1. West Valley High School General Chemistry Mr. Mata
Chapter 11A Gases
West Valley High School
General Chemistry
Mr. Mata

2. The Atmosphere
The atmosphere is the thin layer of gases that surround the earth.
Air is composed of a mixture of gases:
78 % Nitrogen, N2
21 % Oxygen, O2
.1 % Argon, Ar
365 ppm CO2
0 - 4 % water, H2O

3. Physical Properties of Gases
No definite shape or volume:
expand to fill container, take shape of container.
Compressible
increase pressure, decrease volume.
Low Density
air at room temperature and pressure: g/cm3.
Exert uniform pressure on walls of container.
Mix spontaneously and completely.
Diffusion (high conc.  low conc.)

4. Pressure Pressure = force exerted per unit area. P = F/A
Atmospheric Pressure = force exerted by earth’s atmosphere.
Atmospheric Pressure is measured with a barometer.
Aneroid Barometer

5. Measuring Pressure The first device for measuring atmospheric
pressure was developed by Evangelista Torricelli.
during the 17th century.
The device was called a “barometer”.
Baro = “weight”
Meter = “measure”
Mercury Barometer

6. Pressure Units mm of mercury (mm Hg) 1 mm Hg = 1 torr
760 mm Hg = 760 torr = 1 atm
1 atm is 1 atmosphere of pressure, (aka: standard pressure).

7. Robert Boyle (1627-1691) Born into an aristocratic Irish family.
Became interested in medicine, astronomy, chemistry. 
Avid hot air balloonist.

8. Boyle’s Law As pressure of gas increases, volume decreases:
This is an inverse proportion: as one increases the other decreases
P1V1 = P2V2

9. Boyle’s Law Graph
Insert figure 12.9

10. Boyle’s Law Problem #1
A gas has a volume of 5 L at 2 atm pressure and the pressure is increased to 4 atm. Calculate the new volume.
P1V1 = P2V > V2 = P1V1 P2
V2 = (2 atm) (5 L) = 2.5 L
4 atm

11. Boyle’s Law Problem #2
A gas has a pressure of 6.5 atm and a volume of 20 L. Calculate the new pressure if the volume is increased to 80 L.
P1V1 = P2V > P2 = P1V1 V2
P2 = (6.5 atm) (20 L) = 1.6 atm
80 L

12. Jaques Charles (1746-1823) French Physicist.
Conducted the first scientific balloon flight in 1783.

13. CHARLES’ LAW
Volume is directly proportional to temperature
Volume & temperature both increase or both decrease together
V1 = V2
T T2

14. Charles’s Law Graph
Insert figure 12.11

15. Charles’s Problem #1
A certain gas in a closed container has a volume of mL at a temperature of 25oC. If there is no change in pressure, calculate the volume at - 25oC.
First Convert Temperatures to Kelvin.
K = oC + 273
T1 = = 298 K
T2 = = 248 K

16. Charles’s Problem #1 (continued)
T1 = 298 K
T2 = 248 K
V1 = mL
V2 = ?
V1 = V2 -> V2 = V1 T2
T T T1
V2 = (0.567 mL) (248 K) = 0.47 mL
298 K

17. Charles’s Problem #2
A certain gas in a closed container has a volume of 25 mL at a temperature of 23oC. What volume will it occupy at 250 C?
First Convert Temperatures to Kelvin.
K = oC + 273
T1 = = 296 K
T2 = = 523 K

18. Charles’s Problem #2 (continued)
T1 = 296 K
T2 = 523 K
V1 = 25 mL
V2 = ?
V1 = V2 -> V2 = V1 T2
T T T1
V2 = (25 mL) (523 K) = 44.2 mL
296 K