For more information, please see full course syllabus of Physical Chemistry

For more information, please see full course syllabus of Physical Chemistry

### Energy & the First Law 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
- Compression 0:20
- Compression Overview
- Single-stage compression vs. 2-stage Compression
- Multi-stage Compression
- Example I: Compression 14:47
- Example 1: Single-stage Compression
- Example 1: 2-stage Compression
- Example 1: Absolute Minimum
- More on Compression 32:55
- Isothermal Expansion & Compression
- External & Internal Pressure of the System
- Reversible & Irreversible Processes 37:32
- Process 1: Overview
- Process 2: Overview
- Process 1: Analysis
- Process 2: Analysis
- Reversible Process
- Isothermal Expansion and Compression
- Example II: Reversible Isothermal Compression of a Van der Waals Gas 58:10
- Example 2: Reversible Isothermal Compression of a Van der Waals Gas

### Physical Chemistry Online Course

### Transcription: Energy & the First Law III

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

*Today, we are going to continue our discussion of energy and the first law.*0005

*In the lesson, we have talked about the expansion of the gas.*0008

*Today, we are going to talk about the contraction of a gas + some other elements of work and things like that.*0011

*Let us go ahead and jump right on in.*0018

*Compression, as it turns out is going to be exactly the same as expansion in terms of the mathematics, the same equations apply.*0022

*Let us go ahead and recall what these equations are.*0028

*Let me go to blue and see here.*0031

*Compression, the same equations apply.*0035

*If we are expanding a gas under a constant external pressure, the amount of work or the numerical value is the external pressure × the change in volume.*0042

*If we have lots of differential elements, if in fact the pressure is not exactly constant throughout the whole stage,*0055

*if the pressure changes that is constant over a certain differential elements, we have this one which is the general expression v1 to v2, P external dv.*0066

*When we replace, when we take the differential limit as we go to more stages, we get to a point*0080

*where we are actually doing the expansion isothermally right along the isotherm.*0087

*We are not just doing it in 1 stage, 2 stages, or 50 stages.*0093

*We are actually following the isotherm when we do the expansion.*0097

*In that case, the work is going to be v1 to v2, the pressure dv, where we have actually replaced the external pressure*0101

*with the pressure of the system, the internal pressure of the system.*0110

*At that point, we are following the isotherm, the external pressure and the pressure of the system differ only by infinitesimal amount.*0114

*They are essentially equilibrium which is why we justified replacing this with this.*0121

*This is not the same and this is different.*0126

*This is external pressure, this pressure is the internal pressure of the system.*0128

*Let us go ahead and take a look at what a single stage compression looks like, the single stage compression.*0136

*We are taking a gas at a certain volume and we are squeezing it and making the volume smaller.*0149

*The final volume is going to be less than the initial volume.*0155

*A single stage compression and here I’m going to go ahead and draw the Pv diagram for a 2 stage compression so we can actually compare them.*0159

*We already have the background of expansion, the compression is exactly the same.*0171

*It is just reversed.*0175

*The paths says, underneath the isotherm that we are going to be above the isotherm.*0177

*It is going to look something like this.*0182

*We have an isotherm, in this particular case, because we are doing the compression this is our initial state down here.*0186

*It is a higher volume v2 v1.*0194

*We are going to be going from this initial state I’m going to call s1 and s2.*0198

*This is going to be P1 this is going to be P2.*0211

*We start at a certain volume and pressure.*0218

*The external pressure has to be at least as large as the final pressure in order for the compression to take place.*0224

*We know the external pressure has to be bigger than the pressure of the system.*0232

*Otherwise, we would not be able to compress it.*0234

*The external pressure is bigger than the internal pressure, we have to compress it.*0237

*It has to be at least as big as this.*0241

*The external pressure that way, the smallest value of external pressure is going to be that.*0243

*What happens is this, we compress the gas, the volume, and the volume gets smaller.*0249

*As it gets smaller, the pressure rises and rises until it gets to a point where the pressure exactly matches the external pressure.*0259

*The pressure inside the system matches the external pressure.*0270

*The total work done is that right there, work =the external pressure × the change in volume or*0275

*the external pressure is here with this value, that is the height.*0284

*The change in volume is this value, so the work done during this compression, the amount of work that I have to do to compress this gas isothermally*0291

*under constant pressure is the area underneath that graph.*0300

*The same compression but let us do it in 2 stages.*0306

*This was our first path single state compression.*0315

*We have the isotherm, we have state 1, we have state 2, this volume 1, this is volume 2, this is pressure 1, this is pressure 2.*0321

*This time I'm going to compress it, I’m going to use a pressure that is going to be bigger than P1 but less than P2.*0332

*My first external pressure is going to be somewhere around there.*0341

*I'm going to go along this path and I'm going to compress the gas until I hit the volume decreases.*0346

*As the volume decreases, we know the pressure of the system is going to rise until the pressure hits that external value of pressure that is going to stop there.*0357

*I’m going to put more pressure and going to raise the external pressure to P2.*0365

*I’m going to squeeze it some more until the external pressure now matches or the internal pressure matches the external pressure.*0372

*The work that I have done is going to equal this first stage.*0385

*In this 2 stage compression, the amount of work that I have done is the area underneath here, the shaded area.*0408

*Clearly, the work done here is more than the work done here by that amount.*0416

*That is the difference.*0427

*The total work here is the work done in stage 1 + work done in stage 2.*0431

*This is the work in stage 1, this is a work in stage 2.*0438

*These are compressions under constant external pressures.*0443

*In this case, it was 1 external pressure.*0461

*In this case, it was 2 external pressures.*0462

*First here, and then here.*0466

*Clearly, these 2 stage compression requires less work.*0471

*We do not need the numbers, we can see it just from the area.*0489

*This has a lower area than this so it requires less work if I do this compression in 2 stages I do not have to work as hard to actually compress the gas.*0492

*Less work than 1 stage.*0506

*You can see where this is going, a multistage compression 50, 100, 150.*0521

*If I keep going it is going to require less and less and less work.*0528

*A multistage compression would require yet less work.*0540

*A multistage just turning into a calculus course.*0566

*A multistage compression from a higher volume to a lower volume.*0575

*This is volume, this is pressure, it would require less work that we do to compress the gas.*0580

*As we pass to the differential limit, smaller and smaller changes in volume, we find there is a minimum amount of work that*0596

*the surroundings must do to compress the system from an initial state to a final state.*0632

*This is the initial state, this is the final state, there is a minimum amount of work.*0653

*It is work, less work, the more stages I do the compression in, the less work I have to do.*0658

*At some point, the work that I do is the area underneath the graph.*0664

*Underneath all of these, it pass to the differential limit that work is going to achieve a minimum value.*0670

*It is going to be some value below which I cannot go.*0678

*This minimum work is achieved if we compress the gas isothermally.*0683

*If we actually go along the isotherm, we actually go this way.*0715

*Again, if you keep taking, you know this already from calculus along the isotherm.*0721

*We are not going to take that path or that path or this path.*0732

*We are going to take straight, we are going to pass right along the isotherm.*0739

*In this particular case, what you get is work.*0745

*Volume 1 volume 2, Pdv, Ihave replaced external pressure with the internal pressure of the system because*0751

*as we pass along the isotherm, the difference between the external pressure and the internal pressure of the system is infinitesimal.*0760

*It is so small that it really does not matter.*0769

*Therefore, I can replace the external pressure with the internal pressure.*0775

*They are essentially in equilibrium at all times along this path where we have replaced external pressure with the pressure of the system.*0778

*The pressure, internal pressure of the system itself, we do not put the subscript here but this is essentially where it is,*0800

*With the pressure of the system at any given moment along the compression.*0806

*This is the exact same thing as along the expansion.*0818

*When we do the isothermal expansion actually along the isotherm and that is the path we take, we get the maximum amount of work that a gas can do as it expands.*0822

*If we do an isothermal compression along the actual isotherm itself, we get the minimum of amount of work*0834

*that we have to do that the system of the surroundings have to do on the system in order to compress that gas.*0844

*You can do a lot of work to compress a gas or you can do the minimum lot of work to compress a gas.*0853

*You can do a certain amount of work.*0859

*When the gas expands, it can do a certain amount of work but the maximum amount of work that I can do is done when it expands along the isotherm itself.*0862

*That is what is going on here.*0873

*Let us see here, let us go ahead and take a look at an example and see if we can make sense of some of this numerically.*0881

*A little long but it is very important, a lot of things are going to be dealt with here.*0892

*An ideal gas occupies 4 L at 2 atm and 25° C.*0897

*It is compressed to a volume of 1 L, calculate the minimum work needed to compress the gas isothermally by a single stage compression at constant pressure.*0906

*Here is the crucial, then calculate a minimum work required to compress it isothermally in a two stage compression.*0918

*First to a volume of 2 L, then from 2 L to 1 L, the initial status 4 L, the final state is 1 L.*0927

*It is a compression.*0936

*Single stage we are just going from 4 to 1 directly under constant pressure.*0937

*In a 2 stage, we are going to go from 4 to 2 and 2 to 1, under 2 different constant pressures.*0941

*Finally, calculate the absolute minimum.*0948

*The absolute minimum work required to affect the same change of state.*0952

*Let us go ahead and draw the Pv diagram to help us out here.*0959

*We are going to go like this, we have an initial state, we have a final state.*0967

*The initial volume is going to be 4 L and the final volume is going to be 1 L.*0974

*We said that the initial pressure is 2.0 atm and this is going to be our final pressure which we do not know yet.*0982

*Notice that they did not give it to us here.*0993

*I would have to look calculate that, that is not a problem, p1 and v1, p2 and v2.*0995

*We already know how to do that.*0999

*Let us see what we can do, a single stage compression.*1004

*Isothermally just means that the temperature stays the same, the minimum amount of work needed to compress the gas isothermally by a single stage compression.*1010

*In order to compress this gas from this state to this state, from 4 L to 1 L.*1021

*The external pressure has to be bigger than 2 to be able to stop the compression.*1028

*In order to reach the final pressure that was suppose to reach, a volume of 1 L whatever that pressure happens to be, it has to be at least as big as the pressure.*1035

*A single stage, the external pressure has to be at least as big as a final pressure.*1046

*For a single stage, we have the following.*1062

*The minimum work needed means that the external pressure has to be at least as large as the final pressure.*1071

*Let us go ahead and find out what the final pressure has to be.*1104

*P1v1 =P2v2, we are looking for P2.*1109

*Therefore, P2 = P1v1 / v2.*1114

*The initial pressure is 2 atm, the initial volume is 4 L, the final volume is 1 L.*1122

*Therefore, our final pressure has to be 8 atm.*1137

*That means the external pressure has to be at least 8 atm in order to affect that single stage compression.*1142

*The external pressure has to be there in order to do this compression.*1157

*Now that we know what the external pressure is 8 atm, let us go ahead and calculate the work for that.*1162

*Therefore, our work is equal to our external pressure × change in volume.*1170

*Our external pressure is 8 atm and our change in volume final – initial, 1 L -4 L=8 atm × -3 L =-24 L/ atm.*1177

*I’m not going to go ahead and convert it to joules, it is the numerical value that matters.*1199

*We have -24 L atm.*1203

*Let us go ahead and do this in 2 stages.*1208

*A 2 stage compression, for the 2 stage compression let us go ahead and draw that over here.*1214

*The 2 stage compression is going to look like this.*1225

*We have this and we said this is 4 and this is 2 atm, this is 4.0 L, this is going to be a Pf.*1229

*That is our final pressure, it is not a problem.*1244

*A 2 stage it is going to go from 4 L to 2 L and that is going to go from 2 L to 1 L.*1250

*Let us go ahead and put the 1 L here and let us put the 2 L here.*1256

*It is going to look like this.*1262

*That path it is going to take is going that way.*1271

*It is going to compress until it reaches a certain pressure inside and then we are going to have a new pressure.*1273

*It is going to be up here and it is going to compress some more.*1281

*The work done is going to be that right there.*1285

*The work total =work 1 + work 2.*1288

*The area underneath this rectangle, that is going to be work 1, the area underneath this rectangle that is going to be work 2.*1298

*This is going to equal the external pressure × the change in volume for the first stage and the external pressure and the change in volume for the second stage.*1306

*Let us go ahead and find out what these are, put it in, and see if we can come up with some numerical value for that.*1319

*For stage 1, it looks like this.*1326

*Stage 1, I need to find a certain pressure that is going to be bigger than 2 atm but is going to be less than the actual final pressure that we want,*1337

*The pressure that accompanies this 1 L.*1345

*It has to be somewhere in between here, in order to bring it here.*1348

*It has to be that value, the value of the pressure for the first stage has to match the pressure along the isotherm that takes us to 2 L.*1351

*Stage 1, we have P1v1 =P2v2, therefore, P2 =P1v1/ v2.*1363

*The initial is 2.0 atm, I’m going to go ahead and use my units here.*1385

*The initial pressure is 2 atm, the initial volume is 4 L, the final volume in the first stage is going to be 2 L.*1393

*I'm left with 4.0 atm, this is going to be my external pressure for stage 1.*1407

*The change in volume = 2 L -4 L =- 2 L.*1419

*I have the first stage, I have an external of 4 atm, I have a change in volume of 2 L.*1444

*I will do the actual multiplication of the n.*1453

*Let us do stage 2, for stage 2 my external pressure that is equal the external pressure from the first part, the single stage expansion.*1455

*We are still going from an initial state to a final state.*1472

*The final state is 1 L, that 1 L that is just the final state.*1475

*We already calculated what the pressure is going to be, that is going to be 8 atm.*1479

*Because now we need the pressure to be up here, in order to affect this compression, to go from 2 L to 1 L.*1491

*As we go from 2 L to 1 L, the pressure is going to rise until the pressure reaches whatever pressure this is for 1 L, which happens to be 8 atm.*1501

*The δ v of the second stage is equal to 1.0 -2.0 =s-1.0 L so we have that and we have that.*1513

*We can go ahead and calculate our work.*1529

*so the total work for this 2 stage is going to be work 1 + work 2.*1533

*Work 1 = the external pressure for stage 1 is 4 atm and the change in volume is -2.0 L.*1544

*This is going to be 8.0 atm × -1.0 L =-8 + -8 =-16.0 L atm.*1556

*Notice this is less than the single stage.*1575

*The single stage is -24 L atm.*1579

*In other words, the surroundings did 24 L atm of work on the system.*1584

*If we do it in 2 stages, the surroundings does 16 L atm.*1590

*It is dropping, it is confirming everything.*1595

*Let us go ahead and calculate the absolute minimum.*1598

*The absolute minimum is achieved, this is expansion along the isotherm itself, not this way or this way but along the isotherm itself.*1603

*This is expansion along the isotherm and that work is equal to the integral from the initial volume of Pdv.*1623

*This is an ideal gas.*1644

*We will put that in there, work =v1v2 nRT / vdv = when I do this integration and our T is a constant, I pull it out.*1657

*I have to do the integration in a previous lesson or we end up with the following nRT × log of v2 / v1.*1675

*Let us see what we have here, how can we solve this?*1694

*A couple of ways that we can do this, I think I’m actually going to do both ways.*1700

*We know what R is, we know what T is, we need to find what n is.*1712

*We already know what the final volume is 1, we know what the initial volume is 4 L.*1717

*We need to worry about the moles.*1722

*Let us go ahead and find n, Pv =nRT.*1725

*If we rearrange this to solve for n, I get n =Pv/RT.*1728

*The initial pressure is 2.0 atm, the initial volume is.*1754

*This is fine, it is not a problem.*1769

*We have the work equal to this, we do the integration, this is what we want to solve.*1771

*We have v1, v2.*1774

*We have T which is 25℃, the 298 K, we have R which is the 0.08206 if we are dealing with L atm, we need to find the number of moles.*1775

*For Pv =nRT, n =Pv/ RT.*1789

*In terms of Pv, we can either use the initial pressure, the initial volume or the final pressure and the final volume.*1795

*It actually does not matter what it is.*1802

*If you happen to notice, you do not have to go through this to find what n is.*1805

*Let us go back to Pv =nRT, nRT = the pressure × the volume of the system.*1811

*The pressure × the volume P1v1 =P2v2.*1823

*It does not matter, it is equal to P1v1 or I can use P2v2, it does not matter.*1827

*This nRT, this value right here, I can just take P1 × v1 if I need to.*1834

*If I want to do P2v2, that is not a problem.*1841

*In this particular case, nRT = let us use P1v1.*1846

*The initial pressure was 2.0 atm and the initial volume was 4.0 L.*1851

*Therefore, nRT = 8 L atm.*1862

*The work is equal to 8 L atm × nat log.*1872

*The final volume was 1 L, the initial volume was 4 L, when I do this calculation I get -11.1 L atm.*1888

*The single stage compression took 24 L atm of work.*1906

*A 2 stage compression took 16 L atm of work.*1912

*I can take more stages 3, 10, 50, 100 this number is going to keep dropping.*1917

*At a certain point, it is going to hit a minimum.*1922

*That minimum is the area underneath the isotherm from volume 1 to volume 2.*1924

*The minimum amount of work that I have to do in order to compress this gas is 11.1 L atm.*1932

*Notice, because the final volume is less than the initial volume, the sign automatically takes care of itself.*1939

*This -11.1, the amount of work that is done is 11.1.*1947

*It is negative, it is a work done by the surroundings on the system.*1953

*That means energy is transferring as a work from the surroundings to the system.*1957

*This is the minimum amount of work that I have to do with the surroundings has to do in order to compress this gas.*1963

*Let us see what we have got here.*1973

*A gas can expand or be compressed isothermally along many paths.*1978

*When we see that they are doing expansion or a compression isothermally, that does not mean we are doing it along the isotherm.*2004

*We can do it along any path.*2011

*Isothermal just means that we are keeping the temperature the same as it moves from one state to the next, initial to final or any intermediate stage.*2013

*An isothermal expansion or compression is not the same as an isothermal expansion or compression along the isotherm.*2024

*When the isotherm itself is the path, that represents a maximum or minimum quantity, that is the difference.*2032

*When we say isothermal, isothermal just means that it means they are keeping the temperature the same.*2039

*You can follow any path you want.*2044

*When we say isothermally along the isotherm, that represents the absolute minimum and absolute maximum quantities.*2046

*Gas can expand or compressed isothermally along many paths.*2058

*The path that follows the isotherm is the one that gives the maximum work done during expansion.*2065

*It also gives the minimum work required for compression.*2105

*In both cases, the numerical value of the work done is = to the integral from the initial volume to the final volume of the pressure × the differential volume element,*2125

*Where P is the pressure of the system at any point along the transition.*2146

*Along the isotherm, if the path that we actually take is along the isotherm, the external pressure and the internal pressure of the system may differ infinitesimally.*2164

*We are justified in replacing P external by P itself.*2212

*It is very important.*2227

*Let us go ahead and talk about reversible and irreversible processes.*2233

*You are going to hear the word reversible used all the time throughout thermodynamics.*2238

*Let us have a change of pace, let us go ahead and try red.*2248

*Reversible and irreversible processes.*2257

*We are going to subject gas and we are going to subject it to a cyclic process.*2270

*We are going to start at an initial state.*2276

*We are going to take it to a final state.*2278

*We are going to do something else would bring it back to the initial state.*2280

*We are going to do it in two different ways.*2283

*We are going to have some initial state s sub I, which is going to be pressure 1 volume 1.*2288

*It is going to be isothermal so the temperature is going to stay the same.*2297

*We are going to have s2 or s final and this is going to be a pressure 2 volume 2 and T.*2300

*What we are going to do is we are going to get 2 stages.*2311

*We are going to start at stage 1 and go ahead and expand that gas.*2313

*We are going to expand it to a final state.*2319

*What we are going to do is we are going to compress it.*2321

*Under compression, we are going to bring it back to its initial state.*2326

*We are going to do it in 2 different ways.*2329

*Process 1, we are going to do it as a single stage expansion with the external pressure = P2 and we are going to do a single stage compression.*2339

*We are going to bring it back with the external pressure = P1.*2369

*The second process, I will do the same thing.*2376

*I’m going to start with state 1, we are going to take it to state 2.*2381

*We are going to take it back to state 1 but this time the path it would take would follow the isotherm itself.*2383

*The process 2, this is going to be the infinite multistage expansion.*2392

*We know from calculus or from what we just saw, the infinite multistage expansion up and down, that differential infinitesimal amount,*2406

*that is the actual path along the isotherm.*2413

*An infinite multistage expansion and compression then compression along the isotherm.*2415

*Let us see what we got here.*2440

*Let us go ahead and look at process 1.*2443

*Process 1 looks as follows.*2447

*Let me draw it here because I want some room.*2456

*This is state 1, this is the initial state, and this is the final state.*2467

*It is P1 P2 v1 v2, this is v1 v2 P1 P2.*2479

*A single stage expansion going from here to here in a single stage expansion would look like this.*2487

*This is going to be the work done by the expansion.*2501

*A single stage compression, now we go backwards.*2505

*Single stage compression has to follow this path.*2508

*The work of the compression =the work under this whole rectangle.*2516

*The work of the expansion is a work under this rectangle.*2521

*Notice that the work of the compression is a lot bigger than the work of the expansion.*2526

*Therefore, the total work = the work of the compression + the work of the expansion = the work of the compression is going to be P1 × v1 - v2.*2537

*In the compression, this is the final state and this is the initial state.*2557

*It is going to be v1 - v2.*2562

*The work of the expansion is going to be P2 × v2 - v1 v2 - v1.*2566

*I’m going to switch this around, v2 - v1.*2575

*Therefore, this is going to become -P1 × v2 - v1.*2578

*Whenever I switch these, I factor out a negative.*2583

*The factored negative is going to be put there + P2 × v2 - v1.*2586

*I have v2 – v1 v2 – v1.*2593

*This is going to = P2 - P1 × v2 - v1.*2596

*Pressure 2 - pressure 1, pressure 2 - pressure 1, this is negative, v2 - v1 this is positive.*2604

*What I end up is a negative total work done.*2612

*The total work done in a cyclical process I start here, I expand the gas in a single stage.*2621

*The work is this much.*2628

*I compress the gas through the path, the work is this much.*2631

*The total difference in work, the total work done actually ends up being negative.*2637

*Notice, I end up losing that much work.*2641

*The surroundings ends up having to do that much more work in order to compress the gas,*2646

*it does not recover the work that was done along the expansion.*2650

*The total work is negative, mean work has been destroyed in the surroundings.*2657

*Work is negative that means work has flowed from the surroundings to the system.*2677

*It is negative, its float from the surroundings to the system.*2684

*This cyclical process we would expect it to be the same but it is not.*2688

*You end up doing more work in your compression than you get back in the expansion.*2692

*Or you end up doing less work in the expansion but it takes more work to actually compress the gas back to where it was before, back to its initial state.*2697

*the total work of the process is not the same.*2707

*Work has left the surroundings.*2711

*There is a weight in the surroundings that is actually lower than it was before.*2715

*The surroundings has lost work net.*2719

*Let us go ahead and look at process 2.*2727

*Now process 2 looks like this.*2732

*Here is the isotherm, here is the initial state, here is the final state, it is going to be P1, P2, v1, v2.*2741

*We are going to do this along the isotherm.*2755

*We are going to expand along the isotherm and we are going to compress back along the isotherm.*2758

*The work of expansion along the isotherm = the integral from v1 to v2 of Pdv.*2768

*That is this way.*2782

*We will go back to our initial state, the work of compression that is equal to v2 v1.*2784

*We went from v1 to v2, we are going from v2 to v1 of Pdv.*2795

*It is the same, the internal pressure of the system at any point along the compression of expansion so these are the same.*2802

*Therefore, the work total is equal to the work of the expansion + the work of the compression is equal to the integral from v1 to v2 of Pdv + the integral from v2 to v1 of Pdv.*2810

*I’m going to switch this integration and turn this into a negative sign.*2832

*Equals the integral from v1 to v2 of Pdv - the integral v1 v2 of Pdv =0.*2838

*A single stage expansion followed by a single stage compression to go from state 1 to state 2, then back to state 1.*2854

*We ended up with a total loss of work.*2861

*Work is negative it is 0, the surroundings lost work.*2862

*That is done isothermally, we kept the temperature the same.*2868

*In this case, instead of following this path and this path.*2873

*We follow this path during the expansion and this path during the compression.*2881

*We see that the work is actually 0.*2885

*There HAs been no net loss of work by the surroundings.*2889

*The work is 0, this is rather extraordinary.*2893

*No net work has been lost.*2903

*In other words, the system has been restored to its original state which is no different than the first process.*2910

*That process the system has return to its original state.*2931

*In its initial state, we find s1 and s2 and back to s1.*2935

*There is no difference here.*2939

*In both cases, the system has been restored to its original state and the surroundings have been restored to their original state.*2940

*The surroundings have been restored to their original state.*2955

*The work is 0, the surroundings have been restored to their original state.*2959

*We call this irreversible process.*2974

*Any process that restores the surroundings to their original state is irreversible process.*2981

*Any cyclical process that ends up bringing the system back to its original state but where something is not the same in the surroundings, that is an irreversible process.*2988

*Let us go ahead and write this down.*3001

*Reversible process, when a system undergoes a change of state by a specific process that is restored to its original state by the same sequence in reverse,*3003

*then if the surroundings are also restored to their original state, the process is called reversible in both directions.*3007

*When a system undergoes a change a state by a specific process, goes from an initial state to another state and is restored back to its initial state.*3117

*It goes from s1 to s2 and goes from s2 to s1 by following the same sequence of steps that it went from s1 and s2 but in reverse.*3124

*If the surroundings are also restored to their original state, the process is called reversible.*3136

*If during the change, in going from s1 and s2 then back to s1, if the surroundings are not restored to their original state, the process is irreversible.*3144

*Reversibility represents an idealization.*3155

*In this particular case, it represents the expansion along the isotherm.*3159

*It represents the compression along the isotherm.*3164

*When that is the case when we actually perform a process on expansion and a compression, when we do it isothermally along the isotherm,*3168

*there is no net change, there is no loss of work, that is what is important.*3178

*A reversible processes are important.*3183

*You saw that we can actually do the expansion and compression isothermally along any path.*3186

*There are specific paths, mainly one specific path along the isotherm both this way and that way.*3192

*That makes the process reversible where everything is restored exactly to where it was before.*3201

*It is as if nothing happened.*3206

*Reversible processes are important because they represent maximum and minimum quantities or values.*3208

*We already know before we talk about reversible processes, we talked about how if you travel along the isotherm,*3238

*you are going to end up getting the maximum amount of work in the expansion of the gas isothermally*3245

*or the minimum amount of work required for you to actually compress that gas isothermally.*3251

*We are giving it a name, when you are doing this isothermal expansion reversibly, in this case along the isotherm that is when you achieve your maximum and minimum values.*3257

*We will go back to blue, to make it look a lot better.*3274

*The isothermal expansion or compression of a gas along the actual isotherm is the reversible isothermal expansion or compression.*3286

*As opposed to just an isothermal expansion or compression which can happen along any path in a single stage or in multiple stages.*3329

*The maximum work obtained by an expanding gas is irreversible isothermal expansion.*3367

*The minimum work required to compress the gas is the reversible isothermal compression.*3402

*Again, when we say that something is happening isothermally, all we are saying is that the temperature is being kept constant for the process.*3433

*We are not saying that you are actually moving along the isotherm.*3441

*If we mean we want to move along the isotherm, we will say you are explicitly along the isotherm.*3445

*We will just use the term reversible that is what is going on here.*3451

*Isothermal can be any path but reversible, isothermal is one path.*3455

*It is along the isotherm itself.*3460

*It is what gives you maximum and minimum quantities of work.*3462

*Of course, net work again is equal to the integral from an initial volume to a final volume of Pdv.*3467

*Let us go ahead and finish up with an example here.*3485

*One more page, example number 2, the van der Waal’s equation states is a correction to the ideal gas law to account for deviation from ideal behavior in real gases.*3490

*The ideal gas law you know is Pv =nRT or something which is a little more instructive nRT/ v.*3502

*The van der Waal’s gas says that the pressure does equal to nRT =v.*3514

*There is a pressures that have to be make, P = nRT/ v - nb – an²/ v².*3518

*Or a and b are constants that have to do with a particular gas in question.*3527

*We want you to find an expression for the work done in the reversible isothermal compression of a van der Waal’s gas from an initial volume to a final volume.*3533

*Let us go ahead and see what we can do.*3545

*We already know the work is equal to initial v final Pdv.*3547

*We have an expression for P, it is right here.*3563

*Let us go ahead and put that in there.*3566

*It is equal to the integral from vi to to vf of nRT/ v - P – an²/ v² dv.*3570

*Let us go ahead and solve this integral now.*3591

*We will go ahead and separate that out.*3593

*The work is equal the, I’m going to integrate this first term right here and our T is a constant.*3596

*That is v – P, that has to stay in there.*3606

*It is going to = nRT × the integral of v1 to v2 of dv/ v - b – nb.*3609

*V – nb – n² ×the integral from v1 to v2 of dv/ v².*3624

*Dv/ v – nb, this is a logarithm, dv/ v² is v² is the same as v -2 dv.*3637

*You can do that the way you integrate any other thing, what you get is the following.*3646

*I will go to the next page here, work =nRT × log of v2 - nb/ v1 - nb – an² -1 / v.*3652

*v1 to v2 which we will end up with the final expression of nRT × log of a2 - nb/ v1 – nb.*3680

*It comes out here so you will get +an² 1/ v2 -1/ v1.*3695

*There you have that.*3707

*If we are dealing with van der Waal’s gas in a particular situation, we decide to use the van der Waal’s equation instead of the ideal gas equation.*3708

*If we want to find out what the amount of work done by the isothermal reversible expansion or compression of the van der Waal’s gas is, this is the expression for it.*3716

*Thank you so much for joining us here at www.educator.com.*3731

*We will see you next time for further discussion of energy in the first law, bye.*3733

1 answer

Last reply by: Professor Hovasapian

Mon Sep 7, 2015 6:52 AM

Post by Shukree AbdulRashed on September 6, 2015

Just to be clear, with all of these graphs, we're evaluating the work done from the perspective of the system (the gas under the piston)? Thank you.

1 answer

Last reply by: Professor Hovasapian

Tue Jun 23, 2015 1:39 AM

Post by Jinhai Zhang on June 22, 2015

Hello, Professor! In the text you referenced, it has the example of van der waals derivation for reversible isothermal work, and the book it use a - sign in front of the integral, and the answer it derived was -nRTIn(...)...

why it has an negative sign in front, and our discussion has no negative sign in front. is that the textbook you suggested made a mistake?