1. Gas Laws Ideal Gas Law: PV = nRT where P = pressure V = volume
n = moles
T = temperature (K)
R = gas constant
R = atm.L
mol.K
R = L.torr

2. Gas Laws
The ideal gas law is used to describe the behavior of an ideal gas.
Ideal gas: hypothetical gas that obeys kinetic molecular theory and the ideal gas law

3. Gas Laws
The ideal gas law is used in calculations for a specific sample of gas that has a constant T, P, V, and n.
i.e. no changes are being made to the sample of gas
If you know 3 of the 4 variables, you can calculate the other using the ideal gas law.

4. Gas Laws
Example: Calculate the volume of 1.00 mol of an ideal gas at 1.00 atm and 0.00oC.
Given: P = 1.00 atm
n = 1.00 mol
T = 0.00oC + 273
= 273 K
Find: V
PV = nRT

5. Gas Laws PV = nRT Solve for V V = nRT P
V = 1.00 mol x atm.L x 273 K
1.00 atm mol.K
V = 22.4 L

6. Gas Laws
The temperature and pressure used in the previous problem are commonly used to report the properties of gases.
Standard Temperature and Pressure (STP):
0oC and 1 atm.
KNOW THIS!!

7. Gas Laws Molar volume: the volume one mol of a gas occupies (L/mole)
At STP, one mole of an ideal gas has a molar volume of 22.4 L:
22.4 L
1 mol

8. Gas Laws The ideal gas law applies only to ideal gases.
Does not always accurately describe real gases
The molar volumes for many real gases at STP differ slightly from 22.4 L/mol.
In most cases, the differences between ideal gas behavior and real gas behavior is so small that we can ignore it

9. Gas Laws
Example: A weather balloon contains 4.75 moles of He gas. What volume does the gas occupy at an altitude of 4300 m if the temperature is 0oC and the pressure is atm?
Given: P = atm
T = 0oC = 273K
n = 4.75 mol
Find: V
PV = nRT

10. Gas Laws PV = nRT solve for V V = nRT P
V = (4.75 mol) x (273K) x ( atm.L)
(0.595 atm) mol.K
V = 179 L

11. Gas Laws
Example: A used aerosol can contains mol of gas and has a volume of 425 mL. Calculate the pressure in the can if it is accidentally heated to 395oC. (Warning: Don’t do this!!)
Given: V = 425 mL
n = mol
T = 395oC = 668K
Find: P
PV = nRT

12. Gas Laws PV = nRT solve for P P = nRT V
P = ( mol) ( atm.L) (668K)
425 mL mol K
x 1000 mL
1 L
P = 2.23 atm

13. Gas Laws
Example: A tire with an interior volume of 3.50 L contains mol of air at a pressure of 2.49 atm. What is the temperature of the air in the tire in K? In oC?
Given: P = 2.49 atm
V = 3.50 L
n = mol
Find: T (K)
PV = nRT

14. Gas Laws PV = nRT solve for T T = PV nR T = 2.49 atm x 3.50 L x mol.K
0.357 mol atm.L
T = 297 K
T = 297 – 273 = 24oC

15. Gas Laws
The ideal gas law was useful in determining the properties of a specific sample of gas at constant T, P, V, and n.
We often need to know how a change in one (or more) properties impacts the other properties for a sample of a gas.

16. Gas Laws Combined Gas Laws: P1V1 = P2V2 T1 T2
This equation is true when the number of moles of a gas is constant.

17. Gas Laws Special cases for the combined gas law:
At constant temperature (T1 = T2),
P1V1 = P2V2
At constant volume (V1 = V2),
P1 = P2
T1 T2
At constant Pressure (P1 = P2),
V1 = V2

18. Gas Laws
Example: A helium-filled balloon occupies 6.00 L at 19.5oC and atm. What volume will the balloon occupy on top of Pike’s Peak if the pressure is atm and the temperature is constant?
Given: V1 = 6.00L
P1 = atm
T1 = T2 = 19.5oC
P2 = atm
Find: V2
Since T is constant,
P1V1 = P2V2

19. Gas Laws P1 V1 = P2 V2 (0.989 atm) x (6.00 L) = (0.605 atm) x V2
V2 = 9.81 L

20. Gas Laws
Example: Suppose a used aerosol can contains a gas at atm at 23oC. If this can is heated to 425oC, what is the pressure inside the can?
Given: P1 = atm
T1 = 23oC = 296K
T2 = = 698K
V1 = V2
Find: P2
P1V1 = P2V2
T1 T2

21. Gas Laws Since V1= V2: P1 = P2 T1 T2 P2 = 0.989 atm (698K) 296K
P2 = (0.989 atm) x 698K = 2.33 atm
296K

22. Gas Laws--More Applications
The ideal gas equation can be used to determine either the density or the molar mass of a gas.
d = P M RT
Where d = density
P = pressure
T = temperature in K
R = gas constant
M = molar mass

23. Gas Laws--More Applications
The density of a gas depends on:
pressure
temperature
molar mass
At constant temperature and pressure the densities of gas samples are directly proportional to their molar masses:
d = P M RT

24. Gas Laws--More Applications
If molar mass and pressure are held constant, then the density of the gas will decrease with increasing temperature
Hot air rises because density is inversely proportional to temperature
d = P M RT

25. Gas Laws--More Applications
Example: What is the density of helium gas at 1.00 atm and 25oC?
Given: P = 1.00 atm
T = = 298K
Find: d
d = P M RT

26. Gas Laws--More Applications
d = P M RT
M = molar mass of He = 4.00 g/mol
d = (1.00 atm) g mol.K
(298K) 1 mol atm.L
d = g/L
Notice: density of gases is usually in g/L instead of g/mL

27. Gas Laws--More Applications
Example: What is the average molar mass of dry air if it has a density of 1.17 g/L at 21oC and torr?
Given: P = torr
T = = 294K
d = 1.17 g/L
Find: M
d = P M RT

28. Gas Laws--More Applications
d = P M RT
M = dRT P
M = 1.17 g torr.L (294 K)
mol.K torr
M = 29.0 g/mol

29. Gas Laws--More Applications
Understanding the properties of gases is important because gases are often the reactants or products in a chemical reaction.
Often need to calculate the volume of gas produced or consumed during a reaction

30. Gas Laws--More Applications
Moles A B
Molar ratio
grams
Molar mass
Gas
Data A
PA, VA, TA
PV = nRT
Gas
Data B
PB, VB, TB
PV = nRT

31. Gas Laws--More Applications
Example: The air bag in a car is inflated by nitrogen gas formed by the decomposition of NaN3:
2 NaN3(s)  2 Na (s) + 3 N2 (g)
If an inflated air bag has a volume of 36 L and is to be filled with N2 gas at a pressure of 1.15 atm at a temperature of 26oC, how many grams of NaN3 must be decomposed?

32. Gas Laws--More Applications
Moles N2
Moles A
Gas
Data N2
PV = nRT
Molar ratio
Molar ratio
grams
NaN3
grams B
Moles B
Moles
NaN3
Molar mass
Molar mass

33. Gas Laws--More Applications
36L N2, 1.15 atm
26oC
Moles N2
Moles A
PV = nRT
Find
Molar ratio
Molar ratio
Given
grams
NaN3
grams B
Moles
NaN3
Moles B
Molar mass
Molar mass

34. Gas Laws--More Applications
n = PV = (1.15 atm) ( 36 L) mol.K
RT (299 K) atm.L
n = 1.7 mol N2
g NaN3 = 1.7 mol N2 x 2 mol NaN3 x 65.0 g NaN3
3 mol N2 1 mol NaN3
g NaN3 = 74 g NaN3

35. Gas Laws--More Applications
Example: How many mL of oxygen gas can be collected at STP when 1.0 g of KClO3 decomposes:
2 KClO3 (s)  2 KCl (s) O2 (g)

36. Gas Laws--More Applications
mL O2 at STP (1.0atm, 0oC)
Moles O2
Moles A
PV = nRT
Given
Molar ratio
Molar ratio
Find
grams
KClO3
grams B
Moles
KClO3
Moles B
Molar mass
Molar mass

37. Gas Laws--More Applications
Mol O2 = 1.0 g KClO3 x 1 mole KClO3 x 3 mol O2
122.5 g 2 mol KClO3
mol O2 = mol O2
V = nRT = (0.012 mol) atm.L (273K) x 103 mL
P atm mol.K 1 L
V = 270 mL

38. Gas Laws – More Applications
Example: What volume of CO2 at 125oC and 1.15 atm will be produced by the combustion of 1.00 L of C2H6 at 25oC and 1.00 atm?
2 C2H6 (g) O2 (g)  4 CO2 (g) H2O (g)
Given: VC2H6 = 1.00 L
PC2H6 = 1.00 atm
TC2H6 = 25oC = 298 K
PCO2 = 1.15 atm
TCO2 = 125oC = 398 K
Find: VCO2