8.24 moles of CH4, moles of C2H6, and moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)?
Pi = Xi PT
PT = 1.37 atm
0.116
Xpropane = =
Ppropane = x 1.37 atm
= atm

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

57. Collecting a gas “over water”
Gases, since they mix with other gases readily, must be collected in an environment where mixing can not occur. The easiest way to do this is under water because water displaces the air. So when a gas is collected “over water”, that means the container is filled with water and the gas is bubbled through the water into the container. Thus, the pressure inside the container is from the gas AND the water vapor. This is where Dalton’s Law of Partial Pressures becomes useful.

58. Table of Vapor Pressures for Water

59. Solve This!
A student collects some hydrogen gas over water at 20 degrees C and 768 torr. What is the pressure of the gas?
768 torr – 17.5 torr = torr

60. GAS DIFFUSION AND EFFUSION
diffusion is the gradual mixing of molecules of different gases.
effusion is the movement of molecules through a small hole into an empty container.

61. GAS DIFFUSION AND EFFUSION
Graham’s law governs effusion and diffusion of gas molecules.
Rate of effusion is inversely proportional to its molar mass.
Thomas Graham, Professor in Glasgow and London.

62. GAS DIFFUSION AND EFFUSION
Molecules effuse thru holes in a rubber balloon, for example, at a rate (= moles/time) that is
proportional to T
inversely proportional to M.
Therefore, He effuses more rapidly than O2 at same T. He

63. Gas Diffusion relation of mass to rate of diffusion
HCl and NH3 diffuse from opposite ends of tube.
Gases meet to form NH4Cl
HCl heavier than NH3
Therefore, NH4Cl forms closer to HCl end of tube.