Raffi Hovasapian

Entropy Example Problems II

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 2 answers Last reply by: Professor HovasapianSun Nov 25, 2018 5:27 AMPost by Kimberly Davis on November 24, 2018For example II, can we just calc deltaU= Cv dT ? I got the same answer as you did but took a different approach. 3 answers Last reply by: Professor HovasapianTue Oct 2, 2018 5:17 AMPost by Van Anh Do on February 23, 2016For example 1, I think you forgot to calculate dS.

### Entropy Example Problems II

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 0:09
• Example I: Calculate ∆U
• Example I: Calculate Q
• Example I: Calculate Cp
• Example I: Calculate ∆S
• Example II 7:13
• Example II: Calculate W
• Example II: Calculate ∆U
• Example II: Calculate Q
• Example II: Calculate ∆H
• Example II: Calculate ∆S
• Example III 18:47
• Example III: Calculate ∆H
• Example III: Calculate Q
• Example III: Calculate ∆U
• Example III: Calculate W
• Example III: Calculate ∆S
• Example IV 27:57
• Example IV: Diagram
• Example IV: Calculate W
• Example IV: Calculate ∆U
• Example IV: Calculate Q
• Example IV: Calculate ∆H
• Example IV: Calculate ∆S
• Example IV: Summary
• Example V 48:25
• Example V: Diagram
• Example V: Calculate W
• Example V: Calculate ∆U
• Example V: Calculate Q
• Example V: Calculate ∆H
• Example V: Calculate ∆S

### Transcription: Entropy Example Problems II

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

Today, we are going to continue our entropy example problems.0004

Let us jump right on in.0007

For the first problem we are going to look at here is the following.0010

We have 1 mol of an ideal gas initially at 25°C and 1 atm pressure and it is taken to 50°C and 0.55 atm,0014

during this transformation 260 J of work are done on the surroundings.0024

They give us the work that is done during this particular expansion.0029

They want us to calculate the heat δ U, δ H, and δ S for this particular process.0035

Let us see what we can do.0044

We are dealing with an ideal gas so it is going to change a couple of things.0048

We have 25°C to 50°C.0051

Let us go ahead and start with, in this particular problem when you are finding the Q, in this case they gave you W,0057

but in subsequent problems we have to find the work, the change in energy, the change in enthalpy and the change in entropy.0064

There is no systematic way as far as choosing one before the other.0071

Sometimes it is easier to find the energy and then the heat and work.0074

Sometimes it is better to do the work than heat so do not think that you have to go down the line in order, it is not like that.0077

It is what you have at your disposal and whatever strikes your fancy.0083

Let us start off with this one.0088

When we do that let me stick with black.0091

Let us do DU = CV DT again, it is always nice to start with the equations that you know to write them down and take it from there.0094

It is a great review to constantly write down the equations that you need.0104

DDT + DU DV T DV is basic equation as far as the energy of a system in terms of temperature and volume.0108

Now this is an ideal gas so this is automatically goes to 0.0121

All we are dealing with is DU=CV DT and when we integrate this we get δ U =CV × δ T.0127

This is the equation that we are going to use.0143

It looks like we have everything.0144

δ U = CV δ T, CV is 3/2 RN δ T.0147

We can go ahead and put some numbers, we have 3/2, R is 8.314 J/mol-°K and we have 1 mol and we have δ T.0158

δ T is going to be 50 -25 and again Kelvin and Celsius, the δ is the same because the increment is the same so we have just 25°K.0175

°K vs °K and mol vs. mol, energy is going to be in Joules.0188

δ U for this process is 312 J and again I'm hoping that you will actually confirm my arithmetic, I’m notorious for arithmetic mistakes.0195

The energy is taken care of we have 312 J.0205

We know that δ U, let me go ahead and use the differential expression because it is probably the best to begin with.0211

DU = DQ - DW that is the fundamental relationship, that is the first law of thermodynamics.0220

This implies in for a finite change that δ U =Q – W.0230

Again, there is no δ Q δ W, these are Q and W are path functions, this is a state function which is why we have a δ here.0236

When you rearrange this, I get Q = δ U + W.0245

Therefore, Q I have δ U, I have 312 J, I’m going to go ahead and skip the unit.0252

I hope you do not mind just work with numbers.0257

They tell me that 250 J of work are done on the surroundings.0260

Work is set positive if it is being done on the surroundings.0266

Therefore, we have the 260 J.0271

For Q, we end up with 572 J.0276

For this particular expansion, 572 J of heat transpire.0283

Let us go ahead and take a look at δ H.0290

I’m going to go ahead to do this on the next page, it is not a problem I have few pages here.0295

Let us go ahead and write our basic relationship that DH = CP DT + DH DP DP.0299

Again because this is an ideal gas this goes to 0.0315

Basically, all we have is DH =CP DT, nice and straightforward.0319

When we integrate this function, I will integrate this differential equation we end up with CP δ T.0324

This is really nice and straightforward.0331

The only thing we need is they gave us CV, we have to find relationship, we have to find CP.0333

We know the relationship, we know that for an ideal gas CP - CV = Rn.0339

Therefore, CP = Rn + CV.0346

Therefore, CP =Rn + 3/2 Rn CP = Rn × 1 + 3/2.0351

Therefore, our constant pressure heat capacity = 5/2 Rn.0363

Working with an ideal gas is really easy.0368

We have our constant pressure heat capacity now we can just go ahead and put in here.0371

Let us go ahead and write δ H =CP δ T CP is 5/2 Rn and we have δ T.0377

Let us put in some numbers δ H =5/2 × 8.314.0386

Again, I hope you forgive me, I’m going to skip the units.0393

1 mol and δ T is 25°K.0395

When we work out this arithmetic, we end up with 570.0400

No, I’m sorry that was going to be our δ H is going to be 520 J.0406

Let us see what we can do here.0426

Let us go ahead and go to example 2.0431

Example 2, says 1 mol of an ideal gas with the CV = 3 Rn/ 2 was initially in the following state 350°C and 1.5 atm.0438

At constant volume the gas is heated to 450°K, calculate Q, W, δ U, δ H and δ S.0455

They also want us to compare δ S with the value of Q/T.0465

We have 1 mol of an ideal gas, they gave us the constant volume heat capacity 350°K , so we are doing this process at constant volume.0471

These problems are really great because they allow us to actually review our first law issues, the heat, work, and energy, enthalpy, things like that.0483

We just add that final state property which is entropy.0492

Let us see what we have, constant volume is going to be important.0496

These constraints are what changes the equations that we deal with.0501

Constant volume that implies the DV = 0.0505

DW =the external pressure × DV.0513

If DV is 0 that means DW = 0 that means that the work = 0.0520

Constant volume nothing is changing so no work is being done so work is 0.0528

Let us go ahead and see what we can do about DU.0536

Again DU = CV DT + DU DV at constant temperature DV and again because we are dealing with an ideal gas this term goes to 0.0539

It is constant volume so DV goes to 0.0559

Even if this were an ideal gas, in this particular case this would go to 0.0563

In either case, this term goes to 0.0567

What we are left with is DU = CV DT and for a finite difference when we integrate this differential equation we get CV δ T.0569

Let us put in some numbers, we have 3/2, let us go ahead and write it all in, 3/2 Rn δ T and it is going to be 3/2 × 8.314 × 1 mol and0584

the change in temperature 350°K to 450°K, we have 100°K temperature change.0602

Our energy for this process is going to be 1247 J.0611

That is nice.0620

We have W and we have δ U so let us go ahead and find Q.0624

δ U= Q - W and let me just go ahead and just do it this way Q =δ U+ W =δ U which is 1247 + W which is 0.0628

Our Q for this process is also 1247 J.0651

That takes care of the heat, now let us go ahead and take care of the enthalpy.0659

DH =the constant pressure heat capacity × the differential change in temperature or constant pressure heat capacity × temperature +0666

this DH DP the change in enthalpy per unit change in pressure at constant temperature × the change in pressure.0676

But again we are dealing with an ideal gas so this goes to 0.0684

We have DH =CP DT and for integration we have DH = CP δ T.0689

Nice and straightforward, nothing strange is happening here.0700

Again, since we have the 3/2 Rn the constant volume heat capacity was 3/2 Rn because this is an ideal gas our constant pressure heat capacity is going to be 5/2 Rn.0703

It is going to be like that.0715

DH = 5/2 Rn δ T = 5/2 × 8.314 × 1 × 100 we get a δ H for this of 2079 J.0719

That takes care of our δ H.0743

Let us go ahead and see what we can do about δ S.0750

This is what we have been discussing, this particular unit is entropy.0754

We have DS =CV / T DT + nR/ V DV for an ideal gas this is the expression for the entropy or the differential change in entropy of the system.0760

This is happening at constant volume, therefore, this term goes to 0.0783

What we have is just DS =CV/ T DT.0789

We are going to integrate this and what we are going to get is δ S =the integral from temperature 10798

to temperature 2 of CV/ T DT which = CV × the log of T2/T1.0807

CV is a constant so it comes out of the integral which we are left with is the integral of DT/ T, the integral of DT/ T is LN T2/T1.0821

Now I will just put the numbers in.0833

In this particular case, it is going to be 3/2 Rn × the log of temperature 2/ temperature 1.0837

Let us go ahead and DS = 3/2 R is 8.314, the number of moles is 1 log, and we have 450°K/ 350°K.0854

Real quickly here, when you are dealing with temperatures involving the temperature on top of another temperature, you have to work in °K.0875

I know that in previous problems, in this particular problem or either ones where we did Celsius temperature, when we did δ T we just use the 50°C - 25°C.0884

The difference in the δ T is 25°C or 25°K.0898

There we do not have to convert to °K because the increment of Celsius and Kelvin is actually the same.0903

However, for here we cannot go 25°C/ 50°C.0911

The °C do not cancel and the reason is because in the Kelvin temperature is going to be 25 + 273/ 50, + 273 the 273 do not cancel.0919

When you do this, the temperature has to be in °K.0931

In general, we are just avoiding problem, just do everything in °K.0937

In that way, you will never go wrong.0940

I just want to let you know that if the temperature in this particular problem we are given in Celsius,0942

you would not be able to put Celsius on top of Celsius you will get the wrong answer.0946

When we run this, let me go ahead and go back to black here.0952

δ S should be 3.13 J/°K.0956

Since we are dealing with 1 mol, we can just go ahead and put 3.13 J/°K mol or mol°K.0969

It does not matter how you do it because we are talking about 1 mol.0979

It is okay, let us go ahead and put them all there.0984

It also asks us to compare δ S with Q/ T.0987

I do have another page here and that is good.0995

Let us go ahead to the next page.1000

The δ S we just calculated so δ S = 3.13 J/°K mol.1003

The problem with Q/ T is T is not fixed.1014

T goes from 350°K to 450°K so technically we cannot really solve Q/ T so we should leave it alone.1021

However, I’m going to do something different here.1029

I’m going to take T average.1032

I’m going to take the average between 350°K to 450°K instead of just 400°K just for illustrative purposes.1034

1247 J that was the heat for this particular transaction and I'm going to divided by the 400°K.1043

When I do that I will get 3.12 J/°K.1052

Notice that they are almost the same, 3.13 is the actual change in entropy calculated.1058

The Q/ the T average is 3.12.1064

Let me write down this particular note again, we cannot really do Q/ T because T is not fixed.1068

I chose to simply average the temperatures for illustrative purposes.1095

If you are ever asked to do something like this on an exam, it is perfectly fine for you to say this cannot be done.1110

Or if you do it, use T average and make sure you specify that you are actually using an average temperature here.1116

Let us go ahead and do example 3, we have 1 mol of an ideal gas with this particular constant volume heat capacity initially in the following state 350°K, 1.5 atm.1125

This is the same as the previous problem, the initial state is, at constant pressure the gas is heated to 450°K.1140

The previous problem starting with this particular state we took a 450°K under constant volume.1150

We are doing under constant pressure.1157

At constant pressure, this is going to be important part here.1164

I think I’m going to go in red here just for a change of pace, make it a little brighter.1168

Calculate Q, W, δ U, δ H, and δ S and compare δ S with Q/ T.1173

At constant pressure and we are dealing with an ideal gas.1180

At constant pressure we know that δ H = the heat transpired, that is the whole idea behind the constant pressure process.1191

Let us go ahead and take a look at the DH.1200

The DH = CP DT this is one of our equations from the first law + DH DP constant T DP.1205

This is an ideal gas so this term goes to 0 or its constant pressure so this term goes to 0.1222

In either case, it goes to 0.1228

What we have is DH = CP DT and when we integrate that we get δ H =CP δ T.1231

When we do the mathematics I will get δ H =5/2 Rn δ T = 5/2 × 8.314 × 1 mol and the δ T is again 100°K.1244

We get a δ H of 2079 J.1264

At constant pressure we know the δ H = QP.1275

Therefore, we automatically know the Q.1279

The Q under constant pressure, the Q for this particular process is also = 2079 J.1281

It takes care of the heat.1290

Let us see what we can do about δ U.1294

I will do it in this page it is not a problem.1300

I know the δ H =δ U+ δ PV because the definition of enthalpy H =U + PV just apply the δ operator to all of that and you have this.1306

Well δ H =δ U + P δ V because P is constant.1323

Since there is no change in P, you can just go ahead and pull that out.1332

That means δ U =δ H - P δ V or dealing with an ideal gas so PV =nRT.1340

V = nRT/ P therefore δ V δ U =δ H – P.1353

δ V is just V2 – V1 so we have nR T2 / P because it is constant pressure –nR T1/ P.1365

The P cancel and I will get δ U =δ H - nR T2 - T1.1382

I’m just fiddling around with the basic equation that is all it comes down to.1399

We have δ U =δ H which we calculated which was 2079 J – n 1 mol R is 8.314 and δ T which is T2 - T1 is 100.1404

When we end up doing that, we end up with δ U =1247 J.1420

That is the energy of the system spans by 1247 J.1434

δ U =Q – W, therefore W = Q - δ U.1442

Q is 2079 J and δ U is 1247 J.1454

Therefore, we get that the work = 832 J and I'm hoping that you are performing all of my arithmetic.1469

It takes care of the work.1481

We have the work, we have the energy, we have the enthalpy, we have that, let us go ahead and find our δ S.1482

Let me see if I have an extra page.1489

Yes, I do.1491

Let me go ahead and go to the next page.1492

In terms of pressure we have DS.1499

We are express things in terms of temperature and volume, temperature and pressure.1505

In this case because pressure is involved we are going to use this version CP/ T DT - nR/ P DP.1512

The pressure is constant so this term goes to 0.1525

What we have is just DS =CP / T DT and when we integrate the get δ S = it is going to be the integral from T1 to T2 of CP/ T DT.1530

CP is a constant so what comes out from under the integral and which are left with is DS = CP × the log of T2/ T1.1548

When we go ahead and put numbers in here, we go down here a little lower we get 5/2 × 8.314 × 1 mol1563

× the log of 450°K/ 350°K and we get δ S = 5.22 J/°K.1577

Our comparisons so we have δ S = 5.22 J/°K that is our δ S.1597

Again, the T is not fixed so we technically cannot do this.1607

I decided to just go ahead and do a T average.1612

Our Q here is 2079 J and we have an average of 400°K.1616

We get 5.2, it looks like a 6.1625

We have 5.20 J/°K that is our Q/ T.1630

We are just running through our basic equations that is what we want do.1641

We have derived all of these equations and we have set aside specific ones that are important to learn, for example this one regarding entropy.1644

We are just gaining some practice and of course reviewing all of our concept from energy that is why we are comfortable with this.1655

At this stage of the game, we want to be able to understand the thermodynamics but mostly we just want to be able to be comfortable1661

with the mathematics and be able to solve some of the problems that are thrown at us.1668

Let us see what is next, example number 4 I think.1674

Example number 4, the next example 5 and the next couple of examples in the next lesson that we do,1686

they are all going to be based on the same initial state this 350°K, 1.5 atm, an ideal gas with a constant volume heat capacity of 3/2 Rn.1692

What we are going to be doing was we are going to be transforming it from one state to another state under different circumstances.1705

For example 2, we made a transformation from 350 to 450 under constant volume.1711

In the example after that example 3, we do the transformation of 350 to 450 at constant pressure.1719

It is going to be the same starting state, we would be doing something else to it.1726

That is all we are doing, we are just trying to figure out how to deal with these thermodynamic variables under different conditions.1730

1 mol of an ideal gas with the constant volume heat capacity of 3 Rn/ 2 is initially in the following state 350°K. 1.5 atm.1738

The gases expanded and this time isothermally and reversibly until the pressure of the gas drops 2.75 atm.1747

Isothermally the temperature stays the same but the pressure does not 1.5 to 0.75 atm, the gas itself.1760

Calculate these variables and compare δ S with Q/T.1768

Let us see what is going on here.1773

The important thing is isothermal and reversible so this is going to affect the mathematics.1774

Let us go ahead and draw this one real quickly so we know what we are dealing with.1780

This is pressure and this is volume so this is an isotherm.1788

This is state 1 and this is state 2, whatever that happens to be.1795

State 2, we know this is 350°K and it is going to be 1.58 atm.1800

This is going to be 350°K because it is isothermal and it is going to be 0.75 atm.1809

I know it is not the scale and this is not exactly half of it but that is fine.1826

Here we have volume 1 and volume 2, it is expanding.1829

The gas is going from one volume and it is expanding to a larger volume that is what this is.1834

What we are doing is we are keeping the temperature constant so the expansion just not happened randomly.1843

This path or that path it actually happens isothermally, it happens along this path.1848

The temperature stays along this path.1853

Even more than that, it happens reversibly.1856

Reversibly means we absolutely follow this path.1859

In other words, we do not go this way, we do not go that way.1863

We can have an isothermal transformation that is not reversible.1871

That is not a problem, we can go ahead and expand the gas and keep the temperature constant.1875

It follows an isotherm but the actual path that it follows, the work that is done is going to be different depending on the path we follow,1880

this work is a path function.1889

In this case, it is reversible so we are absolutely following this path.1891

What we are doing is we are making a bunch of little of micro changes all the way around.1894

Again this a reversible.1900

Remember what reversible means in terms of mathematics, it means that the external pressure = to the pressure of the system,1901

the internal pressure if you will.1910

The pressure on the outside = the pressure on the inside so the system is always at equilibrium.1913

Any new small change that you make, all you have to do is go back that little small differential change.1920

Essentially, the system is in equilibrium.1925

Mathematically, it means that P external = P.1927

That is what reversible means.1930

Let us go ahead and see what is going on here.1933

I will just this label this so this is an isotherm.1937

It is very important so you can do something isothermally or you can do something isothermally and reversibly.1940

DW = P external × DV that is the basic, that is the definition of work.1950

Because it is reversible, the P external = P so we have DW =P DV.1959

This is an ideal gas so we have PV =nRT volume =nRT/ P.1970

My apologies, I’m going to do this, this is going to be P =nRT/ V.1993

I’m going to have put this into here so I get DW =nRT / V DV then when I integrate this, I end up with the following.1998

I get the integral of DW is W not δ W.2016

I get that the work = nRT × the integral volume 1 to volume 2 of DV/ V= work.2020

When I do this integral, I get nRT × the log of volume 2 / volume 1 so this is my fundamental equation.2043

Let us deal with this.2058

I do not have V2 and V1, what I have is P2 and P1.2060

Again, this is an ideal gas, let me express it this way.2064

I have PV =nRT, I have V =nRT/ P.2067

Therefore, V2 =nRT / P2, V1 =nRT/ P1.2075

This is isothermal so T stays the same, the pressure that is going to be different.2085

V2/ V1, I will go ahead and do on the next page.2093

V2/ V1 = nRT/ PQ / nRT/ P1 you end up with P1/ P2.2102

Therefore, our equation work = nRT Ln of V2/ V1 ends up becoming work = nRT × the log of P1/ P2.2117

We are just fiddling things around, just basic mathematics that is all it is.2138

Now we go ahead and solve it.2143

We have 1 mol, we have 8.314 J/mol-°K, we have a temperature of 350°K, and we have a log of 1.5 atm/ 0.758 atm.2145

This cancels giving us the log of a pure number and we get that the work = 2017 J.2169

That much work is done as the gas expands.2181

The system is doing work on the surroundings, work is positive.2187

It says that it is isothermal.2197

Isothermal implies that δ U =0.2201

However, be careful with this so δ U = 0 and let me tell you why.2208

In this particular case, this is the basic equation DU = CV DT + DU DV at constant temperature DV.2214

The temperature is constant so this term goes to 0.2230

DU DV we are dealing with an ideal gas so this term goes to 0 that is why DU = 0 or δ U =0.2234

You are going to run across a lot of problems in thermodynamics.2245

Most of the problems that you do are going to involve an ideal gas.2248

When you see the word isothermal, you can just automatically set δ U = 0, it is not a problem.2251

However, if you understand that δ U = 0 under isothermal conditions only for an ideal gas.2257

If you are not dealing with an ideal gas this term is not 0, if it is small or whatever happens to be but it is not 0.2263

Isothermal does not automatically mean that δ U is 0.2275

Isothermal for an ideal gas implies that δ U is 0.2279

There are times when we can set something to 0, isothermal δ U= 0, adiabatic Q = 0.2284

It is for an ideal gas, if this is not an ideal gas, if it is a Van Der Waals gas or another kind of gas,2292

or the solid or liquid because this equation works for any system at all this is not to be 0.2301

We cannot just automatically set isothermal means δ U = 0 please remember that.2307

Let us go ahead and finish up here.2314

δ U=Q - W well δ U was 0 and Q the work was 2017 J.2316

Therefore, Q = 2017 J.2329

Let us go ahead and deal with δ H and δ S.2337

Let us do it over here, δ H we said was δ U + δ PV.2342

This is 0, therefore we have δ PV which is nothing more than P2 V2 - P1 V1 which is nothing more than nR T2 – nR T1, which is equal to nR δ T.2356

It is isothermal, the temperature stays the same so this is 0.2381

The δ H for this process is 0.2386

Another way of doing this, this δ H business, we could use the δ H equation DH = CP DT + DH DP at constant T DP,2392

change in temperature δ T, this is an isothermal process so this term goes to 0.2408

For an ideal gas, this term goes to 0.2413

Again, we have DH = 0 which means that δ H =0.2416

Either one of these processes are absolutely fine.2422

You are going to end up getting the same answer.2424

Let us go ahead and take care of the δ S.2428

We have DS = CP/ T DT- nR/ P DP.2431

This is isothermal so that takes this term to 0.2444

We are left with DS = - nR / P DP.2448

We are going to integrate this like we always do and we end up with δ S =.2459

The nR comes out of the integral you going to be left with the integral from DP/ P so what you get is the following.2468

You get - nR × the nat log of pressure 2 / pressure 1.2479

We can go ahead and put the numbers in.2488

This 8.314 × the nat log of pressure 2/pressure 1.2491

Pressure 2 is 0.75 atm, pressure 1 is 1.5 atm, δ S is going to end up equaling -5.76.2498

The log 0.75/ 1.5 is 1/2 , the log of the number less than 1 is negative.2521

This number is negative × the negative gives you a positive.2528

δ S =5.76 J/°K.2534

Let us compare δ S with Q/ T.2544

δ S which is calculated δ S 5.76 J/°K.2547

Let us calculate Q/ T, in this case we can do T at Q/ T because T stays the same, it is just 350°K.2557

In this particular case, it is actually Q reversible because we are doing this process along the reversible paths so Q/ T = Q reversible / T.2565

That is the definition of entropy.2573

The Q reversible we have 2017 J, we have 350°K ends up being 5.76 J/°K.2581

What about 5.76? it is exactly the same because we are dealing with a reversible path.2595

A quick review of what is happening here physically.2608

Notice, how the work and the heat are equal.2611

We ended up with a work and the heat being equal to 2017 J, they ended up being equal.2614

Here is what is going on.2622

As the system expands, it does work on the surroundings.2627

It is W on the surroundings and the amount was I think 2017 J.2638

As it expands, as a gas expands it is going to cool but the problem is we are doing this process isothermally so we do not want it to cool.2651

We are going to keep the temperature the same as it expands.2671

Because the process is to happen isothermally, the system requires energy in the form of heat2676

in order to keep the temperature up so the gas is expanding, the gas wants to cool.2710

It is happening isothermally so something, somehow, energy has to come into the system to keep the temperature from dropping,2715

to keep it from cooling because we want to keep the temperature the same.2719

Because the process happens isothermally, the system requires energy in the form of heat to keep the temperature constant, to keep the T constant.2724

The question is where is this heat going to come from?2739

There is only one source, it has to come from the surroundings.2742

This heat as energy has to come from somewhere.2753

It comes from, it is pulled from the surroundings.2774

Since Q is positive, when it enters the system that is what we get Q =2017 J.2796

What is happening here is the following under isothermal conditions.2814

Work is done on the surroundings, the surroundings get colder.2818

This is our experience with the first law.2832

The surroundings get colder.2835

The gas expands does work on the surroundings 2017 J of work.2841

In the process of expanding the gas inside wants to cool, the system wants to cool.2846

We are doing this process isothermally,2851

In order for it to maintain its temperature, energy has to come into the form of heat.2853

Heat has to come from somewhere so it pulls that heat from the surroundings.2858

As the gas is doing work on the surroundings it is actually pulling heat from the surroundings.2862

When you pull heat from the surrounding, the surroundings are going to get colder.2868

Energy is transferred to the surroundings as work.2879

Energy is taken from the surroundings as heat.2884

That is all that is happening here.2887

Let us see what is going on, let us go to our last example here.2899

Example 5, the same starting conditions, different set of circumstances.2907

1 mol of an ideal gas, constant volume heat capacity 350°K, 1.5 atm, the gas expands isothermally.2913

We are isothermal, against a constant external pressure of 0.75 atm until the pressure of the gas achieves 0.75 atm.2920

The same exact initial conditions.2931

The same exact same exact initial state, same exact final state.2933

The final state is going to be 0.75 atm but now it is not going to expand reversibly.2937

It is going to do it isothermally but that is not going to be reversible.2943

Let us see what this one looks like.2946

We have our pressure, we have our volume, we have our isotherm, we have our state 1, we have our state 2.2953

This is state 1, this is state 2, this is pressure 1 that is the 1.5 atm and this is going to be P2, this is going to be 0.75 atm.2962

1.5 and 0.75 atm.2975

This is volume 1 and this is volume 2.2977

It is going to expand so isothermally the temperature is going to stay the same so we are still going to expand.2980

This is the isotherm but now the particular path that we are going to take is the following.2987

The external pressure is 0.75 so we are going to expand this way.2995

This is the path that we are going to take.3002

It is at constant external pressure so this is the amount of work that is done.3004

The last problem we did isothermal reversible work.3010

We follow this path, we kept the temperature the same but we did along the isotherm itself.3018

There was all this extra.3024

We can only expect that a work in this particular example is going to be less than the work that we had in the previous example.3026

Let us hope it actually turns out that way.3034

It is a different path.3038

The temperature is the same so we can do something isothermally but now we are following a different path so3041

the work is going to be different, the heat is going to be different.3046

Where shall we start?3057

Let us start the same way, the definition of work.3058

The definition of work is external pressure × the change in volume.3062

That means that work = the external pressure because the external pressure is constant, constant external pressure × δ V.3069

We just integrated this differential equation so we get work = P external.3080

This is going to be P external volume 2 - volume 1, we are dealing with an ideal gas therefore V = nRT / P.3092

Volume 2 = nRT/ P2 - nRT/ P1.3103

We get that the work = the external pressure × the nR T.3123

I’m going to pull out the nRT because it is going to be constant.3131

It is going to be 1/ P2 -1/ P1 and now we can go ahead and put our numbers.3133

work = the constant external pressure is 0.75 atm.3142

n is 1 mol, R is 8.314 J/°K.3149

Our temperature is isothermal, it is happening at 350 K.3155

1/ the external pressure P2 is 0.75 -1/1.5.3161

If the arithmetic comes out correct you should have 1455 J.3171

This work is definitely less than the 2017 J.3178

This 1455 accounts for this work was done against a constant external pressure.3182

In the previous problem we did it along the isotherm itself, we did this process reversibly.3187

Therefore, remember reversible processes are important because they represent maxima and minimal, things like that.3192

In the case of an expansion, it is the maximum amount of work that you are going to get from a particular expansion.3201

We are dealing with an ideal gas so isothermal again in the case of an ideal gas, isothermal definitely implies that the change in energy of the system is 0.3212

δ U =Q - W so 0 = Q – W.3226

Therefore, Q = W = 1455 J and it is still happening isothermally.3237

The heat and the work are still the same.3250

The same thing is happening, it is still expanding.3253

It is expanding and wants to get cooler, we are not letting it get cooler, it is doing or forcing it to do isothermally and it has to pull energy from somewhere.3255

It is going to pull it from the surroundings.3263

Work is done on the surroundings in expansion, the surroundings end up getting colder.3264

Let us go ahead and do DH.3273

DH =CP DT + DH DP DP constant temperature that goes to 0, ideal gas that goes to 0.3279

Therefore, our enthalpy change δ H for this process is 0.3296

δ S I’m going to calculate because δ S is a state property.3303

State 1 and state 2, the first path we took was this way.3313

This path now we are going this way.3318

This way and this way.3321

A state property does not depend on the path.3323

Therefore, the change in entropy is still to be 5.76.3327

It is the same number that we got from the previous example J/ K.3333

It is going to be 5.76 J/°K.3338

Heat and work are different because they are not state functions, they are state properties, they are path properties.3342

But because δ S is a state function the path that we taken together does not matter.3349

We can go ahead and use it from the previous example.3353

Let us go ahead and calculate Q and T.3358

Q in this case is 1455 J, temperature is 350°K, in this particular case we get 4.16 J/°K.3360

In this particular case because of a different path the entropy is actually less.3373

It is not the 5.76.3379

You are going to get the same entropy Q/ T and δ S when you run a reversible process because the definition of entropy is DQ reversible / T.3380

This reversible part is very important.3395

Thank you so much for joining us here at www.educator.com.3399

We will see you next time for more example problems, bye.3401

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