1. Behavior of Gases
Chapter 14
Pages

2. 14.1 Properties of Gases Compressibility
The measure of how much the volume of a gas decreases under pressure
Gases are very compressible
Remember liquids and solids do not compress well.
Common use: Air bags in cars
Reaction produces nitrogen gas which inflates air bag. When a wreck happens the bag inflates and the drivers head compresses the gas inside the bag. The energy of the impact is absorbed by the molecules inside the bag, instead of the skull taking the impact.

3. Compression and Kinetic Theory
Kinetic theory explains why gases are compressed more easily than all other states of matter:
Gases are easily compressed because there is a lot of space between particles of that gas.
Remember that the volume that a gas occupies is far greater than its mass.
@ room temperature, the distance between particles is about 10x the diameter of the particle

4. Factors Affecting Gas Pressure
Amount of Gas
Volume
Temperature

5. Amount of Gas
Adding molecules to a closed system (sealed container) greatly effects gas pressure.
The more molecules present, the greater the chance for collisions between molecules, thus a greater increase in pressure.
Take for example a tire that is supposed to be inflated to 35psi. What happens to that tire if you decrease the pressure to 18psi or increase the pressure to 70psi?
For the SI system, take a close container at 100kPa and increase it to 200kPa then the pressure doubles.
Increase it from 100kPa to 300kPa then it triples.
Eventually pressure will become high enough to burst the container.

6. Amount of Gas
If the pressure of the sealed container is lower than the outside air pressure, than molecules will rush into the container if the opened and vice versa.
Aerosol cans (spray paint, hairspray) depend on propelling air from an area of high pressure to an area of lower pressure.
Same thing is true of CO2 cartridges for paintball guns or air rifles.

7. Assignment
Read Section 14.1
Pages

8. Volume You can increase pressure by decreasing volume.
Reducing the amount of space, while keeping the same number of molecules means more collisions and more pressure
The more a gas is compressed, the greater the pressure on the container.
Taking a sealed 1 liter container and reducing the volume to 0.5L or 500mL will double the pressure on the container.

9. Temperature Ever notice certain things “swell” when left in the sun??
Well kinetic theory can explain this, when we increase temperature then we are adding energy to a closed system. This addition of energy speeds particles up, these new higher energy particles impact the sides of the container with greater energy.
Doubling the temperature of a closed system, doubles the pressure.
Some things may eventually reach a pressure too high for the system to contain and they will burst.
Think when you throw a spray paint can in a fire.

10. Assignment
14.1 Review
Pg. 417
Questions 1-6
Due Tomorrow.

11. 14.2: The Gas Laws Boyle’s Law: Pressure and Volume
Charles’ Law: Temperature and Volume
Gay-Lussac’s Law: Pressure and Temperature
The Combined Gas Law
You will need to know these formulas and be able to convert them.

12. Boyle’s Law
If the temperature stays constant, as the pressure of a gas increases the volume decreases.
Boyle’s law states “that for a given mass of gas at a constant temperature, the volume of the gas varies inversely with pressure”
Calculations using Boyle’s Law
Pressure 1 x Volume 1 = Pressure 2 x Volume 2

13. Example of Boyle’s Law Sample Problem 14.1 on page 419
Givens: P1 = 103kPa; V1 = 30.0 L; P2= 25.0 kPa
Unknown: V2
Calculation
103 kPa x 30.0 L = 25.0 kPa x V2
(30 x 103) / 25 = 1.24 x 102 Liters or 124 Liters

14. Practice Problems
Pg. 419
Do Practice Problems 7 and 8.

15. Charles’ Law: Temperature and Volume
As the temperature of an enclosed gas increases, the volume increases, if the pressure is constant.
Charles’ Law states “ that the volume of a fixed mass of gas is directly proportional to its Kelvin temperature if the pressure is kept constant.
Calculations:
Volume 1 / Temp. 1 = Volume 2/ Temp 2

16. Example of Charles’ Law
Pg Sample Problem 14.2
Givens; V1 = 4.0 L; T1 = 24* C; T2 = 58*C
Unknown: V2
Conversions: Celsius to Kelvin = *C +273
Solutions: 4.0L / 297K = V2/ 331K
(4.0 x 331)/ 297 = 4.46L 3

17. Practice Problems
Do Practice problems 9 and 10
Pg. 421 in textbook .

18. Gay-Lussac’s Law: Pressure and Temperature
As the temperature of an enclosed gas increases, the pressure increases, if the volume is constant.
Gay-Lussac’s law states “that the pressure of a gas is directly proportional to the Kelvin temperature if the volume remains constant.
P1/T1 = P2/T2

19. Sample Problem 14.3 Given: P1 = 103kPa; T1 = 25*C; T2 = 928*C
Unknown: P2
Solution:
25*C = 298K
928*C +273 = 1201K
103kPa/298K = P2/1201K
103 x 1201/ 298K
P2 = 415kPa

20. Practice Problems
Do Practice Problems on pg. 423
Questions 11 and 12

21. The Combined Gas Law
The combined gas law allows you to do calculations for situations in which only the amount of gas is fixed.
P1 x V1/ T1 = P2 X V2/ T2
Sample 14.4
Known: V1 = 30L; T1 = 313K; P1 = 153kPa; T2 = 273K; P2 = kPa
Unkown: V2
Solution: V2 = (30 x 153 x 273)/ (101.3 x 313)
V2 = 39.5L

22. Assignment
14.2 Review
Pg. 425
Questions 15-22
Due Monday

23. 14.3: Ideal Gases
The combined gas law allows us to solve problems with 1 of 3 variables: temperature, pressure, volume, because the combined gas law assumes the amount of gas does not change.
However if we want to calculate the # of moles of gas present we need to bring in the variable “n”
We can modify the combined gas law to include “n” and calculate for number of moles:
P1 x V1/ T1 x N1 = P2 x V2/ T2 x N2
This equation’s base is (P x V) / (T x n) is constant. This is true of ideal gases.

24. Ideal Gas Law
The ideal gas constant “R” has a value of 8.31 LxkPa/ Kxmol. It is derived from the previous equation; along with the knowledge that 1 mol of a gas contains STP.
Now if we include all variables we arrive at the Ideal Gas Law.
P x V = n x R x T or more simply PV = nRT
Now lets look at sample problem 14.5 on p. 427
Do practice problems 23 and 24 on p. 427

25. Ideal Gases and Real Gases
An ideal gas is a gas that follows all gas laws at all conditions of pressure and temperature.
This means ideal gases conform to all assumptions of the kinetic theory.
However, there are no gases to which all these assumptions are true, thus what we find are real gases.
Real gases will, however, behave like ideal gases at many different conditions of pressure and temperature.
Real gases differ most from ideal gases at low temperatures and high pressures.
figure on pg. 429 to see an example.

26. Assignment
14.3 Review Pg. 429 Questions

27. 14.4: Gases (Mixtures and Movements)
Dalton’s Law of Partial pressures
At a constant volume and temperature, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the component gases.
We know that as we increase the number of particles we increase pressure due to increased number of particle collisions; this holds true for mixtures of gases as well.
Dalton’s Law Equation:
P(total) = P1 + P2 + P3 ………
All we have to do to calculate this is add up the pressures each individual gas of a mixture exerts on its container.

28. Dalton’s Law cont. Using Dalton’s law Sample Problem 14.6 Solution
Givens: PN2 = 79.10, PCO2 = , Pothers = 0.94, p total =
Unknown: PO2 ??
Solution
101.3 = PO2
PO2 = kPa
Try Practice Problems 31 and 32 on pg. 434

29. Graham’s Law
Graham’s law states that the rate of effusion of a gas is inversely proportional to the square root of the gas’s molar mass.
Diffusion vs. Effusion
Diffusion – tendency of molecules to move toward areas of lower concentration
Effusion – a gas escapes through a tiny hole in its container
Gases of lower molar mass diffuse and effuse faster than those gases of higher molar mass.

30. Graham’s Law
Grahams law makes sense when we look at the relationship between mass, volume, and energy.
Kinetic energy = (Mass x Volume)2
How does it makes sense: Well this means that at a given constant energy any increase in mass is accompanied by a decrease in volume.
2 objects with different masses but the same energy are in motion, the object with the lowest mass will be traveling the fastest.

31. Comparing Effusion Rates
Rate A/ Rate B = sq. rt of Molar mass B/ molar mass A
It’s the principal behind now using nitrogen over helium in balloons
Helium effuses around 3x faster (2.7x to be exact) than Nitrogen does.
This means your balloon will float, but eventually all the Helium will escape through the pores of the balloon.

32. Assignment
14.4 Review
Pg. 436
Questions 33-38
Due tomorrow