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1st Law 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

• ## Related Books

### 1st Law 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 0:11
• Example I: Finding ∆U
• Example I: Finding W
• Example I: Finding Q
• Example I: Finding ∆H
• Example I: Summary
• Example II 21:16
• Example II: Finding W
• Example II: Finding ∆H
• Example II: Finding Q
• Example II: Finding ∆U
• Example III 33:33
• Example III: Finding ∆U, Q & W
• Example III: Finding ∆H
• Example IV 41:50
• Example IV: Finding ∆U
• Example IV: Finding ∆H
• Example V 49:31
• Example V: Finding W
• Example V: Finding ∆U
• Example V: Finding Q
• Example V: Finding ∆H

### Transcription: 1st Law 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 example problems for the first law and for energy and things like that.0004

Let us just jump right on in.0010

Our first example of this lesson is the following.0013

We have 1 mol of Van Der Waals gas at 298 K and it expands isothermally and reversibly from 25 dm³ to 65 dm³.0016

The Van Der Waals constant is 0.366.0029

This is Pascal m⁶/ mol².0034

These are just the units, do not worry about those and B = 0.0429 dm³/ mol.0039

For the gas, the DU DV this should be sub T, my apologies, at constant temperature.0046

The rate of change of energy with respect to volume at temperature = an²/ V².0055

We would like you to find Q, W, Δ U, and δ H for the transformation.0062

There is a lot of information in this particular problem.0067

Things that are important, we have the temperature isothermal and reversible.0070

Remember, reversible means that the pressure external is actually equal the pressure of the system.0077

The change in volume we have the constants and we have this thing so this is a Van Der Waals gas, it is not an ideal gas.0086

For an ideal gas, this DU DV = 0 that term drops out and that expression for the total differential which I will write in just a minute.0095

Let us go ahead and see what we can do.0103

Now the expression I was talking about,0106

Let me do this in blue.0108

The expression I’m talking about is the following.0109

Our basic mathematical expression, what is handful equations that we definitely want to know and use in order to derive everything else.0112

DU = CV DT + DU DV at constant temperature × DV.0123

This is the general expression for the change in energy of a system.0138

For the differential change in energy of the system.0141

We would want the full change which we just integrate this expression.0143

Let us see what we can do to simplify this for us and let the problem itself put constraints on this.0146

This is isothermal.0154

Isothermal means that the temperature is constant.0155

That implies that as DT term = 0 because there is no change in temperature, that means this one drops out.0164

What we are left with is the following.0170

We are left with DU = DU/ DV sub T DV that is going to give us the differential change in energy.0172

Therefore, when we integrate this we end up with a total change in energy for the system =0183

the integral from volume 1 to volume 2 of this DU DV T DV.0193

Fortunately, they give us the DU DV sub T.0202

It is right there and very convenient.0204

This is equal to the integral from the first volume and the second volume and notice we have the first volume and we have the second volume of an² / V² DV.0209

An² comes out, we have V1 / V2 and then we have DV / V.0226

This is what we have to integrate in order to solve this.0235

On notice we started off with this.0236

This is our basic, one of the basic equations that we need to know and we let the problem dmde what sort of constraints, what is going to go to 0.0238

If they said that this under constant volume this would go to 0.0246

This is what is going on here.0250

Once we do that, we end up with the following.0252

This is going to be an² × 1/ V1 - 1/ V2 and when I put these numbers in, that is final.0257

An², a is 0.366, I’m going to go ahead and leave off the units.0272

I hope you guys do not mind.0280

I will leave the units for this first problem it might be nice to see.0284

Pascal m⁶/ mol² × 1 mol², because we are dealing with 1mol of the gas and of course, these are expressed in dm³.0288

We have to express this because this is in meters we have to express the volume in meters.0307

We are dealing with Joules and Pascal so the volume has to be in meters.0311

We have to make the conversion.0314

Volume 1 it is just going to be 25 × 10⁻³.0316

1 dm³ is 10⁻³ m³, basic conversion factor -1/ 65 × 10⁻³.0323

When we do this calculation, we end up with 9.0 J.0332

This is not a strong point but I hope that you actually confirm the arithmetic.0338

The arithmetic is secondary.0345

What is important is this process, getting to this point, this is what is important.0346

Often in a test, for some of the higher end courses you might be asked to calculate and derive.0354

You can stop at the equation.0362

You do not have to do the numbers.0363

9 J that takes care of the change in energy.0366

The change in energy of the system is 9 J.0371

Let us go ahead and see what we can do about the work.0375

Our equation for work is DW = P external DV, again, another one of the basic equations that you have to know.0381

This is the definition for work it is the pressure × change in volume.0392

If this is a reversible process reversible implies that the P external = P.0396

We can write DW = the pressure of the system × the change in volume.0405

This is a Vandrual’s gas, the pressure = nrt /V - nb - an² / V².0412

We put this into here and we integrate, our total work is going to equal the integral from volume 1 to volume 2 of this expression,0423

the P which is nrt / V - nb - an²/ V².0440

Do not let this intimidate you, this is not a difficult and it is a very simple integral.0449

It was just a logarithmic integral.0453

It just looks complicated.0455

Do not to let how something looks intimidate you to stop and look at it.0457

See what it is, it is very simple integration.0461

You can do it by hand or if you like this use mathematical software either mathematical or maple.0463

DV that is it.0471

When you do this integration, you will end up with following.0473

You end up with work = nrt × log V2 - nb/ V1 - nb + an² × 1/ V2 -1 / V1.0475

Notice, in the last problem when we integrated I had 1/ V -1 – 1/ V2.0501

I did that because the integral itself ended up coming out as -1/ V.0509

I just use that negative sign to flip the order.0514

In this case, I took the negative sign out and I dmded to go ahead and make this a positive.0517

It is not a mistake from previous, it just depends on how you want to express it.0523

Again, a particular mathematical equation, at this level you can simplify to any degree you want.0527

The simplification is irrelevant.0533

What is relevant is that you understand what is going on.0535

If you want to leave this as a minus sign, I might put 1/ V1 – 1/ V2.0538

Just look at the same answer is not a problem.0542

When we put all the numbers in, I’m not going to go ahead and write all the numbers in.0547

It is actually a lot.0550

Maybe I should, it is not a problem, we should see everything.0553

1 × 8.314 the temperature is 298 K × log,0557

We have to express the volume in cubic meters.0569

This is going to be 65 × 10⁻³ -nb -1 × 0.064 × 10⁻³.0574

B was expressed in terms of dm³/ mol, we have to express it in terms of dm³/ mol.0586

Thermodynamics, physical chemistry in general, it is a conversion factor that ends up being tedious and toilsome.0594

So 25 × 10⁻³ - 1 × 0.064 × 10⁻³ we have that.0604

It is going to be +a which is 0.366 that is not a problem × 1² × 1/ V2 which is 65 × 10⁻³ -1/ 25 × 10⁻³.0616

When these numbers are floating around, I do hope that you actually confirm my arithmetic.0638

What you will end up with is 2371.3 + -9.0 and you will end up with the work is going to be 2362.3 J.0644

The system is expanding, the gas is expanding isothermally and reversibly.0663

It is doing work on the surroundings which is why the work is positive.0670

Work done on the surroundings is positive that is why the work is positive.0675

We are looking for Q, it looks like we did δ and we did W, let us do Q.0688

We know that δ U= Q – W.0695

Therefore, Q = δ U + W.0698

Therefore, the heat of this reaction = 9 J which is the δ U, the change in energy + 2362.3 J.0705

Therefore, the heat is 2371.3 J.0718

I apologize sometimes I write Joules and sometimes I just put J.0734

Let us go ahead and see what we can do about the,0739

Let us see, with Q let us find δ H.0745

Δ H here is how you want to put it.0750

The definition of H is U + PV.0754

Again, when the basic equation that you start with in this problem is only a handful of equation that you need to know.0759

Everything else should be derived.0765

Do not think that you can memorize a bunch of equations and to apply them to a situation.0767

This is high level science so the problems are not mechanical.0772

It is not just this is what this problem is, you plug the numbers in.0779

It is not going to work that way.0783

There are ton of equations in thermodynamics but there are a handful of basic equations from which the others come.0785

The derivations are not difficult, as far as the problems are concerned.0791

Do not memorize the equations because they will just make you crazy if you try to memorize.0795

So δ H which is what we are looking for = δ U + PV.0800

The δ operator is a linear operator.0807

So δ H linear just means you can distribute operators symbolically.0809

You can distribute that way, distribute that way, which you will end up with is δ U+ δ PV.0814

Well, that is the same as δ H = Δ U.0823

Put the δ PV, that is just P2 V2 - P1 V1.0829

We have to calculate P2 V2 P1 V1 from Van Der Waals equation.0840

Again, the pressure = Van Der Waals equation = nrt / V - nb - an²/ V2.0850

This is going to be a bit tedious as far as a calculation is concerned but it is not a problem.0861

Let us go ahead and calculate P2 V2 first.0867

We will do that one.0869

It is going to be nrt, this first one I just want to make sure to do everything.0872

Over V - nb – an² / V² that is the V2 part, over V2² × V2.0881

When you put all the numbers, you have all the numbers.0896

You have n, R, T, 298.0899

You have the second volume, you have nb.0901

You have all these values.0902

When you put these values in, you end up with 2473.6 J.0904

You do the same for P1 V1.0914

In this case, P1V1 = nrt/ V1 - nb - an² / V1² × V1.0918

When you put those values in make sure to convert volume to cubic meters.0930

In case of 25 × 10⁻³ m³ is 25 dm³.0936

When you do that, you end up with 2467.2 J.0941

Therefore, P2 V2 - P1 V1 = this – that, you end up with 6.4 J that is P2 V2 - P1 V1.0949

This is the δ PV.0966

We can go ahead and calculate our δ H.0969

Δ H = Δ U + δ PV.0973

Δ H where δ U was 9.0 J + 6.4 J for a total of 15.4 J, that is our δ H.0979

This is not the only way to solve this problem but one of the beauties of science itself is the problem they come up with different solutions.0998

I usually just go with the first solution that occurs to me.1009

It might be a little bit longer or shorter, I do not know.1011

By all means, any other solution that you can come up with, you should examine that and you should explore it.1015

And you should use it because that is the way you see the problem.1023

Arithmetic often get in the way of what is going on so I want to pull back a little bit and just go through what is that we did in this particular problem.1029

The first thing that we did is we calculated the energy.1038

We started off with DU = CV DT + DU DV sub T × DV.1041

They said that this was isothermal so this term goes to 0.1053

Again, we start with the basic equation and we derive the relationship that we needed.1058

DU = DU / DV sub T DV.1063

When we integrate that expression, we get the expression for the full energy for the entire change of state1074

which is going to be integral from V1 V2 of this DU DV sub T DV.1080

When you integrate you put the numbers in.1089

That is how we found the energy and then we dmded to go with the work.1092

The definition of work = P external DV.1096

This one is reversible, P external = P.1102

What we end up with is P = the Van Der Waals gas.1107

Nrt/ V - nb – an² / V².1113

We put this expression into here and then we integrate.1119

When we integrate, the work equals the integral from V 1 to V2 of PDV.1125

We put this expression into here and we integrate it and we get the value for the work and we put the numbers in to actually get the work itself.1133

Number 3 was actually pretty straightforward.1143

We used this basic relationship and the differential version is DQ – DW.1145

The integrated version is δ U= Q – W.1157

These are not exact differentials.1162

An exact differential integrates that δ U.1164

These inexact differentials just go to Q and W.1167

We are looking for Q so I just move things around.1173

You already found this and we just found that.1175

Q = Δ U + W and we ended up finding the heat for the particular reaction.1178

How do we do δ H?1189

Δ H = the change in energy which we have + δ PV.1191

Δ PV = P2 V2 - P1 V1.1199

We used the Van Der Waals equation to find this and to find that, make the subtraction, use that and put it into here and solve for δ H.1207

This was the process that we did and it is all based on a handful equations.1217

This equation, right here, that is our first equation.1222

It is one of the fundamental equations that we have to know in order to solve these problems.1225

The second equation right here is the definition of work.1230

This is the total energy of the system.1232

The energy is dependent on temperature and on volume.1235

For this particular problems, isothermal the DT went to 0 so it only depends on the change in volume.1239

This is the definition of work.1245

This is the first law of thermodynamics.1248

The energy of the system = the heat transfer - the work transferred.1250

Of course δ H comes from the definition of enthalpy which is one of the basic equations.1255

Again H = U + PV that is the definition of enthalpy.1261

Excuse me, let us move on to our next problem here.1271

We have a lot of extra pages.1275

Example number 2, so we have 1 mol of an ideal gas and it is confined under pressure P external.1277

This external should actually be down below, I ended up putting on top.1287

P external = P = 250 kl Pascal.1291

Basically, what this says is that the external pressure and the internal pressure are the same.1296

If you have this gas and this piston set up, this with a little weight on top, this is an equilibrium.1300

It is not moving anywhere.1307

In other words, the pressure this way = the pressure that way, that is all these means.1308

So ideal gas is very important.1312

We are provided with pressure 250 kl Pascal, 250,000 kl Pascal.1317

The temperature is changed from 120 to 25°.1322

The temperature of the gas we drop it from 120°C to 25°C.1326

The constant volume heat capacity= 3 Rn/ 2, n is 1.1332

Here we want to calculate the Q, W, δ H.1339

This is a pretty standard problem, the four basic properties of the system.1342

Notice, they say nothing about a change in volume here.1351

Let us see what we can do first.1355

Let us start with work, if it stays in red it is not a problem.1360

I will start with a definition of work DW = P external × DV.1367

The P external is constant, if it is constant when we integrate this, this comes out of the integral sign.1374

When you integrate this, this comes out of the integral sign and what you end up with is work = P external × the change in volume.1382

Let us see what we have got.1399

We have an ideal gas so we have PV = nrt so V volume = nrt / P.1401

We have two different temperatures, we have 120 and 25.1417

The pressure is actually constant so pressure stays the same.1423

For change in volume, let us do this.1428

Let us find, we have the P external we just need a change in volume.1431

This is P external × V2 - V1.1437

I’m going to start with V1, it is going numerical order here.1448

V1 = nr T1/ P which = 1 × 8.314 × temperature 1, the initial temperature is 120.1451

We have 393 K/ 250,000 Pascal.1471

You end up with a value equal to 0.0 1307 and this is going to be in cubic meters.1486

If you are concerned about what is happening here, let us do a quick unit check.1495

We have mol × J/ K mol × K/ Pascal.1502

This is the same as mol × J is N m/ K mol × K and Pascal is a N/ m².1515

Mol cancels mol, K cancels K, N cancels N.1533

This comes up to the top and ends up with cubic meters is what this comes from.1537

This is cubic meters.1542

I’m going to do the same for V2.1544

Volume 2 that is equal to nr T2/ P.1546

We have 1 × 8.314, 25°C which is 298.1551

Try to keep everything straight is going to be the difficult part.1560

This is 250,000 and it is very easy to actually think that 298 here, and put the 393 here by mistake.1563

Again, it is just that is life, lot of numbers floating around.1573

Equals 0.00991 m³.1578

Our work is equal to P external × V2 - V1 = 250,000 Pascal × C V2 is 0.0091 cubic meters -0. 01307 m³ and we end up with Pascal m³ which is J.1586

Our work is going to end up being -790 J.1629

When you drop the temperature of this gas, this gas is going to contract.1637

It is going to fall.1641

The surroundings are going to do work on the system.1643

When the surroundings do work on the system that means work is leading the surroundings.1648

Our consigned convention work is -790 J.1654

790 J is leaving the surroundings and going into the system, that is what this means.1657

Let us go ahead and do our C.1665

Let us go ahead and do our DH next.1671

We have our basic relationship, one of the equations that we have to know.1674

The differential change in enthalpy = the constant pressure heat capacity× the differential temperature + DH DP T DP.1679

The pressure is constant that means DP = 0.1696

Constant pressure, constant P implies the DP = 0.1703

This term goes to 0 so you are left with DH = CP DT.1708

When you integrate this, CP is constant so what comes out which we are left with is the following equation DH = CP δ T.1720

We have δ T that is just going to be the 120 - 25 so we need CP.1735

They did not give us the CP, they gave the CV.1741

They gave the CV = 3/2 Rn and we have a relationship, this is an ideal gas.1744

An ideal gas there is a relationship between the constant pressure heat capacity and the constant volume heat capacity that is the following.1751

CP - CV = Rn.1761

They gave the CV we know Rn and we can find CP and put it in here to find our value.1765

CP = CV + Rn that is equal to 3/2 Rn + 1 Rn.1773

Therefore, the constant pressure heat capacity= 5/2 Rn.1787

And now we can go ahead and put this into here.1791

Therefore, our δ H = 5/2 Rn δ T.1794

We have all the numbers that we need.1801

Δ H = 5/2 × 8.314 × 1 mol × change in temperature final - initial 25°C -120°C.1805

The change in temperature increment, I did not change this to K because the δ T for Celsius and K is the same.1824

The individual things are not the same 393 and this is 298 but 298 -393 is the same as 25 -120.1831

The δ T because all you are doing you are adding 273 so it goes up equally.1840

And when you do this you end up with δ H = -1975 J.1847

Now under constant pressure δ H is actually equal to QP.1858

I will probably write it the other way QP= δ H.1875

In other words the heat happens to equal the enthalpy so we went ahead and took care of that.1878

The Q for this reaction = -1975 J.1884

The only thing we are missing now is δ U.1891

We have a relationship δ U= Q – W.1894

We just found Q, we found W a little bit earlier it is = -1975 J that is the Q - the W which is -790 J.1899

Our change in energy of the system = -1185 J.1915

There we go and what does all this mean?1925

Here is what is happening.1928

We have our system and we have our surroundings Q that was equal to -1975 J.1930

1975 J left the system.1942

Work = -790 J, the surroundings did 790 J of work on the system.1946

In other words, 790 J went into the system.1959

Therefore, the total change in energy 1975 from the systems point of view 1975 J left, 790 J came in.1962

The total net energy, the net loss by the system is 1185 J net loss by the system.1972

That is what is happening here.1988

The surroundings did 790 J of work.,1991

The system in going from 120 to 25 lost 1975.1995

The net loss is the systems.2000

Let us see what we have got here.2007

Let us go to our next example.2010

Example 3, we have 1 mol of ideal gas.2014

We have 1 mol, we have an ideal gas is taken from 10°C to 90°C under constant volume.2017

It is very important.2026

This time, our constrained is under a constant volume.2027

Our constant volume heat capacity is 21 J/ mol K.2030

This is actually a molar heat capacity.2035

21 J/ mol K, calculate Q, W, Δ U, and δ H.2037

Let us see what we have got.2045

Let us see which one do we do first.2056

Constant volume implies that DV = 0.2067

If you remember, under constant volume our Q = δ U.2078

Our equation DU = CV DT + DU DV.2086

Constant volume DV equal 0 but we are dealing with an ideal gas so this is already 0.2106

In either case, that one goes away.2110

Our DU which = QV = CV DT.2113

We just go ahead and integrate this.2123

Therefore, when we integrate this we end up with the following.2127

You end up with δ U= Q = CV δ T.2129

When we integrate this, this is constant so it comes out.2137

The integral of DT is δ T.2139

We have the CV, we have the δ T, so this is really easy.2142

Our DU = CV which we said was 21.0 J/ K and our δ T is 90°C -10°C.2147

And when we run this calculation, we end up with 1680 J.2175

That is δ U= 1680 J also happens to equal Q= 1680 J.2188

We have taken care of the Q and the δ U, let us go ahead and take care of the constant volume.2200

We did that.2211

Let us go ahead and do δ H next.2216

Actually, if we want in this particular case because constant volume, no work is done.2219

PV so work = 0.2235

Let us go ahead and do δ H.2242

Again, we have δ H = Δ U+ δ PV equal to,2244

We have to watch out if a constant volume but V is constant, therefore, let us change this equation a little bit.2263

That means δ H is going to equal Δ U V comes out.2274

V × δ P just the change in pressure.2280

We have δ H or a lot of crazy things floating around.2289

We have δ H = δ U+ V × δ P or ideal gas so PV = nrt.2295

Therefore, the pressure = nrt / V.2307

The δ H = δ U+ V δ P well δ P is P2 - P1.2315

P2 = n R T1/ V nr T2/ P.2328

I'm sorry for V, I’m getting all white variable here.2337

Remember, volume is constant so it stays V.2343

It is not V1 or V2, temperature 2.2344

Nr here is the V and then P1 = nr T1 / V.2348

These cancel out so you get δ H = Δ U+ nr × T2 - T1.2355

We have T2 and we have T1.2365

Therefore, when we put these values in we get δ H = δ U which was 1680 J + 1 mol.2368

R is 8.314 and we have 90°C -10°C.2380

Again the difference in temperature, δ T in Celsius is the same as the δ T in Kelvin.2388

When we run this calculation, we end up with 2345 J is the δ H.2394

Let me actually do these individually.2409

Here we have the 1680, this particular value ends up being 665 J.2412

We took it from 10°C to 90°C so what is going on here is the 1680 J, that much energy comes from the rise in temperature.2428

The 665 J it comes from the rise in pressure.2449

The δ H, the enthalpy is an accounting device.2462

It accounts for the energy change + any change in pressure, volume, work.2465

A change in pressure, change in volume, things like that.2471

1680 if it comes from just the rise in temperature, 665 come from the actual rising pressure because we are keeping at a constant volume.2475

That is what is happening.2483

We keep it at constant volume.2483

It cannot do work this way so the pressure increases inside that increases the energy of the system.2485

You always want to stop, take a look at the numbers, see what they mean in terms of the direction of flow positive or negative,2492

and see if you can actually identify what value is coming from what.2499

Let us move on to example 4, let us see what we have here.2512

Find δ U and δ H for the transformation of 1 mol of ideal gas from 25°C and 1 atm.2518

They give us an initial temperature and initial pressure to a final temperature and a final pressure.2525

The molar heat capacity of the gas CP.2531

This is a CP, CV is 20 + 0.045 J/ mol K.2537

I will make one little correction here.2544

20 + 0.045 × T, T is the variable.2548

In this particular case, the heat capacity is temperature dependent.2553

It starts at 20 but as the temperature rises the heat capacity changes.2557

This is not constant heat capacity, this is something we are going to have to integrate over a change in temperature.2564

I apologize for that mistake so this should be 20 + 0.045 × T.2569

Let me write it here CP = 20.0 + 0.045 T.2577

T is a variable like x.2585

In this particular case you can do Δ U first, you can do δ H first.2589

It does not really matter.2592

Let us go ahead and start off with a basic equation.2594

I’m going to do δ U first.2598

Δ U is equal to CV DT + DU DV T DV.2600

This term goes to 0 because the gas is ideal.2611

An ideal gas the DU DV = 0.2615

What I'm left with is DU = CV DT.2618

When I integrate that, of course I will have to integrate that, I will be left with δ U= the integral from temperature 1 to temperature 2 of CV DT.2628

I do not have CV DT, I have CP DT but it is an ideal gas I know that there is a relationship.2643

I know that CP - CV = Rn.2647

Therefore, CV = CP – Rn.2653

CP = 20.0 + 0.045 T - 8.314 × 1 mol.2664

For my constant volume heat capacity I end up with some 20 - 8.314, I end up with 11.686 + 0.045 T.2682

This is the value that I put in here.2695

Therefore, δ U= the integral from 298 K.2699

These cannot be a Celsius, you have to put these in K.2706

298 K to 225 which is 498 K of the constant volume heat capacity which was 11.686 + 0.045 T DT.2710

I will go ahead and just use math software and if it did it correctly I ended up with 5919 J that is our δ U.2728

Let us go ahead and do a δ H for this.2743

Δ H again, it is up to you how you want to do this.2746

I’m going to do it with a long way with DU and δ PV.2750

You already know that DH = CP DT + DH DP T DP.2755

This is one of the fundamental equations for enthalpy.2769

This is an ideal gas so this goes to 0.2774

Our δ H = CP DT from T1 to T2, this is one way you can actually calculate the enthalpy.2777

I did it another way, I did it with a basic equation by the definition of enthalpy DU + D PV.2790

This is probably the quicker method.2800

This is just one that I happen to have gotten used to so I just tend to automatically default to that one.2802

But either one is fine however you want to see it.2807

This = δ U + P2 V2 - P1 V1 and = Δ U.2812

P2 V2 is an ideal gas PV = nrt.2824

P2 V2 = nr T2 – nr T1.2829

You end up with δ U + nr × T2 - T1.2837

We have T2 and T1 so you end up with δ U which was 5919 + 1 mol ×,2844

I really need to slow this down + 1 × 8.314 × 225 -25 again δ T.2855

225 -25 I can leave that δ H = 7582 J.2870

There you go, nothing strange going on here.2884

P2 V2 P1 V1.2891

This said nothing about volume being constant so despite the fact that this is an ideal gas, P2 V2 - P1 V1 is not 0 because temperature changes.2897

That is what is going on here, be very careful because sometimes you can go with P2 V2 – P1 V1.2913

If other things are the same, if this were isothermal that would not be a problem.2919

If this were isothermal then Q would be constant.2923

P2 V2 would equal nrt and P1 V1 equal nrt.2928

Therefore, P2 V2 – P1 V1 equal 0.2932

But temperature is not constant here so again what the problem says very careful on how you approach it, which is why you do not want to memorize equations.2936

This is an excellent example.2947

If you just memorize the equation you will just plug numbers in without stopping to save yourself and say this is not isothermal situation.2948

If it were an isothermal situation, this term will go to 0.2956

If not isothermal, therefore, this is not 0.2959

You have to run the calculation.2961

You have to be very careful, go slowly.2963

That takes care of that one.2967

Let us go ahead and go to example number 5, it should be our final example for this particular lesson.2969

Let us see.2975

When an ideal gas undergoes an expansion such that the relationship pressure × volume raise to the nth power = a constant2979

or C is a constant and n is greater than one, it is called a reversible poly tropic expansion.2987

We would like to calculate the work for such an expansion of 1 mol of the gas from V1 to V2.2994

When the temperature 1 is 350 K and temperature 2 = 200 K, and n=2.3001

We would like you, given that the constant volume heat capacity= 5/2 Rn or 5n/ 2.3010

We want you to find Q, Δ U, δ H, for this particular expansion.3016

Let us see what we have got.3022

Here is a lot going on here.3023

A new term poly tropic expansion, here we are given a relationship that the pressure × the volume raised to the nth power is actually a constant.3025

Again, this is where the things we just have to dive in and see what is going on.3036

There is only a handful of equations that you know, you want to start with those and see where it takes you.3042

I think I will go ahead and go back to blue for this.3051

Part A, telling me that n = 2, I have PV² = a constant.3055

I know that the pressure is equal to C/ V², C being a constant.3067

My work = I know my basic relationship.3074

Our basic relationship by definition is DW = P DV.3081

Therefore, my work = the integral from V1 to V2 of P DV.3089

I have P here, the relationship is C/ V² so I can just plug this into here and integrate.3097

Work = the integral from one volume to another volume of P DV which P = C / V².3105

This is C/ V² DV which = C × the integral from V1 to V2 of DV / V².3115

And I'm left with C × 1 /V1 -1 / V2 that is great, I have expression for the work = C × 1/ V1.3126

That is what they wanted, calculate the work for such an expansion.3138

However, they gave us temperatures here so it looks like they want an actual numerical value not just some expression.3142

Let us see what we have got.3154

V = C/ nrt so let us see if we can play with this a little bit.3161

We have work = constant × 1/ V1 -1/ V2.3170

We have that P = C/ V², this is PV = nrt.3181

I’m going to go ahead and put this P value of P into here and I end up with C/ V² × V = nrt.3198

This V cancels that V so I’m left with C/ V = nrt.3212

That means V = C/ nrt.3222

That means that V1 is equal to C/ nrt 1 and V2 = C/ nr T2.3231

I can go ahead and put these values into here and here.3244

I got work = C × 1/ C/ nr T1 -1/ C/ nr T2.3248

Work = C × nr T1/ C-1.3270

This is working out really great.3277

C cancels therefore our work = nr × T1 - T2.3280

This is fantastic, we have T1 and T2.3293

We have n and R so work = 1 mol × 8.314 J/ K mol × temperature 1 was 350 K and temperature 2 was 200 K.3295

And therefore, for work for this process is 1247 J.3314

That was fantastic.3321

Work = 1247 J.3324

Let us go ahead and see what we can do with DU.3327

DU= CV DT so δ U = CV δ T well that is equal to 5/2 Rn δ T.3331

Therefore, δ U= 5/2 × 8.314 × 1 mol × δ T.3351

Δ T is final - initial so we have now 200 -350.3366

Therefore, the change in energy = -3118 J.3374

That takes care of the energy.3384

We have a relationship Δ U= Q – W.3386

Therefore, Q = Δ U+ W so Q = δ U which is - 3118 J + 1247 J.3392

Therefore, heat is -874 J.3408

Let us see what we can do with δ H.3417

Δ H = Δ U + δ PV and again this is as my default, I automatically go to this.3421

It is equal to δ U + P2 V2 - P1 V1.3429

We have PV² = a constant therefore P = C / V².3444

Therefore, P2 V2 = C/ V 2² × V2.3453

This V2 cancels one of the V2 so I'm left with C / V2 and P1 V1 analogously is C/ V1.3465

We have δ U + C/ V2 - C/ V1 = δ U + C / nr T2 –C/ nr T1.3487

We end up with δ U+ nr this time T2 - T1 = - 3118 J + 1 × 8.314 × 200 -350 = -3118 J + -1247 J.3517

Our final δ H = -4365 J there we go.3549

A long process and a little tedious and a little involved.3564

But again you are starting off with a basic set of equations and any other information that they gave you that PV ⁺nth power = C.3571

They gave you n = 2.3579

You are going to use that to start fiddling around with it.3582

This relationship holds, this is an ideal gas so this relationship holds.3586

You can start fiddling with these and playing around with the equation.3593

And again, this is not necessarily the only way to do this problem.3598

Maybe there are other ways to do this problem I do not know.3601

I just sort of jump right out on in and did it this way.3603

And now everything seemed work out okay.3606

I will go ahead and leave it like that.3609

Thank you for joining us here at www.educator.com.3612

We will see you next time for a continuation of some more example problems.3614

We want to make sure to get this really nice and tight.3618

Take care, bye.3622

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