1. Dr. Orlando E. Raola Santa Rosa Junior College
Chemistry 1A
General Chemistry
Instructor:
Dr. Orlando E. Raola Santa Rosa Junior College
Chapter 5: Properties of
Gases

2. Gases and the Kinetic Molecular Theory
5.1 An Overview of the Physical States of Matter
5.2 Gas Pressure and Its Measurement
5.3 The Gas Laws and Their Experimental Foundations
5.4 Applications of the Ideal Gas Law

3. The distinction of gases from liquids and solids
Gas volume changes greatly with pressure.
Gas volume changes greatly with temperature.
Gases have relatively low viscosity.
Most gases have relatively low densities under normal conditions.
Gases are miscible.

4. The three states of matter

5. A mercury barometer

6. Units of pressure

7. Sample Problem 5.1
Converting Units of Pressure
PROBLEM:
A geochemist heats a limestone (CaCO3) sample and collects the CO2 released in an evacuated flask attached to a closed-end manometer. After the system comes to room temperature, Dh = mm Hg. Express the CO2 pressure in torr, atmosphere, and kilopascal.
PLAN:
Construct conversion factors to find the other units of pressure.
SOLUTION:
291.4 mmHg
1torr
1 mmHg
= torr
1 atm
760 torr
291.4 torr
= atm
kPa
1 atm
atm
= kPa

8. The relationship between the volume and pressure of a gas.
Boyle’s Law

9. The relationship between the volume and temperature of a gas.
Charles’s Law

10. V a 1 P
Boyle’s Law
n and T are fixed
V x P
= constant
V = constant / P
Charles’s Law
V a T
P and n are fixed V T
= constant
V = constant x T
Amontons’s Law
P a T
V and n are fixed P T
= constant
P = constant x T
V a T P
V = constant x T P PV T
= constant
combined gas law

11. n = chemical amount of gas (mol)
Avogadro’s Law
For a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas
V = a n
a = proportionality constant
V = volume of the gas (m3)
n = chemical amount of gas (mol)

12. (in common units 0.082 atm∙L∙mol-1·K-1)
Ideal Gas Law
An equation of state for a gas.
“state” is the condition of the gas at a given time.
PV = nRT
R = J∙mol-1·K-1
(in common units atm∙L∙mol-1·K-1)

13. Standard molar volume.

14. PV = nRT THE IDEAL GAS LAW IDEAL GAS LAW nRT P PV = nRT or V =
R is the universal gas constant
IDEAL GAS LAW
nRT P
PV = nRT or V =
fixed n and T
fixed n and P
fixed P and T
Boyle’s Law
Charles’s Law
Avogadro’s Law
constant P
V =
constant X n
V =
V =
constant X T

15. Sample Problem 5.2
Applying the Volume-Pressure Relationship
PROBLEM:
Boyle’s apprentice finds that the air trapped in a J tube occupies 24.8 cm3 at 1.12 atm. By adding mercury to the tube, he increases the pressure on the trapped air to 2.64 atm. Assuming constant temperature, what is the new volume of air (in L)?
PLAN:
SOLUTION:
n and T are constant
V1 in cm3
P1 = 1.12 atm
P2 = 2.64 atm
unit conversion
1cm3=1mL
V1 = 24.8 cm3
V2 = unknown
V1 in mL
1 mL
1 cm3 L
103 mL
103 mL=1L
24.8 cm3
= L
V1 in L
gas law calculation
xP1/P2
P1V1
n1T1
P2V2
n2T2
P1V1 = P2V2 =
V2 in L
P1V1 P2
V2 =
1.12 atm
2.46 atm = L = L

16. Sample Problem 5.3
Applying the Temperature-Pressure Relationship
PROBLEM:
A steel tank used for fuel delivery is fitted with a safety valve that opens when the internal pressure exceeds 1.00x103 torr. It is filled with methane at 230C and atm and placed in boiling water at exactly 1000C. Will the safety valve open?
PLAN:
SOLUTION:
P1(atm)
T1 and T2(0C)
P1 = 0.991atm
P2 = unknown
1atm=760torr
K=0C
T1 = 230C
T2 = 1000C
P1(torr)
T1 and T2(K)
P1V1
n1T1
P2V2
n2T2 = P1 T1 P2 T2 =
x T2/T1
P2(torr)
0.991 atm
1 atm
760 torr
= 753 torr
P2 = P1 T2 T1
= 753 torr
373K
296K
= 949 torr

17. Sample Problem 5.4
Applying the Volume-Amount Relationship
PROBLEM:
A scale model of a blimp rises when it is filled with helium to a volume of 55 dm3. When 1.10 mol of He is added to the blimp, the volume is 26.2 dm3. How many more grams of He must be added to make it rise? Assume constant T and P.
PLAN:
We are given initial n1 and V1 as well as the final V2. We have to find n2 and convert it from moles to grams.
n1(mol) of He
SOLUTION:
P and T are constant
x V2/V1
n1 = 1.10 mol
n2 = unknown
P1V1
n1T1
P2V2
n2T2 =
n2(mol) of He
V1 = 26.2 dm3
V2 = 55.0 dm3
subtract n1 V1 n1 V2 n2 =
n2 = n1 V2 V1
mol to be added
x M
n2 = 1.10 mol
55.0 dm3
26.2 dm3
4.003 g He
mol He
g to be added
= 9.24 g He
= 2.31 mol

18. Sample Problem 5.5
Solving for an Unknown Gas Variable at Fixed Conditions
PROBLEM:
A steel tank has a volume of 438 L and is filled with kg of O2. Calculate the pressure of O2 at 210C.
PLAN:
V, T and mass are given. From mass, find amount (n), and use the ideal gas law to find P.
SOLUTION:

19. Relationship between density and molar mass for gases
The density of a gas is directly proportional to its molar mass.
The density of a gas is inversely proportional to the temperature.

20. Sample Problem 5.7
Calculating Gas Density
PROBLEM:
To apply a green chemistry approach, a chemical engineer uses waste CO2 from a manufacturing process, instead of chlorofluorocarbons, as a “blowing agent” in the production of polystyrene containers. Find the density (in g/L) of CO2 and the number of molecules (a) at STP (00C and 1 atm) and (b) at room conditions (20.0C and 1.00 atm).
PLAN:
Density is mass/unit volume; substitute for volume in the ideal gas equation. Since the identity of the gas is known, we can find the molar mass. Convert mass/L to molecules/L with Avogadro’s number. d = RT
M x P
d = mass/volume
PV = nRT
V = nRT/P
SOLUTION:
44.01 g/mol
x 1atm
atm*L
mol*K
0.0821
x K
= 1.96 g/L
d =
(a)
1.96 g L
mol CO2
44.01 g CO2
6.022x1023 molecules
mol
= 2.68x1022 molecules CO2/L

21. Sample Problem 5.6
Calculating Gas Density
continued
d =
44.01 g/mol
x 1 atm
x 293K
atm*L
mol*K
0.0821
(b)
= 1.83 g/L
1.83g L
mol CO2
44.01g CO2
6.022x1023 molecules
mol
= 2.50x1022 molecules CO2/L

22. Determining the molar mass of an unknown volatile liquid.
based on the method of J.B.A. Dumas ( )

23. Sample Problem 5.8
Finding the Molar Mass of a Volatile Liquid
PROBLEM:
An organic chemist isolates a colorless liquid from a petroleum sample. She uses the Dumas method and obtains the following data:
Volume of flask = 213 mL
Mass of flask + gas = g
T = C
Mass of flask = g
P = 754 torr
Calculate the molar mass of the liquid.
PLAN:
Use unit conversions, mass of gas, and density-M relationship.
SOLUTION:
m = ( ) g
= g
0.582 g
atm*L
mol*K
0.0821
373K x
m RT VP x
M = = =
84.4 g/mol
0.213 L x
0.992 atm

24. Relationship between density and molar mass for gases
The density of a gas is directly proportional to its molar mass.
The density of a gas is inversely proportional to the temperature.

25. Sample Problem 5.7
Calculating Gas Density
PROBLEM:
To apply a green chemistry approach, a chemical engineer uses waste CO2 from a manufacturing process, instead of chlorofluorocarbons, as a “blowing agent” in the production of polystyrene containers. Find the density (in g/L) of CO2 and the number of molecules (a) at STP ( K and 1 bar) and (b) at room conditions (20.0C and 1.00 atm).
PLAN:
Density is mass/unit volume; substitute for volume in the ideal gas equation. Since the identity of the gas is known, we can find the molar mass. Convert mass/L to molecules/L with Avogadro’s number. d = RT
M x P
d = mass/volume
PV = nRT
V = nRT/P
SOLUTION:
(a)
1.94 g L
mol CO2
44.01 g CO2
6.022x1023 molecules
mol
= 2.65x1022 molecules CO2/L

26. Sample Problem 5.6
Calculating Gas Density
continued
d =
44.01 g/mol
x 1 atm
x 293K
atm*L
mol*K
0.0821
(b)
= 1.83 g/L
1.83g L
mol CO2
44.01g CO2
6.022x1023 molecules
mol
= 2.50x1022 molecules CO2/L

27. Determining the molar mass of an unknown volatile liquid.
based on the method of J.B.A. Dumas ( )

28. Sample Problem 5.8
Finding the Molar Mass of a Volatile Liquid
PROBLEM:
An organic chemist isolates a colorless liquid from a petroleum sample. She uses the Dumas method and obtains the following data:
Volume of flask = 213 mL
Mass of flask + gas = g
T = C
Mass of flask = g
P = 754 torr
Calculate the molar mass of the liquid.
PLAN:
Use unit conversions, mass of gas, and density-M relationship.
SOLUTION:
m = ( ) g
= g
0.582 g
atm*L
mol*K
0.0821
373K x
m RT VP x
M = = =
84.4 g/mol
0.213 L x
0.992 atm

29. Sample Problem 5.5
Solving for an Unknown Gas Variable at Fixed Conditions
PROBLEM:
A steel tank has a volume of 438 L and is filled with kg of O2. Calculate the pressure of O2 at 210C.
PLAN:
V, T and mass are given. From mass, find amount (n), and use the ideal gas law to find P.
SOLUTION:

30. Mixtures of gases Gases mix homogeneously in any proportions.
Each gas in a mixture behaves as if it were the only gas present.
Dalton’s Law of Partial Pressures
Ptotal = P1 + P2 + P
P1= c1 x Ptotal
where c1 is the mole fraction
c1 = n1
n1 + n2 + n3 +... = n1
ntotal

31. Mole fraction and partial pressure
For each component we define the mole fraction xB
and because of Dalton’s law, we can calculate the partial pressure of each component as

32. Sample Problem 5.9
Applying Dalton’s Law of Partial Pressures
PROBLEM:
In a study of O2 uptake by muscle at high altitude, a physiologist prepares an atmosphere consisting of 79 mol% N2, 17 mol% 16O2, and 4.0 mol% 18O2. (The isotope 18O will be measured to determine the O2 uptake.) The pressure of the mixture is 0.75atm to simulate high altitude. Calculate the mole fraction and partial pressure of 18O2 in the mixture.
PLAN:
Find the c and P from Ptotal and mol% 18O2.
18O2 =
4.0 mol% 18O2
100
mol% 18O2
SOLUTION:
= 0.040 c
18O2
divide by 100
c 18O2
P = c x Ptotal = x 0.75 atm
18O2
= atm
multiply by Ptotal
partial pressure P
18O2

33. Table 5.3 Vapor Pressure of Water (P ) at Different T
H2O
T(0C)
P (torr)
T(0C)
P (torr) 5 10 11 12 13 14 15 16 18 20 22 24 26 28
4.6
6.5
9.2
9.8
10.5
11.2
12.0
12.8
13.6
15.5
17.5
19.8
22.4
25.2
28.3 30 35 40 45 50 55 60 65 70 75 80 85 90 95
100
31.8
42.2
55.3
71.9
92.5
118.0
149.4
187.5
233.7
289.1
355.1
433.6
525.8
633.9
760.0

34. Collecting a water-insoluble gaseous reaction product and determining its pressure.

35. CaC2(s) + 2H2O(l) C2H2(g) + Ca(OH)2(aq)
Sample Problem 5.10
Calculating the Amount of Gas Collected Over Water
PROBLEM:
Acetylene (C2H2), an important fuel in welding, is produced in the laboratory when calcium carbide (CaC2) reaction with water:
CaC2(s) + 2H2O(l) C2H2(g) + Ca(OH)2(aq)
For a sample of acetylene that is collected over water, the total gas pressure (adjusted to barometric pressure) is kPa and the volume is 523 mL. At the temperature of the gas (230C), the vapor pressure of water is kPa. How many grams of acetylene are collected?
PLAN:
The difference in pressures will give us the P for the C2H2. The ideal gas law will allow us to find n. Converting n to grams requires the molar mass, M. P
C2H2
Ptotal P
C2H2
SOLUTION:
= ( ) kPa =95.52 kPa P
n = PV RT
H2O n
C2H2 g
C2H2
x M

36. Sample Problem 5.9
Calculating the Amount of Gas Collected Over Water
continued
0.0203mol
26.04 g C2H2
mol C2H2 
= g C2H2

37. molar ratio from balanced equation
Summary of the stoichiometric relationships among the amount (mol,n) of gaseous reactant or product and the gas variables pressure (P), volume (V), and temperature (T).
amount (mol)
of gas A
amount (mol)
of gas B
P,V,T
of gas A
P,V,T
of gas B
ideal gas law
ideal gas law
molar ratio from balanced equation

38. Sample Problem 5.11
Using Gas Variables to Find Amount of Reactants and Products
PROBLEM:
Dispersed copper in absorbent beds is used to react with oxygen impurities in the ethylene used for producing polyethylene. The beds are regenerated when hot H2 reduces the metal oxide, forming the pure metal and H2O. On a laboratory scale, what volume of H2 at 765 torr and 2250C is needed to reduce 35.5 g of copper(II) oxide?
PLAN:
Since this problem requires stoichiometry and the gas laws, we have to write a balanced equation, use the moles of Cu to calculate mols and then volume of H2 gas.
mass (g) of Cu
SOLUTION:
CuO(s) + H2(g) Cu(s) + H2O(g)
divide by M
mol Cu
63.55 g Cu
1 mol H2
1 mol Cu
35.5 g Cu
= mol H2
mol of Cu
molar ratio x
498K
atm*L
mol*K
0.0821
1.01 atm
0.559 mol H2
= 22.6 L
mol of H2
use known P and T to find V
L of H2

39. Sample Problem 5.12
Using the Ideal Gas Law in a Limiting-Reactant Problem
PROBLEM:
The alkali metals (Group 1) react with the halogens (Group 17) to form ionic metal halides. What mass of potassium chloride forms when L of chlorine gas at atm and 293K reacts with 17.0 g of potassium?
PLAN:
After writing the balanced equation, we use the ideal gas law to find the number of moles of reactants, the limiting reactant and moles of product.
SOLUTION:
2K(s) + Cl2(g) KCl(s)
V = 5.25 L
T = 293K
n = unknown
P = atm PV RT =
0.950 atm
atm*L
mol*K
0.0821 x
293K x
5.25L
n = =
0.207 mol
Cl2
0.207 mol Cl2
2 mol KCl
1 mol Cl2
17.0g
39.10 g K
mol K
= mol
KCl formed =
0.435 mol K
2 mol KCl
2 mol K
Cl2 is the limiting reactant.
0.435 mol K
= mol
KCl formed
74.55 g KCl
mol KCl
0.414 mol KCl
= 30.9 g KCl

40. Problem
Gaseous iodine pentafluoride can be prepared by the reaction between solid iodine and gaseous fluorine. A 5.00-L flask containing 10.0 g of I2 is charged with 10.0 g of F2 and the reaction proceeds until one of the reactants is completely consumed. After the reaction is complete, the temperature in the flask is 125 ºC.
What is the partial pressure of IF5 in the flask?
What is the mole fraction of IF5 in the flask?

41. Postulates of the Kinetic-Molecular Theory
Postulate 1: Particle Volume
Because the volume of an individual gas particle is so small compared to the volume of its container, the gas particles are considered to have mass, but no volume.
Postulate 2: Particle Motion
Gas particles are in constant, random, straight-line motion except when they collide with each other or with the container walls.
Postulate 3: Particle Collisions
Collisions are elastic therefore the total kinetic energy(Ek) of the particles is constant.

42. Distribution of molecular speeds as a function of temperature

43. Distribution of molecular speeds as a function of temperature

44. Molecular description of Boyle’s Law

45. Molecular description of Dalton’s law of partial pressures

46. Molecular description of Charles’s law

47. Molecular description of Avogadro’s law

48. Relationship between molecular speed and mass

49. The meaning of temperature
Absolute temperature is a measure of the average kinetic energy of the molecular random motion

50. Effusion and difussion
Effusion: describes the passage of gas through a small orifice into an evacuated chamber.
Diffusion: describes the mixing of gases. The rate of diffusion is the rate of gas mixing.

51. Effusion:
Diffusion:

52. Effusion and KMT

53. Diffusion through space
distribution of molecular speeds
mean free path
collision frequency

54. √ Sample Problem 5.13 Applying Graham’s Law of Effusion PROBLEM:
Calculate the ratio of the effusion rates of helium and methane (CH4).
PLAN:
The effusion rate is inversely proportional to the square root of the molar mass for each gas. Find the molar mass of both gases and find the inverse square root of their masses.
SOLUTION:
M of CH4 = 16.04g/mol
M of He = 4.003g/mol
CH4 He
rate = √
16.04
4.003
= 2.002

55. (00C and 1 atm) Table 5.4 Molar Volume of Some Common Gases at STP
(L/mol)
Condensation Point
(0C)
Gas He H2 Ne
Ideal gas Ar N2 O2 CO
Cl2
NH3
22.435
22.432
22.422
22.414
22.397
22.396
22.390
22.388
22.184
22.079
-268.9
-252.8
-246.1
---
-185.9
-195.8
-183.0
-191.5
-34.0
-33.4

56. The behavior of several real gases with increasing external pressure

57. Effect of molecular attractions on pressure

58. Effect of molecular volume on measured volume

59. van der Waals equation Gas a (atm·L2·mol-2) b (L·mol-1) Van der Waals
equation for n
moles of a real gas
Gas a (atm·L2·mol-2) b (L·mol-1)
0.034
0.211
1.35
2.32
4.19
0.244
1.39
1.36
6.49
3.59
2.25
4.17
5.46 He Ne Ar Kr Xe H2 N2 O2
Cl2
CO2
CH4
NH3
H2O
0.0237
0.0171
0.0322
0.0398
0.0511
0.0266
0.0391
0.0318
0.0562
0.0427
0.0428
0.0371
0.0305