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Lecture Comments (2)

1 answer

Last reply by: Professor Hovasapian
Fri Feb 26, 2016 4:30 AM

Post by Van Anh Do on February 23 at 07:53:33 PM

For example 1, I think you forgot to calculate dS.

Entropy 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 0:09
    • Example I: Calculate ∆U
    • Example I: Calculate Q
    • Example I: Calculate Cp
    • Example I: Calculate ∆S
  • Example II 7:13
    • Example II: Calculate W
    • Example II: Calculate ∆U
    • Example II: Calculate Q
    • Example II: Calculate ∆H
    • Example II: Calculate ∆S
  • Example III 18:47
    • Example III: Calculate ∆H
    • Example III: Calculate Q
    • Example III: Calculate ∆U
    • Example III: Calculate W
    • Example III: Calculate ∆S
  • Example IV 27:57
    • Example IV: Diagram
    • Example IV: Calculate W
    • Example IV: Calculate ∆U
    • Example IV: Calculate Q
    • Example IV: Calculate ∆H
    • Example IV: Calculate ∆S
    • Example IV: Summary
  • Example V 48:25
    • Example V: Diagram
    • Example V: Calculate W
    • Example V: Calculate ∆U
    • Example V: Calculate Q
    • Example V: Calculate ∆H
    • Example V: Calculate ∆S

Transcription: Entropy Example Problems II

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

Today, we are going to continue our entropy example problems.0004

Let us jump right on in.0007

For the first problem we are going to look at here is the following.0010

We have 1 mol of an ideal gas initially at 25°C and 1 atm pressure and it is taken to 50°C and 0.55 atm,0014

during this transformation 260 J of work are done on the surroundings.0024

They give us the work that is done during this particular expansion.0029

They want us to calculate the heat δ U, δ H, and δ S for this particular process.0035

Let us see what we can do.0044

We are dealing with an ideal gas so it is going to change a couple of things.0048

We have 25°C to 50°C.0051

Let us go ahead and start with, in this particular problem when you are finding the Q, in this case they gave you W, 0057

but in subsequent problems we have to find the work, the change in energy, the change in enthalpy and the change in entropy.0064

There is no systematic way as far as choosing one before the other.0071

Sometimes it is easier to find the energy and then the heat and work.0074

Sometimes it is better to do the work than heat so do not think that you have to go down the line in order, it is not like that.0077

It is what you have at your disposal and whatever strikes your fancy.0083

Let us start off with this one.0088

Let me go ahead and start with energy.0089

When we do that let me stick with black.0091

Let us do DU = CV DT again, it is always nice to start with the equations that you know to write them down and take it from there.0094

It is a great review to constantly write down the equations that you need.0104

DDT + DU DV T DV is basic equation as far as the energy of a system in terms of temperature and volume.0108

Now this is an ideal gas so this is automatically goes to 0.0121

All we are dealing with is DU=CV DT and when we integrate this we get δ U =CV × δ T.0127

This is the equation that we are going to use.0143

It looks like we have everything.0144

δ U = CV δ T, CV is 3/2 RN δ T.0147

We can go ahead and put some numbers, we have 3/2, R is 8.314 J/mol-°K and we have 1 mol and we have δ T.0158

δ T is going to be 50 -25 and again Kelvin and Celsius, the δ is the same because the increment is the same so we have just 25°K.0175

°K vs °K and mol vs. mol, energy is going to be in Joules.0188

δ U for this process is 312 J and again I'm hoping that you will actually confirm my arithmetic, I’m notorious for arithmetic mistakes.0195

The energy is taken care of we have 312 J.0205

We know that δ U, let me go ahead and use the differential expression because it is probably the best to begin with.0211

DU = DQ - DW that is the fundamental relationship, that is the first law of thermodynamics.0220

This implies in for a finite change that δ U =Q – W.0230

Again, there is no δ Q δ W, these are Q and W are path functions, this is a state function which is why we have a δ here.0236

When you rearrange this, I get Q = δ U + W.0245

Therefore, Q I have δ U, I have 312 J, I’m going to go ahead and skip the unit.0252

I hope you do not mind just work with numbers.0257

They tell me that 250 J of work are done on the surroundings.0260

Work is set positive if it is being done on the surroundings.0266

Therefore, we have the 260 J.0271

For Q, we end up with 572 J.0276

For this particular expansion, 572 J of heat transpire.0283

Let us go ahead and take a look at δ H.0290

I’m going to go ahead to do this on the next page, it is not a problem I have few pages here.0295

Let us go ahead and write our basic relationship that DH = CP DT + DH DP DP.0299

Again because this is an ideal gas this goes to 0.0315

Basically, all we have is DH =CP DT, nice and straightforward.0319

When we integrate this function, I will integrate this differential equation we end up with CP δ T.0324

This is really nice and straightforward.0331

The only thing we need is they gave us CV, we have to find relationship, we have to find CP.0333

We know the relationship, we know that for an ideal gas CP - CV = Rn.0339

Therefore, CP = Rn + CV.0346

Therefore, CP =Rn + 3/2 Rn CP = Rn × 1 + 3/2.0351

Therefore, our constant pressure heat capacity = 5/2 Rn.0363

Working with an ideal gas is really easy.0368

We have our constant pressure heat capacity now we can just go ahead and put in here.0371

Let us go ahead and write δ H =CP δ T CP is 5/2 Rn and we have δ T.0377

Let us put in some numbers δ H =5/2 × 8.314.0386

Again, I hope you forgive me, I’m going to skip the units.0393

1 mol and δ T is 25°K.0395

When we work out this arithmetic, we end up with 570.0400

No, I’m sorry that was going to be our δ H is going to be 520 J.0406

Let us see what we can do here.0426

Let us go ahead and go to example 2.0431

Example 2, says 1 mol of an ideal gas with the CV = 3 Rn/ 2 was initially in the following state 350°C and 1.5 atm.0438

At constant volume the gas is heated to 450°K, calculate Q, W, δ U, δ H and δ S.0455

They also want us to compare δ S with the value of Q/T.0465

We have 1 mol of an ideal gas, they gave us the constant volume heat capacity 350°K , so we are doing this process at constant volume.0471

These problems are really great because they allow us to actually review our first law issues, the heat, work, and energy, enthalpy, things like that.0483

We just add that final state property which is entropy.0492

Let us see what we have, constant volume is going to be important.0496

These constraints are what changes the equations that we deal with.0501

Constant volume that implies the DV = 0.0505

DW =the external pressure × DV.0513

If DV is 0 that means DW = 0 that means that the work = 0.0520

Constant volume nothing is changing so no work is being done so work is 0.0528

Let us go ahead and see what we can do about DU.0536

Again DU = CV DT + DU DV at constant temperature DV and again because we are dealing with an ideal gas this term goes to 0.0539

It is constant volume so DV goes to 0.0559

Even if this were an ideal gas, in this particular case this would go to 0.0563

In either case, this term goes to 0.0567

What we are left with is DU = CV DT and for a finite difference when we integrate this differential equation we get CV δ T.0569

Let us put in some numbers, we have 3/2, let us go ahead and write it all in, 3/2 Rn δ T and it is going to be 3/2 × 8.314 × 1 mol and 0584

the change in temperature 350°K to 450°K, we have 100°K temperature change.0602

Our energy for this process is going to be 1247 J.0611

That is nice.0620

We have W and we have δ U so let us go ahead and find Q.0624

δ U= Q - W and let me just go ahead and just do it this way Q =δ U+ W =δ U which is 1247 + W which is 0.0628

Our Q for this process is also 1247 J.0651

That takes care of the heat, now let us go ahead and take care of the enthalpy.0659

DH =the constant pressure heat capacity × the differential change in temperature or constant pressure heat capacity × temperature + 0666

this DH DP the change in enthalpy per unit change in pressure at constant temperature × the change in pressure.0676

But again we are dealing with an ideal gas so this goes to 0.0684

We have DH =CP DT and for integration we have DH = CP δ T.0689

Nice and straightforward, nothing strange is happening here.0700

Again, since we have the 3/2 Rn the constant volume heat capacity was 3/2 Rn because this is an ideal gas our constant pressure heat capacity is going to be 5/2 Rn.0703

It is going to be like that.0715

DH = 5/2 Rn δ T = 5/2 × 8.314 × 1 × 100 we get a δ H for this of 2079 J.0719

That takes care of our δ H.0743

Let us go ahead and see what we can do about δ S.0750

This is what we have been discussing, this particular unit is entropy.0754

We have DS =CV / T DT + nR/ V DV for an ideal gas this is the expression for the entropy or the differential change in entropy of the system.0760

This is happening at constant volume, therefore, this term goes to 0.0783

What we have is just DS =CV/ T DT.0789

We are going to integrate this and what we are going to get is δ S =the integral from temperature 1 0798

to temperature 2 of CV/ T DT which = CV × the log of T2/T1.0807

CV is a constant so it comes out of the integral which we are left with is the integral of DT/ T, the integral of DT/ T is LN T2/T1.0821

Now I will just put the numbers in.0833

In this particular case, it is going to be 3/2 Rn × the log of temperature 2/ temperature 1.0837

Let us go ahead and DS = 3/2 R is 8.314, the number of moles is 1 log, and we have 450°K/ 350°K.0854

Real quickly here, when you are dealing with temperatures involving the temperature on top of another temperature, you have to work in °K.0875

I know that in previous problems, in this particular problem or either ones where we did Celsius temperature, when we did δ T we just use the 50°C - 25°C.0884

The difference in the δ T is 25°C or 25°K.0898

There we do not have to convert to °K because the increment of Celsius and Kelvin is actually the same.0903

However, for here we cannot go 25°C/ 50°C.0911

The °C do not cancel and the reason is because in the Kelvin temperature is going to be 25 + 273/ 50, + 273 the 273 do not cancel.0919

When you do this, the temperature has to be in °K.0931

In general, we are just avoiding problem, just do everything in °K.0937

In that way, you will never go wrong.0940

I just want to let you know that if the temperature in this particular problem we are given in Celsius, 0942

you would not be able to put Celsius on top of Celsius you will get the wrong answer.0946

When we run this, let me go ahead and go back to black here.0952

δ S should be 3.13 J/°K.0956

Since we are dealing with 1 mol, we can just go ahead and put 3.13 J/°K mol or mol°K.0969

It does not matter how you do it because we are talking about 1 mol.0979

It is okay, let us go ahead and put them all there.0984

It also asks us to compare δ S with Q/ T.0987

I do have another page here and that is good.0995

Let us go ahead to the next page.1000

The δ S we just calculated so δ S = 3.13 J/°K mol.1003

The problem with Q/ T is T is not fixed.1014

T goes from 350°K to 450°K so technically we cannot really solve Q/ T so we should leave it alone.1021

However, I’m going to do something different here.1029

I’m going to take T average.1032

I’m going to take the average between 350°K to 450°K instead of just 400°K just for illustrative purposes.1034

1247 J that was the heat for this particular transaction and I'm going to divided by the 400°K.1043

When I do that I will get 3.12 J/°K.1052

Notice that they are almost the same, 3.13 is the actual change in entropy calculated.1058

The Q/ the T average is 3.12.1064

Let me write down this particular note again, we cannot really do Q/ T because T is not fixed.1068

I chose to simply average the temperatures for illustrative purposes.1095

If you are ever asked to do something like this on an exam, it is perfectly fine for you to say this cannot be done. 1110

Or if you do it, use T average and make sure you specify that you are actually using an average temperature here.1116

Let us go ahead and do example 3, we have 1 mol of an ideal gas with this particular constant volume heat capacity initially in the following state 350°K, 1.5 atm.1125

This is the same as the previous problem, the initial state is, at constant pressure the gas is heated to 450°K.1140

The previous problem starting with this particular state we took a 450°K under constant volume.1150

We are doing under constant pressure.1157

At constant pressure, this is going to be important part here.1164

I think I’m going to go in red here just for a change of pace, make it a little brighter.1168

Calculate Q, W, δ U, δ H, and δ S and compare δ S with Q/ T.1173

At constant pressure and we are dealing with an ideal gas.1180

At constant pressure we know that δ H = the heat transpired, that is the whole idea behind the constant pressure process.1191

Let us go ahead and take a look at the DH.1200

The DH = CP DT this is one of our equations from the first law + DH DP constant T DP.1205

This is an ideal gas so this term goes to 0 or its constant pressure so this term goes to 0.1222

In either case, it goes to 0.1228

What we have is DH = CP DT and when we integrate that we get δ H =CP δ T.1231

When we do the mathematics I will get δ H =5/2 Rn δ T = 5/2 × 8.314 × 1 mol and the δ T is again 100°K.1244

We get a δ H of 2079 J.1264

At constant pressure we know the δ H = QP.1275

Therefore, we automatically know the Q.1279

The Q under constant pressure, the Q for this particular process is also = 2079 J.1281

It takes care of the heat.1290

Let us see what we can do about δ U.1294

I will do it in this page it is not a problem.1300

I know the δ H =δ U+ δ PV because the definition of enthalpy H =U + PV just apply the δ operator to all of that and you have this.1306

Well δ H =δ U + P δ V because P is constant.1323

Since there is no change in P, you can just go ahead and pull that out.1332

That means δ U =δ H - P δ V or dealing with an ideal gas so PV =nRT.1340

V = nRT/ P therefore δ V δ U =δ H – P.1353

δ V is just V2 – V1 so we have nR T2 / P because it is constant pressure –nR T1/ P.1365

The P cancel and I will get δ U =δ H - nR T2 - T1.1382

I’m just fiddling around with the basic equation that is all it comes down to.1399

We have δ U =δ H which we calculated which was 2079 J – n 1 mol R is 8.314 and δ T which is T2 - T1 is 100.1404

When we end up doing that, we end up with δ U =1247 J.1420

That is the energy of the system spans by 1247 J.1434

δ U =Q – W, therefore W = Q - δ U.1442

Q is 2079 J and δ U is 1247 J.1454

Therefore, we get that the work = 832 J and I'm hoping that you are performing all of my arithmetic.1469

It takes care of the work.1481

We have the work, we have the energy, we have the enthalpy, we have that, let us go ahead and find our δ S.1482

Let me see if I have an extra page.1489

Yes, I do.1491

Let me go ahead and go to the next page.1492

In terms of pressure we have DS.1499

We are express things in terms of temperature and volume, temperature and pressure.1505

In this case because pressure is involved we are going to use this version CP/ T DT - nR/ P DP.1512

The pressure is constant so this term goes to 0.1525

What we have is just DS =CP / T DT and when we integrate the get δ S = it is going to be the integral from T1 to T2 of CP/ T DT.1530

CP is a constant so what comes out from under the integral and which are left with is DS = CP × the log of T2/ T1.1548

When we go ahead and put numbers in here, we go down here a little lower we get 5/2 × 8.314 × 1 mol 1563

× the log of 450°K/ 350°K and we get δ S = 5.22 J/°K.1577

Our comparisons so we have δ S = 5.22 J/°K that is our δ S.1597

Again, the T is not fixed so we technically cannot do this.1607

I decided to just go ahead and do a T average.1612

Our Q here is 2079 J and we have an average of 400°K.1616

We get 5.2, it looks like a 6.1625

We have 5.20 J/°K that is our Q/ T.1630

We are just running through our basic equations that is what we want do.1641

We have derived all of these equations and we have set aside specific ones that are important to learn, for example this one regarding entropy.1644

We are just gaining some practice and of course reviewing all of our concept from energy that is why we are comfortable with this.1655

At this stage of the game, we want to be able to understand the thermodynamics but mostly we just want to be able to be comfortable 1661

with the mathematics and be able to solve some of the problems that are thrown at us.1668

Let us see what is next, example number 4 I think.1674

Example number 2, example number 3, start with the same state.1682

Example number 4, the next example 5 and the next couple of examples in the next lesson that we do, 1686

they are all going to be based on the same initial state this 350°K, 1.5 atm, an ideal gas with a constant volume heat capacity of 3/2 Rn.1692

What we are going to be doing was we are going to be transforming it from one state to another state under different circumstances.1705

For example 2, we made a transformation from 350 to 450 under constant volume.1711

In the example after that example 3, we do the transformation of 350 to 450 at constant pressure.1719

It is going to be the same starting state, we would be doing something else to it.1726

That is all we are doing, we are just trying to figure out how to deal with these thermodynamic variables under different conditions.1730

1 mol of an ideal gas with the constant volume heat capacity of 3 Rn/ 2 is initially in the following state 350°K. 1.5 atm.1738

The gases expanded and this time isothermally and reversibly until the pressure of the gas drops 2.75 atm.1747

Isothermally the temperature stays the same but the pressure does not 1.5 to 0.75 atm, the gas itself.1760

Calculate these variables and compare δ S with Q/T.1768

Let us see what is going on here.1773

The important thing is isothermal and reversible so this is going to affect the mathematics.1774

Let us go ahead and draw this one real quickly so we know what we are dealing with.1780

This is pressure and this is volume so this is an isotherm.1788

This is state 1 and this is state 2, whatever that happens to be.1795

State 2, we know this is 350°K and it is going to be 1.58 atm.1800

This is going to be 350°K because it is isothermal and it is going to be 0.75 atm.1809

This is your P1, this is your 1.5 and this is your 0.75.1818

I know it is not the scale and this is not exactly half of it but that is fine.1826

Here we have volume 1 and volume 2, it is expanding.1829

The gas is going from one volume and it is expanding to a larger volume that is what this is.1834

What we are doing is we are keeping the temperature constant so the expansion just not happened randomly.1843

This path or that path it actually happens isothermally, it happens along this path.1848

The temperature stays along this path.1853

Even more than that, it happens reversibly.1856

Reversibly means we absolutely follow this path.1859

In other words, we do not go this way, we do not go that way.1863

We can have an isothermal transformation that is not reversible.1871

That is not a problem, we can go ahead and expand the gas and keep the temperature constant.1875

It follows an isotherm but the actual path that it follows, the work that is done is going to be different depending on the path we follow, 1880

this work is a path function.1889

In this case, it is reversible so we are absolutely following this path.1891

What we are doing is we are making a bunch of little of micro changes all the way around.1894

Again this a reversible.1900

Remember what reversible means in terms of mathematics, it means that the external pressure = to the pressure of the system, 1901

the internal pressure if you will.1910

The pressure on the outside = the pressure on the inside so the system is always at equilibrium.1913

Any new small change that you make, all you have to do is go back that little small differential change.1920

Essentially, the system is in equilibrium.1925

Mathematically, it means that P external = P.1927

That is what reversible means.1930

Let us go ahead and see what is going on here.1933

I will just this label this so this is an isotherm.1937

It is very important so you can do something isothermally or you can do something isothermally and reversibly.1940

DW = P external × DV that is the basic, that is the definition of work.1950

Because it is reversible, the P external = P so we have DW =P DV.1959

This is an ideal gas so we have PV =nRT volume =nRT/ P.1970

My apologies, I’m going to do this, this is going to be P =nRT/ V.1993

I’m going to have put this into here so I get DW =nRT / V DV then when I integrate this, I end up with the following.1998

I get the integral of DW is W not δ W.2016

I get that the work = nRT × the integral volume 1 to volume 2 of DV/ V= work.2020

When I do this integral, I get nRT × the log of volume 2 / volume 1 so this is my fundamental equation.2043

Let us deal with this.2058

I do not have V2 and V1, what I have is P2 and P1.2060

Again, this is an ideal gas, let me express it this way.2064

I have PV =nRT, I have V =nRT/ P.2067

Therefore, V2 =nRT / P2, V1 =nRT/ P1.2075

This is isothermal so T stays the same, the pressure that is going to be different.2085

V2/ V1, I will go ahead and do on the next page.2093

V2/ V1 = nRT/ PQ / nRT/ P1 you end up with P1/ P2.2102

Therefore, our equation work = nRT Ln of V2/ V1 ends up becoming work = nRT × the log of P1/ P2.2117

We are just fiddling things around, just basic mathematics that is all it is.2138

Now we go ahead and solve it.2143

We have 1 mol, we have 8.314 J/mol-°K, we have a temperature of 350°K, and we have a log of 1.5 atm/ 0.758 atm.2145

This cancels giving us the log of a pure number and we get that the work = 2017 J.2169

That much work is done as the gas expands.2181

The system is doing work on the surroundings, work is positive.2187

It says that it is isothermal.2197

Isothermal implies that δ U =0.2201

However, be careful with this so δ U = 0 and let me tell you why.2208

In this particular case, this is the basic equation DU = CV DT + DU DV at constant temperature DV.2214

The temperature is constant so this term goes to 0.2230

DU DV we are dealing with an ideal gas so this term goes to 0 that is why DU = 0 or δ U =0.2234

You are going to run across a lot of problems in thermodynamics.2245

Most of the problems that you do are going to involve an ideal gas.2248

When you see the word isothermal, you can just automatically set δ U = 0, it is not a problem.2251

However, if you understand that δ U = 0 under isothermal conditions only for an ideal gas.2257

If you are not dealing with an ideal gas this term is not 0, if it is small or whatever happens to be but it is not 0.2263

Isothermal does not automatically mean that δ U is 0.2275

Isothermal for an ideal gas implies that δ U is 0.2279

There are times when we can set something to 0, isothermal δ U= 0, adiabatic Q = 0.2284

It is for an ideal gas, if this is not an ideal gas, if it is a Van Der Waals gas or another kind of gas, 2292

or the solid or liquid because this equation works for any system at all this is not to be 0.2301

We cannot just automatically set isothermal means δ U = 0 please remember that.2307

Let us go ahead and finish up here.2314

δ U=Q - W well δ U was 0 and Q the work was 2017 J.2316

Therefore, Q = 2017 J.2329

Let us go ahead and deal with δ H and δ S.2337

Let us do it over here, δ H we said was δ U + δ PV.2342

This is 0, therefore we have δ PV which is nothing more than P2 V2 - P1 V1 which is nothing more than nR T2 – nR T1, which is equal to nR δ T.2356

It is isothermal, the temperature stays the same so this is 0.2381

The δ H for this process is 0.2386

Another way of doing this, this δ H business, we could use the δ H equation DH = CP DT + DH DP at constant T DP, 2392

change in temperature δ T, this is an isothermal process so this term goes to 0.2408

For an ideal gas, this term goes to 0.2413

Again, we have DH = 0 which means that δ H =0.2416

Either one of these processes are absolutely fine.2422

You are going to end up getting the same answer.2424

Let us go ahead and take care of the δ S.2428

We have DS = CP/ T DT- nR/ P DP.2431

This is isothermal so that takes this term to 0.2444

We are left with DS = - nR / P DP.2448

We are going to integrate this like we always do and we end up with δ S =.2459

The nR comes out of the integral you going to be left with the integral from DP/ P so what you get is the following.2468

You get - nR × the nat log of pressure 2 / pressure 1.2479

We can go ahead and put the numbers in.2488

This 8.314 × the nat log of pressure 2/pressure 1.2491

Pressure 2 is 0.75 atm, pressure 1 is 1.5 atm, δ S is going to end up equaling -5.76.2498

The log 0.75/ 1.5 is 1/2 , the log of the number less than 1 is negative.2521

This number is negative × the negative gives you a positive.2528

δ S =5.76 J/°K.2534

Let us compare δ S with Q/ T.2544

δ S which is calculated δ S 5.76 J/°K.2547

Let us calculate Q/ T, in this case we can do T at Q/ T because T stays the same, it is just 350°K.2557

In this particular case, it is actually Q reversible because we are doing this process along the reversible paths so Q/ T = Q reversible / T.2565

That is the definition of entropy.2573

The Q reversible we have 2017 J, we have 350°K ends up being 5.76 J/°K.2581

What about 5.76? it is exactly the same because we are dealing with a reversible path.2595

A quick review of what is happening here physically.2608

Notice, how the work and the heat are equal.2611

We ended up with a work and the heat being equal to 2017 J, they ended up being equal.2614

Here is what is going on.2622

As the system expands, it does work on the surroundings.2627

It is W on the surroundings and the amount was I think 2017 J.2638

As it expands, as a gas expands it is going to cool but the problem is we are doing this process isothermally so we do not want it to cool.2651

We are going to keep the temperature the same as it expands.2671

Because the process is to happen isothermally, the system requires energy in the form of heat 2676

in order to keep the temperature up so the gas is expanding, the gas wants to cool.2710

It is happening isothermally so something, somehow, energy has to come into the system to keep the temperature from dropping, 2715

to keep it from cooling because we want to keep the temperature the same.2719

Because the process happens isothermally, the system requires energy in the form of heat to keep the temperature constant, to keep the T constant.2724

The question is where is this heat going to come from?2739

There is only one source, it has to come from the surroundings.2742

This heat as energy has to come from somewhere.2753

It comes from, it is pulled from the surroundings.2774

Since Q is positive, when it enters the system that is what we get Q =2017 J.2796

What is happening here is the following under isothermal conditions.2814

Work is done on the surroundings, the surroundings get colder.2818

This is our experience with the first law.2832

The surroundings get colder.2835

The gas expands does work on the surroundings 2017 J of work.2841

In the process of expanding the gas inside wants to cool, the system wants to cool.2846

We are doing this process isothermally,2851

In order for it to maintain its temperature, energy has to come into the form of heat.2853

Heat has to come from somewhere so it pulls that heat from the surroundings.2858

As the gas is doing work on the surroundings it is actually pulling heat from the surroundings.2862

When you pull heat from the surrounding, the surroundings are going to get colder.2868

Energy is transferred to the surroundings as work.2879

Energy is taken from the surroundings as heat.2884

That is all that is happening here.2887

Let us see what is going on, let us go to our last example here.2899

Example 5, the same starting conditions, different set of circumstances.2907

1 mol of an ideal gas, constant volume heat capacity 350°K, 1.5 atm, the gas expands isothermally.2913

We are isothermal, against a constant external pressure of 0.75 atm until the pressure of the gas achieves 0.75 atm.2920

The same exact initial conditions.2931

The same exact same exact initial state, same exact final state.2933

The final state is going to be 0.75 atm but now it is not going to expand reversibly.2937

It is going to do it isothermally but that is not going to be reversible.2943

Let us see what this one looks like.2946

We have our pressure, we have our volume, we have our isotherm, we have our state 1, we have our state 2.2953

This is state 1, this is state 2, this is pressure 1 that is the 1.5 atm and this is going to be P2, this is going to be 0.75 atm.2962

1.5 and 0.75 atm.2975

This is volume 1 and this is volume 2.2977

It is going to expand so isothermally the temperature is going to stay the same so we are still going to expand.2980

This is the isotherm but now the particular path that we are going to take is the following.2987

The external pressure is 0.75 so we are going to expand this way.2995

This is the path that we are going to take.3002

It is at constant external pressure so this is the amount of work that is done.3004

The last problem we did isothermal reversible work.3010

We follow this path, we kept the temperature the same but we did along the isotherm itself.3018

There was all this extra.3024

We can only expect that a work in this particular example is going to be less than the work that we had in the previous example.3026

Let us hope it actually turns out that way.3034

It is a different path.3038

The temperature is the same so we can do something isothermally but now we are following a different path so 3041

the work is going to be different, the heat is going to be different.3046

Where shall we start?3057

Let us start the same way, the definition of work. 3058

The definition of work is external pressure × the change in volume.3062

That means that work = the external pressure because the external pressure is constant, constant external pressure × δ V.3069

We just integrated this differential equation so we get work = P external.3080

This is going to be P external volume 2 - volume 1, we are dealing with an ideal gas therefore V = nRT / P.3092

Volume 2 = nRT/ P2 - nRT/ P1.3103

We get that the work = the external pressure × the nR T.3123

I’m going to pull out the nRT because it is going to be constant.3131

It is going to be 1/ P2 -1/ P1 and now we can go ahead and put our numbers.3133

work = the constant external pressure is 0.75 atm.3142

n is 1 mol, R is 8.314 J/°K.3149

Our temperature is isothermal, it is happening at 350 K.3155

1/ the external pressure P2 is 0.75 -1/1.5.3161

If the arithmetic comes out correct you should have 1455 J.3171

This work is definitely less than the 2017 J.3178

This 1455 accounts for this work was done against a constant external pressure.3182

In the previous problem we did it along the isotherm itself, we did this process reversibly.3187

Therefore, remember reversible processes are important because they represent maxima and minimal, things like that.3192

In the case of an expansion, it is the maximum amount of work that you are going to get from a particular expansion.3201

We are dealing with an ideal gas so isothermal again in the case of an ideal gas, isothermal definitely implies that the change in energy of the system is 0.3212

δ U =Q - W so 0 = Q – W.3226

Therefore, Q = W = 1455 J and it is still happening isothermally.3237

The heat and the work are still the same.3250

The same thing is happening, it is still expanding.3253

It is expanding and wants to get cooler, we are not letting it get cooler, it is doing or forcing it to do isothermally and it has to pull energy from somewhere.3255

It is going to pull it from the surroundings.3263

Work is done on the surroundings in expansion, the surroundings end up getting colder.3264

Let us go ahead and do DH.3273

DH =CP DT + DH DP DP constant temperature that goes to 0, ideal gas that goes to 0.3279

Therefore, our enthalpy change δ H for this process is 0.3296

δ S I’m going to calculate because δ S is a state property.3303

State 1 and state 2, the first path we took was this way.3313

This path now we are going this way.3318

This way and this way.3321

A state property does not depend on the path.3323

Therefore, the change in entropy is still to be 5.76.3327

It is the same number that we got from the previous example J/ K.3333

It is going to be 5.76 J/°K.3338

Heat and work are different because they are not state functions, they are state properties, they are path properties.3342

But because δ S is a state function the path that we taken together does not matter.3349

We can go ahead and use it from the previous example.3353

Let us go ahead and calculate Q and T.3358

Q in this case is 1455 J, temperature is 350°K, in this particular case we get 4.16 J/°K.3360

In this particular case because of a different path the entropy is actually less.3373

It is not the 5.76.3379

You are going to get the same entropy Q/ T and δ S when you run a reversible process because the definition of entropy is DQ reversible / T.3380

This reversible part is very important.3395

Thank you so much for joining us here at www.educator.com.3399

We will see you next time for more example problems, bye.3401