Raffi Hovasapian

Changes in Energy & State: Constant Pressure

Slide Duration:

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 Changes in Energy & State: Constant Pressure

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Section 2: Energy: Lecture 6 | 43:40 min

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1 answer

Last reply by: Professor Hovasapian
Fri Feb 26, 2016 1:32 AM

Post by Van Anh Do on February 14, 2016

Professor Hovasapian,

How can we have the second term of the equation at 27:45? I thought we got rid of it when we assume at P is constant that's why we only have dH= dQ= (dH/dT)dT then we multiply the dT on both sides to get dQ/dT= Cp? So shouldn't that term be gone if Cp is there?

Thank you! Great lectures!

1 answer

Last reply by: Professor Hovasapian
Sun Sep 13, 2015 9:15 PM

Post by Sam Albert on September 11, 2015

Professor Hovasapian,

I would like to extend my great appreciation to the quality of the explanations that you have given thus far in the Physical Chemistry series. I recently took a full year of physical chemistry and chemical thermodynamics at Oregon State University and struggled immensely. I am in my final year and a half of undergraduate studies and I want you to know that YOU have given me the confidence to tackle this subject matter. Our schooling and education system needs more people like yourself to inspire and encourage struggling students to not give up. Someday I hope to make the impact that you have surely made on math, chemistry, and physics education.

1 answer

Last reply by: Professor Hovasapian
Fri Dec 26, 2014 9:38 PM

Post by tiffany yang on December 26, 2014

Hi Raffi! I was wondering if I get this correctly....
1. All the calculation on enthalpy starts from measuring heat of formation at a specific temperature....which will help us figure out the heat of formation of this molecule at a different temperature
2. the measurement of the heat of formation at that temperature allows us to calculate heat of reaction at that temperature....
3. and figuring out the heat of reaction at that temperature,will help us figure out the heat of reaction at a different temperature.

Is this sort of the purpose of measuring heat of formation ? Thanks Raffi! You're amazing.

Changes in Energy & State: Constant Pressure

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

Changes in Energy & State: Constant Pressure

Important Equations

Example I: Change in Enthalpy (∆H)

Example II: Change in Internal Energy (∆U)

Intro 0:00
Changes in Energy & State: Constant Pressure 0:20

Integrating with Constant Pressure
Defining the New State Function
Heat & Enthalpy of the System at Constant Pressure
Finding ∆U
dH
Constant Pressure Heat Capacity

Important Equations 25:44

Important Equations
Important Equations at Constant Pressure

Example I: Change in Enthalpy (∆H) 28:53
Example II: Change in Internal Energy (∆U) 34:19

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: Changes in Energy & State: Constant Pressure

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

In the previous couple of lessons, we talked about the changes of energy and state and we did them under constant volume.0005

We are going to hold something else constant.0010

We are going to see what happens to the energy of the system when we hold the pressure constant.0013

Let us jump right on in.0018

Most law procedures are conducted under constant pressure conditions.0024

Most law procedures are carried out under conditions of constant pressure, since it is a lot easier that way.0035

One of the experiments that you do end up benched up, you have the constant pressure and the atmospheric pressure.0044

The reaction runs and the system changes pressure.0050

It accommodates that pressure inside the system.0054

It is pretty much going to be the same as the pressure outside.0056

There is always an equilibrium with that.0059

Constant pressure conditions are very simple experimentally.0061

Most law procedures are carried out under constant pressure conditions, under constant P conditions which is atmospheric pressure.0066

The pressure which happens to be the pressure of the system, happens to equal the external pressure, that is all that means.0093

Let us start, we are just playing with mathematics, that is what we are doing.0101

We are hoping that manipulating the mathematical expressions will show us something, will give us something that we recognize.0107

Maybe we will see something that we can move this around.0116

Let us start with, DU =DQ – DW.0121

DW = P external × DV.0134

DU =DQ and again we are working under constant pressure conditions - P DV.0144

Since P is constant, this is just the first law of thermodynamics.0155

It is constant and upon integrating this, we get the following.0163

We get integral from an initial state to a final state of DU = integration from an initial to a final state of DQ sub P - P × the integral from initial state of DV.0177

Pressure is constant, we can pull it out from the integral.0195

We end up with the following.0200

This is going to be, instead of using the δ symbol, I’m not going to write it explicitly.0202

This U as a state functions, so this becomes U2 – U1.0208

This is a path function so it just becomes Q sub P, the heat withdrawn from the system0214

at constant pressure is going to be - P × volume is a state functions so this is going to be V2 - V1.0220

When I distribute this and move it over to the left, I get the following.0231

I get U2 – U1 + P V2 - PV 1 = QP, so far so good.0237

We arrange this and put the U and P together and 2’s together.0251

I get U2 + PV 2 - U1 + PV 1, this is a –U1 not this one and - this one, that is equal to QP.0255

Since P =P1 = P2, the pressure is constant, it does not matter.0285

The initial pressure is the same as the final pressure so we can just call it just P.0291

That is what P is.0296

We can go ahead and put wherever we see a P as associate with the 2, we can put the P20298

and wherever there is a P associated with the subscript 1, we can go ahead and put 1.0304

We get the following.0309

We get U2 + P2V2 – U1 + P1V1 = QP.0313

U is a state function, it is just a function.0336

Pressure is a state function, volume is a state function, so P is a state function, volume is a state function.0348

Therefore, the sum of the state function, the product state function is a state function.0359

This whole thing and this whole thing, they are state functions.0365

This U + PV is a state function.0376

We are going to define a new state function H = U + PV, we call this the enthalpy of the system.0385

The enthalpy of the system is equal to the energy of the system + the pressure of the system × the volume of the system.0413

Because of what we just did mathematically, since U2 + P2V2 - U1 + P1V1 = QP.0425

This is H2, this is H1, what we end up with is δ H =Q sub P.0442

The heat that is transferred from the surroundings in a process that is done under constant pressure = change in enthalpy of the system.0459

This is different from when we did constant volume number that the heat transfer during transformation0471

under conditions of constant volume = the change in energy of the system under conditions of constant pressure.0480

The heat transfer from the surroundings, the heat lost by the surroundings = the change in enthalpy of the system.0488

Enthalpy does the same thing that energy does.0498

Enthalpy does under constant pressure what energy does under constant volume.0503

It is behaving the same way.0507

We had, under constant volume, condition that is equal to change in energy.0509

Here, the heat transfer is equal the change in enthalpy, it is different.0516

Enthalpy is the energy + a little extra.0522

That is what is going on, or when I think about energy = enthalpy - the pressure volume work that the system does.0526

At constant pressure, what is says is at constant pressure the heat withdrawn from the surroundings, in other words the heat it is going into the system.0537

Surroundings = the change in enthalpy of the system.0563

We have δ H = QP, δ U + PV = QP, the δ operator is a linear operate, it distributes.0583

δ U + δ PZ = Q sub P.0605

Since P is constant we can pull it out of there, what we will get is δ U + P δ V = QP.0616

The way of looking at it.0640

Heat effects are usually measured at constant pressure.0643

Again, it is just very easy.0659

Constant pressure is just a natural way things, it was constant pressure, atm pressure.0661

These effects are changes in enthalpy not changes in energy.0672

If we perform a process under constant volume then any heat transfer = change in energy of the system.0693

If we do a process under constant pressure, if we measure some heat transfer, heat change, heat effect under conditions of constant pressure.0700

As the change in the enthalpy of the system, enthalpy is equal the energy + the pressure volume work that is done by the system.0711

It is a little less, if we want a change in energy we have to take the heat and we have to subtract any pressure volume work that is done.0721

We will see what happens in just the second.0729

Sure enough if we want δ U, then δ U + P δ V = Q sub P.0733

If we want δ U, move that over, = Q sub P - P δ V.0752

If we know what the heat lost by the surroundings is and if we know what the pressure is, we know the pressure is the external pressure,0768

the pressure the atm whatever happens to be.0776

If we know a change in volume that the system undergoes, you can find a change in energy.0778

If we know QP which is the heat lost.0786

Sorry if I keep repeating myself, I’m going to do that every time a variable I’m going to say what it is.0798

You need to hear it over and over again.0804

If we know QP which is the heat lost by the surroundings, take the surroundings point of view, that is how we measure changes.0806

Δ V of the system and we can find DU.0819

It is very, very simple.0829

And we will do a problem like that just a little bit.0832

Δ H =QP our basic equation, the differential version is the following.0841

It is the integrated version of finite version, the differential version is D not DU, D is enthalpy DH = DQ sub P.0849

H is a state function.0865

We saw that, it is the combination of state functions.0868

The enthalpy is a state functions so DH is an exact differential.0871

Now we start dealing with the mathematics.0883

We know what exact differential is.0886

If we choose temperature and pressure as the variables for enthalpy, enthalpy is a function of temperature and pressure we can express it as follows.0889

Let us say choosing T and P as our variables, we have that enthalpy = the function of temperature and pressure.0906

Because it is an exact differential, we can express it like this, DH = DH DT under conditions of constant pressure holding the other × DT + DH DP × DP.0931

We have our total differential.0960

If the enthalpy of the system is a function of the temperature and the pressure, a small change in temperature, enthalpy will change by this much.0963

A small change in pressure, the enthalpy will change by this much.0973

If I add these 2 changes, I get the total differential change in enthalpy.0977

If I integrate this expression, I get the total change in enthalpy.0981

This is just mathematical consequences of it being a state function.0985

A state function is an exact differential and an exact differential can be expressed like this, if you choose the variables appropriately.0990

It is your choice as far as the variables are concerned.1000

It is not like when you are doing this a 100 years ago or 150 years ago, it is not like they just knew automatically to check temperature and pressure.1002

You could express this as temperature volume.1011

You can express it as pressure volume.1013

If you want to use an expression as temperature, pressure, and volume, you have three terms here.1015

It is just what you are seeing is the most fruitful work that is come about.1021

It is not like we knew automatically let us just take T and P, we do not.1026

We just start fiddling with it and you see what you get.1030

The derivations that bear fruit, those are the one that end up in the textbooks, that is how it works.1034

DH = this, since P is constant, this DP the change in pressure is 0.1041

Pressure does not change so that DP is 0 that term goes to 0 because that is 0.1057

What you are left with is DH = DH DT sub P × DT, but DH = DQ sub P.1064

That is the equation right there.1086

Here we go again with the heat, DQ sub P = DH DT sub PDT.1093

Or if I bring this down DQ sub P /DT not partial = DH DT sub P.1108

The heat was drawn from the surroundings is related to the temperature change of the system.1130

This is a heat divided by the temperature, this is heat capacity, it is another type of heat capacity.1139

It is the constant pressure heat capacity.1147

This heat capacity is associated with this partial derivative, the change in enthalpy per unit change in temperature.1149

We are associating this derivative with a quantity that we can measure.1157

We can measure how much heat the surrounding loses.1161

We just stick a thermometer in it and see by how much that temperature actually drops.1164

We can find out the change in temperature of the system.1168

We can stick a thermometer in the system and see how much the temperature goes up.1170

When we divide the two, we get the heat capacity.1174

DQ sub P / DT, this ratio is the heat capacity, it is a constant pressure heat capacity of the system.1180

We have a constant pressure heat capacity is defined as DQ P DT, it is defined as the amount of heat that is lost1211

by the surroundings divided by the temperature change of the system.1222

Or if you want to take the systems point of view, the heat gain by the system divided by the temperature change1227

temperature rise of the system, if we preferred to stick with the systems point of view.1234

Again, when we measure heat exchanges we are not measuring the system but we are measuring the surroundings.1240

That is when we say, that happens to be associated with identified with this partial derivative which is the total differential DT sub P.1245

This partial DH DT under constant pressure is now identified with and easily measured quantity heat capacity under constant pressure.1259

I just pull the pressure constant and I measure heat capacity.1286

We now write DH = CP DT + DH DP sub T × DP.1298

We write this, which is the total differential except now because we associated it did this with the heat capacity CP instead of that partial derivative.1325

Let us see what happens.1337

Let me rewrite that again so DH = CPDT + DHDP sub TDP.1340

If the transformation happens at constant pressure there is a general equation, we are not holding anything constant yet.1357

If the transformation happens at constant pressure the DP = 0.1381

This goes to 0, all we are left with is DH = CP DT or if we integrate this we get DH = the integral from1388

temperature 1 temperature 2 of the constant pressure heat capacity × DT.1402

Or if over the temperature range, if CP happens to be constant we can just pull it out of the integral and we get the other version of it,1410

a finite version which is δ H = CP δ T, this is the version that you see in your general chemistry classes.1420

These equations they apply to any system of fixed mass, any system like gas, liquid, solid.1428

Just like the previous equation, in front of a constant volume.1449

This is not just for gases, in general we tend to do the examples for gases but they apply to any system.1453

These are general equations applied at any system of fixed mass, gas, liquid, or solid, as long as there is no phase changed during the process.1461

The equation applied in the system of fixed mass, as long as no phase change takes place or1499

no chemical reaction takes place during this process, during the transformation.1507

If a chemical reaction takes place from the thermodynamic things happening, that is not just this, this is a chemical reaction.1524

This equation holds as long as this is the case.1531

Important equations, let us go ahead and do that.1540

For this particular lesson, the important equations to keep in mind.1550

We will be going through all of these when we do our problems in bulk when we start doing entire lessons of problems.1557

Do not worry, we are going to give a lot of them.1562

We have the general definition of enthalpy, U + PV.1566

It is very important you have to know this definition.1571

We have δ H = δ U + P δ V that is just the finite version of it.1574

We have δ H =QP or the differential version DH = DQP.1585

A state function, exact differential, path function, inexact differential, dealt with the fact that these are the same confuse you.1598

We have the basic mathematical relation DH = DH DT under constant pressure DT + DHDP under constant T and DP which is the same as.1607

That is the final, let us go ahead write the other version, the DH = the constant pressure heat capacity DT + DH DPTDP.1638

At constant pressure, these are the general equations at constant pressure here is what we have.1657

We get DQ sub P = DT sub PDT, in other words DQ sub P = constant pressure heat capacity × DT, DH = CPDT because DQP = DH.1666

And if we integrate this, we get the change in enthalpy of the system = the integral from one temperature1697

to the other temperature of the constant pressure heat capacity × the differential change in temperature.1705

Let us do some problems.1713

During these individual lessons when we are discussing mostly theory, examples are going to be reasonably simple.1717

The more sophisticated or complicated examples, the broader range of them are going to come when we do them in bulk.1723

So no worries, I will not leave you hanging across.1729

Let us see what we have, sample 1.1734

Example 1, the constant pressure molar heat capacity for solid silver is, this in J/ mol K.1740

Be very careful when you are reading these questions, every single word matters.1751

It is not specific heat capacity but a molar heat capacity.1756

If it is a constant volume heat capacity, its constant pressure heat capacity.1759

For sold silver is this, 23.4 + 0.0063 T.1764

Notice, this is a function of temperature which means as the temperature changes the heat capacity changes which is the real behavior of heat capacity.1771

The capacity changes with temperature.1780

It is usually not constant over a large range of temperatures.1781

If 100 g of the silver is taken from 25° C to 900° C at 1 atm pressure, what is the change in enthalpy of this 100 g of silver that is our system?1785

Let us see what we have got.1799

Molar heat capacity, first of all let us go ahead and take care of this 100 g.1802

Let us change it to mol, so we have 100 g × 1 mol silver is 107.9 g.1806

If I did my arithmetic correctly, this will be 0.0927 mol of Ag.1817

I'm not pretty big concern with significant figures, I do not care about that much.1827

Honestly, I do not like to think that important.1831

If 2 or 3 decimal places is usually pretty great.1834

We want the δ H, the δ H, the general equation is.1840

It is the one that we should always fall back on, the integral from temperature 1 to temperature 2 of the constant pressure heat capacity × DT.1847

It is equal to the integral, we are taking it from 25° to 900° C.1859

We are doing integration directly.1866

This constant pressure heat capacity is not constant, it is a function of temperature.1868

Therefore, we cannot just pull it out and use DT.1874

We cannot just use 900 – 25.1876

We have to integrate from 298 which is 25 C to 98 K, 900 is 1173 K, them we go ahead and1878

integrate the 23.4 this is our function, this is our CP 0.4 + 0.0063 TDT.1889

I’m not going to do thee valuation here.1901

Mathematical software - I use maple or mathcad, whatever it is that you need to do the integration.1905

So using the software, I evaluated this and I end up with following value.1913

Δ H I got was 24,529 J/ mol, notice it is just the K that canceled so we are still left with J/ mol, this is the molar heat capacity.1923

We had a 0.927 mol in this particular sample so we multiply this we end up with 22,733 J.1941

In taking this piece of silver from 25° C to 900° C, we have increased the enthalpy of this 100 g of silver by 22,733 J.1957

We said that the enthalpy δ H =δ U + P δ V.1974

In the case of solid, when you heat up a solid, it does expand a little bit.1982

The volume changes, but it changes very little.1986

Therefore, this V δp component, the δv because of δ Vis tiny, we pretty much ignore it.1990

In this particular case, when it comes to liquid and solids the change in enthalpy, we can usually take it to be the same as the change in energy.1999

We cannot do that with a gas.2007

When you raise the temperature of the gas, the volume goes from this to that.2009

The change in volume is huge so the change in energy, the change in enthalpy they are not exactly the same.2013

You will see that in a minute.2019

In the case of the solid or a liquid, because of the change in volume is so small, the change in enthalpy is almost the change in energy.2020

That is what is happening here.2030

Enthalpy is a counting device, it accounts for the change in energy of the system but also accounts for any pressure volume work but the system does,2033

that is why enthalpy and energy are not the same.2044

In a constant volume process because the change in volume is 0, the enthalpy and the energy and the heat are the same.2047

Let us try more examples here.2058

Consider the following reaction, at 25°C under 1 atm pressure.2062

We have the decomposition of ammonia gas to nitrogen gas and to hydrogen gas.2068

2 mol of gas is being converted to 3 mol, 2 mol initial 3 mol total.2074

The δ H for this reaction is 92 kg J.2081

They give you the δ H in this particular problem.2083

Δ H is 92 kg and notice that kg J/ mol it is kg J because it is for the whole process, this is not molar.2086

Find δ U for this reaction as written.2095

What are our equations?2101

We know that the δ H = δ U + P δ V.2103

We want δ U so let us just go ahead and move this over here and we get δ U = δ H - P δ V.2109

If we want δ U, we just have to take the enthalpy and we have to subtract the pressure which is 1 atm, we have it already.2118

So we have δ H, we have the pressure, we just need a change in volume.2126

We just need to find out the change in volume under 1 atm of 2 mol of gas turning into 4 mol of gas.2129

We just the δ V.2137

Let us go ahead and do that.2139

It is very simple.2141

Let us start off with our ideal gas law Pv = nRT.2144

The volume = nRT /P.2152

Therefore, the volume 1, the initial volume = the initial number of mol × RT / P.2159

The initial number of mol is 2 mol.2167

We start off with 2 mol, the system starts off with 2 mol of ammonia gas 0.08206 this is going to be L atm/ mol K.2170

It is taking place at 25° C so it is 298 K and is happening all under pressure 1 atm.2185

Atm atm, mol mol, K K, if I did my arithmetic properly I ended up with something which is 48.91 L.2194

So under these conditions I started off with 48.91 L.2206

As written, the decomposition produces 4 mol of gas, 4 mol of gases could occupy a greater volume.2212

Therefore, let us go and see what the new volume is.2220

The change in state is from volume 1 to volume 2, pressure stays constant 1 atm.2222

Let us find out what volume 2 is.2228

Well, volume 2 is the number of mol for the products × RT / P.2230

We have 4 mol × 0.08206 × 298 K, the temperature stays the same it is still 1 atm and you end up with 97.82 L.2238

The final state has gone to this much, the δ V = V2 – V1 = 97.82 - 48.91.2255

I end up with 48.91 L that is our δ V.2269

We are going to calculate the δ U = δ H - P δ V = 92 kg J which is 92000 J - pressure is 1 atm and we have 48.91 L, we end up with 92,000 J - 48.91 L atm.2275

I’m going to go ahead and convert that.2321

I have got 8.314 J/ 0.8206 L atm and I end up with 92,000 J - 4955 J.2324

I end up with 87,045 J this is δ U.2349

Let us see what I have got here.2362

I find my δ U as 87,045 J.2365

Notice my δ H is 92,000 J.2369

The system gains, the decomposition the change in enthalpy of the system is 92,000 J.2375

The system gains 92,000 J of heat.2386

All of those 92,000 J, 4955 J are converted to work.2391

That much energy is lost by the system in pushing out from 49 L to 98 L.2401

Therefore, what is leftover is the energy of the system 87,045.2411

92,000 J of energy go into the system, 4,955 J leave the system that leaves to work.2416

Work done in pushing away the atm in order to expand the volume to accommodate the fact that 2 mol is turning into 4 mol.2423

When we are dealing with the gas because there is a huge volume difference the 48.91 L, the enthalpy and the energy are not the same.2433

For a liquid and a solid, the energy and enthalpy are pretty much the same because the volume changes a little.2445

For a gas that is not the case, that is what is happening.2451

In other words, it is very important that you understand what is happening physically.2458

In other words, δ H = QP the heat that is lost by the surroundings is the change in enthalpy of the system.2467

The heat that is gained by the system is the change in enthalpy of the system.2479

The system, the reaction the ammonia gains 92 kg J of energy as heat energy but about 5 kg J of this 922485

is used in the volumetric expansion of the system as the system goes from 2 mol to 4 mol.2518

Some of the heat is used as work to push away the atm upon the expansion of this gas.2551

This leaves 87 kg J of energy in the system.2557

Enthalpy and energy are not the same.2567

The enthalpy it accounts for any pressure, volume, work done by the system under conditions of constant pressure.2581

There you go, hope that makes sense.2611

And again, no worries, we will be doing a lot more problems in complete blocks.2613

Thank you so much for joining us here at www.educator.com.2617

We will see you next time, bye.2619

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