1. Gas Laws Boyle’s Law Charle’s law Gay-Lussac’s Law Avogadro’s Law
Dalton’s Law
Henry’s Law

2. How would you calculate the pressure of the gas in this open-end manometer in mm Hg? There is mercury in the tube.
Pgas = h
Pgas = Patm
Pgas = h + Patm
Pgas = h - Patm
Pgas = Patm - h
Pgas =Patm + 2h
gas is more than air by h - so #3

3. How would you calculate the pressure of the gas in this open-end manometer in mm Hg? There is mercury in the tube.
Pgas = h
Pgas = Patm
Pgas = h + Patm
Pgas = h - Patm
Pgas = Patm - h
Pgas =Patm + 2h
gas is more than air by h - so #3

4. How would you calculate the pressure of the gas in this open-end manometer in mm Hg? There is mercury in the tube.
Pgas = h
Pgas = Patm
Pgas = h + Patm
Pgas = h - Patm
Pgas = Patm - h
Pgas =Patm + 2h
gas is more than air by h - so #3

5. How would you calculate the pressure of the gas in this open-end manometer in mm Hg? There is mercury in the tube.
Pgas = h
Pgas = Patm
Pgas = h + Patm
Pgas = h - Patm
Pgas = Patm - h
Pgas =Patm + 2h
gas is more than air by h - so #3

6. How would you calculate the pressure of the gas in this closed-end manometer in mm Hg? There is mercury in the tube.
Pgas = h
Pgas = Patm
Pgas = h + Patm
Pgas = h - Patm
Pgas = Patm - h
Pgas =Patm + 2h
Closed
gas is more than air by h - so #3

7. How would you calculate the pressure of the gas in this closed-end manometer in mm Hg? There is mercury in the tube.
Pgas = h
Pgas = Patm
Pgas = h + Patm
Pgas = h - Patm
Pgas = Patm - h
Pgas =Patm + 2h
Closed
gas is more than air by h - so #3

8. Which graph below best represents the relationship between moles of gas and volume of container? (holding pressure and temperature constant.) 1 2 3 4
moles
Volume

9. Which graph below best represent the relationship between moles of gas and volume of container? (holding pressure and temperature constant.) 1 2 3 4
moles
Volume

10. number of molecules and volume
moles and volume
n & V
increase the number of moles, (while pressure & temp remain constant) and the volume will increase.
This is a direct relationship
n is proportional to V
Avogadro’s Law
k = a constant

11. There is gas in the flask, and more gas is added with the syringe
There is gas in the flask, and more gas is added with the syringe. What would the pressure gauge read in the second picture? ? 5 1 2 3 5 8 2

12. There is gas in the flask, and more gas is added with the syringe
There is gas in the flask, and more gas is added with the syringe. What would the pressure gauge read in the second picture? ? 5 1 2 3 5 8 2

13. number of molecules and pressure
moles and pressure
n & P
increase the number of moles (while maintaining constant volume and temp) and the pressure increases.
this is a direct relationship
n is proportional to P
k = a constant

14. higher than lower than the same as
The height of the piston in the container on the right would be ______ the height of the gas in the container on the left. ??
higher than
lower than
the same as

15. higher than lower than the same as
The height of the piston in the container on the right would be ______ the height of the gas in the container on the left.
higher than
lower than
the same as

16. temperature and volume
T & V
increase the temperature (while pressure remains constant) and the volume increases.
this is a direct relationship
V is proportional to T
Charles’s Law
k = a constant

17. higher than lower than the same as
The pressure of the gas in the container on the right would be ______ the pressure of the gas in the container on the left. (Temperature and moles remain constant.) ??
higher than
lower than
the same as

18. higher than lower than the same as
The pressure of the gas in the container on the right would be ______ the pressure of the gas in the container on the left.
higher than
lower than
the same as

19. pressure and volume P & V
increase the volume (while temp and number of moles remains constant) and the pressure decreases.
this is an inverse relationship
P is inversely proportional to V
Boyle’s Law
k = a constant

20. higher than lower than the same as
The pressure of the gas in the container on the right would be ______ the pressure of the gas in the container on the left. (volume and moles remain constant)
higher than
lower than
the same as

21. higher than lower than the same as
The pressure of the gas in the container on the right would be ______ the pressure of the gas in the container on the left.
higher than
lower than
the same as

22. pressure and temperature
P & T
increase the temperature (while volume and moles remains constant) and the pressure increases.
this is an direct relationship
P is directly proportional to T
atm
Gay-Lussac’s Law
k = a constant

23. Gas Relationships Together
All the k’s are different
Squish the k’s all together, let’s call it kcombined, R
and since R = R
The Combined Gas Law

24. pressure and temperature
Double the temp, does not double the pressure as you might expect.
The temperature must reach 373ºC before the pressure doubles. Why?
1 atm
≠ 2 atm

25. We could graph the following data
Sketch a graph of the following pressure and temp data collected for a tank of helium.
For reasons that might not make sense yet, set the range of your x-axis from - 300ºC to 100ºC
hold the paper landscape
make the graph as large as is reasonable
Temp (ºC)
Pressure (units) 15 30
16.7 50
17.9
100
20.7
-25
13.8
-50
12.3
-140
7.5

26. Absolute Zero As gases cool, they slow down and cause less pressure.
...Slow molecules more = less pressure
...slower still = even less pressure...
...finally the molecules stop, and don’t cause any pressure and can’t go any slower.
this is as slow as the molecules can go, thus this is the coldest the molecules can get.

27. Kelvin the Absolute Temp Scale
Absolute zero, 0 = -273ºC
0 K = -273ºC ºC = K
-273ºC
= 0 K

28. The Combined Gas Law The “Before and After” Gas Law
Consider 0.5 L of gas in sealed syringe in a 50ºC water bath with a pressure of 200 mmHg. The syringe is compressed to 0.1 L held at 50ºC, what will be the new pressure?
2 L of oxygen gas in a sealed, rigid metal cylinder is at 0.80 atm at 25ºC. To what temperature, in Celsius should the gas be changed to reduce the pressure to 0.20 atm?
P1 V1
n1 T1
P2 V2
n2 T2 =
P1 V1
n1 T1
P2 V2
n2 T2 =

29. The Combined Gas Law The “Before and After” Gas Law
Consider 0.5 L of gas in sealed syringe in a 50ºC water bath with a pressure of 200 mmHg. The syringe is compressed to 0.1 L held at 50ºC, what will be the new pressure?
2 L of oxygen gas in a sealed, rigid metal cylinder is at 0.80 atm at 25ºC. To what temperature, in Celsius should the gas be changed to reduce the pressure to 0.20 atm?
P1 V1
n1 T1
P2 V2
n2 T2 =
P1 V1
n1 T1
P2 V2
n2 T2 =

30. Equilibrium Vapor Pressure
“Correcting to STP”
42.0 ml of hydrogen gas was collected at 21.0ºC over water. The barometer read mmHg. Calculate the volume that this amount hydrogen would fill at STP?
Equilibrium Vapor Pressure

31. Equilibrium Vapor Pressure
“Correcting to STP”
42.0 ml of hydrogen gas was collected at 21.0ºC over water. The barometer read mmHg. Calculate the volume that this hydrogen would fill at STP?
Equilibrium Vapor Pressure

32. A sealed container is manipulated such that the absolute temp is doubled, and the pressure is tripled. How will the number of moles of molecules be affected? (Do not use a calculator.)
1 - but if there IMF’s the pressure would not quite be 40
and if there were a reaction, it could be anyone’s guess.
increase
decrease
stay the same 32

33. A sealed container is manipulated such that the absolute temp is doubled, and the pressure is tripled. How will the number of moles of molecules be affected? (Do not use a calculator.)
1 - but if there IMF’s the pressure would not quite be 40
and if there were a reaction, it could be anyone’s guess.
increase
decrease
stay the same
The problem tells you that the container is sealed, thus heat and volume changes must be what is used to change the temp and pressure 33