1. Gas Laws

2. Properties of Gases Variable volume and shape
Expand to occupy volume available
Volume, Pressure, Temperature, and the number of moles present are interrelated
Can be easily compressed
Exert pressure on whatever surrounds them
Easily diffuse into one another

3. Mercury Barometer Used to define and measure atmospheric pressure
On the average at sea level the column of mercury rises to a height of about 760 mm.
This quantity is equal to 1 atmosphere
It is also known as standard atmospheric pressure

4. Pressure Units & Conversions
The above represent some of the more common units for measuring pressure. The standard SI unit is the Pascal or kilopascal. (kPa)

5. Standard Temperature and Pressure
Standard Temperature and Pressure or STP = 0oC or Kelvin and a pressure of 1 atmosphere.
This is used as a reference point when comparing quantities of gases

6. Boyle’s Law
According to Boyle’s Law the pressure and volume of a gas are inversely proportional at constant temperature.
(syringe-P logger)
PV = constant.
P1V1 = P2V2

7. Boyle’s Law A graph of pressure and volume gives an inverse function
A graph of pressure and the reciprocal of volume gives a straight line 7

8. Sample Problem 1:
If the pressure of helium gas in a balloon has a volume of 4.00 dm3 at 210 kPa, what will the pressure be at 2.50 dm3?
P1 V1 = P2 V2
(210 kPa) (4.00 dm3) = P2(2.50 dm3)
P2 = (210 kPa) (4.00 dm3)
(2.50 dm3)
= 340 kPa

9. Charles’ Law
According to Charles’ Law the volume of a gas is proportional to the Kelvin temperature as long as the pressure is constant V = kT (syringe – BB)
Note: The temperature for gas laws must always be expressed
in Kelvin where Kelvin = oC (or 273 to 3 significant digits) V1 = T1 V2 T2

10. Charles’ Law A graph of temperature and volume yields a straight line.
Where this line crosses the x axis (x intercept) is defined as absolute zero

11. Sample Problem 2
A gas sample at 40 oC occupies a volume of 2.32 dm3. If the temperature is increased to 75 oC, what will be the final volume?
V1 = V2
T T2
Convert temperatures to Kelvin. 40oC = 313K
75oC = 348K
2.32 dm3 = V2
313 K K
(313K)( V2) = (2.32 dm3) (348K)
V2 =
2.58 dm3

12. Gay-Lussac’s Law P1 = P2 T2 T1
Gay-Lussac’s Law defines the relationship between pressure and temperature of a gas.
The pressure and temperature of a gas are directly proportional
P = P2 T2 T1 12

13. Sample Problem 3:
The pressure of a gas in a tank is 3.20 atm at 22 oC. If the temperature rises to 60oC, what will be the pressure in the tank?
P1 = P2
T T2
Convert temperatures to Kelvin. 22oC = 295K
60oC = 333K
3.20 atm = P2
295 K K
(295K)( P2) = (3.20 atm)(333K)
P2 =
3.6 atm 13

14. The Combined Gas Law
1. If the amount of the gas is constant, then Boyle’s Charles’ and Gay-Lussac’s Laws can be combined into one relationship
P1 V = P2 V2 T1 T2 14

15. Sample Problem 4:
A gas at 110 kPa and 30 oC fills a container at 2.0 dm3. If the temperature rises to 80oC and the pressure increases to 440 kPa, what is the new volume?
P1V1 = P2V2
T T2
Convert temperatures to Kelvin. 30oC = 303K
80oC = 353K
V2 = V1 P1 T2
P2 T1
= (2.0 dm3) (110 kPa ) (353K)
(440 kPa ) (303 K)
V2 = 0.58 dm3 15

16. Advogadro’s Law
Equal volumes of different gases under the same temperature and pressure contain the same number of particles.
Pa*Va/Ta = na Pb*Vb/Tb = nb
If they occupy same V and T and P are equal, then
Pa*Va/Ta = Pb*Vb/Tb => na = nb 16

17. Ideal Gas Law
Based on the previous laws there are four factors that define the quantity of gas: Volume, Pressure, Kelvin Temperature, and the number of moles of gas present (n).
Putting these all together: PV nT
= Constant = R
The proportionality constant R is known as the
universal gas constant 17

18. Ideal Gas Law PV = nRT V = volume T = Kelvin Temperature
The Ideal gas law is usually written as
PV = nRT
Where P = pressure
V = volume
T = Kelvin Temperature
n = number of moles
The numerical value of R depends on the pressure unit (and perhaps the energy unit) Some common values of R include:
R = dm3 atm mol-1 K-1
= dm3kPa mol-1 K-1 18

19. Sample Problem 5 Example: What volume will 25.0 g O2 occupy
at 20oC and a pressure of atmospheres? :
(25.0 g)
n = = mol
(32.0 g mol-1)
Data
Formula
Calculation
Answer
V =? P = atm; T = ( )K = 293K
R = dm-3 atm mol-1 K-1
PV = nRT so V = nRT/P
V = (0.781 mol)( dm-3 atm mol-1 K-1)(293K)
0.880 atm
V = 21.3 dm3 19