1. Gas Laws
Gases

2. Kelvin Temperature Scale
Used in gas law equations
0 K is referred to as absolute zero – the point where all particle motion stops.
Conversions:
K = 0C
0C = K – 273
Convert 230 C to Kelvin
= 296 K

3. The boiling point of water is 1000 C
The boiling point of water is 1000 C. What is the boiling point in degrees Kelvin?
= 373 K

4. Variables Affecting Gas Behavior
Number of gas particles
Temperature
Pressure
Volume

5. STP Standard Temperature and Pressure
Temp: 0 degrees Celsius or 273 degrees K
Pressure: 1 atm
Equals 760 mm Hg, kPa, 14.7 psi

6. Boyle’s Law
Volume of a given amount of gas at a constant temperature varies inversely with the pressure
When one goes up, the other goes down
(P1)(V1) = (P2)(V2)

7. Boyle’s Law in action

9. Example P1 = 210 kPa V1 = 4.0L P2 = X V2 = 2.5L
A sample of He in a balloon is compressed from 4.0L to 2.5L at a constant temp. If the pressure of the gas is initially 210 kPa, what will the pressure be at 2.5L?
P1 = 210 kPa V1 = 4.0L
P2 = X V2 = 2.5L
(210)(4) = (X)(2.5)
X = 336 kPa

10. Example
Air trapped in a cylinder with piston occupies mL at 1.08 atm. What is the new volume of air when the pressure is increased to 1.43 atm by applying force to the piston?
P1 = 1.08atm V1 = mL
P2 = 1.43atm V2 = X
(1.08)(145.7) = (1.43)(X)
X = mL

11. Charles’s Law
Volume of a given mass of gas is directly proportional to its temp (K) at constant pressure
When one goes up, the other goes up
V1/T1 = V2/T2

12. Charles Law in Action

13. Example
A gas at 40.00C occupies a volume of 2.32L at constant pressure. If the temp is raised to 75.00C, what will the volume be?
V1 = 2.32L T1 = = 313K
V2 = X T2 = = 348K
(2.32)/(313) = (X)/(348)
X = 2.6L

14. Example
A volume of air in a balloon occupies .620L at 250C and constant pressure. If the volume of the balloon decreases to .570L, what was the temp change in the room?
V1 = .620L T1 = = 298K
V2 = .570L T2 = X
(.620)/(298) = (.570)/(X)
X = 273K

15. Lussac’s Law P1/T1 = P2/T2 As temperature goes up, pressure goes up
Pressure of a given mass of gas varies directly with the temp (K) when volume stays constant
When one goes up, the other does too
P1/T1 = P2/T2
As temperature goes up, pressure goes up

17. Example
The pressure of a gas in a tank is 3.20 atm at 22.00C. If the temp rises to 60.00C, what will be the gas pressure in the tank?
P1 = 3.20 atm T1 = 295K
P2 = X T2 = 333K
(3.20)/(295) = (X)/(333)
X = 3.6 atm

18. Combined Gas Law
States relationship between P, V, T for a fixed amount of gas
P is inversely related to V, and directly related to T
V is directly proportional to T
P1V1 = P2V2
T T2

19. Example
A gas at 110kPa and 30.00C fills a container with an initial volume of 2.00L. If the temp is raised to 80.00C and the pressure to 400kPa, what is the new volume?
P1 = 110kPa V1 = 2.0L T1 = 303K
P2 = 400kPa V2 = X T2 = 353K
(110)(2)/(303) = (400)(X)/(353)
X = .58L

20. Example
At 00C and 1.00 atm, a sample of gas occupies 30.0mL. If the temp is increased to 30.00C, and the entire sample transferred to a 20.0mL container, what will be the pressure inside the container?
P1 = 1 atm V1 = 30.0mL T1 = 273K
P2 = X V2 = 20.0mL T2 = 303K
(1)(30)/(273) = (X)(20)/(303)
X = 1.66 atm

21. Avogadro’s Principle
Equal volumes of gas at the same temp and pressure contain equal numbers of particles
How many gas particles are in .166mol He?
.166 mol (6.02 x 1023 particles/mol) = 9.99 x 1022
How many gas particles are in .166mol Ne?

22. Molar Volume
The volume that 1mol of gas occupies at STP (00C and 1atm)
1 mol ANY gas at STP = 22.4L
1 mol of Ar
1 mol of He
1 mol of Xe

23. Example
Calculate the volume that 2000g of methane (CH4) will occupy at STP.
2000g CH4 x 1mol CH4 x L CH4 = L CH4
16.042g CH4 1 mol CH4

24. Example Calculate the volume that .881mol Ar will occupy at STP.
.881 mol Ar x L = 19.7 L Ar
1mol Ar

25. Ideal Gas Law
Describes the behavior of an ideal gas in terms of P, V, T, and # moles (n) of gas present
Ideal Gas
One whose particles take up no space and have no intermolecular attractive forces.
Follows gas laws under all conditions of T and P.
Real Gas
Do not always follow all of the gas laws

26. Ideal Gas Law
Ideal Gas Constant (R) – experimentally determined constant that depends on the units of pressure
PV = nRT
P units may vary
V units are L
n = moles
T unit is K

27. Ideal Gas Constant Units of R Numerical value of R Units of Pressure
L • atm
mol • 0K
0.0821
atm
L • kPa
8.314
kPa
L • mmHg
62.4
mm Hg

28. Example
Calculate the number of moles of gas contained in a 3.0L vessel at 300.0K with a pressure of 1.50atm.
P=1.50atm V=3.0L n=x R= T=300
(1.50)(3) = n(.0821)(300)
n = .18 mol

29. Example
Determine the Celsius temp of 2.59mol of gas contained in a 1.00L vessel at a pressure of 143kPa.
P=143kPa V=1L n= R= T=x
(143)(1) = (2.59)(8.314)(x)
X = 6.64K
6.64K-273 = -2660C

30. Gas Stoichiometry CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) 1 mol = 1 volume
1 vol CH vol O vol CO2
2 vol O vol CO vol H2O

31. Example CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
If you have 6.7L of CH4, how much water vapor can be produced?
6.7L CH4 x 2 vol H2O = 13.4L H2O
1 vol CH4

32. Example CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
Calculate the volume of oxygen needed to completely react with 74.3g CH4
74.3g CH4 x 1mol CH4 x 2 mol O2 x 22.4L O2 = 207.5L O2
16.042g CH4 1mol CH mol O2