1. Gases
Chapter 13 & 14 L

2. I’m worth 1.6 Billion dollars, I don’t need you. You’re fired!
But first
OMG a dollar!
Not negligible
I’m worth 1.6 Billion dollars, I don’t need you. You’re fired!
Negligible

3. Vocabulary pt2 Proportional Good Grades α Happiness
increase in good grades causes an increase in happiness
Decrease in good grades causes a decrease in happiness
Inversely proportional
Coach L singing α 1/Happiness
increase in me singing causes an decrease in happiness
Decrease in me singing causes an increase in happiness

4. Kinetic-molecular theory
Explains why gases behave like they do
6 basic postulates about gases

5. Gases are made of small stuff
Gases are composed of a large number of molecules or atoms
Generally low atomic and molecular weights (bigger molecules form liquids or gases)

6. Gases are… Hello? Most of the volume of a gas is empty space.
That’s why they are easy to compress (to a point)
The volume the actual particle takes up is negligible

7. Gases and pool Always in motion
Move in straight lines, until they collide with the walls of the container or each other
These collisions with the container are called pressure
Behave like pool balls on crisco

8. Gases aren’t attractive… for the most part
There is negligible force of attraction between gas particles or between the walls of the container and the molecules (mostly)

9. Bumper cars, anyone? Collisions are elastic
Elastic Collisions -- none of the energy of a gas particle is lost when it collides with another particle or with the walls of the container.

10. Gases and heat
The temperature of a gas is an average of the kinetic energy of it’s particles
Units: Kelvin (K) = °C

11. Kinetic-molecular theory
A gas that obeys all the postulates of the Kinetic Molecular Theory is called an ideal gas
kinetic energy that is defined as one-half of the product of its mass times its velocity squared.
KE = 1/2 mv2

12. Behaviors of gases
Compression
Expansion

13. Behaviors of a gas
Diffusion
Effusion
Rate of effusion = 1/√molar mass

14. Effusion Rate of effusion α 1/√molar mass For comparing two gases
Ex: a balloon filled with helium will deflate faster then a balloon filled with nitrogen, B/C helium has a smaller molar mass.

15. Pressure Force per unit area
The air in a balloon push against the walls of the balloon equally
Measured with a
barometer (atmospheric pressure)
Manometer (pressure in a container)

16. Units of pressure Pacal (Pa) = 1N/m2 Pounds per square inch (psi)
Millimeters of mercury (mm Hg)
Torr = 1 mm Hg
Atmosphere (atm)
At sea level at 0°C the average air pressure is 760 mm Hg.
AKA 1atm
1 atm = 760 mm Hg = 760 torr = KPa

18. Dalton’s law of partial pressure
In a mix of gases pressure = ∑(pressures in the mix)
Ptotal = Pa + Pb + Pc + Pd…
A sample of oxygen is collected over water at a total pressure of 755 torr at 20°C. Calculate the pressure of oxygen, if water vapor has a partial pressure of 17.5 torr at 20°C.
Ptotal = Poxygen + Pwater vapor
755 torr = Poxygen torr
737.5 torr = Poxygen

19. Practice
What is the partial pressure of hydrogen gas in a mixture of hydrogen and helium if the total pressure is 600 mm Hg and the partial pressure of helium is 439 mm Hg?
Find the total pressure for a mixture that contains four gases with partial pressures of 5.00 KPa, KPa, 3.02 KPa, and 1.20 KPa.
Find the partial pressure of carbon dioxide in a gas mixture with a total pressure of 30.4 KPa if the partial pressures of the other tow gases in the mixture are 16.5 KPa and 3.7 KPa.
161 mm Hg
KPa
10.2 KPa

20. Boyles law Relationship between pressure and volume V α 1/P
If temp. and amount are kept constant
V α 1/P
meaning that as volume decreases pressure increases, and visa versa

21. Boyles law (…continued)
Found that when multiplied P & V produce a constant (K)
PV = K
This is useful in changing P and finding the new V that that gas would occupy.
P1V1 = K
P2V2 = K
P1V1 = P2V2

22. Boyles law (…continued)
P1V1 = P2V2
Ex: you have 5.00 L of helium in a balloon at 1.00 atm, if you decrease the volume to 2.00 L, what would your balloons pressure be?
(5.00L)(1.00atm) = (2.00L)P2
5.00L atm/2.00L = P2
2.5atm = P2

23. Practice
The volume of a gas at 99.0KPa is 300.0mL. If the pressure is increases to 188KPa, what will the new volume be?
The pressure of a sample of helium is a 1.00L container is 0.988atm, what is the new pressure if the sample is placed in a 2.00L container?
A sample of neon gas occupies 0.220L at atm, what ill be it’s volume oat 29.2kPa pressure?
158mL
atm
0.657L

24. Charles’ Law Relationship between temperature and volume V a T
If pressure and amount are kept constant
V a T
Meaning that as temperature increases so does volume

25. Charles’ Law (…continued)
Expressed mathematically the law looks like this:
V = bT –or– V/T = b
If the temperature of a gas with a certain volume (V1) and Kelvin temperature (T1) is changed to a new Kelvin temperature (T2) the volume also changes (V2)
V1/T1 = b
V2/T2 = b
V1/T1 = V2/T2

26. Charles law (…continued)
V1/T1 = V2/T2
Ex: Suppose you live in Alaska, and are outside in the middle of winter where it is -23°C. You inflate a balloon s that it has a volume of 1.00L. You then take your balloon inside where it is 27°C. What is the volume of your balloon now?
(1.00L)/[(-23°C + 273)K] = V2/[(27°C + 273)K]
(1.00L)/250K = V2/300K
1.20L = V2

27. Practice
A gas at 89°C occupies a volume of 0.67L. At what Celsius temperature will the volume increase to 1.12L?
The Celsius temperature of a 3.00L sample of gas is lowered from 80.0°C to 30°C, what will the resulting volume be of this gas?
What is the volume of air in a balloon that occupies 0.620L at 25°C if the temperature is lowered to 0.00°C?
330°C
2.58L
0.57L

28. Gay-Lussac’s Law Relationship between temperature and pressure P a T
If volume and amount are kept constant
P a T
Meaning that as temperature increases so does pressure
Mathematically:
P = kT
P1/T1 = k
P2/T2 = k
P1/T1 = P2/T2

29. The combined gas law
The mathematical combination of all the preceding laws
P1V1 = P2V2
T1 T2
Amount of gas is the only constant

30. The combined gas law (…continued)
Ex: a hot-air balloon initially at 24°C has a volume of 5.41dm3, and a pressure of 101.5KPa. If the air in the balloon is heated to 35°C and the pressure increased to 102.8KPa, what would the new volume be of the balloon?
P1V1 = P2V2
T1 T2
(101.5KPa)(5.41dm3) = (102.8KPa)V2
(24°C + 273) (35°C + 273)
(297K) (308K)

31. The combined gas law (…continued)
Ex: a hot-air balloon initially at 24°C has a volume of 5.41dm3, and a pressure of 101.5KPa. If the air in the balloon is heated to 35°C and the pressure increased to 102.8KPa, what would the new volume be of the balloon?
(101.5KPa)(5.41dm3) = (102.8KPa)V2
(297K) (308K)
(308K)(101.5KPa)(5.41dm3) = (102.8KPa)V2
(297K)
(308K)(101.5KPa)(5.41dm3) = V2
(102.8KPa)(297K)
V2 = 5.54dm3

32. Practice
A helium filled balloon at sea level has a volume of 2.1L at 0.998atm and 36°C. If it is raises to an elevation at which the pressure is 0.900atm and the temperature is 28°C, what will the new volume of the balloon be?
At 0.00°C and 1.00atm, a sample of gas occupies 30.0mL. If the temperature is increased to 30.0°C and the entire sample is transferred to a 20.0mL container, what will the gas pressure be inside the container?
2.3L
1.66atm

33. The ideal gas equation A way of working with all variables PV=nRT
P = pressure (atm)
V = volume (L)
n = moles of gas
T = temperature (K)
R = ideal gas constant (0.0821L atm/K mol)

34. The ideal gas equation (…continued)
Ex: calculate the number of moles of a gas contained in a 3.0L vessel at 300K with a pressure of 1.5atm
PV=nRT
(1.5atm)(3.0L) = n(0.0821L atm/K mol)(300K)
(1.5)(3.0) = n
(0.821)(300)
4.5 = n
246.3
0.18mol = n

35. Practice
At what temperature would 2.10 moles of N2 gas have a pressure of 1.25 atm and in a 25.0 L tank?
What volume is occupied by 5.03 g of O2 at 28°C and a pressure of 0.998atm?
Calculate the pressure in a 212 Liter tank containing 23.3 kg of argon gas at25°C?
181 K
3.9 L
67.3 atm

36. The ideal gas equation (…continued)
Molar mass
Density
Since Molar mass is g/mol (m/M), the ideal gas law can be rewritten for molar mass
PV=nRT
PV=mRT M
M = mRT --or-- M = DRT
PV P
Since density is m/V, the ideal gas law can be rewritten for density
M = mRT PV
M = DRT P
D = MP RT

37. Practice
Molar Mass
Density
What is the molar mass of a pure gas that has a density of 1.40g/L at STP?
31.4g/mol
What is the density of a gas at STP that has a molar mass of 44.0g/mol?
1.96g/L