1. Ch. 16 Sect. 3: Behaviors of Gases & Gas Laws
Boyle’s Law

2. Review
Pressure results from gas particles colliding with the walls of their container
Pressure is force divided by area
High pressure = more collisions
Low pressure = less collisions
Measured in units of:
Pascal’s (Pa)
kilopascals (kPa)
atmospheres (atm)
millimeters mercury (mm Hg)

3. Variables that Affect Gases
There are four variables that we need to consider when looking at the properties of gases:
1. Amount of Gas (number of particles)
2. Pressure (number of collisions in a certain area)
3. Volume (amount of space it takes up)
4. Temperature (average kinetic energy of particles)

4. Robert Boyle (1627-1691, Ireland)
One of the founders of modern chemistry
He made improvements to the construction of an air pump w/the assistance of Robert Hooke; he used this to do experiments on the properties of air
Boyle’s Law originally published in 1662

5. This applies as long as the temperature remains constant as well.
Boyle’s Law
What is the relationship between pressure and volume of a gas?
“The pressure of a fixed amount of gas varies inversely with the volume of that gas” Or
If the pressure goes up, the volume goes down!
This applies as long as the temperature remains constant as well.

6. Explaining Boyle’s Law
If pressure increases, that means the number of collisions must also be increasing.
By decreasing the volume of a container, the number of collisions will increase
This occurs only if the temperature and amount of gas remain constant
Boyle's Law Animation

7. Boyle’s Law In Action
When applied to a real life situation, we find that the pressure multiplied by the volume is always equal to a constant value
The product of the initial volume and pressure will equal the product of the final volume and pressure
P1•V1 = P2•V2

8. Boyle’s Law Practice Problem 1
If a balloon has a volume of 500 mL at a pressure of 1-atm, and is driven into the mountains at a pressure 0.5-atm, what is the new volume of the balloon?
P1 = 1 atm
V1 = 500 mL
P2 = 0.5 atm
V2 = ?
P1•V1 = P2•V2
1 atm • 500 mL = 0.5 atm • V2
1 atm • 500 mL = V2
0.5 atm
1000 mL = V2

9. Boyle’s Law Practice Problem 2
A balloon has a volume of 10.0 L at a pressure of 101 kPa. What will the new pressure be when the volume is increased to 23.5 L?
P1 = 101 kPa
V1 = 10.0 L
P2 = ?
V2 = 23.5 L
P1•V1 = P2•V2
101 kPa • 10.0 L = P2 • 23.5 L
101 kPa • 10.0 L = P2
23.5 L
43.0 kPa= P2

10. More Boyle’s Law Practice
Complete the following table using the Boyle’s Law formula.
P1 (kPa)
V1 (mL)
P2 (kPa)
V2 (mL)
Work 10
100 25 2
500 8 3 50 7 90
180 75 5
1000
250
300

11. More Boyle’s Law Practice Answers
P1 (kPa)
V1 (mL)
P2 (kPa)
V2 (mL) 10
100 25 40 2
500 8
125 3
1.5 50 7 90
3.5
180 15 75 5
200 1
1000 6
250
300