1. Gas Laws
Pressure and Volume (Boyle’s Law) Temperature and Volume (Charles’ Law)

2. P and V Changes
The pressure of a gas is inversely proportional to its volume P1 P2 V1 V2

3. Pressure and Volume
Experiment Pressure Volume P x V
(atm) (L) (atm x L)
↑ ↓
Volume is inversely proportional
_____ to Pressure
_____
P1V = atm x 16 L = atm L (K)
P2V = atm x L = atm L (K)
Boyle's Law: When Temperature and amount of gas remains constant
P x V = k (constant) Pressure x Volume = Constant
P1V1 = k = P2V2
P1V1 = P2V2

4. Use this equation Initial Final P1V1 = P2V2 To find How the volume of a gas changes when its pressure changes. Final V Initial V x Initial P V2 = V1 x P1 P2 or How the pressure of a gas changes when its volume changes. Final P Initial P x Initial V P2 = P1 x V1 V2

5. Freon gas is used in refrigeration systems.
Boyle’s Law Problem #1
The pressure of a gas is inversely proportional to its volume
P1V1 = P2V2
Freon gas is used in refrigeration systems.
What is the final volume (L) of a 1.6 L sample of Freon gas if its initial
pressure of 50 mm Hg is changed to 200 mm of Hg at constant T?
Remember must convert units to atmospheres and liters
Prepare a data table
DATA TABLE
Initial conditions Final conditions
P1 P V1 V2

6. Boyle’s Law Calculation #1
The pressure of a gas is inversely proportional to its volume
P1V1 = P2V2
Freon gas is used in refrigeration systems.
What is the final volume (L) of a 1.6 L sample of Freon gas if its initial
pressure of 50. mm Hg is changed to 200. mm of Hg at constant T?
Remember you must convert units to atmospheres and liters
Convert mm Hg to atm
50 mm Hg (1 atm/760 mm Hg) = atm
200 mm Hg (1 atm/760 mm Hg) = .26 atm
Prepare a data table
DATA TABLE
Initial conditions Final conditions
P1 = atm P2 = .26 atm
V1 = L V = ?

7. 1 Freon-12, CCl2F2, is used in refrigeration systems.
Find New Volume (V2)
1 Freon-12, CCl2F2, is used in refrigeration systems.
What is the final volume (L) of a 1.6 L sample of Freon gas if its initial
pressure of 50. mm Hg is changed to 200. mm of Hg at constant T?
Convert mm Hg to atm
50 mm Hg (1 atm/760 mm Hg) = atm
200 mm Hg (1 atm/760 mm Hg) = .26 atm
Prepare a data table
DATA TABLE
Initial conditions Final conditions
P1 = atm P2 = .26 atm
V1 = L V = ?
Solve for V2: P1V1 = P2V V2 = V1 x P1 P2
V2 = L x atm = L
.26 atm

8. Learning Check #1
Remember you must convert units to atmospheres and liters.
A sample of nitrogen gas has a volume of 6.4 L at a pressure of atm.
What will the new volume be if the pressure is changed to 1.40 atm?
P1V1 = P2V V2 = V1 x P1 P2
Prepare a data table.
There are no units to convert in this problem so you are ready to go.
DATA TABLE
Initial conditions Final conditions
P1 = atm P2 = atm
V1 = L V2 = ? L

9. A sample of nitrogen gas is 6.4 L at a pressure of 0.70 atm.
Solution #1
A sample of nitrogen gas is 6.4 L at a pressure of atm.
What will the new volume be if the pressure is changed to 1.40 atm?
Prepare a data table: DATA TABLE
Initial conditions Final conditions
P1 = atm P2 = 1.4 atm
V1 = L V2 = ?
P1V1 = P2V V2 = V1 x P1 P2
V2 = 6.4 L x atm = 3.2 L
1.40 atm

10. Learning Check #2
A sample of helium gas has a volume of 12.0 L at 600. mm Hg.
What new pressure is needed to change the volume to 36.0 L?
P1V1 = P2V P2 = P1 x V1 V2
Remember you must convert units to atmospheres and liters.
Convert mm Hg to atm
600. mm Hg (1 atm/760 mm Hg) = .789 atm

11. Solution #2 Pressure decrease when volume increases.
A sample of helium gas has a volume of 12.0 L at 600. mm Hg. What new pressure is needed to change the volume to 36.0 L?
Prepare a data table: DATA TABLE
Initial conditions Final conditions
P1 = atm P2 = ? In atm
V1 = L V2 = 36.0L
Remember you must convert units to atmospheres and liters.
.789 atm x L = or x 10-1 atm
36.0 L
Pressure decrease when volume increases.

12. Gas Laws
Pressure and Volume (Boyle’s Law) Temperature and Volume (Charles’ Law)

13. Charles’ Law
V = .250 L V = .175L T = 293 K T = 273K Observe the V and T of the balloons. How does volume change with temperature?

14. Volume and Temperature
The volume of a gas is directly proportional to its absolute (K) temperature
Experiment Volume Temp V/T
(L) (K) (L/K)
_____
_____
V1/ T1 = L / 73K = (b)
V2/ T2 = L / 173K = (b)
Charles’ Law : V = bT
V = b V1 = b = V or V1 = V2
T T T2 T T2

15. You must always convert units to the following:
Charles’ Law: V and T
At constant pressure,
the volume of a gas is directly proportional to its absolute (K) temperature
You must always convert units to the following:
(V) is volume of gas in Liters
(T) is the Temperature in Kelvin V1 = V2
T T2
  Initial Final

16. Learning Check #1 Use Charles’ Law to complete the statements below:
If final Temp. is higher than initial Temp, then
the final Vol. is (greater, or less) than the initial Vol.
If final Volume is less than initial Volume, then
the final Temp. is (higher, or lower) than the initial Temp.
The volume of a gas is directly proportional to its absolute (K) temperature

17. Solution #1
V1 = V2
T T2
1. If final T is higher than initial T, final V is (greater) than the initial V.
2. If final V is less than initial V, final T is (lower) than the initial T.
The volume of a gas is directly proportional to its absolute (K) temperature

18. You must always convert units to the following:
Charles’ Law Problem #1
A balloon has a volume of 785 mL on a fall day when the temperature is 21°C. In the winter, the gas cools to 0°C. What is the new volume of the balloon?
V1 = V2
T T2
You must always convert units to the following:
(V) is volume of gas in Liters
(T) is the Temperature in Kelvin

19. Charles’ Law Calculation #1
Remember you must convert units to Kelvin and liters.
Prepare a data table: DATA TABLE
Initial conditions Final conditions
V1 = ml = .785 L V2 = ? L
T1 = 21 °C + 273= 294 K T2 = 0 °C = 273 K
V1 = V V2 = V1 x T2
T T T1
V2 = .785 L K L or 7.29 x 10-1 L
294 K
Check your answer. If temperature decreases, Volume should decrease

20. Learning Check #2 1) 716K 2) 443K 3) 191K V1 = V2 T2 = T1 x V2
A sample of O2 gas has a volume of 420 mL at a temperature of 18°C. What temperature (K) is needed to change the volume to 640 mL?
1) 716K 2) 443K ) 191K
V1 = V2 T2 = T1 x V2
T T V1

21. Solution #2
A sample of O2 gas has a volume of 420 mL at a temperature of 18°C. What temperature (in K) is needed to change the volume to 640 mL?
Remember you must convert units to Kelvin and liters.
Prepare a data table: DATA TABLE
Initial conditions Final conditions
V1 = ml = L V2 = 640 ml = L
T1 = 18 °C = 291 K T2 = ? K
T2 = 291 K L 443K
.420 L
V1 = V2 T2 = T1 x V2
T T V1