1. CHE 123: General Chemistry I
Chapter 5: Gases
CHE 123: General Chemistry I
Dr. Jerome Williams, Ph.D.
Saint Leo University

2. Overview Gas Laws Ideal Gas Law General Gas Law Example Problems

3. Boyle’s Law Pressure of a gas is inversely proportional to its volume
Robert Boyle (1627–1691)
Pressure of a gas is inversely proportional to its volume
constant T and amount of gas
graph P vs V is curve
graph P vs 1/V is straight line
As P increases, V decreases by the same factor
P x V = constant
P1 x V1 = P2 x V2
Tro: Chemistry: A Molecular Approach, 2/e

4. Boyle’s Experiment, P x V
Tro: Chemistry: A Molecular Approach, 2/e

5. Tro: Chemistry: A Molecular Approach, 2/e

6. Tro: Chemistry: A Molecular Approach, 2/e

7. Boyle’s Law: A Molecular View
Pressure is caused by the molecules striking the sides of the container
When you decrease the volume of the container with the same number of molecules in the container, more molecules will hit the wall at the same instant
This results in increasing the pressure
Tro: Chemistry: A Molecular Approach, 2/e

8. Example 5. 2: A cylinder with a movable piston has a volume of 7
Example 5.2: A cylinder with a movable piston has a volume of 7.25 L at 4.52 atm. What is the volume at 1.21 atm?
Given:
Find:
V1 =7.25 L, P1 = 4.52 atm, P2 = 1.21 atm
V2, L
Conceptual Plan:
Relationships:
P1 ∙ V1 = P2 ∙ V2
V1, P1, P2 V2
Solution:
Check:
because P and V are inversely proportional, when the pressure decreases ~4x, the volume should increase ~4x, and it does
Tro: Chemistry: A Molecular Approach, 2/e

9. Practice – A balloon is put in a bell jar and the pressure is reduced from 782 torr to atm. If the volume of the balloon is now 2.78 x 103 mL, what was it originally?
Tro: Chemistry: A Molecular Approach, 2/e

10. A balloon is put in a bell jar and the pressure is reduced from 782 torr to atm. If the volume of the balloon is now 2.78x 103 mL, what was it originally?
Given:
Find:
V2 =2780 mL, P1 = 782 torr, P2 = atm
V1, mL
Conceptual Plan:
Relationships:
P1 ∙ V1 = P2 ∙ V2 , 1 atm = 760 torr (exactly)
V2, P1, P2 V1
Solution:
Check:
because P and V are inversely proportional, when the pressure decreases ~2x, the volume should increase ~2x, and it does
Tro: Chemistry: A Molecular Approach, 2/e

11. Charles’s Law Volume is directly proportional to temperature
Jacques Charles (1746–1823)
Volume is directly proportional to temperature
constant P and amount of gas
graph of V vs. T is straight line
As T increases, V also increases
Kelvin T = Celsius T + 273
V = constant x T
if T measured in Kelvin
Tro: Chemistry: A Molecular Approach, 2/e

12. If you plot volume vs. temperature for any gas at constant pressure, the points will all fall on a straight line
If the lines are extrapolated back to a volume of “0,” they all show the same temperature, − °C, called absolute zero
Tro: Chemistry: A Molecular Approach, 2/e

13. Example 5. 3: A gas has a volume of 2. 57 L at 0. 00 °C
Example 5.3: A gas has a volume of 2.57 L at 0.00 °C. What was the temperature in Celsius at 2.80 L?
Given:
Find:
V1 =2.57 L, V2 = 2.80 L, t2 = 0.00 °C
t1, K and °C
Conceptual Plan:
Relationships:
V1, V2, T2 T1
Solution:
Check:
because T and V are directly proportional, when the volume decreases, the temperature should decrease, and it does
Tro: Chemistry: A Molecular Approach, 2/e

14. Practice – The temperature inside a balloon is raised from 25
Practice – The temperature inside a balloon is raised from 25.0 °C to °C. If the volume of cold air was 10.0 L, what is the volume of hot air?
Tro: Chemistry: A Molecular Approach, 2/e

15. The temperature inside a balloon is raised from 25. 0 °C to 250. 0 °C
The temperature inside a balloon is raised from 25.0 °C to °C. If the volume of cold air was 10.0 L, what is the volume of hot air?
Given:
Find:
V1 =10.0 L, t1 = 25.0 °C L, t2 = °C
V2, L
Conceptual Plan:
Relationships:
V1, T1, T2 V2
Solution:
Check:
when the temperature increases, the volume should increase, and it does
Tro: Chemistry: A Molecular Approach, 2/e

16. Avogadro’s Law
Amedeo Avogadro (1776–1856)
Volume directly proportional to the number of gas molecules
V = constant x n
constant P and T
more gas molecules = larger volume
Count number of gas molecules by moles
Equal volumes of gases contain equal numbers of molecules
the gas doesn’t matter
Tro: Chemistry: A Molecular Approach, 2/e

17. Example 5. 4:A 0. 225 mol sample of He has a volume of 4. 65 L
Example 5.4:A mol sample of He has a volume of 4.65 L. How many moles must be added to give 6.48 L?
Given:
Find:
V1 = 4.65 L, V2 = 6.48 L, n1 = mol
n2, and added moles
Conceptual Plan:
Relationships:
V1, V2, n1 n2
Solution:
Check:
because n and V are directly proportional, when the volume increases, the moles should increase, and they do
Tro: Chemistry: A Molecular Approach, 2/e

18. Ideal Gas Law
Ideal gases obey an equation incorporating the laws of Charles, Boyle, and Avogadro.
P V = n R T
1 mole of an ideal gas occupies L at STP
STP conditions are K and 1 atm pressure
The gas constant R = L·atm·K–1·mol–1

19. Example 5.6: How many moles of gas are in a basketball with total pressure 24.3 psi, volume of 3.24 L at 25°C?
Given:
Find:
V = 3.24 L, P = 24.3 psi, t = 25 °C
n, mol
Conceptual Plan:
Relationships:
P, V, T, R n
Solution:
Check:
1 mole at STP occupies 22.4 L, because there is a much smaller volume than 22.4 L, we expect less than 1 mole of gas
Tro: Chemistry: A Molecular Approach, 2/e

20. Chapter 5, Unnumbered Figure 1, Page 191

21. Example Problems
Sulfur hexafluoride (SF6) is a colorless, odorless, very unreactive gas. Calculate the pressure (in atm) exerted by 1.82 moles of the gas in a steel vessel of volume 5.43 L at 69.5°C.
Answer = 9.42 atm

22. Example Problems
What is the volume (in liters) occupied by 7.40 g of CO2 at STP?
Answer = 3.77 L

23. Ideal Gas Law: Application
Density and Molar Mass Calculations
Given the following: Note: MM = molar mass
Density d = mass/volume
No. moles n n = mass / MM
Show that MM = dRT / P

24. Figure: UN
Title:
Molar Volume
Caption:
One mole of any gas occupies approximately 22.4 L at standard temperature (273 K) and pressure (1.0 atm).

25. Figure: UN
Title:
Molar density (1)
Caption:
The density in grams/liter can be obtained from molar density by multiplying by the molar mass (M).

26. Figure: UN
Title:
Molar density (2)
Caption:
The density in grams/liter can be obtained from the Ideal Gas Law.

27. Ideal Gas Law: Application Examples
What is the molar mass of a gas with a density of g/L–1 at STP?
What is the density of uranium hexafluoride, UF6, (MM = 352 g/mol) under conditions of STP?
The density of a gaseous compound is 3.38 g/L–1 at 40°C and 1.97 atm. What is its molar mass?

28. Ideal Gas Law: Application Answers
Molar mass = g/mole
Density of UF6 = 15.7 g/mL
Molar mass = g/mole

29. General Gas Law Equation
(P1V1 / T1 )= (P2V2 / T2 )
Only equation that uses two sets of data for V, P, and T.

30. Example Problem
Oxygen gas is normally sold in 49.0 L steel containers at a pressure of atm. What volume would the gas occupy if the pressure was reduced to 1.02 atm and the temperature raised from 20oC to 35oC?