1. Boyle’s Law - Review
P1V1 = P2V2

2. Pressure can be in any units Volume can be in any units
Boyle’s Law
P1V1 = P2V2
Pressure can be in any units
Volume can be in any units

3. Charles’ Law If temperature increases, volume increases
If temperature decreases, volume decreases
This is direct variation (compared to inverse for Boyle’s law)

4. Charles Law Volume can be in any units Temperature must be in Kelvin
V1 = V2
T1 T2
Volume can be in any units
Temperature must be in Kelvin
To convert Celsius to Kelvin, add 273
(K = C )

5. Charles law For example:
The temperature inside my fridge is 4 C. If I place a balloon in my fridge that initially has a temperature of 22 C and a volume of .5 L, what will be the volume of the balloon when it is fully cooled by my refrigerator?

6. Charles’ Law First convert temperatures to Kelvin 4 + 273 = 277K

7. Charles’ Law
Make a list of what you know
V1 =
T1 =
V2 =
T2 =

8. Charles’ Law
V1 = .5 L
T1 =
V2 =
T2 =

9. Charles’ Law
V1 = .5 L
T1 = 295K
V2 =
T2 =

10. Charles’ Law
V1 = .5 L
T1 = 295K
V2 = x
T2 =

11. Charles’ Law
V1 = .5 L
T1 = 295K
V2 = x
T2 = 277K

12. Charles’ Law Put the information into the equation: V1 = V2 T1 T2
.5 = x

13. Charles’ Law Cross multiply to solve for x .5(277) = x(295)
x = .47 Liters

14. Charles’ Law For Example:
A man heats a balloon in the oven. If the balloon initially has a volume of 0.4 L and a temperature of 20 C, what will the volume of the balloon be after he heats it to a temperature of 250 C?

15. Charles’ Law Convert temperatures to Kelvin 20 + 273 = 293 K

16. Charles’ Law
Make a list of what you know
V1 =
T1 =
V2 =
T2 =

17. Charles’ law Make a list of what you know V1 = .4 L T1 = 293 K V2 = x

18. Charles’ law Put the information into the equation: V1 = V2 T1 T2
.4 = x

19. Charles’ law Cross multiply to solve for x .4(523) = x(293)
x = .71 L