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

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

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### 1st Law Example Problems I

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
- Fundamental Equations 0:56
- Work
- Energy (1st Law)
- Definition of Enthalpy
- Heat capacity Definitions
- The Mathematics
- Fundamental Concepts 8:13
- Isothermal
- Adiabatic
- Isobaric
- Isometric
- Ideal Gases
- Example I 12:08
- Example I: Conventions
- Example I: Part A
- Example I: Part B
- Example I: Part C
- Example II: What is the Heat Capacity of the System? 21:49
- Example III: Find Q, W, ∆U & ∆H for this Change of State 24:15
- Example IV: Find Q, W, ∆U & ∆H 31:37
- Example V: Find Q, W, ∆U & ∆H 38:20

### Physical Chemistry Online Course

### Transcription: 1st Law Example Problems I

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

*We have come to the very important parts in these lessons.*0006

*We have come to the beginning of the example problems.*0010

*We have gone through a lot of theory.*0014

*We have done a fair number of derivations that is a lot of mathematics.*0016

*As I promised you in the previous lessons, we re going to start on the example problems and we are going to do a lot of them.*0020

*When I say a lot of them, I mean a lot.*0027

*It is very important that we have good graphs of what is happening particularly with the first law.*0031

*With energy, with the work, and heat, that is going to lay the foundations.*0038

*We do not just handle them mathematically, that is true.*0042

*Of course, we want to because we have to do the problems but we want to understand what is going on.*0046

*That will carry us forward.*0051

*With that, let us go ahead and just jump right on in.*0053

*A lot of you are going to approach this particular class with memorizing a bunch of equations.*0059

*Let me go ahead and tell you right off the bat that is not going to work.*0068

*Memorizing equations will work for a specific set of problems or for a specific type of problem, this is higher science,*0072

*this is more sophisticated science or sophisticated mathematics.*0080

*The idea is to have a reasonable degree of understanding to have a handful of equations at your disposal and let the problem at hand,*0084

*However, it is worded because it is going to be worded and the same thing is going to be worded in 10 different ways.*0092

*You need to be able to extract that information, find out what the problem is asking,*0098

*and use the handful equations that you have at your disposal to derive what you need.*0102

*You are not going to be asked to do a complex derivation that is just for your scientific literacy, the derivations that we went through.*0108

*But as long as you understand what is happening, keep the equation that you learn to a minimum,*0116

*just the fundamental set of equations and that is what I'm going to give you.*0121

*In order to solve all these problems, I'm going to give you the basics of equations that you need to know in order to do all these problems.*0124

*Our fundamental equations, the ones that you want to know are the following.*0133

*Fundamental equations and you can use these for every single problem.*0140

*You can use a subset of them for every single problem but these are the ones that you have to know.*0145

*Fundamental equations.*0156

*Work, we have to know what work is and that is the following.*0163

*I'm going to give almost all of these equations the differential form, the finite version is just a δ in front of it for the state functions, for the path functions.*0170

*Do not stick any δ in front of it.*0177

*The work is equal to the external pressure × change in volume that is the definition of work.*0182

*Now what about energy in the first law?*0191

*Energy in the first law says that change in energy of the system is equal to the heat withdrawn from the surroundings - the work produced in the surrounding.*0195

*That is DQ =- DW.*0220

*The definition of enthalpy is very important, it is equal the energy of the system + the pressure of the system × the volume of the system.*0225

*This is not external pressure, the pressure of the system.*0242

*The definition of the heat capacities is very important.*0247

*The heat capacity definitions we have two of them.*0268

*We have a constant volume heat capacity, the constant volume heat capacity is defined as*0271

*the heat withdrawn from the surroundings divided by the change in temperature of the system.*0278

*When I said heat withdrawn from the surroundings under constant volume, under constant volume*0284

*the heat withdrawn from the surroundings is that heat that goes into the system.*0290

*It is just a questionable point of view.*0293

*It is really important be able to switch back point of views because that means you understand what is happening thermodynamically.*0296

*Qualitatively what is going on is which moving and in which direction.*0302

*That was identified with this partial derivative DU DT.*0307

*If you have a function, if you have the energy and it is a function of temperature and volume,*0314

*the partial derivative of that function with respect the temperature is that heat capacity under constant volume.*0321

*And of course, we have the constant pressure heat capacity.*0327

*The constant pressure heat capacity = DQ DU DT,*0330

*the heat withdrawn from the surroundings under conditions of constant pressure ÷ the change in temperature.*0337

*Again, this is the same that definition of the heat capacity is the same.*0343

*It is heat ÷ temperature.*0347

*This happens to be idea, these three lines for definition.*0350

*This happens to be identified with the enthalpy.*0354

*This is the change in enthalpy per change in temperature.*0358

*Under constant pressure conditions that is how best defined.*0360

*The fundamental equations, you have this one.*0365

*It would do this in red.*0369

*This one you have to know, this one you have to know, and you have to know these two.*0373

*When I say you have to know, you have them as a piece of paper right next to you.*0381

*Eventually, you are going to do enough problems so we are going to memorize them.*0384

*These are the ones that you will memorize not the others.*0387

*Let us go back to blue which you wan to know as far as the mathematics.*0394

*In mathematics we have two relations of a mathematics that you have to know.*0402

*DU = CV DT + DU DV DV this is the general equation.*0410

*The change in energy of the system, if the energy is a function of temperature and volume,*0422

*the differential total change if I change temperature and volume, if I change both it is this.*0430

*And I also have an analogous one for the enthalpy, the DH.*0437

*DH = CP DT + DU DP, constant temperature DP.*0442

*If the enthalpy is a function of temperature and pressure, the change in enthalpy of the system is this + that, if I change both temperature and pressure.*0454

*If they say this is a constant pressure process this goes to 0 because DP = 0.*0463

*If they say it is a constant temperature process DT = 0.*0467

*This goes away.*0471

*This is the general equation.*0473

*You are letting the constraints of the system tell you what is happening, what you need to do these terms.*0474

*Let us go ahead and I will write is the fundamental concepts.*0482

*I will right this on this page.*0488

*Fundamental concept, isothermal, when I say something is isothermal that means it is a constant temperature.*0495

*Mathematically, it means the following.*0515

*It means that the DU is= 0 or finite δ U = 0.*0517

*Isothermal means that the temperature does not change.*0522

*If the temperature does not change, the energy of the system does not change, DU = 0.*0525

*In your equations, you can set DU= 0.*0529

*You can out 0 in for DU.*0533

*Adiabatic means DQ = 0, it means no heat is actually allowed to flow.*0535

*If they say the word insulated that means it is adiabatic DQ = 0 or the finite form δ Q.*0548

*Because heat is a path function, it is not a state function.*0560

*Only state functions have the δ.*0563

*Isobaric, if you happen to see the work that just means constant pressure,*0569

*When I say constant pressure, they will use the word isobaric.*0575

*Mathematically, that means D P = 0.*0583

*If you see isochoric or isometric and the usual you would not use these terms.*0590

*The usually terms for pressure and volume, they usually say constant volume and constant pressure.*0598

*This is constant volume.*0603

*For constant volume, DV = 0, that is the mathematics there.*0604

*Final set, for ideal gases, these are just the basic things that we need to know in order to solve the problems.*0614

*This is what we want to have at our disposal.*0622

*For ideal gases, we know that PV = nr, that is our equation of state from ideal gas.*0625

*We also have that relationship between the constant pressure and constant volume heat capacities that is CP - CV = nr.*0635

*Or in terms of molar, CP mol - CV mol= R.*0645

*Molar just means that you divide everything by n that is all it means.*0653

*From ideal gas, go back to the general equations DU DV under constant temperature = 0 for ideal gas.*0657

*If it is not ideal, it is not going to be 0.*0671

*Actually, in this particular case probably will be 0.*0674

*We want to write a general equation for ideal gas is definitely = 0.*0678

*And for ideal gas is DH DP.*0682

*The change in enthalpy, the change in pressure per unit change in pressure, if I change the pressure by one unit, 1 atm how is the enthalpy going to change?*0686

*0 from ideal gas the enthalpy is not a change of pressure.*0697

*It is not a function of pressure.*0703

*The energy is not a function of volume.*0705

*Energy is a function of temperature.*0708

*The enthalpy is a function of temperature from ideal gas.*0710

*With that, let us just go ahead and jump into this very long process of example problems.*0716

*We are going to do a lot of them like it is very important.*0721

*I want to understand this very well.*0725

*Let us see what we got.*0725

*Our first example is going to be 2 mol of an ideal gas undergo the following changes in state a, b, and c.*0730

*What is the change in temperature for each change in state?*0737

*Molar constant volume heat capacity = 12.5 J/ mol/ K.*0742

*In this case they have given us a molar value.*0748

*They give us that, if we want to keep an eye on units as much as anything else.*0751

*2 mol of ideal gas undergo the following changes in state.*0760

*Let us go ahead and deal with part of a.*0763

*Let us go ahead and recall our conventions before we actually get to the problem.*0766

*Our conventions are the following.*0773

*Heat is positive when heat flows into the system.*0777

*Work is positive when work is produced.*0800

*When you see the work, work is produced that means work is positive.*0811

*Work is produced in the surroundings.*0814

*Work we are looking at something from the point of view.*0817

*In the case of work, we are looking at the surroundings point of view.*0819

*When you see the work produced, it means work is positive in the surroundings.*0823

*If you see work is destroyed that means it is negative.*0827

*That means work is leaving the surroundings.*0830

*Work is going into the system.*0833

*Heat is positive when heat flows into the system.*0835

*That means it is flowing out of the surroundings.*0838

*It is just a question of point of view but this is our convention.*0840

*Let us go ahead and see what we have got.*0845

*We want to know what the temperature changes.*0848

*The equations that we have at our disposal, we are talking about heat and work.*0851

*They want to know change in temperature.*0855

*It looks like we would do something like this.*0858

*We have an ideal gas.*0862

*We know that if we are dealing with an ideal gas we are going to use these equations right here DU =DQ – DW.*0866

*Or we are going to use the finite form because they actually give us the values DU = Q – W.*0880

*We are also going to use the following.*0885

*We are going to use DU = CV DT.*0887

*The second term the DU has this one + the partial derivative, the DU DV this is an ideal gas.*0892

*DU DV = 0 because it = 0 we are left with this equation.*0898

*Or the finite form δ U = CV δ T.*0903

*Will we are looking for changes in temperature so we try to find δ T.*0911

*Therefore, I'm going to solve this for δ T.*0916

*Δ T = DU/ CV.*0919

*I have CV, I just need to find δ U and do the division.*0924

*Let us go ahead and do that.*0929

*Part A, they say that 500 J of heat flows out as heat, that means is flowing out of the system.*0930

*That means it is positive when heat flows into the system.*0941

*If it flows out that means Q is negative so this is -500 J.*0944

*Work they are telling me 150 J of work is destroyed which means work is negative.*0952

*Work = -158 J.*0959

*Δ U = Q - W = -500 -150.*0966

*Therefore, δ U = - 350 J.*0977

*Therefore, δ T is = δ U/ CV.*0986

*This is going to be - 350 J ÷ CV.*0994

*They gave us the molar heat capacity.*1002

*This is 12.5 J/ mol/ K.*1005

*They are 2 mol.*1008

*We do a quick calculation here.*1010

*CV the molar heat capacity it is going to be 12.5,*1013

*Let me just do this on the next page.*1019

*I have 12.5 J/ mol/ K × 2 mol that gives me 25 J/ K.*1023

*Now I can do by problem.*1041

*My δ T in this particular case = -350 J ÷ 25 J/ K.*1043

*J and J cancel and if I get my arithmetic correctly which I’m never quite sure that I do, -40 K.*1054

*That is your answer.*1064

*The system experience is -14°K drop in temperature.*1066

*Let us see here, 500 J of energy flows out of the system and 150 J of energy flows into the system.*1078

*The net loss of energy by the system is 350 J.*1091

*If the system loses energy, the energy of the system is going to drop.*1096

*Part B, part B says that the amount of heat is 400 J, 400 J flows in.*1105

*They also say that 400 J of work is produced, +400.*1117

*Δ U= Q - W = 400 -400 = 0.*1129

*δ T = 0/25 J/ K.*1141

*The change in temperature is 0.*1146

*What happened here is the following.*1148

*You have your system, the boundary, and you have your surroundings 400 J of heat flowed into the system.*1151

*400 J of work was done on the surroundings.*1163

*400 J went this way, 400 J went this way.*1167

*There is no change.*1170

*From the system's point of view 400 came in and 400 left.*1172

*From the surroundings point of view, 400 left and 400 came in.*1177

*Δ U is 0, that is what is happening.*1180

*If δ U is 0 and δ T is 0.*1186

*That is what we want here in this particular case, the temperature change.*1189

*Part C , they are telling me that 0 J flow as heat so Q = 0 and they are telling me that 113 J of work is destroyed.*1194

*Work = -130 J.*1207

*Δ U = Q - W = 0 --130 J.*1212

*Therefore, δ U= 130 J.*1220

*Therefore, δ T =130 J ÷ 2 mol or 12.5 J/ mol/ K gives us 25 J/ K.*1226

*J cancels J and what we are left with is 5.2°K increase.*1234

*System surroundings, no amount of heat flowed.*1252

*Work 113 J was destroyed that means this is negative.*1257

*That means it is flowing out of the surroundings into the system.*1262

*This much work was done.*1269

*If these much work as energy transfers to the system, the energy of the system rises.*1270

*The energy of the system rises, the temperature of the system rises.*1276

*5.2 K change in temperature of the system.*1281

*This is what we are looking at, we are looking at the system.*1287

*Our convention is that heat is positive when it flows into the system.*1290

*Work is positive when it is done on the surroundings or it is negative if it is done on the system.*1296

*It is just a question of perspective.*1305

*Let us look at our next example here.*1310

*Doing a particular change of state of 55 J of work or destroyed and the internal energy of the system increases by 200 J,*1315

*the temperature of the system rises by 12°K.*1324

*What is that heat capacity of the system?*1327

*The heat capacity of the system = DQV / DT or Q / δ T.*1332

*Well they give us the change in temperature that so we have the δ T.*1347

*All we have to do is find Q, this division that will give us that heat capacity of the system.*1353

*We need Q, we know that δ U = Q – W, they tell us that a particular change of state 55 J of work are destroyed.*1360

*They tell us that the internal energy of the system increases by 200 J.*1377

*Δ U = 200 = Q – 55 J of work is destroyed -55.*1381

*Therefore, 200 = Q + 55.*1395

*Our Q is = 145.*1401

*Therefore, let us put that back in there.*1412

*Our constant volume heat capacity is going to be 145 J ÷ 12°K.*1415

*If I did my arithmetic correctly, let us check.*1424

*I never get my arithmetic correct.*1427

*12.1 J/ K that this is our heat capacity.*1430

*Our molar heat capacity, notice there is no mention of a particular moles or anything like that.*1435

*If I want the molar heat capacity I would divide by the number of moles and that would give me J/ K/ mol.*1440

*This is just J/ K.*1445

*Be very careful of the units.*1448

*That was pretty straightforward, nothing too strange here.*1454

*We are starting off easy working our way up.*1457

*3 mol of an ideal gas expand isothermally.*1463

*Isothermally tell us that the temperature is constant.*1466

*Isothermally against a constant pressure 110 kg Pascals for 25 dm³ to 65 dm³.*1474

*We want you to find Q,W, δ U and δ H for this change of state.*1482

*This is a pretty standard problem.*1489

*Find the heat, find the work, find the change in energy, find the change in enthalpy.*1490

*This is a really common problem.*1494

*Some of the problems last for one of the other but again it is nice to ask for all four because that is the relationship.*1497

*This is concerned with work, this and this.*1502

*There is a relationship among these.*1504

*Let us talk about, fundamental concept.*1508

*Isothermal that means the change in temperature = 0.*1512

*If the change in temperature = 0 that means there is no change in energy so that means that DU = 0 or DU = 0.*1522

*I automatically have my DU=0.*1532

*Isothermal means δ U= 0 ideal gas.*1536

*That was nice.*1542

*They gave us a constant pressure 110 km Pascal and the unit of change in volume 25 dm³ to 65 dm³.*1545

*A pressure × a change in volume =work.*1554

*Work = the external pressure × the change in volume, that = 100 × 10³ Pascal × 40 × 10³ m³.*1559

*This is in dm meters, this is km Pascal or Pascal.*1580

*It is a Pascal km that is a Joule.*1585

*We have to make sure working in the right units.*1589

*1 J = 1 Pascal m³.*1591

*We have to make those conversions.*1596

*In this case, 10³ the -3 cancel, 100 × 40 that leaves us with 4000 J.*1598

*The work is 4000 J.*1609

*It is positive that means 4000J of work is done on the surroundings.*1612

*4000 J of work is produced.*1616

*Δ U= Q – W, δ U is 0, isothermal.*1623

*0 = Q – W.*1630

*Therefore, Q = W.*1634

*W is 4000 so Q = 4000.*1639

*There we go we just found Q.*1643

*We have taken care of work, we have taken care of Q, and we have taken care of δ U.*1645

*Now we are almost done.*1652

*All we have to worry about is our δ H.*1653

*Δ H, our recommendation with δ H we will start with the definition of enthalpy.*1655

*Our definition of enthalpy is H = U + PV.*1661

*We want δ H.*1667

*The Δ operator that = δ U + PV and this operating on this.*1669

*The δ operator is linear operator, distributes just regular distribution, consider the symbols they just mean symbolic distribution.*1677

*That = δ U + δ PV.*1685

*We already know what δ U is, δ U is 0.*1695

*Therefore, δ H =δ PV.*1698

*This is not P δ V, this is δ PV.*1704

*It is pressure 2 × volume 2 - pressure 1 × volume 1, that is what δ PV is.*1709

*Δ PV =P2 V2 - P1 V1, this is not P δ V, this is not constant pressure.*1725

*This is the pressure of the system.*1736

*Δ PV is P2V2 – P1 V1.*1742

*This is my ideal gas P1 V1 =P2 V2.*1744

*P1 V1 = P2 V2 because P1/V1 = P2/ V2.*1750

*The temperature is constant so the temperature drops off.*1756

*You are left with this basic equation from general chemistry.*1758

*P2 V2 – P1 V1, if they are equal then δ PV = 0.*1764

*Therefore, δ H = 0.*1771

*Their work is 4000 J, your heat is 4000 J.*1774

*Your change in energy of the system is 0.*1779

*Your change in enthalpy of the system =0.*1782

*That is what is going on here.*1786

*Here is the system, here is the surrounding, 4000 J of work is being produced.*1788

*It is expanding the gas, the system is doing work on the surroundings.*1798

*If energy as work goes this way, if the system does work on the surroundings, the energy of the system is going to decrease.*1806

*The energy of the system decreasing implies that the temperature of the system is going to decrease.*1815

*But this is an isothermal process.*1821

*Isothermal means we are not allowing the temperature to decrease.*1822

*Where is that extra energy coming from in order to make sure that the temperature stays up?*1826

*It is coming from the surroundings.*1832

*Work is being done on the surroundings.*1834

*The surroundings is actually giving back heat to the system, heat is 4000 J.*1836

*Heat is coming into the system that is what is happening here.*1842

*This is what this is saying.*1847

*In this isothermal process, the system does work on the surroundings by pushing against it.*1850

*In the process of pushing against it and doing work its losing energy as work.*1855

*If it loses energy as work, the energy is going to drop but the temperature is going to drop.*1858

*If energy drops, the only way for the temperature to stay the same because we are making sure this happens isothermally,*1863

*it has to pull energy as heat from something else.*1869

*The only other source it has to do that is the surroundings.*1872

*It pulls heat this way so heat goes this way, the temperature stays the same.*1875

*That is what is going on here.*1882

*This is the mathematics and this is what is happening physically.*1884

*I hope that makes sense.*1887

*Let us do some more here.*1893

*We have 3 mol of ideal gas to 30°C expands isothermally and reversibly from 25 dm³ to 65 dm³.*1900

*Find Q, W, δ U, and δ H.*1909

*The expansion is the same, it is going from 25 to 65 dm³.*1911

*The only thing that is different is it is happening isothermally.*1916

*It is an ideal gas but now it is happening reversibly.*1919

*It is not just isothermal, we are not just keeping the temperature constant but we are also doing it reversibly.*1922

*We are moving along the isotherm.*1927

*We are not just going this way.*1930

*Let us go ahead and do this.*1934

*In this particular case, isothermally and reversibly we are going to use this one DW = P external × DV.*1936

*Reversible means that the external pressure = to the pressure of the system.*1947

*The P external = P itself, they are in equilibrium.*1960

*That is what reversible means, the external pressure is = to the pressure of the system.*1963

*Therefore, DW = PDV or we are going to do the derivation itself instead of memorizing equation.*1968

*We are going to do the derivation because we can, we have a mathematics just a simple calculus, PDV.*1978

*Therefore, work = to the integral from V1 to V2 of PDV.*1984

*It equals the integral from V1 to V2 of nrt / V DV because for ideal gas P = nrt / V.*1996

*I’m just substituting this value in for P.*2008

*Our T is a constant so I will pull it out so I get nrt × the integral from V1 to V2 of DV / V.*2012

*I end up with nrt LN of V2 / V1, that is my work.*2022

*And I just put my values in*2031

*I have 3 mol, R is 8.314 J/ mol K.*2033

*This is happening at 30°C.*2045

*It is isothermal because it stays at 30°C.*2047

*This is 303 K.*2050

*In this case I do not have to worry about changing this to meters because the unit cancels 65/25.*2053

*I think this is supposed to be, I think I did my numbers wrong here.*2065

*I think this is supposed to be 75.*2070

*If I’m not mistaken, I think this is supposed to be 75 but any way you can work out the actual arithmetic itself does not really matter.*2075

*All that matters is the process.*2085

*3 mol × log of, I think I’m going to ahead and keep it as written.*2087

*I’m going to do 65/ 25, the dm³ end up cancelling.*2095

*The ratio is what matters so I ended up with 8303 J.*2102

*I can recall this 8303 come from this being 65 of this come from being 75.*2112

*I may have actually been thinking of 75 when I did this problem and put it to my calculator but this is the correct process.*2120

*The number itself is actually not that relevant.*2127

*This takes care of the work.*2131

*So work = 8303 J.*2133

*Isothermal implies that the temperature is constant.*2139

*If the temperature is constant that means there is no change in energy.*2144

*Therefore, isothermal means that δ U = 0.*2148

*If δ U = 0 that means 0 = Q – W, Q = W.*2160

*Therefore, Q = 8303 J.*2178

*Enthalpy is going to be the same as that problem that we just did.*2187

*Let me go over it, H = U + PV.*2193

*Δ H = δ U + PV = δ U + δ PV.*2200

*Δ U = 0 PV, this is an ideal gas so P1 V1 = P2 V2 so that also = 0.*2211

*Therefore, δ H for this process = 0.*2222

*And again this is the pressure of the system not the external pressure.*2226

*The previous example, in example 3, Q = W = 4000 J.*2236

*In this example Q = W = 8303 J.*2245

*This one was just a isothermal against an external pressure 110 km Pascal.*2255

*This one was not just isothermal but it was also reversible.*2262

*It was also reversible, this is the maximum amount of work that this gas can do in expanding from one volume to another.*2274

*The other path that was taken was 4000 J, 8303 J is the maximum amount of work that this gas can do in expanding.*2286

*That is was happening here.*2298

*Example 5, we have 4 mol of an ideal gas.*2304

*They are compressed this time isothermally from 100 L to 20 L by a constant pressure of 7 atm.*2308

*Find Q, W, δ U and δ H.*2316

*This is pretty straightforward.*2319

*We have to take on it by now.*2319

*We have done a couple of problems just like this.*2322

*We are just going to keep doing a lot of problems that are the same just to get a sense of it and to become comfortable with that.*2324

*That is what we want.*2331

*Let us go ahead and do work first.*2333

*Work = the external pressure × the change in volume.*2335

*They give us the change in volume that is just the 120.*2342

*It is final - initial so 20 -100 so it was actually -80.*2346

*The external pressure is 7 atm and -80 L so we end up with -560 L atm.*2352

*-560 let us go ahead and convert this to J.*2370

*-560 L atm × 8.314 J ÷ 0.08206 L atm, that cancels.*2374

*I’m going to go ahead and express in kJ, -56.72 kJ.*2393

*It is an isothermal process.*2399

*We already know that an isothermal process δ U = 0.*2402

*If δ U = 0 that means 0 = Q - W which means that Q = W so Q = -56.7 kJ.*2408

*That takes care of Q and we already work out δ H part.*2421

*Δ H = δ U + δ PV, we are dealing with an ideal gas and this is 0, this is 0.*2427

*Our δ H is 0.*2439

*In this particular case, what is happening is this.*2442

*Here is the system, here is the surroundings, notice in this case that the other work and heat are negative.*2444

*Work is -5.*2454

*Sorry I’m not writing it properly -56.7.*2461

*This is right, I have the next line.*2467

*Everything is good.*2469

*The work done is actually negative, that means work is destroyed.*2478

*It means work is leaving the surroundings 56.7 kJ is moving from the surroundings to the system as work.*2481

*Heat -56.7 kJ.*2490

*Heat is negative when it s leaving the system.*2493

*This is a compression.*2500

*The surroundings are doing work on the system 56.7 kJ of work.*2502

*56.7 kJ energy transfer as work from the surroundings to the system.*2508

*This is an isothermal process which means the temperature have to stay the same.*2513

*If we put energy in the system, the temperature is the one to rise but we want to keep the temperature down.*2518

*We have to take energy away from the system in order to keep the temperature at the same level.*2524

*Heat moves as energy or energy moves as heat from the system to the surroundings.*2530

*Therefore, you get a total of δ U = 0.*2537

*In other words, there is no energy change for the system.*2541

*That is was going on here.*2546

*Let us go ahead and we will stop it with these 5 example problems.*2548

*Certainly there are more example problems to come, quite a few more.*2553

*Thank you for joining us here at www.educator.com.*2557

*We will see you next time, bye.*2559

1 answer

Last reply by: Professor Hovasapian

Wed May 31, 2017 6:31 PM

Post by Vanessa Ralph on May 30 at 05:28:33 PM

Hello,

Thank you for your materials and help. I have a question regarding Example 4.

I originally did my calculation for Pext in the expression dw=Pext dV as P2 = Pext and thus P2 = nRT/V2 = 1.148 atm. Thus w = (1.148 atm)(40L)(101.325 J/L*atm) = 4653.43 J. Why wouldn't this give the correct answer for an ideal gas moving along the isotherm?

Also, I calculated w using your method and received 7,221.207 J as the answer for w = (3)(8.314 J/molK)(303K)ln(65/25)... ?

What am I doing wrong?

Thank you for your time!

1 answer

Last reply by: Professor Hovasapian

Mon Sep 21, 2015 12:50 AM

Post by Shukree AbdulRashed on September 19, 2015

Hello sir. Just a few questions.

I don't see how you arrived at the conslusion that p1v1=p2v2? Any situation where it wont = 0?

Does the "P" in PV=nRT only refer to the external pressure?

Just to clarify, work would be negative in example 4 from perspective of the system?

Is there anyway to solve for pressure in example 4?

for these isothermal problems, if U= Q+ W, if one of Q comes out positive, then W should be negative and vice versa correct?

Thank you.

1 answer

Last reply by: Professor Hovasapian

Mon Sep 21, 2015 12:34 AM

Post by Shukree AbdulRashed on September 19, 2015

Why is it appropriate to use kpA and decimeters for example 3?

1 answer

Last reply by: Professor Hovasapian

Mon Sep 21, 2015 12:24 AM

Post by Shukree AbdulRashed on September 19, 2015

For the equation U= Q - W, can't you just write U = Q + W, then figure out the appropriate signs based on the Point of view of the system?

0 answers

Post by Stuart Nystrom on September 17, 2014

^for example 3

3 answers

Last reply by: Professor Hovasapian

Wed Sep 17, 2014 10:03 PM

Post by Stuart Nystrom on September 17, 2014

external pressure is 110kPa right? Not 100kPa... or am i wrong?