1. Unit 6: Gases and Atmospheric Chemistry
Pressure, Volume, and Temperature
Boyle’s Law

2. Pressure, temperature and volume
Pop can demonstration - Watch this!
A bigger can … - Watch this!

3. Pressure, temperature and volume
Pressure is a measure of force per area.
As the particles strike the walls of their container, they exert a force. The force per area is the pressure of the gas.
Units : The metric unit for pressure is the Pascal:

4. Atmospheric pressure
The pressure of the atmosphere can be measured with a barometer:
760 mm Hg = 760 Torr = 1.00 atm = kPa
Since all of these units are equivalent, we can use them as conversion factors:
e.g. Convert a pressure of 700 mm Hg into kPa:

5. Temperature
Temperature is a measure of the average kinetic energy possessed by the particles of a substance. Celcius (°C) Kelvin (K)
100
373
To convert from °C to K:
TK = TC 25
298
273
To convert from K to °C:
TC = TK
-273
Absolute zero (where all particle motion stops) = 0 K = °C.

6. STANDARD TEMPERATURE AND PRESSURE (STP)
Volume
Units: 1 mL = 1 cm3 1 L = 1000 mL = 1000 cm3
25 cm mL
182 cm L
3.680 L mL
𝑥 1 𝑚𝐿 1 𝑐𝑚 3
= 25
𝑥 1 𝐿 𝑐𝑚 3
= 0.182
𝑥 𝑚𝐿 1 𝐿 =
STANDARD TEMPERATURE AND PRESSURE (STP)
Temperature = 0oC = 273 K
Pressure = kPa

7. Boyle’s Law
If the pressure on a specific amount of gas is increased, the volume will decrease.
 V
 P
 V
 P

8. Boyle’s law
“The volume of a fixed mass of gas is INVERSELY PROPORTIONAL to pressure, provided temperature remains constant.”
𝑉= 𝑘 𝑃 𝑤ℎ𝑒𝑟𝑒 𝑘=𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑉 1 𝑃

9. Boyle’s Law
Since 𝑉= 𝑘 𝑃 , then 𝑃𝑉=𝑘  𝑃 1 𝑉 1 = 𝑃 2 𝑉 2 e.g. A 100 L volume of gas is at a pressure of 32 kPa. If the pressure increases to 44 kPa, what is the final volume?

10. e. g. A 100 L volume of gas is at a pressure of 32 kPa
e.g. A 100 L volume of gas is at a pressure of 32 kPa. If the pressure increases to 44 kPa, what is the final volume? Given: P1 = P2 = V1 = V2 =
32 kpa
44 kpa
Always list given information.
100 L ?
𝑃 1 𝑉 1 = 𝑃 2 𝑉 2
𝑉 2 = 𝑃 1 𝑉 1 𝑃 2 =
32 𝑘𝑃𝑎 (100 𝐿) (44 𝑘𝑃𝑎)
=72.727…𝐿
= 73 L
Rounded to 2 sig. digits.

11. Unit 6: Gases and Atmospheric Chemistry
Charles’ Law and Gay-Lussac’s Law

12. Charles’ Law
“The volume of a fixed mass of gas is DIRECTLY PROPORTIONAL to temperature, provided pressure remains constant.”
𝑉=𝑘𝑇 𝑤ℎ𝑒𝑟𝑒 𝑘=𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑉𝑇

13. Charles’ Law

14. Charles’ Law
Since 𝑉=𝑘𝑇 , then 𝑉 𝑇 =𝑘  𝑉 1 𝑇 1 = 𝑉 2 𝑇 2 e.g. If the volume of a balloon is 10.0L at 0.0°C, what is its volume at room temperature (25°C)?

15. e. g. If the volume of a balloon is 10. 0L at 0
e.g. If the volume of a balloon is 10.0L at 0.0°C, what is its volume at room temperature (25°C)? Given: T1 = T2 = V1 = V2 =
0.0°C = K
25°C = 298 K
Always list given information.
Always convert from Celsius to Kelvin
10.0 L ?
𝑉 1 𝑇 1 = 𝑉 2 𝑇 2
𝑉 2 = 𝑇 2 𝑉 1 𝑇 1 =
298 K (10.0 𝐿) (273.0 𝐾)
= …𝐿
= 10.9 L
Rounded to 3 sig. digits.

16. Gay-Lussac’s law
“The pressure of a fixed mass of gas is DIRECTLY PROPORTIONAL to temperature, provided volume remains constant.”
P =𝑘𝑇 𝑤ℎ𝑒𝑟𝑒 𝑘=𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
P 𝑇

17. Gay-Lussac’s Law

18. Gay-Lussac’s Law
Since 𝑃=𝑘𝑇 , then 𝑃 𝑇 =𝑘  𝑃 1 𝑇 1 = 𝑃 2 𝑇 2 e.g. A can of hairspray contains a gas which exerts a pressure of 900 kPa at 25°C. If the can can only withstand a maximum of 1800 kPa, at what temperature will it explode?

19. e.g. A can of hairspray contains a gas which exerts a pressure of 900 kPa at 25°C. If the can can only withstand a maximum of 1800 kPa, at what temperature will it explode? Given: T1 = T2 = P1 = P2 =
25°C = 298K ?
Always list given information.
Always convert from Celsius to Kelvin.
900 kPa
1800 kPa
𝑃 1 𝑇 1 = 𝑃 2 𝑇 2
𝑇 2 = 𝑃 2 𝑇 1 𝑃 1 =
1800 kPa (298K ) (900 𝑘𝑃𝑎)
=596 K
=323 °C

20. The Combined Gas Law

21. Combining the gas laws P1V1 P2V2 T1 T2 = V1 T1 = V2 T2 P1 T1 = P2 T2
So far we have seen three gas laws:
Robert Boyle
Jacques Charles
Joseph Louis Gay-Lussac V1 T1 = V2 T2 P1 T1 = P2 T2
P1V1 =
P2V2
These are all subsets of a more encompassing law, the combined gas law:
P1V1 P2V2
T T2 =

22. Combined Gas Law Equations:P1 =
P2T1V2
T2V1 P2 =
P1T2V1
T1V2 T1 =
P1T2V1
P2V2 T2 =
P2T1V2
P1V1 V1 =
P2T1V2
T2P1 V2 =
P1T2V1
P2T1

23. Ex. 1) A 10. 0L volume of gas is at 27°C and 101. 3 kPa
Ex.1) A 10.0L volume of gas is at 27°C and kPa. What would be its volume if the gas was at 0°C and 50 kPa? Given: T1 = T2 = P1 = P2 = V1 = V2 =
27°C = 300K
0°C = 273K
Always list given information.
Always convert from Celsius to Kelvin.
101.3 kPa
50 kPa
10.0 L ?
𝑃 1 𝑉 1 𝑇 1 = 𝑃 2 𝑉 2 𝑇 2
𝑉 2 = 𝑃 1 𝑉 1 𝑇 2 𝑃 2 𝑇 1 =
(101.3 𝑘𝑃𝑎)(10.0 𝐿)(273𝐾) (50 𝑘𝑃𝑎)(300𝐾)
= 𝐿
≐18 L

24. (87.5 𝑘𝑃𝑎)(820 𝑚𝐿)(270𝐾) (1.05 𝑎𝑡𝑚)(1010 𝑚𝐿)
Ex.2) 1010 mL of a gas is at 1.05 atm and -3°C. What would the temperature be if the pressure is changed to 87.5 kPa and the volume become 820 mL? Given: T1 = T2 = P1 = P2 = V1 = V2 =
-3°C = 270K ?
1.05 atm
87.5 kPa
1010 mL
820 mL
𝑃 1 =1.05 𝑎𝑡𝑚 × 𝑘𝑃𝑎 1.00 𝑎𝑡𝑚
≐106 𝑘𝑃𝑎
𝑃 1 𝑉 1 𝑇 1 = 𝑃 2 𝑉 2 𝑇 2
𝑇 2 = 𝑃 2 𝑉 2 𝑇 1 𝑃 1 𝑉 1 =
(87.5 𝑘𝑃𝑎)(820 𝑚𝐿)(270𝐾) (1.05 𝑎𝑡𝑚)(1010 𝑚𝐿)
= …𝐾
≐181𝐾
If the original question uses °C you must answer in the same units.
≐−92°C

25. (1.0 𝑎𝑡𝑚)(820 𝑚𝐿)(270𝐾) (1.05 𝑎𝑡𝑚)(1010 𝑚𝐿)
Ex.3) A solid steel cylinder is full of gas at a temperature of 25°C and pressure of 1.6 atm. When the gas was released into a balloon under STP conditions the volume was 865 L. What was the original volume of the cylinder? Given: T1 = T2 = P1 = P2 = V1 = V2 =
25°C = 298K
273K
1.6 atm
101.3 kPa = 1.0 atm ?
865 L
𝑃 1 𝑉 1 𝑇 1 = 𝑃 2 𝑉 2 𝑇 2
𝑉 1 = 𝑃 2 𝑉 2 𝑇 1 𝑃 1 𝑇 2 =
(1.0 𝑎𝑡𝑚)(820 𝑚𝐿)(270𝐾) (1.05 𝑎𝑡𝑚)(1010 𝑚𝐿)
= …𝐿
≐590 𝐿

26. Ex. 4) A 2000 L volume of gas at STP was compressed into 0
Ex.4) A 2000 L volume of gas at STP was compressed into L at 35°C. What would the final pressure be in the cylinder? Given: T1 = T2 = P1 = P2 = V1 = V2 =
273K
35°C = 308K
101.3 kPa ?
2000 L
0.500 L
𝑃 1 𝑉 1 𝑇 1 = 𝑃 2 𝑉 2 𝑇 2
𝑃 2 = 𝑃 1 𝑉 1 𝑇 2 𝑉 2 𝑇 1 =
(101.3 𝑘𝑃𝑎)(2000 𝐿)(308𝐾) (0.500 𝐿)(273𝐾)
= …𝑘𝑃𝑎
≐ 𝑘𝑃𝑎

27. Ex. 5) A balloon filled with 0
Ex.5) A balloon filled with mol of helium at 30°C and a pressure of 1.0 atm occupies a volume of 0.75 L and has a density of g/L. What would the density of the helium gas be if the balloon was placed in the freezer at -10°C and a pressure of 2.0 atm? Given: T1 = T2 = P1 = P2 = V1 = V2 = nHe = D1= D2 = mHe=
30°C = 303K
-10°C = 263K
1.0 atm
2.0 atm
0.75 L ?
mol
Not important
0.161 g/L ?
𝑃 1 𝑉 1 𝑇 1 = 𝑃 2 𝑉 2 𝑇 2
𝑉 2 = 𝑃 1 𝑉 1 𝑇 2 𝑃 2 𝑉 2 =
(1.0 𝑎𝑡𝑚)(0.75 𝐿)(263𝐾) (2.0 𝑎𝑡𝑚)(303𝐾)
= …L
≐0.33 𝐿
𝐷 2 = 𝑚 𝐻𝑒 𝑉 2 =
0.12 𝑔 0.33 𝐿 ≐0.37 𝑔/𝐿