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1st Law Example Problems III

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 0:15
    • Example I: Finding the Final Temperature
    • Example I: Finding Q
    • Example I: Finding ∆U
    • Example I: Finding W
    • Example I: Finding ∆H
  • Example II 11:27
    • Example II: Finding the Final Temperature
    • Example II: Finding ∆U
    • Example II: Finding W & Q
    • Example II: Finding ∆H
  • Example III 24:38
    • Example III: Finding the Final Temperature
    • Example III: Finding W, ∆U, and Q
    • Example III: Finding ∆H
  • Example IV 29:23
    • Example IV: Finding ∆U, W, and Q
    • Example IV: Finding ∆H
  • Example V 32:24
    • Example V: Finding the Final Temperature
    • Example V: Finding ∆U
    • Example V: Finding W
    • Example V: First Way of Finding ∆H
    • Example V: Second Way of Finding ∆H

Transcription: 1st Law Example Problems III

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

We are going to continue our example problems on energy and the first law.0005

Today, we are going to spend some time on adiabatic transformations. 0009

Let us go ahead and get started.0014

The first problem example number 1, we have 1 mol of ideal gas, so we are definitely dealing with an ideal gas here, 0018

that is subjected to a single stage adiabatic compression at a constant external pressure of 1.3 mega Pascal.0025

Initially, the gas is at 30°C and 0.2 mega Pascal.0035

The final pressure of the gas is 1.3 mega Pascal which is the external pressure that we are putting on it.0041

Calculate the final temperature of the gas Q, W, δ U, and δ H, the constant volume heat capacity = 3/2 Rn or if 3Rn/2.0049

This is an adiabatic compression, it does not say reversible it just says adiabatic.0062

Remember, we differentiate it, it can be something can be isothermal and it can be isothermal and reversible.0068

It can be adiabatic and it can be adiabatic and reversible.0074

Those are not the same.0076

Reversible does not necessarily with the situation.0079

The irreversible compression or expansion is a very specific thing.0083

An adiabatic and isothermal are very specific things.0087

Let see what we can do.0090

Now because this is adiabatic, it implies that DQ = 0.0092

The first law which is DU = DQ - DW this goes to 0 so what we are left with is that DU = - DW.0104

If we find δ U we know what W is or if we find W, we automatically know what δ U is.0116

An integrated form, this is δ U = - W.0123

Again it is –W not –δ W, I keep repeating myself, work is a path function not a state function.0127

Δ when it integrates it integrates just what quantity and a quantity depends on the path that you take.0135

State functions only depend on the beginning and end that is why I write Δ, the path to them is irrelevant.0141

Let us go ahead and see what we got here.0148

Let us do a direct derivation of everything so DU = CV DT + DU DV sub T × DV this is the basic equation for the δ U for this one.0151

This is an ideal gas so this term goes to 0 so the energy δ U = CV DT.0171

Let me see, DW= P external × DV.0178

Therefore, W = P external because it is a constant external pressure, P external × δ V.0193

We can go ahead and put these values in for here.0205

Here we have δ U = CV δ T, when we integrated version so we have this and we have that with that or left with the following.0210

DU which is CV δ T = - W which is P external δ V.0223

We have CV T2 - T1 again, one of the first things we want find is the final temperature T2 = - P external × V2 - V1.0238

This is an ideal gas so PV = nrt V = nrt/ P.0254

P is constant therefore, V2 = nr T2/P V1 = nr T1/ P.0265

We have CV ×T2 - T1 = - P external × nr T2/ P2.0272

I'm sorry the external pressure is constant not the pressure of the gas, my apologies.0289

So this is nr T2/ P2 –nr T1/ P1.0297

Yes again, my apologies this is both the temperature and the pressure change.0304

The pressure goes from 0.12 to 1.3 mega Pascal and the temperature is at 30 and it has a final temperature.0311

Both of those are variable.0322

This is the equation that we get.0324

Let us see.0328

What should I do?0348

Should I go and put the numbers in now or should I have and go ahead?0349

I had CV, this is 3/2 Rn so we have 3/2 Rn × T2 - T1.0354

Let us simplify as much as we can a little bit more.0363

The Rn comes out so we end up with - Rn P external × T2/ P2 - T1/ P1.0367

The R and n cancels the R and n so we are left with of 3/2 × T2 - T1 = - P external × T2/ P2 - T1/ P1.0382

What we are looking for is T2.0405

That is what we are looking for, now we can put the values in.0409

We have 3/2 × T2 - the temperature 1 so 303, I think it was 30°C if I’m not mistaken.0413

303 K, again this is what is important.0426

The numbers themselves are actually irrelevant.0429

The final pressure or the external pressure is 1.30 × 10⁶ Pascal.0432

The temperature 2 / the pressure 2, the pressure to the final pressure the system was 1.30 × 10⁶ Pascal - temperature 1 0443

which is of 303 K/ 0.2 × 10⁶ Pascal.0454

We have to convert to Pascal and when we do this, when we do the algebra, you should end up with the temperature 2 = 970 K or 697°C.0462

I will go ahead and run to the algebraic process.0482

That takes care of the temperature.0485

Well, because this process is adiabatic, the Q = 0 because it is adiabatic.0487

That is easy.0492

Let us see here, δ U, we have Q, should we do work or let us go ahead and do δ U.0499

Δ U = CV × δ T and now that we have the final temperature, we know what the δ T is.0510

Δ U = 3/2 Rn δ T which is 3/2 × 8.314 × 1 × 970 K - 303 K.0519

We end up with a change in energy of the system 8318 J.0539

This is an adiabatic process so the work = -δ U.0549

From the first law remember, δ U = Q – W.0557

It is adiabatic so Q is 0 so δ U= - W or W= -δ U.0562

The work = -8318 J.0568

Let us go ahead and deal with δ H.0577

I suppose I can do it one more page here.0580

It is fine let us do it in the next page.0590

My default is just go automatically this one, it does not have to be like this.0594

You can do the δ H = the constant pressure heat capacity C sub P × δ T, that is probably faster and easier than this.0599

But again, I just tend to do like this δ U + δ PV H = δ U + this is an ideal gas Pv = nrt = δ nrt.0612

Δ H = δ U = nr δ T.0628

Δ H = 8318 J + 1 ×8.314 × a change in temperature which is 970 - 303 and you are left with δ H of 13,863 J or 13.8 kJ.0646

However, you want to express it.0674

Adiabatic process just runs the equation.0677

Let us see what another example brings.0684

Example number 2, we have 1 mol of ideal gas initially at 30°C and 0.2 mega Pascal and it is compressed adiabatically and reversibly.0689

We have adiabatic and reversible.0703

Not just adiabatic, the first example was adiabatic, now it is adiabatic and reversible to a final pressure of 1.3 mega Pascal.0707

The numbers are the same, 30°C, 0.2 initial, 1.3 final.0714

Find the final temperature of the gas Q, W, δ U, and δ H, molar constant volume heat capacity is 3R/ 2.0719

Notice the n is missing, molar means that this is CV with a line over it = 3R/ 2.0729

CV with a line over it is CV/ n = 3/2 R.0740

Therefore, the heat capacity itself a constant volume is 3/2 Rn.0747

It was the same, it just depends.0752

If you see molar that means that it CV ÷ n.0754

It is just moving n over.0757

Let us see what we can do.0760

We have the same sort of thing.0762

Let us see, P external DV.0765

It is going to be the situation we have an adiabatic so we know that DQ = 0.0771

If DQ = 0 and we know the DU = - DW.0779

Therefore, we have DU which is going to = CV DT and – DW.0791

That is equal to - P external DV, this is reversible, reversible means that P external = P.0804

It means that at any time along the compression or expansion the outside pressure = the inside pressure, they are in equilibrium.0812

We are going along an adiabat.0819

This = - PDV, well P is an ideal gas P is no longer constant.0823

P = nrt/ V so I need to put this into here so I get CV DT = -nrt / V DV.0835

Let us see here, what do I do next?0854

Let us go ahead and separate the variables.0857

I will divide by T so I'm going to get CV DT / T = - nr/ V DV.0858

We have been to this process when we are discussing adiabatic theoretically.0871

I'm just running through the derivation again.0876

We do not really need to go through the derivation again but I think it is nice to go through it because in that way it locks it in your mind and 0880

it confirms the idea that when we solve these problems we only want to start with a handful of equations.0888

And with those handfuls of equations, derive what comes next, that is what is important,0894

that we would have to memorize some equation which we are going to get to in just a minute.0899

When we go ahead and now that variables are separated, we have DT here and we have DV here, and V and T, we can integrate both sides.0904

This is a constant so what comes out so you end up with.0912

When you do this integration this comes out as a constant.0917

What this becomes is CV × the integral from T1 to T2 of DT/ T = - nr × the integral from V1 to V2 of DV/ V.0921

When you do this integration you end up with the following.0938

You end up with CV × log of T2/ T1 = -nr × nat log of V2/ V1.0941

This is the fundamental equation for adiabatic and reversible expansion or compression of the gas in a change of state.0962

This is the relationship CV × LN of T2/ T1 =-nr LN V2 V1 for an ideal gas.0976

We want to find the final temperature.0986

We have the initial temperature 30°C so we have this one.0990

We do not really know what the volume is but we know what the pressures are.1000

Let us go ahead and manipulate this equation and see what we can come up with.1005

And this is sort of the exciting part as far as mathematical derivation is concerned.1010

Hopefully, you can follow it.1015

You will be able to follow it, there is nothing strange here.1017

I’m going to start rewriting this equation.1020

It is going to be a little bit of mathematical play here.1022

I’m going to bring this up, just properties of logarithms.1024

Log of T2/ T1 ⁺CV = log of V2/ V1 ⁻n/R.1027

Let us see, this is equal to nat log of,1046

I can do this either way I suppose it does not really matter.1055

Let us see if we can cancel log first or V, let us go ahead and do it that way.1058

This is the log of something, the logs go away and what you are left with is, in CV we already know what CV is.1062

CV = 3 Rn/3/2 Rn.1071

What we are left with is T2/ T1³/2 Rn = V2/ V1 ⁻Rn.1076

The Rn goes with Rn so what you end up with is T2/ T1³/2 = V2/ V1⁻¹ which is T2/ T1³/2 = V1/ V2.1090

I just took the reciprocal of that.1115

Let us see where we are.1118

Let us go ahead and do across multiplication.1120

We end up with is T2³/2 V2 = T1³/2 V1.1122

This is an ideal gas, we have PV = nrt so V = nrt/ P.1136

We have T2³/2 V2 is nr T2/ P2 T1³/2 V1 = nr T1/ P1.1146

The nr cancels with the nr.1171

The T2 and T2 this becomes T2⁵/2/ P2 = T1 to 5/2 / P1.1173

I had everything, for initial pressure I have the final pressure, I have the initial temperature, I just need to go ahead and solve for that.1186

I have the following expression.1194

This is equivalent to, let me rewrite it over here so that I have more room.1198

I have T⁵/2 / P2, my final pressure is 1.30 × 10⁶ Pascal = T1 which is 303 K/ 0.20 × 10⁶ Pascal.1203

When I go ahead and solve for this I end up with a temperature of 641 K.1221

There you go, it is all based on that initial equations that you know and the rest is just mathematical manipulation and sort of recognizing.1229

Remembering that P= nrt we are dealing with ideal gas.1239

Do not get me wrong, I’m suggesting that these problems are simple, there is a lot going on.1244

But again, this is higher level science and you are talking about a junior senior course.1249

And this is the hardest kind of problem that you will get on quizzes and tests.1253

A lot will be more straightforward and simple.1261

Probably not necessarily needing to see this kind of derivation.1264

But it is nice to know the derivation is there for you.1269

That is a good feeling to do something like this and be able to follow problem like this.1271

We have the final temperature 641 K and we had the initial temperature which is 30°C or 303 K.1277

Let us see what we can do next.1285

Let us go ahead and do energy DU = CV DT or the integrated form DU = CV δ T.1287

We have δ T now so this is really great.1299

We have 3/2 Rn δ T.1302

We have R, we have n which is one we have δ T which is 641 -303 so this = 3/2 × 8.314 × 1 × 641 -303.1307

Our change in energy for the system ends up being 4215 J.1324

Again, I hope you will check my arithmetic here.1329

That is the change in energy.1332

This is an adiabatic process so the work = - the change in energy.1335

Therefore, the work = -4215 J that takes care of the work.1341

Adiabatic so Q = 0 and δ H.1354

You can do δ H via the integral process.1360

You can either do that now you have δ T you can go ahead and do δ H = CP δ T.1363

You have CV but you know there is a relationship an ideal gas CP - CV = Rn.1370

You can go ahead and put the CV in and solve for CP.1376

Put it back in here and do the multiplication because now you have the δ T or my default δ H = δ U + δ PV = δ U + δ this is nrt.1379

The ideal gas PV = nrt so = δ U + nr × T2 - T1 = 4215 J + 1 × 8.314 × 641 -303.1398

You are left with a δ H of 7025 J.1431

That is all that is going on.1438

You have a certain amount of this energy.1443

I like the δ H because I like actually seeing where each value comes from.1445

This comes from a change in temperature, this comes from a change in volume and pressure, whatever it is it is going on, it is an accounting.1451

That is what is enthalpy is, it is an accounting of how much comes from the energy, how much comes from pressure volume work, things like that.1461

Again it does not really matter, you can use this one and you are going to end up with the same number.1470

Let us see example number 3, in an adiabatic expansion of 1 mol of an ideal gas.1476

Again, we are talking about an ideal gas initially of 30°C so the initial temperature is 30°C, 1400 J of work is produced.1487

This time to give us the work.1497

If the constant volume heat capacity is 5/2 Rn calculate the final temperature.1499

We want the final temperature, we want Q, W, δ U and δ H.1505

This was going to be an adiabatic expansion.1511

In adiabatic expansion there is going to be a cooling.1513

We are looking for a final temperature which is going to be less than 30°C or less than 303 K.1516

I just want to keep that in mind because again you get so caught up with the actual problem itself and doing it you stop to see if the value that you got made sense.1523

Hopefully, they make sense.1533

Let us see here, let us just start with our regular.1538

We have δ U= Q - W this is an adiabatic expansion and Q = 0.1542

Therefore, δ U = - W.1551

Δ U = CV δ T = - W.1556

We are looking for the final temperature so here δ T = - W/ CV because they gave us the work.1564

It is right here so δ T = that, δ T they gave us the work is -1400 J and they gave us the constant volume heat capacity which is 5/2 × 8.314 × 1.1572

Our δ T ends up being -67.4°C or -67.4 K because again that δ T is the same.1594

Δ T = T2 - T1, therefore T2 = T1 + δ T, T2 = T1 which was 303 K + -67.4 K.1609

I end up with a final temperature of 235.6 K or -37.4°C.1632

There you go, sure enough it did end up cooling from 303 to 235.6.1642

We ended up with a final temperature of 235.6.1651

We already calculated the work for this.1656

Let me see if I need an extra page.1658

I will just go ahead with an extra page, it is not a problem.1661

They gave us the work, the work was 1400 J.1664

Δ U= - the work so = -1400 J this is an easy problem.1669

A Q is adiabatic so it is equal 0, nothing is lost or gained as heat and δ H.1679

δ H =δ U + δ PV δ, H = δ U + nr δ T, δ H = -1400 δ U is -1400 + n which is 1, R which is 8.314 and δ T which is -67.4.1687

You end up with the δ H of -1960 J.1719

There you go, the system ends up losing energy in terms of a temperature drop and 1727

it ends up losing energy in terms of the expansion itself, the pressure volume work that is done.1738

The total loss, the total change in enthalpy of the system is -1960 J.1746

Let us see what example 4 has, I need one page for this one.1760

1 mol of ideal gas is compressed adiabatically, its temperature rises from 30 to 60, its CV is that, find Q, W, δ U and δ H for this transformation.1773

We know we have an adiabatic, let us start off with our basic.1786

You just have to dive in.1795

We have a handful of equations, pick a couple of them and see where they go.1799

That is what it was going on.1802

We have the DU = CV DT + DU DV DV this term goes to 0 because we are dealing with the ideal gas.1805

Therefore, we have just DU = CV DT or the integrated form δ U =CV × δ T.1820

This is great, they already gave us both temperatures.1833

The δ T is 30, we have the CV is right there, this is an easy problem.1835

δ U= 5/2 Rn × δ T which is 30°, when you put R which is 8.314 and for n you end up with δ U = 624 J.1840

There you go, that is the δ U.1860

This is adiabatic, it implies that this is Q = 0.1863

Adiabatic means that work =-δ U.1874

Δ U = Q - W this goes to 0 so δ U = - W or W = - δ U which means that work = -624 J.1878

And now we just need δ H.1894

Again, δ H so this is getting old at this point + δ PV = δ U + nr δ T δ H = 624 J + 1 × 8.314 × 30 you end up with the δ H = 873 J.1897

Nice and easy, really straightforward.1932

They give us the two temperatures.1934

There was a lot of extra work to do.1935

Let us see what is next, I think this is going to be the final problem of this particular lesson.1940

Let me see here, an ideal gas whose constant volume molar heat capacity = 5 R/2 or 5/2 R is expanded adiabatically.1948

Adiabatic = 0, against an external pressure of 2 atm until it triples in volume.1967

This time it tells us it triples in volume, the initial temperature is 25°C and the initial pressure is 7 atm.1977

The final temperature of the gas Q, W, δ U, δ H per mol gas for this transformation.1988

Let us take a look at what is going on here.1995

Let us do this carefully.1997

Our volume, let me do this in purple.1999

Let us see how green looks.2008

Our volume is going from volume 1 to triple volume, three volume 1.2015

This is initial and this is final.2021

Our initial pressure going to a final pressure, the initial pressure is actually 7 atm and this is going from some T1 to T2.2023

DU = DQ - DW it is adiabatic so this goes to 0.2040

Therefore, DU = - DW.2050

Let us see here DU = CV DT and DW.2056

DW = P external DV so - P external DV, there you go.2072

Let us see here, CV is constant that is this.2084

CV = 5/2 R therefore CV = 5/2 Rn.2093

This n goes over there like before.2101

CV with a line over the CV/ n was moved here you get the CV.2106

The external pressure is 2 atm that is constant.2113

This is a constant also so they actually come out of the integral so we are left with is CV δ T.2122

When we integrate both of these, we end up with CV δ T = - P external δ V.2132

We have of 5/2 Rn that CV δ T is T2 - T1 the external pressure is 2 atm.2146

Δ V = V2 - V1.2162

V2 = 3V1 - V1 so δ V = 2V1.2170

This is an ideal gas V = nrt / P.2182

Therefore, V1 = nr T1/ P1 so this is going to be 2 atm δ V.2190

I will just go ahead and put this in δ V is 2V1 × 2V1 so 5/2 Rn × T2 - T1 = -2 atm × 2V1 V1 = nr T1/ P1.2207

This is going to be 2 nr T1/ P1.2232

Let us see what happens next, the R and n will cancel the R and n.2241

We end up with something a little bit simpler here.2247

We are going to be left with 5/2 × T2 - T1 = -2 × 2T1/ P1.2250

Let us move to the next page here, I’m going to write that so I have 5/2 × T2 which I'm looking for, -298.2272

I think the initial temperature was 25 to 298 = -2 atm × 2 × 298/ 7 atm.2283

In this particular case, notice I did not change atm to Pascal.2300

The reason I did not change atm to Pascal is because the unit atmosphere and atmosphere cancel.2304

If I convert this to Pascal and this to Pascal, the conversion is linear.2311

It is just converting this, multiplying this by a number, multiplying this by a number to convert a Pascal.2315

The numbers cancel, the units cancel.2322

If units cancel you do not have to do the conversion.2324

In this case, the atmosphere and atmosphere cancel, I can go ahead and leave the numbers the same.2327

I do not have to convert it to the appropriate unit so that is all you have to watch out for.2332

When I do this, let me go ahead and actually write this out.2339

5/2 T2 -745 = when I multiply this out I get -170.3= 5/2 T2 = 574.7.2344

And I get the final temperature of 230 K.2360

It expanded, it is cooling.2369

Δ U = CV × δ T which = 5/2 Rn × 230 -298.2372

I have a value of R and n, when I do this, when I put that values in and solve.2388

It is equal to 5/2 × 8.314 × 1=230 -298.2393

I end up with δ U = -1413 J.2409

There you go, work = - δ U.2416

Therefore, work = 1413 J.2422

The system is expanding, it is doing work, it is pushing the surroundings.2434

It is doing work on the surroundings, work is positive.2438

Change in energy is expanding, it cooled from 298 to 230.2443

It cooled, it lost energy.2449

It lost energy which is why this 1413 is negative.2452

Let us go ahead and do δ H.2465

I’m going to do δ H both ways just so you actually see it.2467

Both ways of finding δ H.2477

I’m going to do it my default way δ H = δ U + δ PV = δ U + nr δ T I get -1413 J + 1 × 8.314 × 230 -298.2483

I end up with δ H of -1979 J.2518

Let us do the other way.2531

This was my default way where I actually separated it in terms of energy and work.2536

One of the fundamental equations you have to know DH = CP DT + DH DP under constant temperature × the change in pressure.2543

For ideal gas, this term DH DP T is 0 therefore, this term goes to 0.2558

What we are left with is DH = CP DT or because the CP is constant we are left with δ H = CP × δ T.2565

I have δ T is 230 – 298, I just need CP. I have a relationship this is an ideal gas CP - CV = RN.2581

CP = CV + Rn = 5/2 Rn + Rn = 7/2 Rn CP = 7/2 Rn.2593

Therefore, δ H = CP which is 7/2 Rn × δ T which is T2 - T1= 7/2 × 8.314 × 1 × a change in temperature which is 230 K -298 K.2611

And what you end up getting is -1979 J, just like before.2635

You might be off by a couple of decimal points, it is not really a big deal.2645

There you go, two different ways to do the δ H.2647

This particular way with δ H happens to be what I sort of have fallen into, this is my default.2650

I like to see the energy term and the work term, here this is sort of the direct way of doing it.2656

The numbers are the same.2664

It is totally your choice whatever you are comfortable with.2666

Thank you so much for joining us here at www.educator.com.2670

We will see you next time, take care, bye. 2672