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

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

### 1st Law Example Problems II

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Example I 0:11
- Example I: Finding ∆U
- Example I: Finding W
- Example I: Finding Q
- Example I: Finding ∆H
- Example I: Summary
- Example II 21:16
- Example II: Finding W
- Example II: Finding ∆H
- Example II: Finding Q
- Example II: Finding ∆U
- Example III 33:33
- Example III: Finding ∆U, Q & W
- Example III: Finding ∆H
- Example IV 41:50
- Example IV: Finding ∆U
- Example IV: Finding ∆H
- Example V 49:31
- Example V: Finding W
- Example V: Finding ∆U
- Example V: Finding Q
- Example V: Finding ∆H

### Physical Chemistry Online Course

### Transcription: 1st Law 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 example problems for the first law and for energy and things like that.*0004

*Let us just jump right on in.*0010

*Our first example of this lesson is the following.*0013

*We have 1 mol of Van Der Waals gas at 298 K and it expands isothermally and reversibly from 25 dm³ to 65 dm³.*0016

*The Van Der Waals constant is 0.366.*0029

*This is Pascal m⁶/ mol².*0034

*These are just the units, do not worry about those and B = 0.0429 dm³/ mol.*0039

*For the gas, the DU DV this should be sub T, my apologies, at constant temperature.*0046

*The rate of change of energy with respect to volume at temperature = an²/ V².*0055

*We would like you to find Q, W, Δ U, and δ H for the transformation.*0062

*There is a lot of information in this particular problem.*0067

*Things that are important, we have the temperature isothermal and reversible.*0070

*Remember, reversible means that the pressure external is actually equal the pressure of the system.*0077

*The change in volume we have the constants and we have this thing so this is a Van Der Waals gas, it is not an ideal gas.*0086

*For an ideal gas, this DU DV = 0 that term drops out and that expression for the total differential which I will write in just a minute.*0095

*Let us go ahead and see what we can do.*0103

*Now the expression I was talking about,*0106

*Let me do this in blue.*0108

*The expression I’m talking about is the following.*0109

*Our basic mathematical expression, what is handful equations that we definitely want to know and use in order to derive everything else.*0112

*DU = CV DT + DU DV at constant temperature × DV.*0123

*This is the general expression for the change in energy of a system.*0138

*For the differential change in energy of the system.*0141

*We would want the full change which we just integrate this expression.*0143

*Let us see what we can do to simplify this for us and let the problem itself put constraints on this.*0146

*This is isothermal.*0154

*Isothermal means that the temperature is constant.*0155

*That implies that as DT term = 0 because there is no change in temperature, that means this one drops out.*0164

*What we are left with is the following.*0170

*We are left with DU = DU/ DV sub T DV that is going to give us the differential change in energy.*0172

*Therefore, when we integrate this we end up with a total change in energy for the system =*0183

*the integral from volume 1 to volume 2 of this DU DV T DV.*0193

*Fortunately, they give us the DU DV sub T.*0202

*It is right there and very convenient.*0204

*This is equal to the integral from the first volume and the second volume and notice we have the first volume and we have the second volume of an² / V² DV.*0209

*An² comes out, we have V1 / V2 and then we have DV / V.*0226

*This is what we have to integrate in order to solve this.*0235

*On notice we started off with this.*0236

*This is our basic, one of the basic equations that we need to know and we let the problem dmde what sort of constraints, what is going to go to 0.*0238

*If they said that this under constant volume this would go to 0.*0246

*This is what is going on here.*0250

*Once we do that, we end up with the following.*0252

*This is going to be an² × 1/ V1 - 1/ V2 and when I put these numbers in, that is final.*0257

*An², a is 0.366, I’m going to go ahead and leave off the units.*0272

*I hope you guys do not mind.*0280

*I will leave the units for this first problem it might be nice to see.*0284

*Pascal m⁶/ mol² × 1 mol², because we are dealing with 1mol of the gas and of course, these are expressed in dm³.*0288

*We have to express this because this is in meters we have to express the volume in meters.*0307

*We are dealing with Joules and Pascal so the volume has to be in meters.*0311

*We have to make the conversion.*0314

*Volume 1 it is just going to be 25 × 10⁻³.*0316

*1 dm³ is 10⁻³ m³, basic conversion factor -1/ 65 × 10⁻³.*0323

*When we do this calculation, we end up with 9.0 J.*0332

*This is not a strong point but I hope that you actually confirm the arithmetic.*0338

*The arithmetic is secondary.*0345

*What is important is this process, getting to this point, this is what is important.*0346

*Often in a test, for some of the higher end courses you might be asked to calculate and derive.*0354

*You can stop at the equation.*0362

*You do not have to do the numbers.*0363

*9 J that takes care of the change in energy.*0366

*The change in energy of the system is 9 J.*0371

*Let us go ahead and see what we can do about the work.*0375

*Our equation for work is DW = P external DV, again, another one of the basic equations that you have to know.*0381

*This is the definition for work it is the pressure × change in volume.*0392

*If this is a reversible process reversible implies that the P external = P.*0396

*We can write DW = the pressure of the system × the change in volume.*0405

*This is a Vandrual’s gas, the pressure = nrt /V - nb - an² / V².*0412

*We put this into here and we integrate, our total work is going to equal the integral from volume 1 to volume 2 of this expression,*0423

*the P which is nrt / V - nb - an²/ V².*0440

*Do not let this intimidate you, this is not a difficult and it is a very simple integral.*0449

*It was just a logarithmic integral.*0453

*It just looks complicated.*0455

*Do not to let how something looks intimidate you to stop and look at it.*0457

*See what it is, it is very simple integration.*0461

*You can do it by hand or if you like this use mathematical software either mathematical or maple.*0463

*DV that is it.*0471

*When you do this integration, you will end up with following.*0473

*You end up with work = nrt × log V2 - nb/ V1 - nb + an² × 1/ V2 -1 / V1.*0475

*Notice, in the last problem when we integrated I had 1/ V -1 – 1/ V2.*0501

*I did that because the integral itself ended up coming out as -1/ V.*0509

*I just use that negative sign to flip the order.*0514

*In this case, I took the negative sign out and I dmded to go ahead and make this a positive.*0517

*It is not a mistake from previous, it just depends on how you want to express it.*0523

*Again, a particular mathematical equation, at this level you can simplify to any degree you want.*0527

*The simplification is irrelevant.*0533

*What is relevant is that you understand what is going on.*0535

*If you want to leave this as a minus sign, I might put 1/ V1 – 1/ V2.*0538

*Just look at the same answer is not a problem.*0542

*When we put all the numbers in, I’m not going to go ahead and write all the numbers in.*0547

*It is actually a lot.*0550

*Maybe I should, it is not a problem, we should see everything.*0553

*1 × 8.314 the temperature is 298 K × log,*0557

*We have to express the volume in cubic meters.*0569

*This is going to be 65 × 10⁻³ -nb -1 × 0.064 × 10⁻³.*0574

*B was expressed in terms of dm³/ mol, we have to express it in terms of dm³/ mol.*0586

*Thermodynamics, physical chemistry in general, it is a conversion factor that ends up being tedious and toilsome.*0594

*So 25 × 10⁻³ - 1 × 0.064 × 10⁻³ we have that.*0604

*It is going to be +a which is 0.366 that is not a problem × 1² × 1/ V2 which is 65 × 10⁻³ -1/ 25 × 10⁻³.*0616

*When these numbers are floating around, I do hope that you actually confirm my arithmetic.*0638

*What you will end up with is 2371.3 + -9.0 and you will end up with the work is going to be 2362.3 J.*0644

*The system is expanding, the gas is expanding isothermally and reversibly.*0663

*It is doing work on the surroundings which is why the work is positive.*0670

*Work done on the surroundings is positive that is why the work is positive.*0675

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

*We are looking for Q, it looks like we did δ and we did W, let us do Q.*0688

*We know that δ U= Q – W.*0695

*Therefore, Q = δ U + W.*0698

*Therefore, the heat of this reaction = 9 J which is the δ U, the change in energy + 2362.3 J.*0705

*Therefore, the heat is 2371.3 J.*0718

*I apologize sometimes I write Joules and sometimes I just put J.*0734

*Let us go ahead and see what we can do about the,*0739

*Let us see, with Q let us find δ H.*0745

*Δ H here is how you want to put it.*0750

*The definition of H is U + PV.*0754

*Again, when the basic equation that you start with in this problem is only a handful of equation that you need to know.*0759

*Everything else should be derived.*0765

*Do not think that you can memorize a bunch of equations and to apply them to a situation.*0767

*This is high level science so the problems are not mechanical.*0772

*It is not just this is what this problem is, you plug the numbers in.*0779

*It is not going to work that way.*0783

*There are ton of equations in thermodynamics but there are a handful of basic equations from which the others come.*0785

*The derivations are not difficult, as far as the problems are concerned.*0791

*Do not memorize the equations because they will just make you crazy if you try to memorize.*0795

*So δ H which is what we are looking for = δ U + PV.*0800

*The δ operator is a linear operator.*0807

*So δ H linear just means you can distribute operators symbolically.*0809

*You can distribute that way, distribute that way, which you will end up with is δ U+ δ PV.*0814

*Well, that is the same as δ H = Δ U.*0823

*Put the δ PV, that is just P2 V2 - P1 V1.*0829

*We have to calculate P2 V2 P1 V1 from Van Der Waals equation.*0840

*Again, the pressure = Van Der Waals equation = nrt / V - nb - an²/ V2.*0850

*This is going to be a bit tedious as far as a calculation is concerned but it is not a problem.*0861

*Let us go ahead and calculate P2 V2 first.*0867

*We will do that one.*0869

*It is going to be nrt, this first one I just want to make sure to do everything.*0872

*Over V - nb – an² / V² that is the V2 part, over V2² × V2.*0881

*When you put all the numbers, you have all the numbers.*0896

*You have n, R, T, 298.*0899

*You have the second volume, you have nb.*0901

*You have all these values.*0902

*When you put these values in, you end up with 2473.6 J.*0904

*You do the same for P1 V1.*0914

*In this case, P1V1 = nrt/ V1 - nb - an² / V1² × V1.*0918

*When you put those values in make sure to convert volume to cubic meters.*0930

*In case of 25 × 10⁻³ m³ is 25 dm³.*0936

*When you do that, you end up with 2467.2 J.*0941

*Therefore, P2 V2 - P1 V1 = this – that, you end up with 6.4 J that is P2 V2 - P1 V1.*0949

*This is the δ PV.*0966

*We can go ahead and calculate our δ H.*0969

*Δ H = Δ U + δ PV.*0973

*Δ H where δ U was 9.0 J + 6.4 J for a total of 15.4 J, that is our δ H.*0979

*This is not the only way to solve this problem but one of the beauties of science itself is the problem they come up with different solutions.*0998

*I usually just go with the first solution that occurs to me.*1009

*It might be a little bit longer or shorter, I do not know.*1011

*By all means, any other solution that you can come up with, you should examine that and you should explore it.*1015

*And you should use it because that is the way you see the problem.*1023

*Arithmetic often get in the way of what is going on so I want to pull back a little bit and just go through what is that we did in this particular problem.*1029

*The first thing that we did is we calculated the energy.*1038

*We started off with DU = CV DT + DU DV sub T × DV.*1041

*They said that this was isothermal so this term goes to 0.*1053

*Again, we start with the basic equation and we derive the relationship that we needed.*1058

*DU = DU / DV sub T DV.*1063

*When we integrate that expression, we get the expression for the full energy for the entire change of state*1074

*which is going to be integral from V1 V2 of this DU DV sub T DV.*1080

*When you integrate you put the numbers in.*1089

*That is how we found the energy and then we dmded to go with the work.*1092

*The definition of work = P external DV.*1096

*This one is reversible, P external = P.*1102

*What we end up with is P = the Van Der Waals gas.*1107

*Nrt/ V - nb – an² / V².*1113

*We put this expression into here and then we integrate.*1119

*When we integrate, the work equals the integral from V 1 to V2 of PDV.*1125

*We put this expression into here and we integrate it and we get the value for the work and we put the numbers in to actually get the work itself.*1133

*Number 3 was actually pretty straightforward.*1143

*We used this basic relationship and the differential version is DQ – DW.*1145

*The integrated version is δ U= Q – W.*1157

*These are not exact differentials.*1162

*An exact differential integrates that δ U.*1164

*These inexact differentials just go to Q and W.*1167

*We are looking for Q so I just move things around.*1173

*You already found this and we just found that.*1175

*Q = Δ U + W and we ended up finding the heat for the particular reaction.*1178

*How do we do δ H?*1189

*Δ H = the change in energy which we have + δ PV.*1191

*Δ PV = P2 V2 - P1 V1.*1199

*We used the Van Der Waals equation to find this and to find that, make the subtraction, use that and put it into here and solve for δ H.*1207

*This was the process that we did and it is all based on a handful equations.*1217

*This equation, right here, that is our first equation.*1222

*It is one of the fundamental equations that we have to know in order to solve these problems.*1225

*The second equation right here is the definition of work.*1230

*This is the total energy of the system.*1232

*The energy is dependent on temperature and on volume.*1235

*For this particular problems, isothermal the DT went to 0 so it only depends on the change in volume.*1239

*This is the definition of work.*1245

*This is the first law of thermodynamics.*1248

*The energy of the system = the heat transfer - the work transferred.*1250

*Of course δ H comes from the definition of enthalpy which is one of the basic equations.*1255

*Again H = U + PV that is the definition of enthalpy.*1261

*Excuse me, let us move on to our next problem here.*1271

*We have a lot of extra pages.*1275

*Example number 2, so we have 1 mol of an ideal gas and it is confined under pressure P external.*1277

*This external should actually be down below, I ended up putting on top.*1287

*P external = P = 250 kl Pascal.*1291

*Basically, what this says is that the external pressure and the internal pressure are the same.*1296

*If you have this gas and this piston set up, this with a little weight on top, this is an equilibrium.*1300

*It is not moving anywhere.*1307

*In other words, the pressure this way = the pressure that way, that is all these means.*1308

*So ideal gas is very important.*1312

*We are provided with pressure 250 kl Pascal, 250,000 kl Pascal.*1317

*The temperature is changed from 120 to 25°.*1322

*The temperature of the gas we drop it from 120°C to 25°C.*1326

*The constant volume heat capacity= 3 Rn/ 2, n is 1.*1332

*Here we want to calculate the Q, W, δ H.*1339

*This is a pretty standard problem, the four basic properties of the system.*1342

*Notice, they say nothing about a change in volume here.*1351

*Let us see what we can do first.*1355

*Let us start with work, if it stays in red it is not a problem.*1360

*I will start with a definition of work DW = P external × DV.*1367

*The P external is constant, if it is constant when we integrate this, this comes out of the integral sign.*1374

*When you integrate this, this comes out of the integral sign and what you end up with is work = P external × the change in volume.*1382

*Let us see what we have got.*1399

*We have an ideal gas so we have PV = nrt so V volume = nrt / P.*1401

*We have two different temperatures, we have 120 and 25.*1417

*The pressure is actually constant so pressure stays the same.*1423

*For change in volume, let us do this.*1428

*Let us find, we have the P external we just need a change in volume.*1431

*This is P external × V2 - V1.*1437

*I’m going to start with V1, it is going numerical order here.*1448

*V1 = nr T1/ P which = 1 × 8.314 × temperature 1, the initial temperature is 120.*1451

*We have 393 K/ 250,000 Pascal.*1471

*You end up with a value equal to 0.0 1307 and this is going to be in cubic meters.*1486

*If you are concerned about what is happening here, let us do a quick unit check.*1495

*We have mol × J/ K mol × K/ Pascal.*1502

*This is the same as mol × J is N m/ K mol × K and Pascal is a N/ m².*1515

*Mol cancels mol, K cancels K, N cancels N.*1533

*This comes up to the top and ends up with cubic meters is what this comes from.*1537

*This is cubic meters.*1542

*I’m going to do the same for V2.*1544

*Volume 2 that is equal to nr T2/ P.*1546

*We have 1 × 8.314, 25°C which is 298.*1551

*Try to keep everything straight is going to be the difficult part.*1560

*This is 250,000 and it is very easy to actually think that 298 here, and put the 393 here by mistake.*1563

*Again, it is just that is life, lot of numbers floating around.*1573

*Equals 0.00991 m³.*1578

*Our work is equal to P external × V2 - V1 = 250,000 Pascal × C V2 is 0.0091 cubic meters -0. 01307 m³ and we end up with Pascal m³ which is J.*1586

*Our work is going to end up being -790 J.*1629

*When you drop the temperature of this gas, this gas is going to contract.*1637

*It is going to fall.*1641

*The surroundings are going to do work on the system.*1643

*When the surroundings do work on the system that means work is leading the surroundings.*1648

*Our consigned convention work is -790 J.*1654

*790 J is leaving the surroundings and going into the system, that is what this means.*1657

*Let us go ahead and do our C.*1665

*Let us go ahead and do our DH next.*1671

*We have our basic relationship, one of the equations that we have to know.*1674

*The differential change in enthalpy = the constant pressure heat capacity× the differential temperature + DH DP T DP.*1679

*The pressure is constant that means DP = 0.*1696

*Constant pressure, constant P implies the DP = 0.*1703

*This term goes to 0 so you are left with DH = CP DT.*1708

*When you integrate this, CP is constant so what comes out which we are left with is the following equation DH = CP δ T.*1720

*We have δ T that is just going to be the 120 - 25 so we need CP.*1735

*They did not give us the CP, they gave the CV.*1741

*They gave the CV = 3/2 Rn and we have a relationship, this is an ideal gas.*1744

*An ideal gas there is a relationship between the constant pressure heat capacity and the constant volume heat capacity that is the following.*1751

*CP - CV = Rn.*1761

*They gave the CV we know Rn and we can find CP and put it in here to find our value.*1765

*CP = CV + Rn that is equal to 3/2 Rn + 1 Rn.*1773

*Therefore, the constant pressure heat capacity= 5/2 Rn.*1787

*And now we can go ahead and put this into here.*1791

*Therefore, our δ H = 5/2 Rn δ T.*1794

*We have all the numbers that we need.*1801

*Δ H = 5/2 × 8.314 × 1 mol × change in temperature final - initial 25°C -120°C.*1805

*The change in temperature increment, I did not change this to K because the δ T for Celsius and K is the same.*1824

*The individual things are not the same 393 and this is 298 but 298 -393 is the same as 25 -120.*1831

*The δ T because all you are doing you are adding 273 so it goes up equally.*1840

*And when you do this you end up with δ H = -1975 J.*1847

*Now under constant pressure δ H is actually equal to QP.*1858

*I will probably write it the other way QP= δ H.*1875

*In other words the heat happens to equal the enthalpy so we went ahead and took care of that.*1878

*The Q for this reaction = -1975 J.*1884

*The only thing we are missing now is δ U.*1891

*We have a relationship δ U= Q – W.*1894

*We just found Q, we found W a little bit earlier it is = -1975 J that is the Q - the W which is -790 J.*1899

*Our change in energy of the system = -1185 J.*1915

*There we go and what does all this mean?*1925

*Here is what is happening.*1928

*We have our system and we have our surroundings Q that was equal to -1975 J.*1930

*1975 J left the system.*1942

*Work = -790 J, the surroundings did 790 J of work on the system.*1946

*In other words, 790 J went into the system.*1959

*Therefore, the total change in energy 1975 from the systems point of view 1975 J left, 790 J came in.*1962

*The total net energy, the net loss by the system is 1185 J net loss by the system.*1972

*That is what is happening here.*1988

*The surroundings did 790 J of work.,*1991

*The system in going from 120 to 25 lost 1975.*1995

*The net loss is the systems.*2000

*Let us see what we have got here.*2007

*Let us go to our next example.*2010

*Example 3, we have 1 mol of ideal gas.*2014

*We have 1 mol, we have an ideal gas is taken from 10°C to 90°C under constant volume.*2017

*It is very important.*2026

*This time, our constrained is under a constant volume.*2027

*Our constant volume heat capacity is 21 J/ mol K.*2030

*This is actually a molar heat capacity.*2035

*21 J/ mol K, calculate Q, W, Δ U, and δ H.*2037

*Let us see what we have got.*2045

*Let us see which one do we do first.*2056

*Constant volume implies that DV = 0.*2067

*If you remember, under constant volume our Q = δ U.*2078

*Our equation DU = CV DT + DU DV.*2086

*Constant volume DV equal 0 but we are dealing with an ideal gas so this is already 0.*2106

*In either case, that one goes away.*2110

*Our DU which = QV = CV DT.*2113

*We just go ahead and integrate this.*2123

*Therefore, when we integrate this we end up with the following.*2127

*You end up with δ U= Q = CV δ T.*2129

*When we integrate this, this is constant so it comes out.*2137

*The integral of DT is δ T.*2139

*We have the CV, we have the δ T, so this is really easy.*2142

*Our DU = CV which we said was 21.0 J/ K and our δ T is 90°C -10°C.*2147

*And when we run this calculation, we end up with 1680 J.*2175

*That is δ U= 1680 J also happens to equal Q= 1680 J.*2188

*We have taken care of the Q and the δ U, let us go ahead and take care of the constant volume.*2200

*We did that.*2211

*Let us go ahead and do δ H next.*2216

*Actually, if we want in this particular case because constant volume, no work is done.*2219

*PV so work = 0.*2235

*What about that one?*2240

*Let us go ahead and do δ H.*2242

*Again, we have δ H = Δ U+ δ PV equal to,*2244

*We have to watch out if a constant volume but V is constant, therefore, let us change this equation a little bit.*2263

*That means δ H is going to equal Δ U V comes out.*2274

*V × δ P just the change in pressure.*2280

*Let me rewrite that again on this page.*2287

*We have δ H or a lot of crazy things floating around.*2289

*We have δ H = δ U+ V × δ P or ideal gas so PV = nrt.*2295

*Therefore, the pressure = nrt / V.*2307

*The δ H = δ U+ V δ P well δ P is P2 - P1.*2315

*P2 = n R T1/ V nr T2/ P.*2328

*I'm sorry for V, I’m getting all white variable here.*2337

*Remember, volume is constant so it stays V.*2343

*It is not V1 or V2, temperature 2.*2344

*Nr here is the V and then P1 = nr T1 / V.*2348

*These cancel out so you get δ H = Δ U+ nr × T2 - T1.*2355

*We have T2 and we have T1.*2365

*Therefore, when we put these values in we get δ H = δ U which was 1680 J + 1 mol.*2368

*R is 8.314 and we have 90°C -10°C.*2380

*Again the difference in temperature, δ T in Celsius is the same as the δ T in Kelvin.*2388

*When we run this calculation, we end up with 2345 J is the δ H.*2394

*Let me actually do these individually.*2409

*Here we have the 1680, this particular value ends up being 665 J.*2412

*We took it from 10°C to 90°C so what is going on here is the 1680 J, that much energy comes from the rise in temperature.*2428

*The 665 J it comes from the rise in pressure.*2449

*The δ H, the enthalpy is an accounting device.*2462

*It accounts for the energy change + any change in pressure, volume, work.*2465

*A change in pressure, change in volume, things like that.*2471

*1680 if it comes from just the rise in temperature, 665 come from the actual rising pressure because we are keeping at a constant volume.*2475

*That is what is happening.*2483

*We keep it at constant volume.*2483

*It cannot do work this way so the pressure increases inside that increases the energy of the system.*2485

*You always want to stop, take a look at the numbers, see what they mean in terms of the direction of flow positive or negative,*2492

*and see if you can actually identify what value is coming from what.*2499

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

*Let us move on to example 4, let us see what we have here.*2512

*Find δ U and δ H for the transformation of 1 mol of ideal gas from 25°C and 1 atm.*2518

*They give us an initial temperature and initial pressure to a final temperature and a final pressure.*2525

*The molar heat capacity of the gas CP.*2531

*This is a CP, CV is 20 + 0.045 J/ mol K.*2537

*I will make one little correction here.*2544

*20 + 0.045 × T, T is the variable.*2548

*In this particular case, the heat capacity is temperature dependent.*2553

*It starts at 20 but as the temperature rises the heat capacity changes.*2557

*This is not constant heat capacity, this is something we are going to have to integrate over a change in temperature.*2564

*I apologize for that mistake so this should be 20 + 0.045 × T.*2569

*Let me write it here CP = 20.0 + 0.045 T.*2577

*T is a variable like x.*2585

*In this particular case you can do Δ U first, you can do δ H first.*2589

*It does not really matter.*2592

*Let us go ahead and start off with a basic equation.*2594

*I’m going to do δ U first.*2598

*Δ U is equal to CV DT + DU DV T DV.*2600

*This term goes to 0 because the gas is ideal.*2611

*An ideal gas the DU DV = 0.*2615

*What I'm left with is DU = CV DT.*2618

*When I integrate that, of course I will have to integrate that, I will be left with δ U= the integral from temperature 1 to temperature 2 of CV DT.*2628

*I do not have CV DT, I have CP DT but it is an ideal gas I know that there is a relationship.*2643

*I know that CP - CV = Rn.*2647

*Therefore, CV = CP – Rn.*2653

*CP = 20.0 + 0.045 T - 8.314 × 1 mol.*2664

*For my constant volume heat capacity I end up with some 20 - 8.314, I end up with 11.686 + 0.045 T.*2682

*This is the value that I put in here.*2695

*Therefore, δ U= the integral from 298 K.*2699

*These cannot be a Celsius, you have to put these in K.*2706

*298 K to 225 which is 498 K of the constant volume heat capacity which was 11.686 + 0.045 T DT.*2710

*I will go ahead and just use math software and if it did it correctly I ended up with 5919 J that is our δ U.*2728

*Let us go ahead and do a δ H for this.*2743

*Δ H again, it is up to you how you want to do this.*2746

*I’m going to do it with a long way with DU and δ PV.*2750

*You already know that DH = CP DT + DH DP T DP.*2755

*This is one of the fundamental equations for enthalpy.*2769

*This is an ideal gas so this goes to 0.*2774

*Our δ H = CP DT from T1 to T2, this is one way you can actually calculate the enthalpy.*2777

*I did it another way, I did it with a basic equation by the definition of enthalpy DU + D PV.*2790

*This is probably the quicker method.*2800

*This is just one that I happen to have gotten used to so I just tend to automatically default to that one.*2802

*But either one is fine however you want to see it.*2807

*This = δ U + P2 V2 - P1 V1 and = Δ U.*2812

*P2 V2 is an ideal gas PV = nrt.*2824

*P2 V2 = nr T2 – nr T1.*2829

*You end up with δ U + nr × T2 - T1.*2837

*We have T2 and T1 so you end up with δ U which was 5919 + 1 mol ×,*2844

*I really need to slow this down + 1 × 8.314 × 225 -25 again δ T.*2855

*225 -25 I can leave that δ H = 7582 J.*2870

*There you go, nothing strange going on here.*2884

*P2 V2 P1 V1.*2891

*This said nothing about volume being constant so despite the fact that this is an ideal gas, P2 V2 - P1 V1 is not 0 because temperature changes.*2897

*That is what is going on here, be very careful because sometimes you can go with P2 V2 – P1 V1.*2913

*If other things are the same, if this were isothermal that would not be a problem.*2919

*If this were isothermal then Q would be constant.*2923

*P2 V2 would equal nrt and P1 V1 equal nrt.*2928

*Therefore, P2 V2 – P1 V1 equal 0.*2932

*But temperature is not constant here so again what the problem says very careful on how you approach it, which is why you do not want to memorize equations.*2936

*This is an excellent example.*2947

*If you just memorize the equation you will just plug numbers in without stopping to save yourself and say this is not isothermal situation.*2948

*If it were an isothermal situation, this term will go to 0.*2956

*If not isothermal, therefore, this is not 0.*2959

*You have to run the calculation.*2961

*You have to be very careful, go slowly.*2963

*That takes care of that one.*2967

*Let us go ahead and go to example number 5, it should be our final example for this particular lesson.*2969

*Let us see.*2975

*When an ideal gas undergoes an expansion such that the relationship pressure × volume raise to the nth power = a constant*2979

*or C is a constant and n is greater than one, it is called a reversible poly tropic expansion.*2987

*We would like to calculate the work for such an expansion of 1 mol of the gas from V1 to V2.*2994

*When the temperature 1 is 350 K and temperature 2 = 200 K, and n=2.*3001

*We would like you, given that the constant volume heat capacity= 5/2 Rn or 5n/ 2.*3010

*We want you to find Q, Δ U, δ H, for this particular expansion.*3016

*Let us see what we have got.*3022

*Here is a lot going on here.*3023

*A new term poly tropic expansion, here we are given a relationship that the pressure × the volume raised to the nth power is actually a constant.*3025

*Again, this is where the things we just have to dive in and see what is going on.*3036

*There is only a handful of equations that you know, you want to start with those and see where it takes you.*3042

*Let us go ahead and start with part A.*3048

*I think I will go ahead and go back to blue for this.*3051

*Part A, telling me that n = 2, I have PV² = a constant.*3055

*I know that the pressure is equal to C/ V², C being a constant.*3067

*My work = I know my basic relationship.*3074

*Our basic relationship by definition is DW = P DV.*3081

*Therefore, my work = the integral from V1 to V2 of P DV.*3089

*I have P here, the relationship is C/ V² so I can just plug this into here and integrate.*3097

*Work = the integral from one volume to another volume of P DV which P = C / V².*3105

*This is C/ V² DV which = C × the integral from V1 to V2 of DV / V².*3115

*And I'm left with C × 1 /V1 -1 / V2 that is great, I have expression for the work = C × 1/ V1.*3126

*That is what they wanted, calculate the work for such an expansion.*3138

*However, they gave us temperatures here so it looks like they want an actual numerical value not just some expression.*3142

*Let us see what we have got.*3154

*V = C/ nrt so let us see if we can play with this a little bit.*3161

*We have work = constant × 1/ V1 -1/ V2.*3170

*We have that P = C/ V², this is PV = nrt.*3181

*I’m going to go ahead and put this P value of P into here and I end up with C/ V² × V = nrt.*3198

*This V cancels that V so I’m left with C/ V = nrt.*3212

*That means V = C/ nrt.*3222

*That means that V1 is equal to C/ nrt 1 and V2 = C/ nr T2.*3231

*I can go ahead and put these values into here and here.*3244

*I got work = C × 1/ C/ nr T1 -1/ C/ nr T2.*3248

*Work = C × nr T1/ C-1.*3270

*This is working out really great.*3277

*C cancels therefore our work = nr × T1 - T2.*3280

*This is fantastic, we have T1 and T2.*3293

*We have n and R so work = 1 mol × 8.314 J/ K mol × temperature 1 was 350 K and temperature 2 was 200 K.*3295

*And therefore, for work for this process is 1247 J.*3314

*That was fantastic.*3321

*Work = 1247 J.*3324

*Let us go ahead and see what we can do with DU.*3327

*DU= CV DT so δ U = CV δ T well that is equal to 5/2 Rn δ T.*3331

*I had all information.*3349

*Therefore, δ U= 5/2 × 8.314 × 1 mol × δ T.*3351

*Δ T is final - initial so we have now 200 -350.*3366

*Therefore, the change in energy = -3118 J.*3374

*That takes care of the energy.*3384

*We have a relationship Δ U= Q – W.*3386

*Therefore, Q = Δ U+ W so Q = δ U which is - 3118 J + 1247 J.*3392

*Therefore, heat is -874 J.*3408

*Let us see what we can do with δ H.*3417

*Δ H = Δ U + δ PV and again this is as my default, I automatically go to this.*3421

*It is equal to δ U + P2 V2 - P1 V1.*3429

*We have PV² = a constant therefore P = C / V².*3444

*Therefore, P2 V2 = C/ V 2² × V2.*3453

*This V2 cancels one of the V2 so I'm left with C / V2 and P1 V1 analogously is C/ V1.*3465

*We have δ U + C/ V2 - C/ V1 = δ U + C / nr T2 –C/ nr T1.*3487

*We end up with δ U+ nr this time T2 - T1 = - 3118 J + 1 × 8.314 × 200 -350 = -3118 J + -1247 J.*3517

*Our final δ H = -4365 J there we go.*3549

*A long process and a little tedious and a little involved.*3564

*But again you are starting off with a basic set of equations and any other information that they gave you that PV ⁺nth power = C.*3571

*They gave you n = 2.*3579

*You are going to use that to start fiddling around with it.*3582

*This relationship holds, this is an ideal gas so this relationship holds.*3586

*You can start fiddling with these and playing around with the equation.*3593

*And again, this is not necessarily the only way to do this problem.*3598

*Maybe there are other ways to do this problem I do not know.*3601

*I just sort of jump right out on in and did it this way.*3603

*And now everything seemed work out okay.*3606

*I will go ahead and leave it like that.*3609

*Thank you for joining us here at www.educator.com.*3612

*We will see you next time for a continuation of some more example problems.*3614

*We want to make sure to get this really nice and tight.*3618

*Take care, bye.*3622

1 answer

Last reply by: Professor Hovasapian

Fri Feb 26, 2016 4:12 AM

Post by Van Anh Do on February 21, 2016

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

1 answer

Last reply by: Professor Hovasapian

Wed Oct 28, 2015 12:50 AM

Post by Chanelle Brown on October 27, 2015

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

1 answer

Last reply by: Professor Hovasapian

Mon Sep 21, 2015 12:54 AM

Post by Shukree AbdulRashed on September 19, 2015

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

0 answers

Post by Shukree AbdulRashed on September 19, 2015

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

1 answer

Last reply by: Professor Hovasapian

Sun Dec 7, 2014 6:21 PM

Post by Carly Sisk on December 7, 2014

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