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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### Three Miscellaneous Example Problems

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 0:41
- Part A: Calculating ∆H
- Part B: Calculating ∆S
- Example II 24:39
- Part A: Final Temperature of the System
- Part B: Calculating ∆S
- Example III 46:49

### Physical Chemistry Online Course

### Transcription: Three Miscellaneous Example Problems

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

*Today, I want to throw in these 3 miscellaneous problems.*0004

*It is not like they have another place in the previous problem sets.*0009

*In fact, they actually technically belong to the entropy section but I want talk about few other things before I brought these back.*0014

*They have to do with a little bit of what we discussed previously.*0023

*They have to do with heat transfer, changes in phase, things like that.*0027

*In some sense, it is a little bit of everything.*0031

*The third example problem is strictly an entropy problem but I just thought I collect them here like this.*0033

*Let us get started.*0040

*The first example is on a fully insulated flask 40 g of liquid water at 25°C is added to 25 g of ice at -7°C,*0044

*What is the final state of the system given the following data and pressure is held constant during this process.*0056

*The constant pressure heat capacity for liquid water 4.18, constant pressure heat capacity for solid water, ice is 2.09 and the δ H of fusion is 334 J/ g.*0063

*In addition to finding the final state calculate δ H and δ S for this transformation.*0079

*The final state of system, what they are asking for is we have 40 g of liquid water and we have 25 g of ice.*0084

*Some of the ice is going to melt and it is going to melt it because the liquid water is going to end up cooling.*0094

*We want to find out what is the final temperature of the system?*0103

*How much ice is there and how much liquid water is there?*0110

*That is what they mean by what is the final state of the system.*0112

*Let us just jump on in and see what we can do.*0116

*A lot of this has to do with a lot of heat capacity that we have talked about earlier but*0119

*it actually harkens back to what it is that you learn in general chemistry.*0124

*Remember Q = MC δ T and the amount of heat that something loses is the amount of heat that something else gains.*0128

*In this particular case, the amount of energy lost upon cooling of the water is going to be gained by the ice and that is going to end up melting.*0136

*We are going to reach some final equilibrium temperature that is going to be some mixture of water and ice.*0144

*Maybe it is going to be all ice or maybe it is going to be all water.*0149

*We do not know, that is what we need to find out.*0152

*Let us see what we have.*0157

*Once again, we know that liquid water is going to cool.*0160

*We know that the ice is going to warm up and it is going to melt.*0162

*The first thing I'm going to do is find out how much energy is required to bring the 25 g of ice from 7°C up to 0°C.*0168

*The reason that I happen to pick the ice first just to see it, is just that I notice that since the ice was at -7, it is not that far from 0.*0180

*I want to see how much energy and it is only 25 g, where is the water itself you have actually 40 g and it is at 25°C.*0191

*The heat capacity of liquid water is higher than the heat capacity of the ice.*0200

*Chances are just qualitatively looking at the problem,*0205

*I'm probably going to be able to raise all of the ice to 0 without dropping the liquid water temperature too far.*0211

*That is what my intuition tells me just based on the data that I have.*0219

*Let us see if that is actually true.*0223

*It does not really matter where you start, whether it is with the ice or with the water.*0225

*You just have to start somewhere and inch your way, converge on some final temperature and final state.*0229

*Remember from general chemistry we had this Q = MC δ T.*0237

*The amount of energy lost or gained by something = the mass of something× the specific heat × the change in temperature.*0245

*This δ G gives us the change in temperature of that particular substance.*0254

*To find out how much energy is required to bring 25 g of ice to 0°C, here is what we do.*0262

*Our Q is going to equal the mass which is 25 g of ice and the heat capacity of the ice which is 2.09 J/ g/°C.*0272

*°C and °K, I do not give it to you in J/ GK.*0287

*The difference of degree for °C and °K is the same so it does not matter whether I use °C or °K.*0294

*I'm going from -7 to 0°C.*0302

*It is going to be δ T of 7°.*0305

*I’m just going to write it as 7°C.*0310

*When I go ahead and multiply and cancel units, I'm going to end up with 365.75 J.*0313

*This is how much energy is required to bring the 7 g of ice from -7° all the way up to 0°C.*0323

*Energy has to come from somewhere.*0331

*The energy is going to come from the water actually cooling down.*0333

*Let us see what this affect has on the liquid water because that is where the energy is being pulled from.*0338

*This is going to be Q = MC δ T but this time we are going to be looking for our δ T for water.*0349

*Our Q is going to be 365.75 J and I’m going to skip the units, I hope you do not mind.*0357

*We have 40 g of water and a specific heat of water is 4.18 J/ g/°C.*0364

*And we want to know what δ T is.*0374

*When we only do this math, nice and simple arithmetic, we get 2.19°C.*0375

*That means the water cools from 25°C it drops down by 2.19°C.*0382

*That amount of energy that is lost by the water in cooling, that energy goes towards bringing the ice from -7 to 0.*0390

*That is what is happening here, that is what I have done.*0398

*This is an exchange of energy.*0401

*Let us see where it is that we are.*0404

*The water is now at 22.81°C.*0417

*It started at 25, it dropped by 2.19, now it is at 22.81°C.*0426

*We have 40 g of liquid water at 22.81°C and we still have a 25 g of ice.*0433

*Except now, the ice is at 0°C.*0455

*We bought the solid ice from -7 to 0 but we have melted the ice yet.*0459

*This is very important.*0464

*Just because something comes to 0°C does not mean that it starts to melt or the other way around.*0467

*It does not mean that the liquid water starts to freeze.*0472

*That is the δ H of fusion, there is a certain amount of energy that is required to make the conversion from one phase to the other.*0475

*We do not have 25 g of ice, we do not have 25 g of water to 0, we have 25 g of ice at 0°C.*0483

*Let us see what we can do about the water at 22.81°C.*0499

*I have got to write it a little bit better.*0503

*Water at 22.81°C, in order to take that to water at 0°C, we need a certain amount of energy.*0516

*Let us calculate that energy.*0534

*Once again, Q = MC δ T.*0536

*Q = I have 40 g of water, 4.18 is its heat capacity J/ g/°C.*0539

*I want my difference in temperature to be 22.81.*0553

*I want to take it from 22.81 down to 0°C.*0556

*When I do this calculation, it tells me that I require 3113.8 J.*0561

*Water is going to lose 3813.8 J in going from 22.1° down to 0, that heat is going to go into the ice at 0°C*0572

*and convert it from solid ice at 0 to liquid water at 0.*0586

*It is going to cause the ice to melt, this is where we use the δ H of fusion.*0591

*Let us see how much ice actually melts.*0596

*Once again, let me go blue because I like the blue.*0600

*We are going to take this 3813.8 J and we are going to multiply it by the reciprocal of the δ H of fusion.*0614

*1 g and 334 J, 334 J/ g is the δ H of fusion, that is how much energy it takes to either melt ice at 0°C and convert it to liquid water at 0,*0634

*Or it is the amount of energy that is going to be lost by liquid water at 0 and then convert that to ice.*0650

*When I do this, I end up with 11.42 g.*0661

*What is this 11.42 g?*0674

*11.42 g is the amount of ice at 0°C converted to liquid water.*0677

*Let us recap what is that we have done in terms of a nice picture.*0702

*We had liquid water at 25°C, I will just go ahead and put a 0 mark right there.*0705

*This is 0°C and then we have solid water at -7°C.*0713

*This H₂O liquid and this is H₂O solid.*0720

*A certain amount of heat was lost by the water in cooling and drop it down to 22.81°.*0730

*This amount of heat was enough to bring the ice of 20°C but still ice.*0738

*Now from the 22.81 there is a certain amount of heat that is going to be lost upon the water actually cooling down to 0°C.*0746

*Water at 0°C.*0755

*This amount of heat that is lost by the water in cooling, that amount of heat,*0758

*since the ice is now already at 0°C it is going to convert a certain amount of ice into liquid water.*0762

*That is the 11.42 g, that is our final state.*0771

*Here is where we are now.*0775

*Let us see, I have 40 g of H₂O liquid at 0°C, that is the initial 40 that drop down to 0.*0777

*I have 11.42 g of H₂O liquid at 0°C is the water that came from the ice that melted.*0796

*That gives me a total of 51.42 g of liquid water*0811

*I'm left with 25 g of ice initially -11.42 that had melted, that leaves me with 13.58 g of ice or solid water at 0°C.*0822

*This is our final state.*0844

*Our final state, once everything comes to equilibrium I have 51.42 g of liquid water at 0°C and I have 13.58 g of ice at 0°C.*0846

*That is what is going on, that is our final state.*0861

*Let us go ahead and do the δ H and δ S.*0865

*Any time you see the word insulated that means no heat is allowed to come in or go out.*0870

*It means that Q = 0.*0879

*Since, the pressure is constant, since P is constant we know that under conditions of constant pressure the δ H = Q so δ H= 0.*0888

*There is no δ H, there is no enthalpy transaction in this particular process.*0908

*Let us go ahead and talk about the δ S.*0915

*The δ S for this process is going to be composed of three parts.*0919

*It is going to be the δ S total.*0923

*It is going to be equal to the change in entropy of the first phase,*0926

*that means 40 g of H₂O liquid at 25°C taken to 40 g of H₂O liquid at 0°C.*0931

*The first part of the total entropy is going to be the entropy change in taking the 40 g of liquid water at 25*0950

*and converting it to 40 g of liquid water at 0.*0956

*To that we are going to add δ S2, to that is going to be the change in entropy associated with taking 25 g of H₂O solid at -7°C and converting it to 25 g of H₂O solid at 0°C.*0961

*We have to account for each and every single transformation, that is what we are doing.*0986

*To that we are going to add the δ S3,*0991

*We have 11.42 g of H₂O solid at 0°C converted to 11.42 g of H₂O liquid at 0°C.*1006

*There are three things that happen here.*1034

*Liquid water went from 25 to 0, there is a certain entropy change associated with that, that is δ S1.*1036

*25 g of ice went from -7° to 0 that is δ S2.*1043

*Out of that 25 g, 11.42 g of that was converted to liquid water and there is a change in entropy associate with that.*1056

*The total change in entropy, I just have to add them all up.*1067

*Let us see what we can do.*1070

*Let us do δ S1, let us always go back to our basic equations, always start with your basic fundamental equation.*1074

*From there, derive what you need.*1088

*DS = for conditions of constant temperature and pressure CP/ T DT – V Α DP and*1092

*that is our basic equation for calculating changes in entropy when the temperature of the pressure are constant.*1107

*In this particular case, this is great.*1113

*They tell us that the pressure is constant and that mean DP = 0.*1116

*This is equal to 0 because the pressure is constant.*1121

*This is not hard, most of these problems are actually very simple as long as you to start with the basic equation.*1126

*You want to be consistent, always start at the same place and then take it from there.*1131

*You might go this way or this way, but always start at the same place.*1136

*That means, once I integrate this I have δ S1 = the integral from T1 to T2 of CP/ T DT.*1141

*The CP is constant and this is liquid water that we are talking about so it was going to be 4.18 × the integral of T1 to T2 of DT/ T = 4.18 × the log of T2/ T1.*1156

*δ S1 = 4.18 × the nat log of the final temperature was 0°C, 273.*1177

*The initial temperature was 25°C, 298.*1186

*We have to work in °K and when we do that I get -0.366 J/ g/°K.*1190

*This is 4.18 J/ g/°K.*1208

*It is J/ g, I have 40 g and I still have to multiply this by the 40.*1211

*This is the part that you have to be very careful of.*1217

*It is really important to keep track of your units.*1220

*We have this many J/ g/°K.*1224

*We have 40 g so let me write that down.*1228

*We have 40 g of water so 40 g × -0.3,*1233

*There are numbers everywhere, 366 J/ g/°K or J/ g/°C.*1245

*And we end up with the δ S1 =- 14.65 J/°K.*1253

*I'm hoping that you confirm my arithmetic.*1262

*Our δ S1 = -1465 J/°K.*1265

*Let us do δ S2, it is going to be the same thing.*1271

*I mean, except now the heat capacity is going to be 2.09 instead of 4.18 because now we are talking about the ice going from -7 up to 0.*1275

*We have 2.09 × the nat log, final temperature is 273, initial temperature was 266.*1287

*We end up with 0.054 J/ g/ °K.*1299

*We have 25 g of ice so we have 25 g × 0.054 J/ g °K.*1307

*We are left with δ S2 of +1.36 J/ °K, that is δ S2.*1318

*Let me go ahead and do the next one on the next page.*1331

*We have δ S3, this one had to do with the melting of the ice, the converting of the 11.42 g of ice to 11.42 g of liquid water.*1334

*We are going to use the δ H of fusion.*1351

*The δ S3 remember that is the δ H of fusion divided by the temperature, that was the entropy change when there is a phase change.*1355

*During the phase change, the temperature is constant.*1366

*Because the temperature is constant we can go ahead and use the δ H.*1368

*This is how to calculate the entropy for phase changes.*1373

*The δ H of fusion is 334 J/ g and it is happening at 0℃ which is 273 °K.*1378

*We end up with 1.22 J/ g °K.*1381

*We have 11.42 g of ice as being converted 11.42 × 1.22 J/ g °K.*1398

*We end up with 13.97 J/ °K, that is δ S3.*1409

*We are going from ice to liquid water.*1419

*Entropy is rising, it is positive.*1422

*We just put them all together.*1426

*Our δ S total for this particular process is - 14.65 + 1.36 + 13.97 and we get a δ S total equal to 0.68 J/ °K.*1430

*A very practical problem because you are putting something at a certain temperature, something else at another temperature,*1457

*and there is going to be some final state that is going to be involved here.*1463

*Let us see what we have got now.*1472

*Let us go to example 2.*1476

*This is going to be the same sort of thing except that now we would be dealing with steam and liquid water.*1481

*It is going to be handled precisely the same way.*1486

*We are going to combine our intuition based on heat capacities which is going to absorb more heat.*1489

*Without change in temperature, it is going to be liquid water.*1496

*You are going to see how much you have of each and decide where you want to start the problem.*1498

*25 g of steam at 100°C and 310 g of liquid water at 25°C are combined in a fully insulated flask.*1503

*We are fully insulated so again Q = 0 so δ H is going to equal 0.*1514

*Given the following data, pressure is constant, we have the heat capacity for liquid water.*1521

*We have the heat capacity for the gas.*1527

*Notice 1.86 is not very high, it is a bit surprising given that for water.*1529

*The δ H of vaporization is 2257 J / g.*1535

*The δ H of vaporization, this is the amount of heat necessary to take 1 g of liquid water from the liquid phase to the gaseous phase, still at 100°C.*1540

*Or it is the amount of heat that is lost by 1 g of steam in transforming from 1 g of steam to 1 g of liquid water at 100°C.*1553

*They want us to know, what is the final temperature of the system and what phase or phases are present?*1566

*Same thing as before, what is the final state?*1570

*It is just a different way of asking it.*1572

*They want you to calculate δ S for this transformation.*1573

*We already know δ H is going to be 0 because we are at constant pressure and this is a fully insulated flask.*1576

*Let us see what we can do.*1583

*Let us deal with the 25 g of steam first.*1587

*I have got Q = MC δ T.*1591

*I got 25 g of steam at 1.86 J/ g/°C × 30°C.*1595

*What I'm calculating here is the amount of energy that is going to be required*1606

*in order to take the steam at 130°C and bring it down to steam at 100°C.*1609

*This is still steam, it has not converted yet.*1617

*This multiplication, I end up with 1395 J.*1620

*This is the total amount of energy has lost by steam in going from H₂O gas at 130°C to H₂O gas at 100°C.*1626

*Let us see what this heat lost, this 1395 J.*1647

*If that heat is lost, that heat is going to end up going into the water so the water temperature is going to rise.*1652

*Let us see what is the rise in water temperature is.*1659

*1395 J = we have 310 g of water, its heat capacity is 4.18 J/ g/°C.*1664

*And we are going to multiply that by δ T.*1676

*When I calculate δ T, I get δ T = 1.08°C.*1680

*The water rises by only 1.08°C.*1686

*We have 310 g of H₂O liquid at 25°C and we just took it to 310 g H₂O liquid at 26.08°C.*1693

*Now the water is at 26.08°C.*1714

*Let us see what is next, let us find out how much energy is lost now that the steam is at 0°C.*1719

*We have 25 g of steam at 100°C and we have water at 26.08°C.*1732

*They still need to find some in between temperature that is going to be the final temperature, the equilibrium state.*1742

*The steam has to cool a little bit further but now it is steam at 100°C.*1749

*It cannot drop any temperature now it is going to undergo a phase change once it hits 100.*1756

*Now we are going to have to use the δ H of vaporization.*1761

*We have 25 g of steam and the δ H of vaporization is 2257 J/ g.*1767

*We have 56,425 J.*1777

*This 25 g of steam at 100°C, when it converts to liquid water at 100°C it gives up 56,425 J of energy.*1783

*It is going to give that energy to the liquid water, the 310 g.*1796

*This amount of energy is going to raise the temperature of water yet some more from the 26.08.*1801

*Let us find out what the new temperature is.*1807

*Let us just say what this is here, write it down.*1813

*For H₂O gas at 100°C converted to H₂O liquid at 100°C.*1819

*This is the amount of heat lost by the 25 g of steam.*1835

*Let us see what this does to the water.*1840

*Now, we have 56,425 J = 310 g 4.18 J/ g/°C × δ T.*1842

*Our δ T is going to be 43.54°C.*1858

*Let us do this on the next page.*1870

*We are at 26.08° and we just raised the temperature at another 43.54°C.*1876

*Now, my temperature is 69.62°C.*1889

*310 g of liquid water is at 69.62°C.*1904

*We also have 25 g of liquid water at 100°C.*1918

*They are both liquid water but at very different temperatures.*1934

*This 25 g came from the condensation of the steam.*1936

*This 310 g at 69.62 just came from the normal liquid water that is rising the temperature.*1941

*We have water at 100°C, we have water at 69.62°C.*1948

*This water is going to cool, this water is going to warm.*1966

*They are going to reach some final temperature.*1969

*We need to find what that is.*1973

*The heat lost by the water at 100°C as it cools, is the heat gained by the water at 69.62 as it warms up.*1977

*That is the whole idea, heat out heat in.*1987

*We have Q lost = - Q gained.*1992

*This negative sign is very important.*1999

*The heats are equal.*2002

*When I add them, they add to 0.*2003

*This mathematic this ends up being a negative sign, very important.*2005

*The heat that is lost by one substance, in this case, water at a certain temperature, is negative of the heat gain by the other.*2009

*One is gaining and one is losing.*2016

*Let us write it out.*2020

*The heat lost that is going to be the MC δ T of the water at 100 and the heat gain is going to be the MC δ T of the water at 69.62.*2023

*Let us write that out, MC δ T.*2034

*I’m going to write M1 C1 δ T = M2 C2 δ T to let you know we are talking about two different masses.*2041

*In this particular case, it is not going to be two different heat capacities because now they are both water but the masses ore different.*2052

*We have got 25 g of water, the heat capacity is 4.18 and the change in temperature is temperature final – initial.*2059

*This initial temperature is 100°C for the 25 g = negative.*2075

*I have 310 g of water, its heat capacity is 4.18 and now it is going to be temperature final.*2083

*Its initial temperature is 69.62, this is the equation.*2091

*I'm looking for a final temperature.*2097

*We have done this from general chemistry.*2100

*Let us see what the arithmetic looks like.*2105

*The 4.18 go away and I'm going to be left with 25 TF - 2500 = -310 TF + 21,396.2.*2109

*I should have 335 TF = 23,896.2.*2132

*I have a temperature final of 71.33°C, this is what I wanted, the final temperature.*2144

*We have 335 g of H₂O liquid, the 310 g + 25 g of liquid water at 71.33°C.*2157

*This is our final state, that is what happened here.*2177

*The steam cooled, the water rose in temperature.*2181

*The steam converted to liquid water at 100 and that conversion, the heat that was lost by that conversion,*2185

*one of the 310 g of water lifting the temperature to the 69.62.*2192

*We have water at 100 and water at 69.62.*2198

*We are going to find some medium temperature.*2201

*I’m going to write it in between them.*2203

*Again, 25 g and 310 g there is a big difference.*2204

*There is only one from 69 to 71.*2208

*This one is from 100 down to 71, that is the whole idea.*2211

*Let us go ahead and calculate the δ S for this process.*2218

*δ S this time is going to be 4 different things that happened.*2223

*δ S total = we have δ S1.*2227

*This is the conversion of 25 g of gaseous water at 130° taken to 25 g of H₂O gas at 100 °.*2232

*To that we are going to add the δ S2, the second process we are going to take this 25 g of H₂O gas at 100°C.*2253

*And we are going to convert it to 25 g of H₂O liquid at 100°C.*2266

*And we have another process, δ S3 is going to be taking 25 g of H₂O liquid at 100°C and converting it to 25 g of H₂O liquid at 71.33°C.*2277

*Then the final entropy change is going to be 310 g of H₂O liquid at 25°C and 310 g of liquid water at 71.33°C.*2305

*25 g of steam at 130 went to 25 g of steam to 100, there is an entropy change associated with that.*2337

*25 g of steam at 100 went to 25 g of liquid water at 100, there is entropy change associated with that.*2344

*+ 25 g of H₂O liquid at 100 and now this 25 g of H₂O liquid at 71.33, there is an entropy change associated with the temperature change.*2351

*And this one we have 310 g of H₂O liquid at 25 going to 310 g of H₂O at 71.33.*2360

*There are 4 parts to this entropy.*2368

*Let us go ahead and calculate them.*2370

*Let us see here, what should I do.*2374

*The general equation once again, DS = CP/ T DT – V Α DP, this goes to 0 because pressure is constant.*2377

*Therefore, our general equation for change in entropy upon a temperature change*2390

*is going to be the integral of CP/ T DT from temperature 1 to temperature 2.*2394

*And for a phase change, our entropy is going to be δ H of vaporization ÷ T.*2405

*For 3 of these, our temperature changes we are going to use this one.*2412

*One of these is going to be a phase change, we are going to use this one.*2416

*Let us do it.*2419

*The δ S1 = 1.86 we are taking it from 403 which is 130 to 373 which is 100.*2424

*Which is going to be DT/ T and we are going to get 1.86 × the nat log of 373/ 403.*2438

*We end up with -0.144 J/ g/ °K.*2451

*We have 25 g of steam so 25 × 0.144 J/ g/ °K gives us - 3.60 J/ °K this is δ S1.*2459

*δ S2 is going to equal the,*2482

*This was the phase change 1.*2485

*We have the δ H of vaporization/ T which id equal 2257 J/ g ÷ 373 °K because the transformation took place at 100°C.*2487

*We get 6.05 J/ g/ °K.*2504

*We are changing 25 g of it so g/ °K and we end up with -151.27 J/ °K.*2511

*It is negative, there is a negative sign here but it is negative because it is dropping.*2532

*I'm actually going from a gaseous phase to a solid phase, entropy decreases.*2537

*You can use your qualitative understanding.*2544

*As things go from solid, liquid, to gas, the entropy is going to be positive.*2547

*As things go from gas to liquid to solid, the entropy is going to be negative.*2551

*These are just the absolute values, these are the actual numbers.*2555

*The magnitudes of the energy changes of things like that.*2559

*But I put negative sign here, the entropy is negative because we are going from a gaseous phase to a liquid phase, in more ordered phase.*2562

*Our entropy change is negative.*2569

*By all means, use the qualitative.*2571

*If there is ever a place where you want a qualitative analysis, it is in thermodynamics.*2574

*δ S3, it is going to equal 4.18 × the integral 373 to 344.33 DT/ T = 4.18 × the nat log of the final temperature which is 344.33 °K / 373.*2580

*We end up with -0.334 J/ g °K.*2610

*Once again, we have 25 g of this -0.334 J/ g °K and you get -8.36 J/ °K, this is δ S3.*2617

*In this particular case, the negative that shows up automatically based on the mathematics because we are actually doing integration.*2639

*As integration ends up being a logarithm.*2645

*And when you have a logarithm in the quotient and the quotient is less than one, you will get a negative number.*2649

*This negative number is already taken care of.*2655

*We do not have to worry about it being qualitative.*2657

*We have got one more δ S here.*2663

*We have our δ S4 and that is going to equal the 4.18 because now are talking about water × the nat log .*2665

*And we went from 298 °, we went to the 344.33 °, and tat is going to equal 0.604.*2678

*This is positive because we are rising in temperature.*2691

*Temperature increases, entropy increases.*2694

*J/ g °K and this time we have 310 g of the water that we raised from 25 to 78 or whatever it was.*2699

*× 0.604 J/ g °K and end up with 187.3 J/ °K.*2710

*Our δ S total = -3.60 - 151.27 - 8.36 + 187.3.*2726

*Our δ S total for this particular transformation = 24.02 J/ °K, a net increase.*2746

*There you go, a long problem, a little bit tedious but nothing altogether difficult in terms of what is actually happening.*2758

*Just take it one piece at a time.*2767

*Any time you are dealing with any sort of a mixture of two things, one is at a certain temperature and one is at another temperature,*2771

*the heat is lost by one thing is going to be the heat that is gained by the other thing.*2782

*Particularly, when we are talking about something that is going to be in a completely insulated flask which is often how we run these things.*2785

*We do not have to worry about any heat being transferred in or out of the combined system.*2791

*There you go.*2797

*Let us round this out with one final example and it does have to do with entropy.*2801

*Let us see what it says.*2807

*Given the following expression DS = CP/ T DT – V A DP.*2812

*We already know what this expression is, this is one of general expressions for finding the change in entropy.*2817

*This should be S.*2822

*Finding the entropy under conditions of constant temperature and constant pressure.*2826

*Given the following expression, calculate the decrease in temperature that occurs when 1 mol of water*2832

*at 25°C and 1100 atm is brought adiabatically and reversibly to a pressure of 1 atm.*2837

*Assume that K, the coefficient of compressibility = 0.*2847

*That is extra information, do not worry about that.*2851

*You do not need to do anything with it.*2852

*You have the following data for water.*2855

*The molar volume of water is 18 cm³/ mol, constant pressure heat capacity this time is expressed in J/ mol °K,*2858

*75.3 J/ °K and Α the coefficient of thermal expansion is 2.07 × 10⁻⁴/ atm.*2867

*Let us see what they are asking us here.*2880

*I have 1 mol of water it is at 25°C but its pressure is 1100°,*2882

*Under conditions of adiabatically and reversibly, I drop the pressure to 1 atm.*2892

*Adiabatic means there is no change in the Q = 0.*2899

*In other words, there is no heat is allowed to flow.*2904

*I know that when I drop the pressure from 1100 to 1 atm, my intuition tells me that the temperature is going to drop.*2907

*The temperature is going to drop precisely because no heat is allowed to flow, this is the adiabatic part.*2915

*No heat is allowed to flow into the system, in order to keep the temperature up.*2921

*The temperature is definitely going to drop.*2926

*Remember that an adiabatic process is when you increase the pressure, it is the hugest,*2928

*the largest drop in temperature happens in an adiabatic reversible process.*2935

*We want to know what that pressure drop is.*2941

*In other words, starting at 25°C if I do not allow any heat to flow, if I conduct this reversibly and very slowly,*2943

*adiabatically what is my final temperature going to be when my pressure in the system is now 1 atm, that is what this is asking.*2952

*Let us do it, adiabatic and reversible, that means that DQ reversible = 0.*2960

*Adiabatic Q is 0, we now we just put a reversible because it is telling us that it is actually reversible.*2985

*DS by definition = DQ reversible/ T.*2993

*If DQ reversible is 0 it implies that DS = 0.*2998

*There is no entropy change for this process.*3005

*We have an equation 0 = CP/ T DT – V Α DP.*3007

*Move is over, I get CP/ T DT = V Α DP.*3021

*Integrate this expression, the heat capacity is going to be constant so I end up with the following.*3031

*CP × the integral from T1 to T2 of DT/ T = V Α × the integral from P1 to P2 of DP.*3038

*I’m just taking the expression and found out that DS = 0.*3055

*I set it equal to 0, I rearranged it and I integrate this expression.*3058

*My final equation is going to be.*3062

*I get CP × the nat log of the final temperature/ the initial temperature = V Α δ P.*3070

*I have everything that I need, now I need to put it in terms of units that match and everything should be fine.*3084

*Let us go ahead and deal with the molar volume first.*3091

*I have 18 cm³/ mol = 18 × 10⁻³ dm³/ mol.*3094

*A cubic decimeter is a liter.*3108

*If there is 1 unit that you should know, 1 liter is 1 cubic decimeter.*3111

*We get 18 × 10⁻³ L/ mol, that just means that 1 mol of liquid water occupies 18 × 10⁻³ L.*3115

*Volume is it L and pressure is at atmosphere.*3131

*We are going to end up with something in L atm and we need to convert it to J.*3134

*Let us go ahead and take care of the V DP first.*3140

*I have got V × δ P.*3145

*The volume is 18 × 10⁻³ L/ mol.*3150

*The δ P was final - initial so it is going to be 1 atm - 1100 atm.*3167

*V δ P when I actually multiply this out I get - 19.782 L atm/ mol.*3177

*In order to convert it, I multiply by 8.314 J ÷ 0.08206 L/ atm.*3191

*And I end up with - 2004 J/ mol.*3203

*V × DP = V × DP = -2004.*3212

*I just have to convert to J because I’m working J, because heat capacity is in Joules.*3221

*I need to make sure that my units match.*3226

*Let us go to the next page.*3233

*I have my equation CP LN TF/ TI = VΑ DP.*3236

*I have got 75.3 × the nat log of the final temperature.*3252

*That is what I want, I want the final temperature.*3261

*They wanted to know what the temperature drop was.*3262

*TF I started off at 298 = Α V DP.*3265

*Α is 2.07 × 10⁻⁴ and V DP was – 2004.*3275

*And so I get 75.3 × LN of TF/ 298 = - 0.415.*3287

*I go ahead and I will do the division and exponentiation.*3301

*Let me write that out LN of TF/ 298 = -0.00551.*3306

*When I exponentiate, e to this power, e to that power.*3317

*TF/ 298 = 0.9945 and I get the final temperature of 296.36.*3326

*And therefore, δ T = final temperature 298.*3345

*I think it is a little bit of an arithmetic problem here.*3364

*The final temperature, final – initial will give me something positive, it is going to be something negative.*3366

*I apologize if I calculated some of it a little bit of arithmetic issue here.*3372

*I’m going to go ahead and leave this.*3376

*The process is correct but I seem to have done the multiplication wrong.*3378

*It is from what I see on the table.*3382

*This is correct and now if you just go ahead and multiply this × that.*3384

*I'm sorry, I apologize.*3394

*I wrote 273 on the paper, it should be 298.*3395

*This was correct.*3400

*We have 296.36 and our δ T = 296.36 – 298.*3402

*I'm sorry I thought it was 0 so I had 273 on my paper.*3416

*Our δ T -1.64°C or °K, I will just go ahead and write °C and there we go, this is our temperature drop.*3420

*Just use what is given to you in the problem.*3437

*You have a general equation because it is adiabatic and reversible, there is no change in entropy.*3439

* DS= 0, rearrange the equation and just put the numbers in.*3444

*The only thing you have to watch out for in this problem is the units.*3449

*Cm³/ mol you have to change that to L because the pressures and atmospheres.*3455

*That is just my particular preference.*3462

*You can work in cm³ and work in Pa, it does not really matter.*3464

*Just what you are comfortable with.*3468

*I’m comfortable with L atm, convert it to J.*3470

*Once I’m in Joules, I’m going to go ahead and run in the problem.*3473

*There you have it, a change in temperature is -1.64°C .*3477

*Thank you so much for joining us here at www.educator.com.*3482

*We will see you next time, bye.*3485

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