Raffi Hovasapian

The Relationship Between Cp & Cv

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Table of Contents

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

Physical Chemistry The Relationship Between Cp & Cv

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The Relationship Between Cp & Cv

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Intro

The Relationship Between Cp & Cv

When Cp is Greater than Cv

Constant P Process: 3 Parts

Define : γ = (Cp/Cv)

Intro 0:00
The Relationship Between Cp & Cv 0:21

For a Constant Volume Process No Work is Done
For a Constant Pressure Process ∆V ≠ 0, so Work is Done
The Relationship Between Cp & Cv: For an Ideal Gas
The Relationship Between Cp & Cv: In Terms of Molar heat Capacities
Heat Capacity Can Have an Infinite # of Values
The Relationship Between Cp & Cv

When Cp is Greater than Cv 17:13

2nd Term
1st Term

Constant P Process: 3 Parts 22:36

Part 1
Part 2
Part 3

Define : γ = (Cp/Cv) 28:06

For Gases
For Liquids
For an Ideal Gas

Physical Chemistry Online Course

Section 1: Classical Thermodynamics Preliminaries
	The Ideal Gas Law	46:05
	Math Lesson 1: Partial Differentiation	46:02
Section 2: Energy
	Energy & the First Law I	1:06:45
	Energy & the First Law II	1:06:33
	Energy & the First Law III	1:02:17
	Changes in Energy & State: Constant Volume	1:04:39
	Joule's Experiment	16:50
	Changes in Energy & State: Constant Pressure	43:40
	The Relationship Between Cp & Cv	32:23
	The Joule Thompson Experiment	39:15
	Adiabatic Changes of State	35:52
Section 3: Energy Example Problems
	1st Law Example Problems I	42:40
	1st Law Example Problems II	1:00:23
	1st Law Example Problems III	44:34
	Thermochemistry Example Problems	59:07
Section 4: Entropy
	Entropy	49:16
	Math Lesson II	33:59
	Entropy As a Function of Temperature & Volume	54:37
	Entropy as a Function of Temperature & Pressure	31:18
	Summary of Entropy So Far	23:06
	Entropy Changes for an Ideal Gas	25:42
Section 5: Entropy Example Problems
	Entropy Example Problems I	43:39
	Entropy Example Problems II	56:44
	Entropy Example Problems III	57:06
Section 6: Entropy and Probability
	Entropy & Probability I	54:35
	Entropy & Probability II	35:05
Section 7: Spontaneity, Equilibrium, and the Fundamental Equations
	Spontaneity & Equilibrium I	28:42
	Spontaneity & Equilibrium II	34:38
	The Fundamental Equations of Thermodynamics	30:50
	The General Thermodynamic Equations of State	34:06
	Properties of the Helmholtz & Gibbs Energies	39:18
	The Entropy of the Universe & the Surroundings	19:40
Section 8: Free Energy Example Problems
	Free Energy Example Problems I	54:16
	Free Energy Example Problems II	31:17
	Free Energy Example Problems III	45:00
	Three Miscellaneous Example Problems	58:05
Section 9: Equation Review for Thermodynamics
	Looking Back Over Everything: All the Equations in One Place	25:20
Section 10: Quantum Mechanics Preliminaries
	Complex Numbers	34:25
	Probability & Statistics	59:57
	SchrÓ§dinger Equation & Operators	42:05
	SchrÓ§dinger Equation as an Eigenvalue Problem	30:26
	The Plausibility of the SchrÓ§dinger Equation	21:34
Section 11: The Particle in a Box
	The Particle in a Box Part I	56:22
	The Particle in a Box Part II	45:24
	The Particle in a Box Part III	48:43
Section 12: Postulates and Principles of Quantum Mechanics
	The Postulates & Principles of Quantum Mechanics, Part I	46:18
	The Postulates & Principles of Quantum Mechanics, Part II	39:28
	The Postulates & Principles of Quantum Mechanics, Part III	35:32
	The Postulates & Principles of Quantum Mechanics, Part IV	29:55
Section 13: Postulates and Principles Example Problems, Including Particle in a Box
	Example Problems I	54:25
	Example Problems II	46:58
	Example Problems III	44:11
Section 14: The Harmonic Oscillator
	The Harmonic Oscillator I	35:33
	The Harmonic Oscillator II	43:04
	The Harmonic Oscillator III	26:30
Section 15: The Rigid Rotator
	The Rigid Rotator I	41:10
	The Rigid Rotator II	30:32
	The Rigid Rotator III	35:19
Section 16: Oscillator and Rotator Example Problems
	Example Problems I	33:48
	Example Problems II	46:43
Section 17: The Hydrogen Atom
	The Hydrogen Atom I	40:00
	The Hydrogen Atom II	35:58
	The Hydrogen Atom III	36:18
	The Hydrogen Atom IV	33:55
	The Hydrogen Atom V: Where We Are	51:53
	The Hydrogen Atom VI	51:53
	The Hydrogen Atom VII	34:29
Section 18: Hydrogen Atom Example Problems
	Hydrogen Atom Example Problems I	43:49
	The Hydrogen Atom Example Problems II	1:01:57
	The Hydrogen Atom Example Problems III	48:33
	The Hydrogen Atom Example Problems IV	48:33
	The Hydrogen Atom Example Problems V	48:33
Section 19: Spin Quantum Number and Atomic Term Symbols
	Spin Quantum Number: Term Symbols I	59:18
	Spin Quantum Number: Term Symbols II	34:54
	Spin Quantum Number: Term Symbols III	38:03
	Term Symbols & Atomic Spectra	57:49
Section 20: Term Symbols Example Problems
	Example Problems I	1:01:20
	Example Problems II	56:34
Section 21: Equation Review for Quantum Mechanics
	Quantum Mechanics: All the Equations in One Place	18:24
Section 22: Molecular Spectroscopy
	Spectroscopic Overview: Which Equation Do I Use & Why	50:02
	Vibration-Rotation	59:47
	Vibration-Rotation Interaction	46:22
	The Non-Rigid Rotator	29:24
	The Anharmonic Oscillator	30:53
	Electronic Transitions	1:01:33
Section 23: Molecular Spectroscopy Example Problems
	Example Problems I	33:47
	Example Problems II	1:01:05
	Example Problems III	33:31
Section 24: Statistical Thermodynamics
	Statistical Thermodynamics: The Big Picture	1:01:15
	Statistical Thermodynamics: The Various Partition Functions I	47:23
	Statistical Thermodynamics: The Various Partition Functions II	54:09
Section 25: Statistical Thermodynamics Example Problems
	Example Problems I	48:32
	Example Problems II	57:30

Transcription: The Relationship Between Cp & Cv

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

In the last couple of lessons we have introduced this thing called the constant volume heat capacity C sub V.0005

And we have introduced this thing called the constant pressure heat capacity C sub P.0011

Let us see if we can elucidate the relationship between the two.0016

First of all, know that numerically the constant pressure heat capacity is greater than the constant volume heat capacity.0024

The qualitative reason for that is, the intuitive reason is this, for a constant volume process no work is done because the change in volume is 0.0032

Remember, work is pressure times the change in volume.0043

The change in volume is 0, the work is 0.0045

Therefore, all of the heat transfer during the process goes just straight into the chaotic motion of the molecules of the system.0049

All of the heat transfer during the process accounts for the energy change that is why we have that0059

the heat transfer during constant volume process = a change in energy.0066

All of the heat goes to change in energy.0073

For constant pressure process, the change in volume is not 0 so work is done.0077

Therefore, the heat transfer to the process, some of this energy that is transferred as heat is used to perform work.0083

Heat goes in and a part of that heat is used to do work in the expansion leaving δ U to be less than expected.0098

Thus, the change in energy is equal to the heat transfer which is similar to this - the energy lost as pressure volume work.0107

The change in energy is going to be less than heat transferred or is for constant volume process, all of the heat accounts for the energy change.0117

If the change in energy is less than expected, the change in temperature is going to be0127

less than expected because energy and temperature are the same essentially.0131

That is what energy is a function of temperature so if energy is less that means the temperature of the system is less.0137

In order to achieve the same unit change in temperature, the same 1° increase in temperature,0145

more heat has to be transferred during the constant pressure process.0151

The more heat for the same change in temperature means a higher heat capacity.0156

The constant volume heat capacity, a certain amount of heat is transferred, it is all energy.0162

In a constant pressure process, a certain amount of heat is transferred of that heat, some of it is actually used to expand the gas.0167

My energy is less, my temperature is less.0176

In order to achieve the same change in temperature, I have to put more heat in.0179

Therefore, my constant pressure heat capacity is greater than my constant volume heat capacity.0184

Under conditions of constant pressure, I have to put more heat to achieve the same 1° change in temperature.0190

It is the qualitative reason so you should be able to qualitatively, intuitively, that is why it is true.0199

Let us go ahead and do a little bit of work here.0207

For an ideal gas, in other words let us find out what this relationship between CP and CV actually is.0212

Let us find an equation.0219

For an ideal gas it is really simple, there are several ways to get this.0221

I just decided to take what I consider to be the easiest, quickest approach.0230

Let us begin with the definition enthalpy.0235

Begin with the enthalpy is equal to the energy + PV.0240

For ideal gas, we know that PV = nRT so we can just go ahead and substitute that into that and we get that H =U + nRT.0252

Let us go ahead and take the derivative of both sides with respect to T.0265

We get DH DT = DU DT = nR.0269

We know what DH DT is DHDT is the constant pressure heat capacity, the change in enthalpy per unit change in temperature.0279

We know what DUDT is, the change in the energy vs. the unit change in temperature, this is the constant volume heat capacity.0288

Let us keep our = and + separate here, that is our relationship.0304

For an ideal gas, its constant pressure heat capacity is equal to its constant volume heat capacity + the number of mol × R or you can write it this way CP - CV = nr.0311

In terms of molar heat capacities, molar quantities is per mol, which means just divide everything by n, dividing everything by the number of moles.0331

When I divide everything by n, what I get is the following.0341

In terms of molar heat capacities, CP this is J/ K.0346

When we divide by mol we get J/ K/ mol, the one that we are used to seeing.0360

In terms of molar heat capacity, divide everything by n and what you get is CP/ N + CD/ n = R.0366

All molar variables just put a line over them.0386

I’m missing a bunch of + and - signs here, - there you go - CV = R, these are our relations right here.0395

You can think of it this way, it does not matter.0407

Any time you see a line over something it is a molar, that means that you just divide everything by n, the number of mol.0409

There you go.0415

To for an ideal gas, this is the relationship the constant pressure heat capacity is equal to the constant volume heat capacity + nr or CP - CV = nr.0416

Either one is fine depending on how you want to use it.0427

In general, if the system has a differential change in temperature that is associated with a change of state,0433

if it is going from state 1 to state 2, there is some change in temperature for the system.0463

Then Q, the amount of heat transfer, the amount withdrawn from the surroundings0474

then Q can have an infinite number of values and because heat Q is a path function.0483

Temperature is a state function.0510

If I want to take something from 10° C to 25° C, the 25° change I can do a whole bunch of ways.0512

I can just go straight to it, I can go off to 100 and other 25, I can go down to 100 then back to 15 then to 25.0520

The heat transfer as the heat changes the temperatures itself is a state function.0529

Q was a path function.0538

Therefore, the heat transfer can have an infinite number of values going from state 1 of one temperature to state 2 of different temperature.0540

There are a bunch of ways that heat can transfer.0548

The heat transfer can have an infinite number of values.0551

Heat capacity is defined as the change in heat per a change in temperature.0559

If the quantities for heat have an infinite number of values, the truth is for a given change in temperature0570

there are an infinite number of heat capacities.0577

Heat capacity can also have an infinite number of values for a particular temperature change because0584

they are a bunch of paths in order to get from one temperature to another.0600

Only two heat capacities are generally important, heat capacity under constant pressure and heat capacity under constant volume.0606

When we put these two constraints on there, those are the two that matter not the infinite variety of them.0621

In another words, we want to go from temperature 1 to temperature 2 under constant volume.0629

We want to go from temperature 1 to temperature 2 under constant pressure.0633

In both of those cases, there is heat capacity associated with both of the paths, with both of those processes.0637

Only 2 heat capacities are generally important.0643

But it is important for you to know again, that heat capacity there is an infinite number of them depending on the path that you take.0650

It depends on the heat transfer.0656

The heat capacity is defined as the heat transfer per unit change in temperature.0658

If there is an infinite number of heat, heat has to be transferred as an infinite number of heat capacities CP and CV.0662

There you go, constant pressure and constant volume heat capacity.0677

Let us go back.0681

Let us return to our mathematics, DU = DU DT VDT.0684

Let us see if I can keep it all straight, DU/ DV constant TDV.0700

We also have DU is equal to DQ - DW which is equal to DQ - P external DV.0708

The definition of work is pressure × volume.0728

We have our equation DQ - P external DV = this side is going to be DU/ DV constant V × DT + DU/ DV constant T × DV.0734

Let us see what we can do.0760

At constant pressure under conditions of constant P, the P of the system is equal to P external.0763

I put P over here and move it over the other side and I get the following.0783

I get DQ sub P because now we are at under constant pressure, DQ without the subscript it just means that heat transfer.0788

DQ sub P is under conditions of constant pressure is equal to DUDT sub V × DT was DUDV under TDV + P DV.0797

I just put this into here and moved it over to that side.0824

I’m going to go ahead and I have a DV here and here.0828

Let me combine some terms.0831

I have got DQP is equal to CVDT because the DUDT is equal to the constant volume heat capacity + DU DV T + P × DV.0834

Now I'm going to divide DT.0857

If I divide everything by DT, I get the following.0864

I get DQP DT this is just mathematical manipulation.0868

It is just people playing with symbols.0877

They did not know where they are going when they are doing it, they just started doing it.0882

It looks like we know what we are doing because we have the result of their fruitful work.0886

We have CV, DT DT is just 1 + DU DV sub T + P × DV DT under constant pressure conditions.0894

What it is that we have got, DQ DP DQ PDT this is a constant pressure heat capacity.0929

We have constant pressure heat capacity = constant volume heat capacity + DU DV T + P × DV.0938

I’m trying to keep this all straight is not easy.0960

There you go, this is the relationship that we are looking for.0963

This expresses a relationship between the constant pressure heat capacity and the constant volume heat capacity.0966

The constant volume heat capacity + something accounts for the constant pressure heat capacity.0974

Let us talk about what this something is.0981

You can move the CV over there to write in a different way.0984

I'm going to go ahead and actually multiply and distribute this over this and over this and write it this way.0989

CP =, let me go ahead and first to bring the CV over for our discussion.0998

I'm going to get DU DV under constant temperature conditions × DVDT, our constant pressure + P × DVDT under constant pressure conditions.1010

The amount by which CP is greater than CV has two components.1036

This is one component that is the other component.1054

I’m going to the second term first.1064

Let me actually rewrite it here so I have on top.1067

I have got CP - CV = DU/ DV under conditions of constant pressure × DVDT under conditions of constant pressure + PDVDT conditions of constant pressure.1071

That is the second term first.1091

This term right here, the second term.1094

This is the work that the system does on the surroundings per unit change in temperature at constant pressure, constant P.1096

Just take a look at the units.1127

We have pressure × volume divided by temperature.1129

Pressure × volume is work per unit temperature.1136

First term, the first term is a little stranger.1143

It is a little harder to wrap your mind around this one.1148

In this particular case, I’m going to ask to take it off.1150

This is the change in energy per unit change in volume under conditions of constant pressure × a change in volume1154

or the change in temperature under conditions of constant pressure.1161

This is the energy required to pull molecules apart against intermolecular attractive forces.1166

This term right here, this accounts for the amount of energy it takes to actually pull molecules apart against attractive forces1207

and separate them by a certain distance.1217

In the active separating at a certain distance, the gas is going to expand.1219

If you are taking a molecule, collection of molecules, are pulling further way from other molecules the gas is going to expand.1223

That expansion is going to do work on the surroundings.1230

Some of the energy is lost in actually pushing the atmosphere away as the gas expands.1234

That is the difference between the two heat capacities, two of the terms of the energy that transfers heat.1241

Some of the energy goes to pull the molecules apart against intermolecular forces.1251

Some of the energy goes to doing pressure and the process of pulling them apart against the atmosphere that you are pushing.1258

The rest of it is leftover, that is your CV component, that is the part that goes straight into the random motion of the molecules.1266

The temperature changes accounted by this term not by this term and not by this term.1275

We will say more about it in just a second.1281

Let me rewrite this way so CP = CV + DU DV constant pressure DVDT, what makes sense is the amount of energy to have to put the system1286

in order to change its volume or separate, that is what change in volume is.1310

You are separating gas molecules and then the amount of separation per unit change in temperature.1315

It is the amount of energy that takes per change in temperature was accompanied by a change in volume.1323

In other words, the amount of energy I have to take in order to take this molecule and pull them apart because it requires energy,1329

because of the inter molecular attraction.1335

Under constant pressure + the pressure × the change in volume over the change in temperature at constant pressure.1339

For a constant pressure process, for constant P process, the heat was transferred per unit T change in temp per unit T change.1350

The heat transferred per unit change in T, in other words the CP heat capacity is divided into three parts.1393

Part 1, did not do this in order for the equation is concerned.1419

So one part pulls the molecules apart.1426

Let us slow down, pulls the molecules apart and separates them.1440

Another part, does work on the surroundings in the process of separation, that is what we call expansion is.1452

The third part, is the actual increase in chaotic motion.1488

In other words the δ U, only this part reflects the temperature change.1508

Let me write the equation again, the constant pressure heat capacity is equal to the constant volume heat capacity + DU DV1531

under T and this is DVD T under P + P × DVDT under P.1549

The constant pressure heat capacity is divided into three parts, the amount of heat transfer per unit change in temperature.1562

One part of it does work on the surroundings.1571

One part pulls the molecules apart and separates them, that is this part.1573

That is the amount of energy it takes to actually pull molecules apart against their intermolecular forces.1579

Another part does work on the surroundings in the process of separation.1588

In the process it pulled the energy it takes to pull it apart and separate them, that separation, that pushing or pulling molecules away from each other,1591

in order to pull it this way, I have to push it against the atmosphere.1600

That ends up doing work on the surroundings.1604

What is left over the third part is the actual increase in the chaotic motion.1607

This is what accounts for the δ U, this is what accounts for the temperature increase.1612

This is not change temperature.1618

This is where the temperature change happens.1622

Therefore, in order to achieve the same amount of temperature increase as a constant volume process, the constant pressure process has to put all this.1625

You have to put this much + this much + this much, more heat is required in order to achieve1634

the same 1° temperature rise which is the definition of heat capacity.1642

This is the relationship.1646

It is very important.1648

Again, there is nothing here that is counterintuitive.1650

You get it now, you understand where this is coming from.1652

Under a constant volume process, I have to put a certain amount of heat under constant pressure process,1656

a certain heat is transferred but all the heat that is transferred some of the heat has to go to pulling this molecules apart.1662

In the process of pulling them apart, some of that heat has to be converted to work to actually push the atmosphere away so that we can pull the molecules apart.1670

The rest of it just goes to the energy of the system.1678

That is what is happening here.1682

Let us go ahead and finish up with a couple of definitions.1688

Excuse me, we are going to define this thing called gamma.1691

Let us make it a little bit better, you will see it every so often.1700

Gamma is equal to, it is a ratio of the constant pressure heat capacity to the constant volume heat capacity.1705

This is going to be greater than 1.1712

You knew this already because CP is greater than CV.1714

For gases, the difference between CP and CV is significant, of course it is, because gases expand.1719

For liquids and solids, because the change in volume is so small, it is not nearly 0.1748

It is small but we do not say it is close to 0 because it is reasonably significant.1771

Because the change in volume is so small CP is approximately equal to CV.1776

Tabulated values, the values that you see for heat capacity for liquid and solids in your books and in all the table that you read, that constant pressure heat capacities.1783

Tabulated values for liquids and solids are constant pressure heat capacities and the reason is it is very easy to measure.1798

We just do it under atmosphere conditions because these are easy to measure experimentally.1818

Not quite so easy to do constant volume for a liquid or solid.1836

This is kind of messy actually.1844

For an ideal gas, as we said the CP - CV = nr or for molar CP - CV = R, that is just heat capacity per mol.1847

This is actually good approximation for real gas as well.1873

When you are dealing with the real gas and if you are given the constant pressure heat capacity and1891

you need the constant volume heat capacity, just go ahead and solve this equation.1897

That is fine, for all practical purposes.1902

Real gases under conditions of low pressure and high temperature they behave ideally which is why we use the Pv=nRT unless, we are doing really precise work.1906

For a real gas, that is a good approximation for a real gas also.1916

There you have it, there is the relationship between the constant pressure heat capacity and the constant volume heat capacity.1923

Who know that there was a relationship?1929

You know you probably never thought that there is actually an infinite number of heat capacities but there are.1931

There are only two we are concerned with constant pressure and constant volume.1935

Thank you so much for joining us here at www.educator.com.1940

We will see you next time, bye.1942

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ISBN: 0935702997

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