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Lecture Comments (9)

1 answer

Last reply by: Professor Hovasapian
Fri Feb 26, 2016 4:12 AM

Post by Van Anh Do on February 21 at 10:14:38 AM

For example 4, can we find Q using CpdT by converting the given Cv into Cp and then find dU using dU= Q - w?

1 answer

Last reply by: Professor Hovasapian
Wed Oct 28, 2015 12:50 AM

Post by Chanelle Brown on October 27, 2015

For example 1, you have the integral of [an^2]*(1/v^2)dV going from V1 to V2 is equal to [an^2]*{[1/(V1)] - [1/(V2)]}. Why isn't this integral equal to [an^2]*{[1/(V2)] - [1/(V1)]}?

1 answer

Last reply by: Professor Hovasapian
Mon Sep 21, 2015 12:54 AM

Post by Shukree AbdulRashed on September 19, 2015

Does delta U = Q-W only work when your point of view is the surroundings? I seem to come up with different values when I use U= Q + W and evaluate the problems from the perspective of the system. Thank you.

0 answers

Post by Shukree AbdulRashed on September 19, 2015

For Example #1, why is b=.064 x 10^-3? Shouldn't it be 0.049 dm^3/mol?

1 answer

Last reply by: Professor Hovasapian
Sun Dec 7, 2014 6:21 PM

Post by Carly Sisk on December 7, 2014

Just to clarify, the only time delta U equals zero under isothermal conditions is with an ideal gas?

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 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

Let us go ahead and see about the next phase here.0681

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 state 1074

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

What about that one?2240

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

Let me rewrite that again on this page.2287

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

This process will really help you clarify what is happening physically.2504

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 constant 2979

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

Let us go ahead and start with part A.3048

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

I had all information.3349

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

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