  Raffi Hovasapian

Three Miscellaneous Example Problems

Slide Duration:

Section 1: Classical Thermodynamics Preliminaries
The Ideal Gas Law

46m 5s

Intro
0:00
Course Overview
0:16
Thermodynamics & Classical Thermodynamics
0:17
Structure of the Course
1:30
The Ideal Gas Law
3:06
Ideal Gas Law: PV=nRT
3:07
Units of Pressure
4:51
Manipulating Units
5:52
Atmosphere : atm
8:15
Millimeter of Mercury: mm Hg
8:48
SI Unit of Volume
9:32
SI Unit of Temperature
10:32
Value of R (Gas Constant): Pv = nRT
10:51
Extensive and Intensive Variables (Properties)
15:23
Intensive Property
15:52
Extensive Property
16:30
Example: Extensive and Intensive Variables
18:20
Ideal Gas Law
19:24
Ideal Gas Law with Intensive Variables
19:25
Graphing Equations
23:51
Hold T Constant & Graph P vs. V
23:52
Hold P Constant & Graph V vs. T
31:08
Hold V Constant & Graph P vs. T
34:38
Isochores or Isometrics
37:08
More on the V vs. T Graph
39:46
More on the P vs. V Graph
42:06
Ideal Gas Law at Low Pressure & High Temperature
44:26
Ideal Gas Law at High Pressure & Low Temperature
45:16
Math Lesson 1: Partial Differentiation

46m 2s

Intro
0:00
Math Lesson 1: Partial Differentiation
0:38
Overview
0:39
Example I
3:00
Example II
6:33
Example III
9:52
Example IV
17:26
Differential & Derivative
21:44
What Does It Mean?
21:45
Total Differential (or Total Derivative)
30:16
Net Change in Pressure (P)
33:58
General Equation for Total Differential
38:12
Example 5: Total Differential
39:28
Section 2: Energy
Energy & the First Law I

1h 6m 45s

Intro
0:00
Properties of Thermodynamic State
1:38
Big Picture: 3 Properties of Thermodynamic State
1:39
Enthalpy & Free Energy
3:30
Associated Law
4:40
Energy & the First Law of Thermodynamics
7:13
System & Its Surrounding Separated by a Boundary
7:14
In Other Cases the Boundary is Less Clear
10:47
State of a System
12:37
State of a System
12:38
Change in State
14:00
Path for a Change in State
14:57
Example: State of a System
15:46
Open, Close, and Isolated System
18:26
Open System
18:27
Closed System
19:02
Isolated System
19:22
Important Questions
20:38
Important Questions
20:39
Work & Heat
22:50
Definition of Work
23:33
Properties of Work
25:34
Definition of Heat
32:16
Properties of Heat
34:49
Experiment #1
42:23
Experiment #2
47:00
More on Work & Heat
54:50
More on Work & Heat
54:51
Conventions for Heat & Work
1:00:50
Convention for Heat
1:02:40
Convention for Work
1:04:24
Schematic Representation
1:05:00
Energy & the First Law II

1h 6m 33s

Intro
0:00
The First Law of Thermodynamics
0:53
The First Law of Thermodynamics
0:54
Example 1: What is the Change in Energy of the System & Surroundings?
8:53
Energy and The First Law II, cont.
11:55
The Energy of a System Changes in Two Ways
11:56
Systems Possess Energy, Not Heat or Work
12:45
Scenario 1
16:00
Scenario 2
16:46
State Property, Path Properties, and Path Functions
18:10
Pressure-Volume Work
22:36
When a System Changes
22:37
Gas Expands
24:06
Gas is Compressed
25:13
Pressure Volume Diagram: Analyzing Expansion
27:17
What if We do the Same Expansion in Two Stages?
35:22
Multistage Expansion
43:58
General Expression for the Pressure-Volume Work
46:59
Upper Limit of Isothermal Expansion
50:00
Expression for the Work Done in an Isothermal Expansion
52:45
Example 2: Find an Expression for the Maximum Work Done by an Ideal Gas upon Isothermal Expansion
56:18
Example 3: Calculate the External Pressure and Work Done
58:50
Energy & the First Law III

1h 2m 17s

Intro
0:00
Compression
0:20
Compression Overview
0:34
Single-stage compression vs. 2-stage Compression
2:16
Multi-stage Compression
8:40
Example I: Compression
14:47
Example 1: Single-stage Compression
14:47
Example 1: 2-stage Compression
20:07
Example 1: Absolute Minimum
26:37
More on Compression
32:55
Isothermal Expansion & Compression
32:56
External & Internal Pressure of the System
35:18
Reversible & Irreversible Processes
37:32
Process 1: Overview
38:57
Process 2: Overview
39:36
Process 1: Analysis
40:42
Process 2: Analysis
45:29
Reversible Process
50:03
Isothermal Expansion and Compression
54:31
Example II: Reversible Isothermal Compression of a Van der Waals Gas
58:10
Example 2: Reversible Isothermal Compression of a Van der Waals Gas
58:11
Changes in Energy & State: Constant Volume

1h 4m 39s

Intro
0:00
Recall
0:37
State Function & Path Function
0:38
First Law
2:11
Exact & Inexact Differential
2:12
Where Does (∆U = Q - W) or dU = dQ - dU Come from?
8:54
Cyclic Integrals of Path and State Functions
8:55
Our Empirical Experience of the First Law
12:31
∆U = Q - W
18:42
Relations between Changes in Properties and Energy
22:24
Relations between Changes in Properties and Energy
22:25
Rate of Change of Energy per Unit Change in Temperature
29:54
Rate of Change of Energy per Unit Change in Volume at Constant Temperature
32:39
Total Differential Equation
34:38
Constant Volume
41:08
If Volume Remains Constant, then dV = 0
41:09
Constant Volume Heat Capacity
45:22
Constant Volume Integrated
48:14
Increase & Decrease in Energy of the System
54:19
Example 1: ∆U and Qv
57:43
Important Equations
1:02:06
Joule's Experiment

16m 50s

Intro
0:00
Joule's Experiment
0:09
Joule's Experiment
1:20
Interpretation of the Result
4:42
The Gas Expands Against No External Pressure
4:43
Temperature of the Surrounding Does Not Change
6:20
System & Surrounding
7:04
Joule's Law
10:44
More on Joule's Experiment
11:08
Later Experiment
12:38
Dealing with the 2nd Law & Its Mathematical Consequences
13:52
Changes in Energy & State: Constant Pressure

43m 40s

Intro
0:00
Changes in Energy & State: Constant Pressure
0:20
Integrating with Constant Pressure
0:35
Defining the New State Function
6:24
Heat & Enthalpy of the System at Constant Pressure
8:54
Finding ∆U
12:10
dH
15:28
Constant Pressure Heat Capacity
18:08
Important Equations
25:44
Important Equations
25:45
Important Equations at Constant Pressure
27:32
Example I: Change in Enthalpy (∆H)
28:53
Example II: Change in Internal Energy (∆U)
34:19
The Relationship Between Cp & Cv

32m 23s

Intro
0:00
The Relationship Between Cp & Cv
0:21
For a Constant Volume Process No Work is Done
0:22
For a Constant Pressure Process ∆V ≠ 0, so Work is Done
1:16
The Relationship Between Cp & Cv: For an Ideal Gas
3:26
The Relationship Between Cp & Cv: In Terms of Molar heat Capacities
5:44
Heat Capacity Can Have an Infinite # of Values
7:14
The Relationship Between Cp & Cv
11:20
When Cp is Greater than Cv
17:13
2nd Term
18:10
1st Term
19:20
Constant P Process: 3 Parts
22:36
Part 1
23:45
Part 2
24:10
Part 3
24:46
Define : γ = (Cp/Cv)
28:06
For Gases
28:36
For Liquids
29:04
For an Ideal Gas
30:46
The Joule Thompson Experiment

39m 15s

Intro
0:00
General Equations
0:13
Recall
0:14
How Does Enthalpy of a System Change Upon a Unit Change in Pressure?
2:58
For Liquids & Solids
12:11
For Ideal Gases
14:08
For Real Gases
16:58
The Joule Thompson Experiment
18:37
The Joule Thompson Experiment Setup
18:38
The Flow in 2 Stages
22:54
Work Equation for the Joule Thompson Experiment
24:14
Insulated Pipe
26:33
Joule-Thompson Coefficient
29:50
Changing Temperature & Pressure in Such a Way that Enthalpy Remains Constant
31:44
Joule Thompson Inversion Temperature
36:26
Positive & Negative Joule-Thompson Coefficient
36:27
Joule Thompson Inversion Temperature
37:22
Inversion Temperature of Hydrogen Gas
37:59

35m 52s

Intro
0:00
0:10
0:18
Work & Energy in an Adiabatic Process
3:44
Pressure-Volume Work
7:43
Adiabatic Changes for an Ideal Gas
9:23
Adiabatic Changes for an Ideal Gas
9:24
Equation for a Fixed Change in Volume
11:20
Maximum & Minimum Values of Temperature
14:20
18:08
18:09
21:54
22:34
Fundamental Relationship Equation for an Ideal Gas Under Adiabatic Expansion
25:00
More on the Equation
28:20
Important Equations
32:16
32:17
Reversible Adiabatic Change of State Equation
33:02
Section 3: Energy Example Problems
1st Law Example Problems I

42m 40s

Intro
0:00
Fundamental Equations
0:56
Work
2:40
Energy (1st Law)
3:10
Definition of Enthalpy
3:44
Heat capacity Definitions
4:06
The Mathematics
6:35
Fundamental Concepts
8:13
Isothermal
8:20
8:54
Isobaric
9:25
Isometric
9:48
Ideal Gases
10:14
Example I
12:08
Example I: Conventions
12:44
Example I: Part A
15:30
Example I: Part B
18:24
Example I: Part C
19:53
Example II: What is the Heat Capacity of the System?
21:49
Example III: Find Q, W, ∆U & ∆H for this Change of State
24:15
Example IV: Find Q, W, ∆U & ∆H
31:37
Example V: Find Q, W, ∆U & ∆H
38:20
1st Law Example Problems II

1h 23s

Intro
0:00
Example I
0:11
Example I: Finding ∆U
1:49
Example I: Finding W
6:22
Example I: Finding Q
11:23
Example I: Finding ∆H
16:09
Example I: Summary
17:07
Example II
21:16
Example II: Finding W
22:42
Example II: Finding ∆H
27:48
Example II: Finding Q
30:58
Example II: Finding ∆U
31:30
Example III
33:33
Example III: Finding ∆U, Q & W
33:34
Example III: Finding ∆H
38:07
Example IV
41:50
Example IV: Finding ∆U
41:51
Example IV: Finding ∆H
45:42
Example V
49:31
Example V: Finding W
49:32
Example V: Finding ∆U
55:26
Example V: Finding Q
56:26
Example V: Finding ∆H
56:55
1st Law Example Problems III

44m 34s

Intro
0:00
Example I
0:15
Example I: Finding the Final Temperature
3:40
Example I: Finding Q
8:04
Example I: Finding ∆U
8:25
Example I: Finding W
9:08
Example I: Finding ∆H
9:51
Example II
11:27
Example II: Finding the Final Temperature
11:28
Example II: Finding ∆U
21:25
Example II: Finding W & Q
22:14
Example II: Finding ∆H
23:03
Example III
24:38
Example III: Finding the Final Temperature
24:39
Example III: Finding W, ∆U, and Q
27:43
Example III: Finding ∆H
28:04
Example IV
29:23
Example IV: Finding ∆U, W, and Q
25:36
Example IV: Finding ∆H
31:33
Example V
32:24
Example V: Finding the Final Temperature
33:32
Example V: Finding ∆U
39:31
Example V: Finding W
40:17
Example V: First Way of Finding ∆H
41:10
Example V: Second Way of Finding ∆H
42:10
Thermochemistry Example Problems

59m 7s

Intro
0:00
Example I: Find ∆H° for the Following Reaction
0:42
Example II: Calculate the ∆U° for the Reaction in Example I
5:33
Example III: Calculate the Heat of Formation of NH₃ at 298 K
14:23
Example IV
32:15
Part A: Calculate the Heat of Vaporization of Water at 25°C
33:49
Part B: Calculate the Work Done in Vaporizing 2 Mols of Water at 25°C Under a Constant Pressure of 1 atm
35:26
Part C: Find ∆U for the Vaporization of Water at 25°C
41:00
Part D: Find the Enthalpy of Vaporization of Water at 100°C
43:12
Example V
49:24
Part A: Constant Temperature & Increasing Pressure
50:25
Part B: Increasing temperature & Constant Pressure
56:20
Section 4: Entropy
Entropy

49m 16s

Intro
0:00
Entropy, Part 1
0:16
Coefficient of Thermal Expansion (Isobaric)
0:38
Coefficient of Compressibility (Isothermal)
1:25
Relative Increase & Relative Decrease
2:16
More on α
4:40
More on κ
8:38
Entropy, Part 2
11:04
Definition of Entropy
12:54
Differential Change in Entropy & the Reversible Path
20:08
State Property of the System
28:26
Entropy Changes Under Isothermal Conditions
35:00
Recall: Heating Curve
41:05
Some Phase Changes Take Place Under Constant Pressure
44:07
Example I: Finding ∆S for a Phase Change
46:05
Math Lesson II

33m 59s

Intro
0:00
Math Lesson II
0:46
Let F(x,y) = x²y³
0:47
Total Differential
3:34
Total Differential Expression
6:06
Example 1
9:24
More on Math Expression
13:26
Exact Total Differential Expression
13:27
Exact Differentials
19:50
Inexact Differentials
20:20
The Cyclic Rule
21:06
The Cyclic Rule
21:07
Example 2
27:58
Entropy As a Function of Temperature & Volume

54m 37s

Intro
0:00
Entropy As a Function of Temperature & Volume
0:14
Fundamental Equation of Thermodynamics
1:16
Things to Notice
9:10
Entropy As a Function of Temperature & Volume
14:47
Temperature-dependence of Entropy
24:00
Example I
26:19
Entropy As a Function of Temperature & Volume, Cont.
31:55
Volume-dependence of Entropy at Constant Temperature
31:56
Differentiate with Respect to Temperature, Holding Volume Constant
36:16
Recall the Cyclic Rule
45:15
Summary & Recap
46:47
Fundamental Equation of Thermodynamics
46:48
For Entropy as a Function of Temperature & Volume
47:18
The Volume-dependence of Entropy for Liquids & Solids
52:52
Entropy as a Function of Temperature & Pressure

31m 18s

Intro
0:00
Entropy as a Function of Temperature & Pressure
0:17
Entropy as a Function of Temperature & Pressure
0:18
Rewrite the Total Differential
5:54
Temperature-dependence
7:08
Pressure-dependence
9:04
Differentiate with Respect to Pressure & Holding Temperature Constant
9:54
Differentiate with Respect to Temperature & Holding Pressure Constant
11:28
Pressure-Dependence of Entropy for Liquids & Solids
18:45
Pressure-Dependence of Entropy for Liquids & Solids
18:46
Example I: ∆S of Transformation
26:20
Summary of Entropy So Far

23m 6s

Intro
0:00
Summary of Entropy So Far
0:43
Defining dS
1:04
Fundamental Equation of Thermodynamics
3:51
Temperature & Volume
6:04
Temperature & Pressure
9:10
Two Important Equations for How Entropy Behaves
13:38
State of a System & Heat Capacity
15:34
Temperature-dependence of Entropy
19:49
Entropy Changes for an Ideal Gas

25m 42s

Intro
0:00
Entropy Changes for an Ideal Gas
1:10
General Equation
1:22
The Fundamental Theorem of Thermodynamics
2:37
Recall the Basic Total Differential Expression for S = S (T,V)
5:36
For a Finite Change in State
7:58
If Cv is Constant Over the Particular Temperature Range
9:05
Change in Entropy of an Ideal Gas as a Function of Temperature & Pressure
11:35
Change in Entropy of an Ideal Gas as a Function of Temperature & Pressure
11:36
Recall the Basic Total Differential expression for S = S (T, P)
15:13
For a Finite Change
18:06
Example 1: Calculate the ∆S of Transformation
22:02
Section 5: Entropy Example Problems
Entropy Example Problems I

43m 39s

Intro
0:00
Entropy Example Problems I
0:24
Fundamental Equation of Thermodynamics
1:10
Entropy as a Function of Temperature & Volume
2:04
Entropy as a Function of Temperature & Pressure
2:59
Entropy For Phase Changes
4:47
Entropy For an Ideal Gas
6:14
Third Law Entropies
8:25
Statement of the Third Law
9:17
Entropy of the Liquid State of a Substance Above Its Melting Point
10:23
Entropy For the Gas Above Its Boiling Temperature
13:02
Entropy Changes in Chemical Reactions
15:26
Entropy Change at a Temperature Other than 25°C
16:32
Example I
19:31
Part A: Calculate ∆S for the Transformation Under Constant Volume
20:34
Part B: Calculate ∆S for the Transformation Under Constant Pressure
25:04
Example II: Calculate ∆S fir the Transformation Under Isobaric Conditions
27:53
Example III
30:14
Part A: Calculate ∆S if 1 Mol of Aluminum is taken from 25°C to 255°C
31:14
Part B: If S°₂₉₈ = 28.4 J/mol-K, Calculate S° for Aluminum at 498 K
33:23
Example IV: Calculate Entropy Change of Vaporization for CCl₄
34:19
Example V
35:41
Part A: Calculate ∆S of Transformation
37:36
Part B: Calculate ∆S of Transformation
39:10
Entropy Example Problems II

56m 44s

Intro
0:00
Example I
0:09
Example I: Calculate ∆U
1:28
Example I: Calculate Q
3:29
Example I: Calculate Cp
4:54
Example I: Calculate ∆S
6:14
Example II
7:13
Example II: Calculate W
8:14
Example II: Calculate ∆U
8:56
Example II: Calculate Q
10:18
Example II: Calculate ∆H
11:00
Example II: Calculate ∆S
12:36
Example III
18:47
Example III: Calculate ∆H
19:38
Example III: Calculate Q
21:14
Example III: Calculate ∆U
21:44
Example III: Calculate W
23:59
Example III: Calculate ∆S
24:55
Example IV
27:57
Example IV: Diagram
29:32
Example IV: Calculate W
32:27
Example IV: Calculate ∆U
36:36
Example IV: Calculate Q
38:32
Example IV: Calculate ∆H
39:00
Example IV: Calculate ∆S
40:27
Example IV: Summary
43:41
Example V
48:25
Example V: Diagram
49:05
Example V: Calculate W
50:58
Example V: Calculate ∆U
53:29
Example V: Calculate Q
53:44
Example V: Calculate ∆H
54:34
Example V: Calculate ∆S
55:01
Entropy Example Problems III

57m 6s

Intro
0:00
Example I: Isothermal Expansion
0:09
Example I: Calculate W
1:19
Example I: Calculate ∆U
1:48
Example I: Calculate Q
2:06
Example I: Calculate ∆H
2:26
Example I: Calculate ∆S
3:02
Example II: Adiabatic and Reversible Expansion
6:10
Example II: Calculate Q
6:48
Example II: Basic Equation for the Reversible Adiabatic Expansion of an Ideal Gas
8:12
Example II: Finding Volume
12:40
Example II: Finding Temperature
17:58
Example II: Calculate ∆U
19:53
Example II: Calculate W
20:59
Example II: Calculate ∆H
21:42
Example II: Calculate ∆S
23:42
Example III: Calculate the Entropy of Water Vapor
25:20
Example IV: Calculate the Molar ∆S for the Transformation
34:32
Example V
44:19
Part A: Calculate the Standard Entropy of Liquid Lead at 525°C
46:17
Part B: Calculate ∆H for the Transformation of Solid Lead from 25°C to Liquid Lead at 525°C
52:23
Section 6: Entropy and Probability
Entropy & Probability I

54m 35s

Intro
0:00
Entropy & Probability
0:11
Structural Model
3:05
Recall the Fundamental Equation of Thermodynamics
9:11
Two Independent Ways of Affecting the Entropy of a System
10:05
Boltzmann Definition
12:10
Omega
16:24
Definition of Omega
16:25
Energy Distribution
19:43
The Energy Distribution
19:44
In How Many Ways can N Particles be Distributed According to the Energy Distribution
23:05
Example I: In How Many Ways can the Following Distribution be Achieved
32:51
Example II: In How Many Ways can the Following Distribution be Achieved
33:51
Example III: In How Many Ways can the Following Distribution be Achieved
34:45
Example IV: In How Many Ways can the Following Distribution be Achieved
38:50
Entropy & Probability, cont.
40:57
More on Distribution
40:58
Example I Summary
41:43
Example II Summary
42:12
Distribution that Maximizes Omega
42:26
If Omega is Large, then S is Large
44:22
Two Constraints for a System to Achieve the Highest Entropy Possible
47:07
What Happened When the Energy of a System is Increased?
49:00
Entropy & Probability II

35m 5s

Intro
0:00
Volume Distribution
0:08
Distributing 2 Balls in 3 Spaces
1:43
Distributing 2 Balls in 4 Spaces
3:44
Distributing 3 Balls in 10 Spaces
5:30
Number of Ways to Distribute P Particles over N Spaces
6:05
When N is Much Larger than the Number of Particles P
7:56
Energy Distribution
25:04
Volume Distribution
25:58
Entropy, Total Entropy, & Total Omega Equations
27:34
Entropy, Total Entropy, & Total Omega Equations
27:35
Section 7: Spontaneity, Equilibrium, and the Fundamental Equations
Spontaneity & Equilibrium I

28m 42s

Intro
0:00
Reversible & Irreversible
0:24
Reversible vs. Irreversible
0:58
Defining Equation for Equilibrium
2:11
Defining Equation for Irreversibility (Spontaneity)
3:11
TdS ≥ dQ
5:15
Transformation in an Isolated System
11:22
Transformation in an Isolated System
11:29
Transformation at Constant Temperature
14:50
Transformation at Constant Temperature
14:51
Helmholtz Free Energy
17:26
Define: A = U - TS
17:27
Spontaneous Isothermal Process & Helmholtz Energy
20:20
Pressure-volume Work
22:02
Spontaneity & Equilibrium II

34m 38s

Intro
0:00
Transformation under Constant Temperature & Pressure
0:08
Transformation under Constant Temperature & Pressure
0:36
Define: G = U + PV - TS
3:32
Gibbs Energy
5:14
What Does This Say?
6:44
Spontaneous Process & a Decrease in G
14:12
Computing ∆G
18:54
Summary of Conditions
21:32
Constraint & Condition for Spontaneity
21:36
Constraint & Condition for Equilibrium
24:54
A Few Words About the Word Spontaneous
26:24
Spontaneous Does Not Mean Fast
26:25
Putting Hydrogen & Oxygen Together in a Flask
26:59
Spontaneous Vs. Not Spontaneous
28:14
Thermodynamically Favorable
29:03
Example: Making a Process Thermodynamically Favorable
29:34
Driving Forces for Spontaneity
31:35
Equation: ∆G = ∆H - T∆S
31:36
Always Spontaneous Process
32:39
Never Spontaneous Process
33:06
A Process That is Endothermic Can Still be Spontaneous
34:00
The Fundamental Equations of Thermodynamics

30m 50s

Intro
0:00
The Fundamental Equations of Thermodynamics
0:44
Mechanical Properties of a System
0:45
Fundamental Properties of a System
1:16
Composite Properties of a System
1:44
General Condition of Equilibrium
3:16
Composite Functions & Their Differentiations
6:11
dH = TdS + VdP
7:53
dA = -SdT - PdV
9:26
dG = -SdT + VdP
10:22
Summary of Equations
12:10
Equation #1
14:33
Equation #2
15:15
Equation #3
15:58
Equation #4
16:42
Maxwell's Relations
20:20
Maxwell's Relations
20:21
Isothermal Volume-Dependence of Entropy & Isothermal Pressure-Dependence of Entropy
26:21
The General Thermodynamic Equations of State

34m 6s

Intro
0:00
The General Thermodynamic Equations of State
0:10
Equations of State for Liquids & Solids
0:52
More General Condition for Equilibrium
4:02
General Conditions: Equation that Relates P to Functions of T & V
6:20
The Second Fundamental Equation of Thermodynamics
11:10
Equation 1
17:34
Equation 2
21:58
Recall the General Expression for Cp - Cv
28:11
For the Joule-Thomson Coefficient
30:44
Joule-Thomson Inversion Temperature
32:12
Properties of the Helmholtz & Gibbs Energies

39m 18s

Intro
0:00
Properties of the Helmholtz & Gibbs Energies
0:10
Equating the Differential Coefficients
1:34
An Increase in T; a Decrease in A
3:25
An Increase in V; a Decrease in A
6:04
We Do the Same Thing for G
8:33
Increase in T; Decrease in G
10:50
Increase in P; Decrease in G
11:36
Gibbs Energy of a Pure Substance at a Constant Temperature from 1 atm to any Other Pressure.
14:12
If the Substance is a Liquid or a Solid, then Volume can be Treated as a Constant
18:57
For an Ideal Gas
22:18
Special Note
24:56
Temperature Dependence of Gibbs Energy
27:02
Temperature Dependence of Gibbs Energy #1
27:52
Temperature Dependence of Gibbs Energy #2
29:01
Temperature Dependence of Gibbs Energy #3
29:50
Temperature Dependence of Gibbs Energy #4
34:50
The Entropy of the Universe & the Surroundings

19m 40s

Intro
0:00
Entropy of the Universe & the Surroundings
0:08
Equation: ∆G = ∆H - T∆S
0:20
Conditions of Constant Temperature & Pressure
1:14
Reversible Process
3:14
Spontaneous Process & the Entropy of the Universe
5:20
Tips for Remembering Everything
12:40
Verify Using Known Spontaneous Process
14:51
Section 8: Free Energy Example Problems
Free Energy Example Problems I

54m 16s

Intro
0:00
Example I
0:11
Example I: Deriving a Function for Entropy (S)
2:06
Example I: Deriving a Function for V
5:55
Example I: Deriving a Function for H
8:06
Example I: Deriving a Function for U
12:06
Example II
15:18
Example III
21:52
Example IV
26:12
Example IV: Part A
26:55
Example IV: Part B
28:30
Example IV: Part C
30:25
Example V
33:45
Example VI
40:46
Example VII
43:43
Example VII: Part A
44:46
Example VII: Part B
50:52
Example VII: Part C
51:56
Free Energy Example Problems II

31m 17s

Intro
0:00
Example I
0:09
Example II
5:18
Example III
8:22
Example IV
12:32
Example V
17:14
Example VI
20:34
Example VI: Part A
21:04
Example VI: Part B
23:56
Example VI: Part C
27:56
Free Energy Example Problems III

45m

Intro
0:00
Example I
0:10
Example II
15:03
Example III
21:47
Example IV
28:37
Example IV: Part A
29:33
Example IV: Part B
36:09
Example IV: Part C
40:34
Three Miscellaneous Example Problems

58m 5s

Intro
0:00
Example I
0:41
Part A: Calculating ∆H
3:55
Part B: Calculating ∆S
15:13
Example II
24:39
Part A: Final Temperature of the System
26:25
Part B: Calculating ∆S
36:57
Example III
46:49
Section 9: Equation Review for Thermodynamics
Looking Back Over Everything: All the Equations in One Place

25m 20s

Intro
0:00
Work, Heat, and Energy
0:18
Definition of Work, Energy, Enthalpy, and Heat Capacities
0:23
Heat Capacities for an Ideal Gas
3:40
Path Property & State Property
3:56
Energy Differential
5:04
Enthalpy Differential
5:40
Joule's Law & Joule-Thomson Coefficient
6:23
Coefficient of Thermal Expansion & Coefficient of Compressibility
7:01
Enthalpy of a Substance at Any Other Temperature
7:29
Enthalpy of a Reaction at Any Other Temperature
8:01
Entropy
8:53
Definition of Entropy
8:54
Clausius Inequality
9:11
Entropy Changes in Isothermal Systems
9:44
The Fundamental Equation of Thermodynamics
10:12
Expressing Entropy Changes in Terms of Properties of the System
10:42
Entropy Changes in the Ideal Gas
11:22
Third Law Entropies
11:38
Entropy Changes in Chemical Reactions
14:02
Statistical Definition of Entropy
14:34
Omega for the Spatial & Energy Distribution
14:47
Spontaneity and Equilibrium
15:43
Helmholtz Energy & Gibbs Energy
15:44
Condition for Spontaneity & Equilibrium
16:24
Condition for Spontaneity with Respect to Entropy
17:58
The Fundamental Equations
18:30
Maxwell's Relations
19:04
The Thermodynamic Equations of State
20:07
Energy & Enthalpy Differentials
21:08
Joule's Law & Joule-Thomson Coefficient
21:59
Relationship Between Constant Pressure & Constant Volume Heat Capacities
23:14
One Final Equation - Just for Fun
24:04
Section 10: Quantum Mechanics Preliminaries
Complex Numbers

34m 25s

Intro
0:00
Complex Numbers
0:11
Representing Complex Numbers in the 2-Dimmensional Plane
0:56
2:35
Subtraction of Complex Numbers
3:17
Multiplication of Complex Numbers
3:47
Division of Complex Numbers
6:04
r & θ
8:04
Euler's Formula
11:00
Polar Exponential Representation of the Complex Numbers
11:22
Example I
14:25
Example II
15:21
Example III
16:58
Example IV
18:35
Example V
20:40
Example VI
21:32
Example VII
25:22
Probability & Statistics

59m 57s

Intro
0:00
Probability & Statistics
1:51
Normalization Condition
1:52
Define the Mean or Average of x
11:04
Example I: Calculate the Mean of x
14:57
Example II: Calculate the Second Moment of the Data in Example I
22:39
Define the Second Central Moment or Variance
25:26
Define the Second Central Moment or Variance
25:27
1st Term
32:16
2nd Term
32:40
3rd Term
34:07
Continuous Distributions
35:47
Continuous Distributions
35:48
Probability Density
39:30
Probability Density
39:31
Normalization Condition
46:51
Example III
50:13
Part A - Show that P(x) is Normalized
51:40
Part B - Calculate the Average Position of the Particle Along the Interval
54:31
Important Things to Remember
58:24
Schrӧdinger Equation & Operators

42m 5s

Intro
0:00
Schrӧdinger Equation & Operators
0:16
Relation Between a Photon's Momentum & Its Wavelength
0:17
Louis de Broglie: Wavelength for Matter
0:39
Schrӧdinger Equation
1:19
Definition of Ψ(x)
3:31
Quantum Mechanics
5:02
Operators
7:51
Example I
10:10
Example II
11:53
Example III
14:24
Example IV
17:35
Example V
19:59
Example VI
22:39
Operators Can Be Linear or Non Linear
27:58
Operators Can Be Linear or Non Linear
28:34
Example VII
32:47
Example VIII
36:55
Example IX
39:29
Schrӧdinger Equation as an Eigenvalue Problem

30m 26s

Intro
0:00
Schrӧdinger Equation as an Eigenvalue Problem
0:10
Operator: Multiplying the Original Function by Some Scalar
0:11
Operator, Eigenfunction, & Eigenvalue
4:42
Example: Eigenvalue Problem
8:00
Schrӧdinger Equation as an Eigenvalue Problem
9:24
Hamiltonian Operator
15:09
Quantum Mechanical Operators
16:46
Kinetic Energy Operator
19:16
Potential Energy Operator
20:02
Total Energy Operator
21:12
Classical Point of View
21:48
Linear Momentum Operator
24:02
Example I
26:01
The Plausibility of the Schrӧdinger Equation

21m 34s

Intro
0:00
The Plausibility of the Schrӧdinger Equation
1:16
The Plausibility of the Schrӧdinger Equation, Part 1
1:17
The Plausibility of the Schrӧdinger Equation, Part 2
8:24
The Plausibility of the Schrӧdinger Equation, Part 3
13:45
Section 11: The Particle in a Box
The Particle in a Box Part I

56m 22s

Intro
0:00
Free Particle in a Box
0:28
Definition of a Free Particle in a Box
0:29
Amplitude of the Matter Wave
6:22
Intensity of the Wave
6:53
Probability Density
9:39
Probability that the Particle is Located Between x & dx
10:54
Probability that the Particle will be Found Between o & a
12:35
Wave Function & the Particle
14:59
Boundary Conditions
19:22
What Happened When There is No Constraint on the Particle
27:54
Diagrams
34:12
More on Probability Density
40:53
The Correspondence Principle
46:45
The Correspondence Principle
46:46
Normalizing the Wave Function
47:46
Normalizing the Wave Function
47:47
Normalized Wave Function & Normalization Constant
52:24
The Particle in a Box Part II

45m 24s

Intro
0:00
Free Particle in a Box
0:08
Free Particle in a 1-dimensional Box
0:09
For a Particle in a Box
3:57
Calculating Average Values & Standard Deviations
5:42
Average Value for the Position of a Particle
6:32
Standard Deviations for the Position of a Particle
10:51
Recall: Energy & Momentum are Represented by Operators
13:33
Recall: Schrӧdinger Equation in Operator Form
15:57
Average Value of a Physical Quantity that is Associated with an Operator
18:16
Average Momentum of a Free Particle in a Box
20:48
The Uncertainty Principle
24:42
Finding the Standard Deviation of the Momentum
25:08
Expression for the Uncertainty Principle
35:02
Summary of the Uncertainty Principle
41:28
The Particle in a Box Part III

48m 43s

Intro
0:00
2-Dimension
0:12
Dimension 2
0:31
Boundary Conditions
1:52
Partial Derivatives
4:27
Example I
6:08
The Particle in a Box, cont.
11:28
Operator Notation
12:04
Symbol for the Laplacian
13:50
The Equation Becomes…
14:30
Boundary Conditions
14:54
Separation of Variables
15:33
Solution to the 1-dimensional Case
16:31
Normalization Constant
22:32
3-Dimension
28:30
Particle in a 3-dimensional Box
28:31
In Del Notation
32:22
The Solutions
34:51
Expressing the State of the System for a Particle in a 3D Box
39:10
Energy Level & Degeneracy
43:35
Section 12: Postulates and Principles of Quantum Mechanics
The Postulates & Principles of Quantum Mechanics, Part I

46m 18s

Intro
0:00
Postulate I
0:31
Probability That The Particle Will Be Found in a Differential Volume Element
0:32
Example I: Normalize This Wave Function
11:30
Postulate II
18:20
Postulate II
18:21
Quantum Mechanical Operators: Position
20:48
Quantum Mechanical Operators: Kinetic Energy
21:57
Quantum Mechanical Operators: Potential Energy
22:42
Quantum Mechanical Operators: Total Energy
22:57
Quantum Mechanical Operators: Momentum
23:22
Quantum Mechanical Operators: Angular Momentum
23:48
More On The Kinetic Energy Operator
24:48
Angular Momentum
28:08
Angular Momentum Overview
28:09
Angular Momentum Operator in Quantum Mechanic
31:34
The Classical Mechanical Observable
32:56
Quantum Mechanical Operator
37:01
Getting the Quantum Mechanical Operator from the Classical Mechanical Observable
40:16
Postulate II, cont.
43:40
Quantum Mechanical Operators are Both Linear & Hermetical
43:41
The Postulates & Principles of Quantum Mechanics, Part II

39m 28s

Intro
0:00
Postulate III
0:09
Postulate III: Part I
0:10
Postulate III: Part II
5:56
Postulate III: Part III
12:43
Postulate III: Part IV
18:28
Postulate IV
23:57
Postulate IV
23:58
Postulate V
27:02
Postulate V
27:03
Average Value
36:38
Average Value
36:39
The Postulates & Principles of Quantum Mechanics, Part III

35m 32s

Intro
0:00
The Postulates & Principles of Quantum Mechanics, Part III
0:10
Equations: Linear & Hermitian
0:11
Introduction to Hermitian Property
3:36
Eigenfunctions are Orthogonal
9:55
The Sequence of Wave Functions for the Particle in a Box forms an Orthonormal Set
14:34
Definition of Orthogonality
16:42
Definition of Hermiticity
17:26
Hermiticity: The Left Integral
23:04
Hermiticity: The Right Integral
28:47
Hermiticity: Summary
34:06
The Postulates & Principles of Quantum Mechanics, Part IV

29m 55s

Intro
0:00
The Postulates & Principles of Quantum Mechanics, Part IV
0:09
Operators can be Applied Sequentially
0:10
Sample Calculation 1
2:41
Sample Calculation 2
5:18
Commutator of Two Operators
8:16
The Uncertainty Principle
19:01
In the Case of Linear Momentum and Position Operator
23:14
When the Commutator of Two Operators Equals to Zero
26:31
Section 13: Postulates and Principles Example Problems, Including Particle in a Box
Example Problems I

54m 25s

Intro
0:00
Example I: Three Dimensional Box & Eigenfunction of The Laplacian Operator
0:37
Example II: Positions of a Particle in a 1-dimensional Box
15:46
Example III: Transition State & Frequency
29:29
Example IV: Finding a Particle in a 1-dimensional Box
35:03
Example V: Degeneracy & Energy Levels of a Particle in a Box
44:59
Example Problems II

46m 58s

Intro
0:00
Review
0:25
Wave Function
0:26
Normalization Condition
2:28
Observable in Classical Mechanics & Linear/Hermitian Operator in Quantum Mechanics
3:36
Hermitian
6:11
Eigenfunctions & Eigenvalue
8:20
Normalized Wave Functions
12:00
Average Value
13:42
If Ψ is Written as a Linear Combination
15:44
Commutator
16:45
Example I: Normalize The Wave Function
19:18
Example II: Probability of Finding of a Particle
22:27
Example III: Orthogonal
26:00
Example IV: Average Value of the Kinetic Energy Operator
30:22
Example V: Evaluate These Commutators
39:02
Example Problems III

44m 11s

Intro
0:00
Example I: Good Candidate for a Wave Function
0:08
Example II: Variance of the Energy
7:00
Example III: Evaluate the Angular Momentum Operators
15:00
Example IV: Real Eigenvalues Imposes the Hermitian Property on Operators
28:44
Example V: A Demonstration of Why the Eigenfunctions of Hermitian Operators are Orthogonal
35:33
Section 14: The Harmonic Oscillator
The Harmonic Oscillator I

35m 33s

Intro
0:00
The Harmonic Oscillator
0:10
Harmonic Motion
0:11
Classical Harmonic Oscillator
4:38
Hooke's Law
8:18
Classical Harmonic Oscillator, cont.
10:33
General Solution for the Differential Equation
15:16
Initial Position & Velocity
16:05
Period & Amplitude
20:42
Potential Energy of the Harmonic Oscillator
23:20
Kinetic Energy of the Harmonic Oscillator
26:37
Total Energy of the Harmonic Oscillator
27:23
Conservative System
34:37
The Harmonic Oscillator II

43m 4s

Intro
0:00
The Harmonic Oscillator II
0:08
Diatomic Molecule
0:10
Notion of Reduced Mass
5:27
Harmonic Oscillator Potential & The Intermolecular Potential of a Vibrating Molecule
7:33
The Schrӧdinger Equation for the 1-dimensional Quantum Mechanic Oscillator
14:14
Quantized Values for the Energy Level
15:46
Ground State & the Zero-Point Energy
21:50
Vibrational Energy Levels
25:18
Transition from One Energy Level to the Next
26:42
Fundamental Vibrational Frequency for Diatomic Molecule
34:57
Example: Calculate k
38:01
The Harmonic Oscillator III

26m 30s

Intro
0:00
The Harmonic Oscillator III
0:09
The Wave Functions Corresponding to the Energies
0:10
Normalization Constant
2:34
Hermite Polynomials
3:22
First Few Hermite Polynomials
4:56
First Few Wave-Functions
6:37
Plotting the Probability Density of the Wave-Functions
8:37
Probability Density for Large Values of r
14:24
Recall: Odd Function & Even Function
19:05
More on the Hermite Polynomials
20:07
Recall: If f(x) is Odd
20:36
Average Value of x
22:31
Average Value of Momentum
23:56
Section 15: The Rigid Rotator
The Rigid Rotator I

41m 10s

Intro
0:00
Possible Confusion from the Previous Discussion
0:07
Possible Confusion from the Previous Discussion
0:08
Rotation of a Single Mass Around a Fixed Center
8:17
Rotation of a Single Mass Around a Fixed Center
8:18
Angular Velocity
12:07
Rotational Inertia
13:24
Rotational Frequency
15:24
Kinetic Energy for a Linear System
16:38
Kinetic Energy for a Rotational System
17:42
Rotating Diatomic Molecule
19:40
Rotating Diatomic Molecule: Part 1
19:41
Rotating Diatomic Molecule: Part 2
24:56
Rotating Diatomic Molecule: Part 3
30:04
Hamiltonian of the Rigid Rotor
36:48
Hamiltonian of the Rigid Rotor
36:49
The Rigid Rotator II

30m 32s

Intro
0:00
The Rigid Rotator II
0:08
Cartesian Coordinates
0:09
Spherical Coordinates
1:55
r
6:15
θ
6:28
φ
7:00
Moving a Distance 'r'
8:17
Moving a Distance 'r' in the Spherical Coordinates
11:49
For a Rigid Rotator, r is Constant
13:57
Hamiltonian Operator
15:09
Square of the Angular Momentum Operator
17:34
Orientation of the Rotation in Space
19:44
Wave Functions for the Rigid Rotator
20:40
The Schrӧdinger Equation for the Quantum Mechanic Rigid Rotator
21:24
Energy Levels for the Rigid Rotator
26:58
The Rigid Rotator III

35m 19s

Intro
0:00
The Rigid Rotator III
0:11
When a Rotator is Subjected to Electromagnetic Radiation
1:24
Selection Rule
2:13
Frequencies at Which Absorption Transitions Occur
6:24
Energy Absorption & Transition
10:54
Energy of the Individual Levels Overview
20:58
Energy of the Individual Levels: Diagram
23:45
Frequency Required to Go from J to J + 1
25:53
Using Separation Between Lines on the Spectrum to Calculate Bond Length
28:02
Example I: Calculating Rotational Inertia & Bond Length
29:18
Example I: Calculating Rotational Inertia
29:19
Example I: Calculating Bond Length
32:56
Section 16: Oscillator and Rotator Example Problems
Example Problems I

33m 48s

Intro
0:00
Equations Review
0:11
Energy of the Harmonic Oscillator
0:12
Selection Rule
3:02
3:27
Harmonic Oscillator Wave Functions
5:52
Rigid Rotator
7:26
Selection Rule for Rigid Rotator
9:15
Frequency of Absorption
9:35
Wave Numbers
10:58
Example I: Calculate the Reduced Mass of the Hydrogen Atom
11:44
Example II: Calculate the Fundamental Vibration Frequency & the Zero-Point Energy of This Molecule
13:37
Example III: Show That the Product of Two Even Functions is even
19:35
Example IV: Harmonic Oscillator
24:56
Example Problems II

46m 43s

Intro
0:00
Example I: Harmonic Oscillator
0:12
Example II: Harmonic Oscillator
23:26
Example III: Calculate the RMS Displacement of the Molecules
38:12
Section 17: The Hydrogen Atom
The Hydrogen Atom I

40m

Intro
0:00
The Hydrogen Atom I
1:31
Review of the Rigid Rotator
1:32
Hydrogen Atom & the Coulomb Potential
2:50
Using the Spherical Coordinates
6:33
Applying This Last Expression to Equation 1
10:19
13:26
Angular Equation
15:56
Solution for F(φ)
19:32
Determine The Normalization Constant
20:33
Differential Equation for T(a)
24:44
Legendre Equation
27:20
Legendre Polynomials
31:20
The Legendre Polynomials are Mutually Orthogonal
35:40
Limits
37:17
Coefficients
38:28
The Hydrogen Atom II

35m 58s

Intro
0:00
Associated Legendre Functions
0:07
Associated Legendre Functions
0:08
First Few Associated Legendre Functions
6:39
s, p, & d Orbital
13:24
The Normalization Condition
15:44
Spherical Harmonics
20:03
Equations We Have Found
20:04
Wave Functions for the Angular Component & Rigid Rotator
24:36
Spherical Harmonics Examples
25:40
Angular Momentum
30:09
Angular Momentum
30:10
Square of the Angular Momentum
35:38
Energies of the Rigid Rotator
38:21
The Hydrogen Atom III

36m 18s

Intro
0:00
The Hydrogen Atom III
0:34
Angular Momentum is a Vector Quantity
0:35
The Operators Corresponding to the Three Components of Angular Momentum Operator: In Cartesian Coordinates
1:30
The Operators Corresponding to the Three Components of Angular Momentum Operator: In Spherical Coordinates
3:27
Z Component of the Angular Momentum Operator & the Spherical Harmonic
5:28
Magnitude of the Angular Momentum Vector
20:10
Classical Interpretation of Angular Momentum
25:22
Projection of the Angular Momentum Vector onto the xy-plane
33:24
The Hydrogen Atom IV

33m 55s

Intro
0:00
The Hydrogen Atom IV
0:09
The Equation to Find R( r )
0:10
Relation Between n & l
3:50
The Solutions for the Radial Functions
5:08
Associated Laguerre Polynomials
7:58
1st Few Associated Laguerre Polynomials
8:55
Complete Wave Function for the Atomic Orbitals of the Hydrogen Atom
12:24
The Normalization Condition
15:06
In Cartesian Coordinates
18:10
Working in Polar Coordinates
20:48
Principal Quantum Number
21:58
Angular Momentum Quantum Number
22:35
Magnetic Quantum Number
25:55
Zeeman Effect
30:45
The Hydrogen Atom V: Where We Are

51m 53s

Intro
0:00
The Hydrogen Atom V: Where We Are
0:13
Review
0:14
Let's Write Out ψ₂₁₁
7:32
Angular Momentum of the Electron
14:52
Representation of the Wave Function
19:36
28:02
Example: 1s Orbital
28:34
33:46
1s Orbital: Plotting Probability Densities vs. r
35:47
2s Orbital: Plotting Probability Densities vs. r
37:46
3s Orbital: Plotting Probability Densities vs. r
38:49
4s Orbital: Plotting Probability Densities vs. r
39:34
2p Orbital: Plotting Probability Densities vs. r
40:12
3p Orbital: Plotting Probability Densities vs. r
41:02
4p Orbital: Plotting Probability Densities vs. r
41:51
3d Orbital: Plotting Probability Densities vs. r
43:18
4d Orbital: Plotting Probability Densities vs. r
43:48
Example I: Probability of Finding an Electron in the 2s Orbital of the Hydrogen
45:40
The Hydrogen Atom VI

51m 53s

Intro
0:00
The Hydrogen Atom VI
0:07
Last Lesson Review
0:08
Spherical Component
1:09
Normalization Condition
2:02
Complete 1s Orbital Wave Function
4:08
1s Orbital Wave Function
4:09
Normalization Condition
6:28
Spherically Symmetric
16:00
Average Value
17:52
Example I: Calculate the Region of Highest Probability for Finding the Electron
21:19
2s Orbital Wave Function
25:32
2s Orbital Wave Function
25:33
Average Value
28:56
General Formula
32:24
The Hydrogen Atom VII

34m 29s

Intro
0:00
The Hydrogen Atom VII
0:12
p Orbitals
1:30
Not Spherically Symmetric
5:10
Recall That the Spherical Harmonics are Eigenfunctions of the Hamiltonian Operator
6:50
Any Linear Combination of These Orbitals Also Has The Same Energy
9:16
Functions of Real Variables
15:53
Solving for Px
16:50
Real Spherical Harmonics
21:56
Number of Nodes
32:56
Section 18: Hydrogen Atom Example Problems
Hydrogen Atom Example Problems I

43m 49s

Intro
0:00
Example I: Angular Momentum & Spherical Harmonics
0:20
Example II: Pair-wise Orthogonal Legendre Polynomials
16:40
Example III: General Normalization Condition for the Legendre Polynomials
25:06
Example IV: Associated Legendre Functions
32:13
The Hydrogen Atom Example Problems II

1h 1m 57s

Intro
0:00
Example I: Normalization & Pair-wise Orthogonal
0:13
Part 1: Normalized
0:43
Part 2: Pair-wise Orthogonal
16:53
Example II: Show Explicitly That the Following Statement is True for Any Integer n
27:10
Example III: Spherical Harmonics
29:26
Angular Momentum Cones
56:37
Angular Momentum Cones
56:38
Physical Interpretation of Orbital Angular Momentum in Quantum mechanics
1:00:16
The Hydrogen Atom Example Problems III

48m 33s

Intro
0:00
Example I: Show That ψ₂₁₁ is Normalized
0:07
Example II: Show That ψ₂₁₁ is Orthogonal to ψ₃₁₀
11:48
Example III: Probability That a 1s Electron Will Be Found Within 1 Bohr Radius of The Nucleus
18:35
Example IV: Radius of a Sphere
26:06
Example V: Calculate <r> for the 2s Orbital of the Hydrogen-like Atom
36:33
The Hydrogen Atom Example Problems IV

48m 33s

Intro
0:00
Example I: Probability Density vs. Radius Plot
0:11
Example II: Hydrogen Atom & The Coulombic Potential
14:16
Example III: Find a Relation Among <K>, <V>, & <E>
25:47
Example IV: Quantum Mechanical Virial Theorem
48:32
Example V: Find the Variance for the 2s Orbital
54:13
The Hydrogen Atom Example Problems V

48m 33s

Intro
0:00
Example I: Derive a Formula for the Degeneracy of a Given Level n
0:11
Example II: Using Linear Combinations to Represent the Spherical Harmonics as Functions of the Real Variables θ & φ
8:30
Example III: Using Linear Combinations to Represent the Spherical Harmonics as Functions of the Real Variables θ & φ
23:01
Example IV: Orbital Functions
31:51
Section 19: Spin Quantum Number and Atomic Term Symbols
Spin Quantum Number: Term Symbols I

59m 18s

Intro
0:00
Quantum Numbers Specify an Orbital
0:24
n
1:10
l
1:20
m
1:35
4th Quantum Number: s
2:02
Spin Orbitals
7:03
Spin Orbitals
7:04
Multi-electron Atoms
11:08
Term Symbols
18:08
Russell-Saunders Coupling & The Atomic Term Symbol
18:09
Example: Configuration for C
27:50
Configuration for C: 1s²2s²2p²
27:51
Drawing Every Possible Arrangement
31:15
Term Symbols
45:24
Microstate
50:54
Spin Quantum Number: Term Symbols II

34m 54s

Intro
0:00
Microstates
0:25
We Started With 21 Possible Microstates
0:26
³P State
2:05
Microstates in ³P Level
5:10
¹D State
13:16
³P State
16:10
²P₂ State
17:34
³P₁ State
18:34
³P₀ State
19:12
9 Microstates in ³P are Subdivided
19:40
¹S State
21:44
Quicker Way to Find the Different Values of J for a Given Basic Term Symbol
22:22
Ground State
26:27
Hund's Empirical Rules for Specifying the Term Symbol for the Ground Electronic State
27:29
Hund's Empirical Rules: 1
28:24
Hund's Empirical Rules: 2
29:22
Hund's Empirical Rules: 3 - Part A
30:22
Hund's Empirical Rules: 3 - Part B
31:18
Example: 1s²2s²2p²
31:54
Spin Quantum Number: Term Symbols III

38m 3s

Intro
0:00
Spin Quantum Number: Term Symbols III
0:14
Deriving the Term Symbols for the p² Configuration
0:15
Table: MS vs. ML
3:57
¹D State
16:21
³P State
21:13
¹S State
24:48
J Value
25:32
Degeneracy of the Level
27:28
When Given r Electrons to Assign to n Equivalent Spin Orbitals
30:18
p² Configuration
32:51
Complementary Configurations
35:12
Term Symbols & Atomic Spectra

57m 49s

Intro
0:00
Lyman Series
0:09
Spectroscopic Term Symbols
0:10
Lyman Series
3:04
Hydrogen Levels
8:21
Hydrogen Levels
8:22
Term Symbols & Atomic Spectra
14:17
Spin-Orbit Coupling
14:18
Selection Rules for Atomic Spectra
21:31
Selection Rules for Possible Transitions
23:56
Wave Numbers for The Transitions
28:04
Example I: Calculate the Frequencies of the Allowed Transitions from (4d) ²D →(2p) ²P
32:23
Helium Levels
49:50
Energy Levels for Helium
49:51
Transitions & Spin Multiplicity
52:27
Transitions & Spin Multiplicity
52:28
Section 20: Term Symbols Example Problems
Example Problems I

1h 1m 20s

Intro
0:00
Example I: What are the Term Symbols for the np¹ Configuration?
0:10
Example II: What are the Term Symbols for the np² Configuration?
20:38
Example III: What are the Term Symbols for the np³ Configuration?
40:46
Example Problems II

56m 34s

Intro
0:00
Example I: Find the Term Symbols for the nd² Configuration
0:11
Example II: Find the Term Symbols for the 1s¹2p¹ Configuration
27:02
Example III: Calculate the Separation Between the Doublets in the Lyman Series for Atomic Hydrogen
41:41
Example IV: Calculate the Frequencies of the Lines for the (4d) ²D → (3p) ²P Transition
48:53
Section 21: Equation Review for Quantum Mechanics
Quantum Mechanics: All the Equations in One Place

18m 24s

Intro
0:00
Quantum Mechanics Equations
0:37
De Broglie Relation
0:38
Statistical Relations
1:00
The Schrӧdinger Equation
1:50
The Particle in a 1-Dimensional Box of Length a
3:09
The Particle in a 2-Dimensional Box of Area a x b
3:48
The Particle in a 3-Dimensional Box of Area a x b x c
4:22
The Schrӧdinger Equation Postulates
4:51
The Normalization Condition
5:40
The Probability Density
6:51
Linear
7:47
Hermitian
8:31
Eigenvalues & Eigenfunctions
8:55
The Average Value
9:29
Eigenfunctions of Quantum Mechanics Operators are Orthogonal
10:53
Commutator of Two Operators
10:56
The Uncertainty Principle
11:41
The Harmonic Oscillator
13:18
The Rigid Rotator
13:52
Energy of the Hydrogen Atom
14:30
Wavefunctions, Radial Component, and Associated Laguerre Polynomial
14:44
Angular Component or Spherical Harmonic
15:16
Associated Legendre Function
15:31
Principal Quantum Number
15:43
Angular Momentum Quantum Number
15:50
Magnetic Quantum Number
16:21
z-component of the Angular Momentum of the Electron
16:53
Atomic Spectroscopy: Term Symbols
17:14
Atomic Spectroscopy: Selection Rules
18:03
Section 22: Molecular Spectroscopy
Spectroscopic Overview: Which Equation Do I Use & Why

50m 2s

Intro
0:00
Spectroscopic Overview: Which Equation Do I Use & Why
1:02
Lesson Overview
1:03
Rotational & Vibrational Spectroscopy
4:01
Frequency of Absorption/Emission
6:04
Wavenumbers in Spectroscopy
8:10
Starting State vs. Excited State
10:10
Total Energy of a Molecule (Leaving out the Electronic Energy)
14:02
Energy of Rotation: Rigid Rotor
15:55
Energy of Vibration: Harmonic Oscillator
19:08
Equation of the Spectral Lines
23:22
Harmonic Oscillator-Rigid Rotor Approximation (Making Corrections)
28:37
Harmonic Oscillator-Rigid Rotor Approximation (Making Corrections)
28:38
Vibration-Rotation Interaction
33:46
Centrifugal Distortion
36:27
Anharmonicity
38:28
Correcting for All Three Simultaneously
41:03
Spectroscopic Parameters
44:26
Summary
47:32
Harmonic Oscillator-Rigid Rotor Approximation
47:33
Vibration-Rotation Interaction
48:14
Centrifugal Distortion
48:20
Anharmonicity
48:28
Correcting for All Three Simultaneously
48:44
Vibration-Rotation

59m 47s

Intro
0:00
Vibration-Rotation
0:37
What is Molecular Spectroscopy?
0:38
Microwave, Infrared Radiation, Visible & Ultraviolet
1:53
Equation for the Frequency of the Absorbed Radiation
4:54
Wavenumbers
6:15
Diatomic Molecules: Energy of the Harmonic Oscillator
8:32
Selection Rules for Vibrational Transitions
10:35
Energy of the Rigid Rotator
16:29
Angular Momentum of the Rotator
21:38
Rotational Term F(J)
26:30
Selection Rules for Rotational Transition
29:30
Vibration Level & Rotational States
33:20
Selection Rules for Vibration-Rotation
37:42
Frequency of Absorption
39:32
Diagram: Energy Transition
45:55
Vibration-Rotation Spectrum: HCl
51:27
Vibration-Rotation Spectrum: Carbon Monoxide
54:30
Vibration-Rotation Interaction

46m 22s

Intro
0:00
Vibration-Rotation Interaction
0:13
Vibration-Rotation Spectrum: HCl
0:14
Bond Length & Vibrational State
4:23
Vibration Rotation Interaction
10:18
Case 1
12:06
Case 2
17:17
Example I: HCl Vibration-Rotation Spectrum
22:58
Rotational Constant for the 0 & 1 Vibrational State
26:30
Equilibrium Bond Length for the 1 Vibrational State
39:42
Equilibrium Bond Length for the 0 Vibrational State
42:13
Bₑ & αₑ
44:54
The Non-Rigid Rotator

29m 24s

Intro
0:00
The Non-Rigid Rotator
0:09
Pure Rotational Spectrum
0:54
The Selection Rules for Rotation
3:09
Spacing in the Spectrum
5:04
Centrifugal Distortion Constant
9:00
Fundamental Vibration Frequency
11:46
Observed Frequencies of Absorption
14:14
Difference between the Rigid Rotator & the Adjusted Rigid Rotator
16:51
21:31
Observed Frequencies of Absorption
26:26
The Anharmonic Oscillator

30m 53s

Intro
0:00
The Anharmonic Oscillator
0:09
Vibration-Rotation Interaction & Centrifugal Distortion
0:10
Making Corrections to the Harmonic Oscillator
4:50
Selection Rule for the Harmonic Oscillator
7:50
Overtones
8:40
True Oscillator
11:46
Harmonic Oscillator Energies
13:16
Anharmonic Oscillator Energies
13:33
Observed Frequencies of the Overtones
15:09
True Potential
17:22
HCl Vibrational Frequencies: Fundamental & First Few Overtones
21:10
Example I: Vibrational States & Overtones of the Vibrational Spectrum
22:42
Example I: Part A - First 4 Vibrational States
23:44
Example I: Part B - Fundamental & First 3 Overtones
25:31
Important Equations
27:45
Energy of the Q State
29:14
The Difference in Energy between 2 Successive States
29:23
Difference in Energy between 2 Spectral Lines
29:40
Electronic Transitions

1h 1m 33s

Intro
0:00
Electronic Transitions
0:16
Electronic State & Transition
0:17
Total Energy of the Diatomic Molecule
3:34
Vibronic Transitions
4:30
Selection Rule for Vibronic Transitions
9:11
More on Vibronic Transitions
10:08
Frequencies in the Spectrum
16:46
Difference of the Minima of the 2 Potential Curves
24:48
Anharmonic Zero-point Vibrational Energies of the 2 States
26:24
Frequency of the 0 → 0 Vibronic Transition
27:54
Making the Equation More Compact
29:34
Spectroscopic Parameters
32:11
Franck-Condon Principle
34:32
Example I: Find the Values of the Spectroscopic Parameters for the Upper Excited State
47:27
Table of Electronic States and Parameters
56:41
Section 23: Molecular Spectroscopy Example Problems
Example Problems I

33m 47s

Intro
0:00
Example I: Calculate the Bond Length
0:10
Example II: Calculate the Rotational Constant
7:39
Example III: Calculate the Number of Rotations
10:54
Example IV: What is the Force Constant & Period of Vibration?
16:31
Example V: Part A - Calculate the Fundamental Vibration Frequency
21:42
Example V: Part B - Calculate the Energies of the First Three Vibrational Levels
24:12
Example VI: Calculate the Frequencies of the First 2 Lines of the R & P Branches of the Vib-Rot Spectrum of HBr
26:28
Example Problems II

1h 1m 5s

Intro
0:00
Example I: Calculate the Frequencies of the Transitions
0:09
Example II: Specify Which Transitions are Allowed & Calculate the Frequencies of These Transitions
22:07
Example III: Calculate the Vibrational State & Equilibrium Bond Length
34:31
Example IV: Frequencies of the Overtones
49:28
Example V: Vib-Rot Interaction, Centrifugal Distortion, & Anharmonicity
54:47
Example Problems III

33m 31s

Intro
0:00
Example I: Part A - Derive an Expression for ∆G( r )
0:10
Example I: Part B - Maximum Vibrational Quantum Number
6:10
Example II: Part A - Derive an Expression for the Dissociation Energy of the Molecule
8:29
Example II: Part B - Equation for ∆G( r )
14:00
Example III: How Many Vibrational States are There for Br₂ before the Molecule Dissociates
18:16
Example IV: Find the Difference between the Two Minima of the Potential Energy Curves
20:57
Example V: Rotational Spectrum
30:51
Section 24: Statistical Thermodynamics
Statistical Thermodynamics: The Big Picture

1h 1m 15s

Intro
0:00
Statistical Thermodynamics: The Big Picture
0:10
Our Big Picture Goal
0:11
Partition Function (Q)
2:42
The Molecular Partition Function (q)
4:00
Consider a System of N Particles
6:54
Ensemble
13:22
Energy Distribution Table
15:36
Probability of Finding a System with Energy
16:51
The Partition Function
21:10
Microstate
28:10
Entropy of the Ensemble
30:34
Entropy of the System
31:48
Expressing the Thermodynamic Functions in Terms of The Partition Function
39:21
The Partition Function
39:22
Pi & U
41:20
Entropy of the System
44:14
Helmholtz Energy
48:15
Pressure of the System
49:32
Enthalpy of the System
51:46
Gibbs Free Energy
52:56
Heat Capacity
54:30
Expressing Q in Terms of the Molecular Partition Function (q)
59:31
Indistinguishable Particles
1:02:16
N is the Number of Particles in the System
1:03:27
The Molecular Partition Function
1:05:06
Quantum States & Degeneracy
1:07:46
Thermo Property in Terms of ln Q
1:10:09
Example: Thermo Property in Terms of ln Q
1:13:23
Statistical Thermodynamics: The Various Partition Functions I

47m 23s

Intro
0:00
Lesson Overview
0:19
Monatomic Ideal Gases
6:40
Monatomic Ideal Gases Overview
6:42
Finding the Parition Function of Translation
8:17
Finding the Parition Function of Electronics
13:29
Example: Na
17:42
Example: F
23:12
Energy Difference between the Ground State & the 1st Excited State
29:27
The Various Partition Functions for Monatomic Ideal Gases
32:20
Finding P
43:16
Going Back to U = (3/2) RT
46:20
Statistical Thermodynamics: The Various Partition Functions II

54m 9s

Intro
0:00
Diatomic Gases
0:16
Diatomic Gases
0:17
Zero-Energy Mark for Rotation
2:26
Zero-Energy Mark for Vibration
3:21
Zero-Energy Mark for Electronic
5:54
Vibration Partition Function
9:48
When Temperature is Very Low
14:00
When Temperature is Very High
15:22
Vibrational Component
18:48
Fraction of Molecules in the r Vibration State
21:00
Example: Fraction of Molecules in the r Vib. State
23:29
Rotation Partition Function
26:06
Heteronuclear & Homonuclear Diatomics
33:13
Energy & Heat Capacity
36:01
Fraction of Molecules in the J Rotational Level
39:20
Example: Fraction of Molecules in the J Rotational Level
40:32
Finding the Most Populated Level
44:07
Putting It All Together
46:06
Putting It All Together
46:07
Energy of Translation
51:51
Energy of Rotation
52:19
Energy of Vibration
52:42
Electronic Energy
53:35
Section 25: Statistical Thermodynamics Example Problems
Example Problems I

48m 32s

Intro
0:00
Example I: Calculate the Fraction of Potassium Atoms in the First Excited Electronic State
0:10
Example II: Show That Each Translational Degree of Freedom Contributes R/2 to the Molar Heat Capacity
14:46
Example III: Calculate the Dissociation Energy
21:23
Example IV: Calculate the Vibrational Contribution to the Molar heat Capacity of Oxygen Gas at 500 K
25:46
Example V: Upper & Lower Quantum State
32:55
Example VI: Calculate the Relative Populations of the J=2 and J=1 Rotational States of the CO Molecule at 25°C
42:21
Example Problems II

57m 30s

Intro
0:00
Example I: Make a Plot of the Fraction of CO Molecules in Various Rotational Levels
0:10
Example II: Calculate the Ratio of the Translational Partition Function for Cl₂ and Br₂ at Equal Volume & Temperature
8:05
Example III: Vibrational Degree of Freedom & Vibrational Molar Heat Capacity
11:59
Example IV: Calculate the Characteristic Vibrational & Rotational temperatures for Each DOF
45:03
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• ## Related Books 1 answer Last reply by: Professor HovasapianTue Nov 27, 2018 4:49 AMPost by Kimberly Davis on November 26, 2018In the section Three miscellaneous example problems, example I, you wrote the Cp in term of J/g KELVIN unit, but you solved the problem as J/g Celsius. Does it  matter which unit I use for this? Thank you!

### Three Miscellaneous Example Problems

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Example I 0:41
• Part A: Calculating ∆H
• Part B: Calculating ∆S
• Example II 24:39
• Part A: Final Temperature of the System
• Part B: Calculating ∆S
• Example III 46:49

### Transcription: Three Miscellaneous Example Problems

Hello and welcome back to www.educator.com and welcome back to Physical Chemistry.0000

Today, I want to throw in these 3 miscellaneous problems.0004

It is not like they have another place in the previous problem sets.0009

In fact, they actually technically belong to the entropy section but I want talk about few other things before I brought these back.0014

They have to do with a little bit of what we discussed previously.0023

They have to do with heat transfer, changes in phase, things like that.0027

In some sense, it is a little bit of everything.0031

The third example problem is strictly an entropy problem but I just thought I collect them here like this.0033

Let us get started.0040

The first example is on a fully insulated flask 40 g of liquid water at 25°C is added to 25 g of ice at -7°C,0044

What is the final state of the system given the following data and pressure is held constant during this process.0056

The constant pressure heat capacity for liquid water 4.18, constant pressure heat capacity for solid water, ice is 2.09 and the δ H of fusion is 334 J/ g.0063

In addition to finding the final state calculate δ H and δ S for this transformation.0079

The final state of system, what they are asking for is we have 40 g of liquid water and we have 25 g of ice.0084

Some of the ice is going to melt and it is going to melt it because the liquid water is going to end up cooling.0094

We want to find out what is the final temperature of the system?0103

How much ice is there and how much liquid water is there?0110

That is what they mean by what is the final state of the system.0112

Let us just jump on in and see what we can do.0116

A lot of this has to do with a lot of heat capacity that we have talked about earlier but0119

it actually harkens back to what it is that you learn in general chemistry.0124

Remember Q = MC δ T and the amount of heat that something loses is the amount of heat that something else gains.0128

In this particular case, the amount of energy lost upon cooling of the water is going to be gained by the ice and that is going to end up melting.0136

We are going to reach some final equilibrium temperature that is going to be some mixture of water and ice.0144

Maybe it is going to be all ice or maybe it is going to be all water.0149

We do not know, that is what we need to find out.0152

Let us see what we have.0157

Once again, we know that liquid water is going to cool.0160

We know that the ice is going to warm up and it is going to melt.0162

The first thing I'm going to do is find out how much energy is required to bring the 25 g of ice from 7°C up to 0°C.0168

The reason that I happen to pick the ice first just to see it, is just that I notice that since the ice was at -7, it is not that far from 0.0180

I want to see how much energy and it is only 25 g, where is the water itself you have actually 40 g and it is at 25°C.0191

The heat capacity of liquid water is higher than the heat capacity of the ice.0200

Chances are just qualitatively looking at the problem,0205

I'm probably going to be able to raise all of the ice to 0 without dropping the liquid water temperature too far.0211

That is what my intuition tells me just based on the data that I have.0219

Let us see if that is actually true.0223

It does not really matter where you start, whether it is with the ice or with the water.0225

You just have to start somewhere and inch your way, converge on some final temperature and final state.0229

Remember from general chemistry we had this Q = MC δ T.0237

The amount of energy lost or gained by something = the mass of something× the specific heat × the change in temperature.0245

This δ G gives us the change in temperature of that particular substance.0254

To find out how much energy is required to bring 25 g of ice to 0°C, here is what we do.0262

Our Q is going to equal the mass which is 25 g of ice and the heat capacity of the ice which is 2.09 J/ g/°C.0272

°C and °K, I do not give it to you in J/ GK.0287

The difference of degree for °C and °K is the same so it does not matter whether I use °C or °K.0294

I'm going from -7 to 0°C.0302

It is going to be δ T of 7°.0305

I’m just going to write it as 7°C.0310

When I go ahead and multiply and cancel units, I'm going to end up with 365.75 J.0313

This is how much energy is required to bring the 7 g of ice from -7° all the way up to 0°C.0323

Energy has to come from somewhere.0331

The energy is going to come from the water actually cooling down.0333

Let us see what this affect has on the liquid water because that is where the energy is being pulled from.0338

This is going to be Q = MC δ T but this time we are going to be looking for our δ T for water.0349

Our Q is going to be 365.75 J and I’m going to skip the units, I hope you do not mind.0357

We have 40 g of water and a specific heat of water is 4.18 J/ g/°C.0364

And we want to know what δ T is.0374

When we only do this math, nice and simple arithmetic, we get 2.19°C.0375

That means the water cools from 25°C it drops down by 2.19°C.0382

That amount of energy that is lost by the water in cooling, that energy goes towards bringing the ice from -7 to 0.0390

That is what is happening here, that is what I have done.0398

This is an exchange of energy.0401

Let us see where it is that we are.0404

The water is now at 22.81°C.0417

It started at 25, it dropped by 2.19, now it is at 22.81°C.0426

We have 40 g of liquid water at 22.81°C and we still have a 25 g of ice.0433

Except now, the ice is at 0°C.0455

We bought the solid ice from -7 to 0 but we have melted the ice yet.0459

This is very important.0464

Just because something comes to 0°C does not mean that it starts to melt or the other way around.0467

It does not mean that the liquid water starts to freeze.0472

That is the δ H of fusion, there is a certain amount of energy that is required to make the conversion from one phase to the other.0475

We do not have 25 g of ice, we do not have 25 g of water to 0, we have 25 g of ice at 0°C.0483

Let us see what we can do about the water at 22.81°C.0499

I have got to write it a little bit better.0503

Water at 22.81°C, in order to take that to water at 0°C, we need a certain amount of energy.0516

Let us calculate that energy.0534

Once again, Q = MC δ T.0536

Q = I have 40 g of water, 4.18 is its heat capacity J/ g/°C.0539

I want my difference in temperature to be 22.81.0553

I want to take it from 22.81 down to 0°C.0556

When I do this calculation, it tells me that I require 3113.8 J.0561

Water is going to lose 3813.8 J in going from 22.1° down to 0, that heat is going to go into the ice at 0°C0572

and convert it from solid ice at 0 to liquid water at 0.0586

It is going to cause the ice to melt, this is where we use the δ H of fusion.0591

Let us see how much ice actually melts.0596

Once again, let me go blue because I like the blue.0600

We are going to take this 3813.8 J and we are going to multiply it by the reciprocal of the δ H of fusion.0614

1 g and 334 J, 334 J/ g is the δ H of fusion, that is how much energy it takes to either melt ice at 0°C and convert it to liquid water at 0,0634

Or it is the amount of energy that is going to be lost by liquid water at 0 and then convert that to ice.0650

When I do this, I end up with 11.42 g.0661

What is this 11.42 g?0674

11.42 g is the amount of ice at 0°C converted to liquid water.0677

Let us recap what is that we have done in terms of a nice picture.0702

We had liquid water at 25°C, I will just go ahead and put a 0 mark right there.0705

This is 0°C and then we have solid water at -7°C.0713

This H₂O liquid and this is H₂O solid.0720

A certain amount of heat was lost by the water in cooling and drop it down to 22.81°.0730

This amount of heat was enough to bring the ice of 20°C but still ice.0738

Now from the 22.81 there is a certain amount of heat that is going to be lost upon the water actually cooling down to 0°C.0746

Water at 0°C.0755

This amount of heat that is lost by the water in cooling, that amount of heat,0758

since the ice is now already at 0°C it is going to convert a certain amount of ice into liquid water.0762

That is the 11.42 g, that is our final state.0771

Here is where we are now.0775

Let us see, I have 40 g of H₂O liquid at 0°C, that is the initial 40 that drop down to 0.0777

I have 11.42 g of H₂O liquid at 0°C is the water that came from the ice that melted.0796

That gives me a total of 51.42 g of liquid water0811

I'm left with 25 g of ice initially -11.42 that had melted, that leaves me with 13.58 g of ice or solid water at 0°C.0822

This is our final state.0844

Our final state, once everything comes to equilibrium I have 51.42 g of liquid water at 0°C and I have 13.58 g of ice at 0°C.0846

That is what is going on, that is our final state.0861

Let us go ahead and do the δ H and δ S.0865

Any time you see the word insulated that means no heat is allowed to come in or go out.0870

It means that Q = 0.0879

Since, the pressure is constant, since P is constant we know that under conditions of constant pressure the δ H = Q so δ H= 0.0888

There is no δ H, there is no enthalpy transaction in this particular process.0908

The δ S for this process is going to be composed of three parts.0919

It is going to be the δ S total.0923

It is going to be equal to the change in entropy of the first phase,0926

that means 40 g of H₂O liquid at 25°C taken to 40 g of H₂O liquid at 0°C.0931

The first part of the total entropy is going to be the entropy change in taking the 40 g of liquid water at 250950

and converting it to 40 g of liquid water at 0.0956

To that we are going to add δ S2, to that is going to be the change in entropy associated with taking 25 g of H₂O solid at -7°C and converting it to 25 g of H₂O solid at 0°C.0961

We have to account for each and every single transformation, that is what we are doing.0986

To that we are going to add the δ S3,0991

We have 11.42 g of H₂O solid at 0°C converted to 11.42 g of H₂O liquid at 0°C.1006

There are three things that happen here.1034

Liquid water went from 25 to 0, there is a certain entropy change associated with that, that is δ S1.1036

25 g of ice went from -7° to 0 that is δ S2.1043

Out of that 25 g, 11.42 g of that was converted to liquid water and there is a change in entropy associate with that.1056

The total change in entropy, I just have to add them all up.1067

Let us see what we can do.1070

Let us do δ S1, let us always go back to our basic equations, always start with your basic fundamental equation.1074

From there, derive what you need.1088

DS = for conditions of constant temperature and pressure CP/ T DT – V Α DP and1092

that is our basic equation for calculating changes in entropy when the temperature of the pressure are constant.1107

In this particular case, this is great.1113

They tell us that the pressure is constant and that mean DP = 0.1116

This is equal to 0 because the pressure is constant.1121

This is not hard, most of these problems are actually very simple as long as you to start with the basic equation.1126

You want to be consistent, always start at the same place and then take it from there.1131

You might go this way or this way, but always start at the same place.1136

That means, once I integrate this I have δ S1 = the integral from T1 to T2 of CP/ T DT.1141

The CP is constant and this is liquid water that we are talking about so it was going to be 4.18 × the integral of T1 to T2 of DT/ T = 4.18 × the log of T2/ T1.1156

δ S1 = 4.18 × the nat log of the final temperature was 0°C, 273.1177

The initial temperature was 25°C, 298.1186

We have to work in °K and when we do that I get -0.366 J/ g/°K.1190

This is 4.18 J/ g/°K.1208

It is J/ g, I have 40 g and I still have to multiply this by the 40.1211

This is the part that you have to be very careful of.1217

It is really important to keep track of your units.1220

We have this many J/ g/°K.1224

We have 40 g so let me write that down.1228

We have 40 g of water so 40 g × -0.3,1233

There are numbers everywhere, 366 J/ g/°K or J/ g/°C.1245

And we end up with the δ S1 =- 14.65 J/°K.1253

I'm hoping that you confirm my arithmetic.1262

Our δ S1 = -1465 J/°K.1265

Let us do δ S2, it is going to be the same thing.1271

I mean, except now the heat capacity is going to be 2.09 instead of 4.18 because now we are talking about the ice going from -7 up to 0.1275

We have 2.09 × the nat log, final temperature is 273, initial temperature was 266.1287

We end up with 0.054 J/ g/ °K.1299

We have 25 g of ice so we have 25 g × 0.054 J/ g °K.1307

We are left with δ S2 of +1.36 J/ °K, that is δ S2.1318

Let me go ahead and do the next one on the next page.1331

We have δ S3, this one had to do with the melting of the ice, the converting of the 11.42 g of ice to 11.42 g of liquid water.1334

We are going to use the δ H of fusion.1351

The δ S3 remember that is the δ H of fusion divided by the temperature, that was the entropy change when there is a phase change.1355

During the phase change, the temperature is constant.1366

Because the temperature is constant we can go ahead and use the δ H.1368

This is how to calculate the entropy for phase changes.1373

The δ H of fusion is 334 J/ g and it is happening at 0℃ which is 273 °K.1378

We end up with 1.22 J/ g °K.1381

We have 11.42 g of ice as being converted 11.42 × 1.22 J/ g °K.1398

We end up with 13.97 J/ °K, that is δ S3.1409

We are going from ice to liquid water.1419

Entropy is rising, it is positive.1422

We just put them all together.1426

Our δ S total for this particular process is - 14.65 + 1.36 + 13.97 and we get a δ S total equal to 0.68 J/ °K.1430

A very practical problem because you are putting something at a certain temperature, something else at another temperature,1457

and there is going to be some final state that is going to be involved here.1463

Let us see what we have got now.1472

Let us go to example 2.1476

This is going to be the same sort of thing except that now we would be dealing with steam and liquid water.1481

It is going to be handled precisely the same way.1486

We are going to combine our intuition based on heat capacities which is going to absorb more heat.1489

Without change in temperature, it is going to be liquid water.1496

You are going to see how much you have of each and decide where you want to start the problem.1498

25 g of steam at 100°C and 310 g of liquid water at 25°C are combined in a fully insulated flask.1503

We are fully insulated so again Q = 0 so δ H is going to equal 0.1514

Given the following data, pressure is constant, we have the heat capacity for liquid water.1521

We have the heat capacity for the gas.1527

Notice 1.86 is not very high, it is a bit surprising given that for water.1529

The δ H of vaporization is 2257 J / g.1535

The δ H of vaporization, this is the amount of heat necessary to take 1 g of liquid water from the liquid phase to the gaseous phase, still at 100°C.1540

Or it is the amount of heat that is lost by 1 g of steam in transforming from 1 g of steam to 1 g of liquid water at 100°C.1553

They want us to know, what is the final temperature of the system and what phase or phases are present?1566

Same thing as before, what is the final state?1570

It is just a different way of asking it.1572

They want you to calculate δ S for this transformation.1573

We already know δ H is going to be 0 because we are at constant pressure and this is a fully insulated flask.1576

Let us see what we can do.1583

Let us deal with the 25 g of steam first.1587

I have got Q = MC δ T.1591

I got 25 g of steam at 1.86 J/ g/°C × 30°C.1595

What I'm calculating here is the amount of energy that is going to be required1606

in order to take the steam at 130°C and bring it down to steam at 100°C.1609

This is still steam, it has not converted yet.1617

This multiplication, I end up with 1395 J.1620

This is the total amount of energy has lost by steam in going from H₂O gas at 130°C to H₂O gas at 100°C.1626

Let us see what this heat lost, this 1395 J.1647

If that heat is lost, that heat is going to end up going into the water so the water temperature is going to rise.1652

Let us see what is the rise in water temperature is.1659

1395 J = we have 310 g of water, its heat capacity is 4.18 J/ g/°C.1664

And we are going to multiply that by δ T.1676

When I calculate δ T, I get δ T = 1.08°C.1680

The water rises by only 1.08°C.1686

We have 310 g of H₂O liquid at 25°C and we just took it to 310 g H₂O liquid at 26.08°C.1693

Now the water is at 26.08°C.1714

Let us see what is next, let us find out how much energy is lost now that the steam is at 0°C.1719

We have 25 g of steam at 100°C and we have water at 26.08°C.1732

They still need to find some in between temperature that is going to be the final temperature, the equilibrium state.1742

The steam has to cool a little bit further but now it is steam at 100°C.1749

It cannot drop any temperature now it is going to undergo a phase change once it hits 100.1756

Now we are going to have to use the δ H of vaporization.1761

We have 25 g of steam and the δ H of vaporization is 2257 J/ g.1767

We have 56,425 J.1777

This 25 g of steam at 100°C, when it converts to liquid water at 100°C it gives up 56,425 J of energy.1783

It is going to give that energy to the liquid water, the 310 g.1796

This amount of energy is going to raise the temperature of water yet some more from the 26.08.1801

Let us find out what the new temperature is.1807

Let us just say what this is here, write it down.1813

For H₂O gas at 100°C converted to H₂O liquid at 100°C.1819

This is the amount of heat lost by the 25 g of steam.1835

Let us see what this does to the water.1840

Now, we have 56,425 J = 310 g 4.18 J/ g/°C × δ T.1842

Our δ T is going to be 43.54°C.1858

Let us do this on the next page.1870

We are at 26.08° and we just raised the temperature at another 43.54°C.1876

Now, my temperature is 69.62°C.1889

310 g of liquid water is at 69.62°C.1904

We also have 25 g of liquid water at 100°C.1918

They are both liquid water but at very different temperatures.1934

This 25 g came from the condensation of the steam.1936

This 310 g at 69.62 just came from the normal liquid water that is rising the temperature.1941

We have water at 100°C, we have water at 69.62°C.1948

This water is going to cool, this water is going to warm.1966

They are going to reach some final temperature.1969

We need to find what that is.1973

The heat lost by the water at 100°C as it cools, is the heat gained by the water at 69.62 as it warms up.1977

That is the whole idea, heat out heat in.1987

We have Q lost = - Q gained.1992

This negative sign is very important.1999

The heats are equal.2002

This mathematic this ends up being a negative sign, very important.2005

The heat that is lost by one substance, in this case, water at a certain temperature, is negative of the heat gain by the other.2009

One is gaining and one is losing.2016

Let us write it out.2020

The heat lost that is going to be the MC δ T of the water at 100 and the heat gain is going to be the MC δ T of the water at 69.62.2023

Let us write that out, MC δ T.2034

I’m going to write M1 C1 δ T = M2 C2 δ T to let you know we are talking about two different masses.2041

In this particular case, it is not going to be two different heat capacities because now they are both water but the masses ore different.2052

We have got 25 g of water, the heat capacity is 4.18 and the change in temperature is temperature final – initial.2059

This initial temperature is 100°C for the 25 g = negative.2075

I have 310 g of water, its heat capacity is 4.18 and now it is going to be temperature final.2083

Its initial temperature is 69.62, this is the equation.2091

I'm looking for a final temperature.2097

We have done this from general chemistry.2100

Let us see what the arithmetic looks like.2105

The 4.18 go away and I'm going to be left with 25 TF - 2500 = -310 TF + 21,396.2.2109

I should have 335 TF = 23,896.2.2132

I have a temperature final of 71.33°C, this is what I wanted, the final temperature.2144

We have 335 g of H₂O liquid, the 310 g + 25 g of liquid water at 71.33°C.2157

This is our final state, that is what happened here.2177

The steam cooled, the water rose in temperature.2181

The steam converted to liquid water at 100 and that conversion, the heat that was lost by that conversion,2185

one of the 310 g of water lifting the temperature to the 69.62.2192

We have water at 100 and water at 69.62.2198

We are going to find some medium temperature.2201

I’m going to write it in between them.2203

Again, 25 g and 310 g there is a big difference.2204

There is only one from 69 to 71.2208

This one is from 100 down to 71, that is the whole idea.2211

Let us go ahead and calculate the δ S for this process.2218

δ S this time is going to be 4 different things that happened.2223

δ S total = we have δ S1.2227

This is the conversion of 25 g of gaseous water at 130° taken to 25 g of H₂O gas at 100 °.2232

To that we are going to add the δ S2, the second process we are going to take this 25 g of H₂O gas at 100°C.2253

And we are going to convert it to 25 g of H₂O liquid at 100°C.2266

And we have another process, δ S3 is going to be taking 25 g of H₂O liquid at 100°C and converting it to 25 g of H₂O liquid at 71.33°C.2277

Then the final entropy change is going to be 310 g of H₂O liquid at 25°C and 310 g of liquid water at 71.33°C.2305

25 g of steam at 130 went to 25 g of steam to 100, there is an entropy change associated with that.2337

25 g of steam at 100 went to 25 g of liquid water at 100, there is entropy change associated with that.2344

+ 25 g of H₂O liquid at 100 and now this 25 g of H₂O liquid at 71.33, there is an entropy change associated with the temperature change.2351

And this one we have 310 g of H₂O liquid at 25 going to 310 g of H₂O at 71.33.2360

There are 4 parts to this entropy.2368

Let us go ahead and calculate them.2370

Let us see here, what should I do.2374

The general equation once again, DS = CP/ T DT – V Α DP, this goes to 0 because pressure is constant.2377

Therefore, our general equation for change in entropy upon a temperature change2390

is going to be the integral of CP/ T DT from temperature 1 to temperature 2.2394

And for a phase change, our entropy is going to be δ H of vaporization ÷ T.2405

For 3 of these, our temperature changes we are going to use this one.2412

One of these is going to be a phase change, we are going to use this one.2416

Let us do it.2419

The δ S1 = 1.86 we are taking it from 403 which is 130 to 373 which is 100.2424

Which is going to be DT/ T and we are going to get 1.86 × the nat log of 373/ 403.2438

We end up with -0.144 J/ g/ °K.2451

We have 25 g of steam so 25 × 0.144 J/ g/ °K gives us - 3.60 J/ °K this is δ S1.2459

δ S2 is going to equal the,2482

This was the phase change 1.2485

We have the δ H of vaporization/ T which id equal 2257 J/ g ÷ 373 °K because the transformation took place at 100°C.2487

We get 6.05 J/ g/ °K.2504

We are changing 25 g of it so g/ °K and we end up with -151.27 J/ °K.2511

It is negative, there is a negative sign here but it is negative because it is dropping.2532

I'm actually going from a gaseous phase to a solid phase, entropy decreases.2537

You can use your qualitative understanding.2544

As things go from solid, liquid, to gas, the entropy is going to be positive.2547

As things go from gas to liquid to solid, the entropy is going to be negative.2551

These are just the absolute values, these are the actual numbers.2555

The magnitudes of the energy changes of things like that.2559

But I put negative sign here, the entropy is negative because we are going from a gaseous phase to a liquid phase, in more ordered phase.2562

Our entropy change is negative.2569

By all means, use the qualitative.2571

If there is ever a place where you want a qualitative analysis, it is in thermodynamics.2574

δ S3, it is going to equal 4.18 × the integral 373 to 344.33 DT/ T = 4.18 × the nat log of the final temperature which is 344.33 °K / 373.2580

We end up with -0.334 J/ g °K.2610

Once again, we have 25 g of this -0.334 J/ g °K and you get -8.36 J/ °K, this is δ S3.2617

In this particular case, the negative that shows up automatically based on the mathematics because we are actually doing integration.2639

As integration ends up being a logarithm.2645

And when you have a logarithm in the quotient and the quotient is less than one, you will get a negative number.2649

This negative number is already taken care of.2655

We do not have to worry about it being qualitative.2657

We have got one more δ S here.2663

We have our δ S4 and that is going to equal the 4.18 because now are talking about water × the nat log .2665

And we went from 298 °, we went to the 344.33 °, and tat is going to equal 0.604.2678

This is positive because we are rising in temperature.2691

Temperature increases, entropy increases.2694

J/ g °K and this time we have 310 g of the water that we raised from 25 to 78 or whatever it was.2699

× 0.604 J/ g °K and end up with 187.3 J/ °K.2710

Our δ S total = -3.60 - 151.27 - 8.36 + 187.3.2726

Our δ S total for this particular transformation = 24.02 J/ °K, a net increase.2746

There you go, a long problem, a little bit tedious but nothing altogether difficult in terms of what is actually happening.2758

Just take it one piece at a time.2767

Any time you are dealing with any sort of a mixture of two things, one is at a certain temperature and one is at another temperature,2771

the heat is lost by one thing is going to be the heat that is gained by the other thing.2782

Particularly, when we are talking about something that is going to be in a completely insulated flask which is often how we run these things.2785

We do not have to worry about any heat being transferred in or out of the combined system.2791

There you go.2797

Let us round this out with one final example and it does have to do with entropy.2801

Let us see what it says.2807

Given the following expression DS = CP/ T DT – V A DP.2812

We already know what this expression is, this is one of general expressions for finding the change in entropy.2817

This should be S.2822

Finding the entropy under conditions of constant temperature and constant pressure.2826

Given the following expression, calculate the decrease in temperature that occurs when 1 mol of water2832

at 25°C and 1100 atm is brought adiabatically and reversibly to a pressure of 1 atm.2837

Assume that K, the coefficient of compressibility = 0.2847

That is extra information, do not worry about that.2851

You do not need to do anything with it.2852

You have the following data for water.2855

The molar volume of water is 18 cm³/ mol, constant pressure heat capacity this time is expressed in J/ mol °K,2858

75.3 J/ °K and Α the coefficient of thermal expansion is 2.07 × 10⁻⁴/ atm.2867

Let us see what they are asking us here.2880

I have 1 mol of water it is at 25°C but its pressure is 1100°,2882

Under conditions of adiabatically and reversibly, I drop the pressure to 1 atm.2892

Adiabatic means there is no change in the Q = 0.2899

In other words, there is no heat is allowed to flow.2904

I know that when I drop the pressure from 1100 to 1 atm, my intuition tells me that the temperature is going to drop.2907

The temperature is going to drop precisely because no heat is allowed to flow, this is the adiabatic part.2915

No heat is allowed to flow into the system, in order to keep the temperature up.2921

The temperature is definitely going to drop.2926

Remember that an adiabatic process is when you increase the pressure, it is the hugest,2928

the largest drop in temperature happens in an adiabatic reversible process.2935

We want to know what that pressure drop is.2941

In other words, starting at 25°C if I do not allow any heat to flow, if I conduct this reversibly and very slowly,2943

adiabatically what is my final temperature going to be when my pressure in the system is now 1 atm, that is what this is asking.2952

Let us do it, adiabatic and reversible, that means that DQ reversible = 0.2960

Adiabatic Q is 0, we now we just put a reversible because it is telling us that it is actually reversible.2985

DS by definition = DQ reversible/ T.2993

If DQ reversible is 0 it implies that DS = 0.2998

There is no entropy change for this process.3005

We have an equation 0 = CP/ T DT – V Α DP.3007

Move is over, I get CP/ T DT = V Α DP.3021

Integrate this expression, the heat capacity is going to be constant so I end up with the following.3031

CP × the integral from T1 to T2 of DT/ T = V Α × the integral from P1 to P2 of DP.3038

I’m just taking the expression and found out that DS = 0.3055

I set it equal to 0, I rearranged it and I integrate this expression.3058

My final equation is going to be.3062

I get CP × the nat log of the final temperature/ the initial temperature = V Α δ P.3070

I have everything that I need, now I need to put it in terms of units that match and everything should be fine.3084

Let us go ahead and deal with the molar volume first.3091

I have 18 cm³/ mol = 18 × 10⁻³ dm³/ mol.3094

A cubic decimeter is a liter.3108

If there is 1 unit that you should know, 1 liter is 1 cubic decimeter.3111

We get 18 × 10⁻³ L/ mol, that just means that 1 mol of liquid water occupies 18 × 10⁻³ L.3115

Volume is it L and pressure is at atmosphere.3131

We are going to end up with something in L atm and we need to convert it to J.3134

Let us go ahead and take care of the V DP first.3140

I have got V × δ P.3145

The volume is 18 × 10⁻³ L/ mol.3150

The δ P was final - initial so it is going to be 1 atm - 1100 atm.3167

V δ P when I actually multiply this out I get - 19.782 L atm/ mol.3177

In order to convert it, I multiply by 8.314 J ÷ 0.08206 L/ atm.3191

And I end up with - 2004 J/ mol.3203

V × DP = V × DP = -2004.3212

I just have to convert to J because I’m working J, because heat capacity is in Joules.3221

I need to make sure that my units match.3226

Let us go to the next page.3233

I have my equation CP LN TF/ TI = VΑ DP.3236

I have got 75.3 × the nat log of the final temperature.3252

That is what I want, I want the final temperature.3261

They wanted to know what the temperature drop was.3262

TF I started off at 298 = Α V DP.3265

Α is 2.07 × 10⁻⁴ and V DP was – 2004.3275

And so I get 75.3 × LN of TF/ 298 = - 0.415.3287

I go ahead and I will do the division and exponentiation.3301

Let me write that out LN of TF/ 298 = -0.00551.3306

When I exponentiate, e to this power, e to that power.3317

TF/ 298 = 0.9945 and I get the final temperature of 296.36.3326

And therefore, δ T = final temperature 298.3345

I think it is a little bit of an arithmetic problem here.3364

The final temperature, final – initial will give me something positive, it is going to be something negative.3366

I apologize if I calculated some of it a little bit of arithmetic issue here.3372

I’m going to go ahead and leave this.3376

The process is correct but I seem to have done the multiplication wrong.3378

It is from what I see on the table.3382

This is correct and now if you just go ahead and multiply this × that.3384

I'm sorry, I apologize.3394

I wrote 273 on the paper, it should be 298.3395

This was correct.3400

We have 296.36 and our δ T = 296.36 – 298.3402

I'm sorry I thought it was 0 so I had 273 on my paper.3416

Our δ T -1.64°C or °K, I will just go ahead and write °C and there we go, this is our temperature drop.3420

Just use what is given to you in the problem.3437

You have a general equation because it is adiabatic and reversible, there is no change in entropy.3439

DS= 0, rearrange the equation and just put the numbers in.3444

The only thing you have to watch out for in this problem is the units.3449

Cm³/ mol you have to change that to L because the pressures and atmospheres.3455

That is just my particular preference.3462

You can work in cm³ and work in Pa, it does not really matter.3464

Just what you are comfortable with.3468

I’m comfortable with L atm, convert it to J.3470

Once I’m in Joules, I’m going to go ahead and run in the problem.3473

There you have it, a change in temperature is -1.64°C .3477

Thank you so much for joining us here at www.educator.com.3482

We will see you next time, bye.3485

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