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

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

## Discussion

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### Entropy 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
- Entropy Example Problems I
- Fundamental Equation of Thermodynamics
- Entropy as a Function of Temperature & Volume
- Entropy as a Function of Temperature & Pressure
- Entropy For Phase Changes
- Entropy For an Ideal Gas
- Third Law Entropies
- Statement of the Third Law
- Entropy of the Liquid State of a Substance Above Its Melting Point
- Entropy For the Gas Above Its Boiling Temperature
- Entropy Changes in Chemical Reactions
- Entropy Change at a Temperature Other than 25°C
- Example I
- Part A: Calculate ∆S for the Transformation Under Constant Volume
- Part B: Calculate ∆S for the Transformation Under Constant Pressure
- Example II: Calculate ∆S fir the Transformation Under Isobaric Conditions
- Example III
- Part A: Calculate ∆S if 1 Mol of Aluminum is taken from 25°C to 255°C
- Part B: If S°₂₉₈ = 28.4 J/mol-K, Calculate S° for Aluminum at 498 K
- Example IV: Calculate Entropy Change of Vaporization for CCl₄
- Example V

- Intro 0:00
- Entropy Example Problems I 0:24
- Fundamental Equation of Thermodynamics
- Entropy as a Function of Temperature & Volume
- Entropy as a Function of Temperature & Pressure
- Entropy For Phase Changes
- Entropy For an Ideal Gas
- Third Law Entropies
- Statement of the Third Law
- Entropy of the Liquid State of a Substance Above Its Melting Point
- Entropy For the Gas Above Its Boiling Temperature
- Entropy Changes in Chemical Reactions
- Entropy Change at a Temperature Other than 25°C
- Example I 19:31
- Part A: Calculate ∆S for the Transformation Under Constant Volume
- Part B: Calculate ∆S for the Transformation Under Constant Pressure
- Example II: Calculate ∆S fir the Transformation Under Isobaric Conditions 27:53
- Example III 30:14
- Part A: Calculate ∆S if 1 Mol of Aluminum is taken from 25°C to 255°C
- Part B: If S°₂₉₈ = 28.4 J/mol-K, Calculate S° for Aluminum at 498 K
- Example IV: Calculate Entropy Change of Vaporization for CCl₄ 34:19
- Example V 35:41
- Part A: Calculate ∆S of Transformation
- Part B: Calculate ∆S of Transformation

### Physical Chemistry Online Course

### Transcription: Entropy Example Problems I

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

*Today, we are going to start off with our example problems for this particular entropy sections.*0005

*We have done a lot of discussion about entropy, a lot of mathematical derivations and now we just want to become familiar with doing problems.*0010

*We are going to do a lot of them just like we did for energy.*0018

*Let us go ahead and get started.*0022

*Again, before I jump into the problems, I want you to do the same thing that I did with energy which was stop and*0025

*take a look at what it is that we want to bring to the table as far as our tools and what we need to know.*0031

*I will discuss a little bit of that in the last few lessons, specially when we did a summary of entropy but I would like to do it again*0036

*just so we have a list of equations that we want to refer to immediately instead of pouring through a whole bunch of derivations and things.*0045

*What we need to know in order to solve entropy problems?*0054

*One of the things that you need to know is the fundamental equation of thermodynamics.*0058

*In general, the fundamental equation of thermodynamics is not necessarily going to be used for problem solving but it is a nice thing to be able to know.*0063

*The fundamental equation of thermodynamics is DS = 1/ T DU + P RT PV that is the fundamental equation of thermodynamics.*0077

*It is derived from the first and second laws.*0103

*First law being DU = DQ - DW and the second law which is the definition of entropy DS = DQ reversible/ T.*0107

*Definitely that equation you want to know.*0121

*For entropy as a function of temperature and volume, this with what you want to know.*0129

*You want to know the total differential expression which is always easily derived.*0136

*You just take the derivative so it is going to be DS/ DT constant V DT + DS DV under constant T DV.*0140

*The actual equation that you want to bring to the table is DS = CV/ T DT + A/ K DV this is the equation that you want to know just like for energy.*0155

*Absolutely, you have to know this most problems begin with this equation or the next one that I’m going to write.*0175

*In the case of dealing with entropy where entropy is going to be a function of temperature and pressure,*0180

*If pressure is mentioned in the problem we have DS = DS DT under constant pressure DT + DS DP under constant temperature DP.*0186

*The actual derivation that we did was the following DS = CP/ T DT - volume × the coefficient of thermal expansion × DP.*0209

*This is the equation, this is the other one that you want to bring to the table.*0225

*Of course, A and K we want to know those definitions.*0230

*A that was equal to 1/ V × , DV DT under constant pressure and K which is the coefficient of compressibility is -1/ V and*0235

*it is a change in volume per unit change in pressure under conditions of constant temperature.*0251

*These are some of the things that we want to bring to the table when we start thinking about these problems.*0261

*We also want to remember the relationship for an ideal gas, the relationship between the constant pressure heat capacity and*0268

*the constant volume heat capacity which was CP - CV = Rn this is for an ideal gas.*0274

*For phase changes, for problems involving phase changes.*0289

*We have a change in entropy for the process of vaporization = the change in enthalpy for the process of vaporization divided by*0297

*the boiling temperature at which the vaporization takes place.*0307

*The change in entropy of fusion or melting, or the back and forth with solid liquid, liquid solid,*0312

*that is equal to the δ H of fusion divided by the temperature of melting.*0318

*For phase change, solid liquid, liquid to gas, or the other way, coming back the other way gas to liquid, liquid to solid,*0325

*these are the equations that we use when we talk about entropy.*0333

*In going from a solid to a liquid the entropy increases because it is becoming more disorder.*0336

*There is more ways for the molecules to move around.*0341

*In going from a liquid to a gas the entropy rises again, it rises a lot because that now you are going from a condensed state liquid to a gas state.*0343

*There is a whole bunch of room for gas molecules to bounce around it so the entropy rises a lot.*0353

*In going from gas to a liquid, entropy decreases.*0361

*In going from liquid to a solid, the entropy decrease, not a lot but it does decrease.*0364

*Because again you are going from a liquid state which is more disordered to a solid state which is more orderly.*0366

*For an ideal gas, when we are dealing with problems that have an ideal gas you can use the previous equations or*0376

*you can use this set of equations, it is totally up to you.*0384

*For S = S a function of temperature and volume we had DS = CV / T DT + nR/ V DV which implies that DS/ DV under constant T = nR/ V which is positive.*0388

*For entropy of an ideal gas as a function of temperature and pressure we have a differential expression DS = CP/ T DT - nR/ P DP*0419

*which implies that DS/ DP under constant temperature is going to equal - nR/ P.*0435

*Integrating these two, we get δ S = CV Ln T2/ T1 + nR LN of V2/ V1 δ S = CP × LN of T2/ T1 - nR × log of P2/ P1.*0446

*This equation and this equation if you like these two are the differential expressions.*0495

*Third law in entropy, the third law of entropy, the entropy at a given temperature is equal to the integral from 0 to T CP/ T DT.*0507

*This is the entropy of a solid at temperature T and pressure P.*0532

*Let us go ahead in the statement of the third law.*0558

*Let us go ahead and write that down again real quickly.*0566

*It says the entropy of a pure perfectly crystalline solid substance at 0°K is 0.*0569

*The entropy of a pure crystal as 0°K is 0.*0603

*If I want to know what the entropy is at any other temperature, I just take the integral from 0 to that particular °K temperature*0607

*of the constant pressure heat capacity divided by the temperature DT.*0616

*This is the definition of the third law of entropy.*0620

*In computing the entropy of the liquid state of this particular solid, the liquid state of the substance about its melting point,*0629

*in other words I have taken a solid at a particular entropy, it is melted and it is a little higher temperature.*0662

*The entropy of that will include the entropy of the solid, the melting process and then the rise in temperature as a liquid.*0671

*It was entropy at temperature T = the entropy of the solid which is this thing CP/ T DT this standard just means 1 atm pressure.*0682

*When you see the circle there, you are used to thinking the standard temperature pressure , it is not standard temperature,*0701

*this circle is just standard pressure which is 1 atm.*0707

*That is really what it means because now we have come to the point where it is more sophisticated.*0711

*We do not have to stick with 298 K or 25°C but generally we stick with 1 atm.*0717

*These third law of entropy, these are the values that you actually see in your tables of thermodynamic data.*0724

*They are calculated with this particular integral.*0730

*It is the entropy of the solid + the entropy of the change of the melting which is δ H of fusion divided by the melting temperature + from going*0734

*from the melting temperature to another temperature above that, it is going to be integral of the melting temperature to T ×*0749

*the heat capacity of the liquid phase/ T DT, this is the solid phase.*0756

*I’m just accounting for all the phase changes.*0765

*The entropy of the solid that is given from here, the rise of entropy that come from melting, and the rise in entropy that come from taking it*0768

*from the melting temperature to a higher temperature which is not quite yet its boiling temperature.*0776

*For the gas if it actually boils, for the gas at its boiling temperature is the exact same thing,*0783

*just a couple of more terms for the gas above its boiling temperature.*0796

*The total entropy for the gas = 0 to the melting temperature of the constant pressure heat capacity of the solid/ T DT + the entropy change*0809

*that accompanies the melting + the temperature change from the melting temperature*0832

*to the boiling temperature of the liquid phase constant pressure heat capacity divided by T DT.*0841

*This is just the expression of this depending on the temperature range + the entropy change that accompanies the vaporization*0849

*as the liquid goes from liquid to the gaseous state which is going to be isothermal process +*0860

*any extra rise in temperature of the boiling temperature up to my final temperature.*0871

*This time I'm going to use the heat capacity of the gaseous phase / T DT this is it.*0876

*If I want to calculate the third law entropy of the solid, if I want to calculate the entropy of liquid,*0884

*if I want to calculate the entropy of the gas of that substance, this is what I use.*0890

*It implies that I have to know what the heat capacities are, I have to know what the δ H are, what the boiling temperature is,*0896

*what the melting temperature is, all of these things are readily available in tables all over the place.*0903

*It is very easy to get all this information for a solid, for the liquid phase, for the gas phase.*0907

*Let me make this clear, this says gas so this is the G.*0915

*For entropy changes in chemical reactions.*0923

*For problems that involve entropy changes in chemical reaction the δ S = the sum of the entropy, the standard entropy of the products*0927

*including these coefficients - the sum of the standard entropies for the reactants.*0951

*We are looking this up in a table of thermodynamic data, these are tables of third law entropies.*0959

*Just look them up in a table of the data and just take the sum of the products - the sum of the reactants and you have the δ S of that process.*0966

*When we go ahead and write including isometric coefficients and this is nothing that you have done a 1000 times in general chemistry.*0978

*Final thing is the change in entropy of a particular chemical process at a temperature other than 25°C or*0992

*other than the temperature at which you happen to know the change in entropy which for tables of thermodynamic data happened to be 25°C or 290°K.*1004

*In general, it is going to be that particular entropy at some initial temperature + T0 to T of the change that δ of the heat capacities products – reactants.*1015

*Where δ CP is the sum of the heat capacities for the products - the sum of the heat capacities for the reactants*1039

*including the volumetric coefficients because heat capacity is an extensive property.*1064

*It depends on how much is there.*1072

*If I have 1 mol twice that is twice the heat capacity.*1072

*Regarding this, this is valid as long as none of the reactants or products undergoes a phase change in that particular temperature range from T0 to T.*1091

*In this particular temperature range, the reactants and products they cannot change their phase,*1132

*they cannot go from solid to liquid, liquid to gas, things like that, then this is valid.*1136

*If I know the δ S of a particular chemical reaction at a given temperature T 0, I can calculate what the entropy would be at any other temperature.*1142

*Let us do some problems.*1155

*This first set is going to be reasonably simple and straightforward.*1158

*We just want to get comfortable with them.*1161

*That is what we do, we get comfortable with the set of concepts by doing problems.*1163

*That is the only way to do it.*1169

*Let us go ahead and start.*1172

*Let us go ahead and do this in red.*1177

*A temperature of 2 mol and an ideal gas, so I have 2 mol of an ideal gas is raised from 25°C to 225°C, the constant volume heat capacity is 3 Rn/ 2.*1180

*They want us to calculate δ S for this transformation under constant volume so we are dealing with constant volume.*1193

*We are dealing with an ideal gas so that is always nice.*1199

*Let us go ahead and see what we can do here.*1204

*I have an ideal gas, I had two things that I can do.*1207

*I can go ahead and use the equations for the ideal gas which is absolutely fine or I can go back to my original equations,*1211

*the equations that I apply to every system and let the problem tell me how to handle it.*1218

*Me, personally, it is just a habit of mine to always fall back on the basic mathematical equations*1223

*that apply to all systems and let the problem tell me where to go from there.*1229

*I'm going to go ahead and do it that way.*1234

*Part A, I’m going to start off with my basic equation.*1237

*This is the temperature changes constant volume so this is the temperature volume issue.*1239

*DS = CV/ T DT + A/ K DV.*1246

*It is telling me the constant volume so DV is 0 so this whole term goes to 0 because V is constant.*1256

*This is beautiful.*1265

*I’m left with is this so I have DS = CV/ T DT well δ S, if I integrate this expression = the integral from T1 to T2 of CV/ T DT.*1266

*This constant volume heat capacity, I have it is 3 Rn/ 2 and it is constant so I can pull that out.*1287

*It ends up being CV × T1 T2 DT/ T.*1294

*I end up getting δ S = CV × LN of T2/ T1 which I would have gotten any way if I had just used the equation for the ideal gas directly.*1305

*It does not matter, this is just my habit, you do not have to do it this way.*1321

*3/2 Rn/ 2 so 3/2 × 8.314 J/mol-°K × n which is the number of moles which we happen to have 2 mol and 2 mol × the nat log of 225 and 25.*1329

*This is going to be 498°K ÷ 298°K and when I do this I end up with the following δ S = 12.81 J/°K because mol and mol cancel.*1351

*Nice and simple, the entropy of this gas from 25 to 225°C with this particular constant volume heat capacity,*1369

*I have actually increased the entropy, I have increased disorder like 12.81 J/°K.*1382

*That is all that is going on here, temperature increase δ S is positive, everything is good.*1387

*If we end up with a -δ S there would be a problem here so we know that we made some sort of a mistake.*1394

*What I did, take a look at this 498 ÷ 298 and I'm hoping that you guys recognize this*1400

*but sometimes things like this actually slip through the cracks, I do want to mention that.*1407

*°K temperature must be used and here is why, in this case °K must be used and the reason is the following.*1413

*Even though, if I took 225°C/ 25°C you know the °C cancels °C, it is fine.*1428

*225 or 25 is a pure number, you can take the log of a pure number.*1439

*However, 225 or 25 = 9 that 9 does not equal 498/ 298 = 1.67 that is the difference.*1445

*The reason is the following, 225 + 273/ 25 + 273, the 273 still not cancel because they cannot cancel these numbers are not the same.*1461

*Even the units cancel, it looks like you can use that, you cannot use the Celsius temperature but you have to use the Kelvin temperature.*1481

*This is not linear.*1487

*If I multiply it by something then we cancel.*1490

*In this particular case, temperature is not linear, you are actually adding a term, these terms, this does not happen so you have to use the 498 to 298.*1494

*Let us go ahead and do part B.*1506

*In this particular case, we use constant pressure well DS = CP/ T DT - VA DP.*1511

*In this particular case it is going to be the pressure that is constant so that term goes to 0 because P is constant.*1524

*We have the same thing, we end up with just this expression right here.*1532

*When we integrate that expression we get δ S = the integral from temperature 1 to temperature 2 of CP / T DT.*1536

*Let us go ahead and see what the relationship is.*1548

*They gave us CV they said that CV = 3/2 Rn.*1549

*We know the relationship, this is an ideal gas so we know that CP - CV = Rn.*1556

*Therefore, CP = Rn - CV CP = Rn + 3/2 Rn CP = Rn × 1 + 3/2.*1566

*I get CP = 5/2 Rn there we go, that is what I put in to here.*1590

*Therefore, I get δ S and while I do the integration I get δ S = 5/2 Rn temperature 1 to temperature 2 of DT / T = 5/2 Rn LN of T2/ T1.*1596

*I have δ S = 5/2 × 8.314 J/°K × 2 mol × log of 498/ 298.*1622

*I end up with δ S = 21.35 J/°K.*1643

*If I hold the pressure constant and I change the temperature from 25°C to 225°C, I create more disorder.*1655

*That idea if I held the volume constant.*1662

*That is all this is saying.*1666

*Let us take a look at example 2, we have a monatomic solid is measured to have a constant pressure heat capacity of 3.25 Rn,*1671

*we have 2.5 mol of this solid and it is taken from 298°K to 548°K, calculate δ S for this transformation under isobaric conditions.*1683

*Isobaric means constant pressure.*1693

*You can see all these terms, you can see constant pressure, you can see isobaric, you are going to see CP,*1697

*you are going to see molar heat capacity, heat capacity, you just need to interpret what they mean.*1701

*This is constant pressure so we want to deal with, we have DS = CP/ T DT – VA DP.*1706

*That is the general equation.*1725

*Again, I would like to begin with a general equation.*1726

*Let us see, isobaric conditions constant pressure which means DP= 0 so that falls out.*1730

*We are left with δ S = CP × integral from T1 to T2 of DT/ T which is equal to CP × the LN of T2/ T1.*1735

*This is starting to become very simple I hope, δ S = 3.25 Rn × 8.314 J/ mol °K.*1749

*I have 2.5 mol of the solid , mol cancels mol log.*1765

*I’m taking it from 298 to 548 so it is going to be 548 ÷ 298.*1773

*What I get is δ S =241.15 J/ °K.*1782

*The biggest problem you should have what this is one of arithmetic.*1790

*Increase in temperature implies an increase in entropy because a change in entropy is positive and δ is final – initial.*1795

*Let us see what we have got.*1809

*We have the constant pressure heat capacity per mol for aluminum is 20.7 + 12.4 × 10⁻³ × T.*1818

*In this particular case, the constant pressure heat capacity is not constant, it is actually a function of temperature.*1826

*As the temperature rises the heat capacity rises.*1834

*Here we cannot pull it out from under the integral sign.*1837

*We have to leave it there and do the integration but it is pretty straightforward, it is just polynomials with a degree of 1.*1841

*Calculate δ S if 1 mol of aluminum is taken from 25 to 225 ℃.*1849

*In part B, they say if the standard entropy at 298°K is 20.4 J/mol-°K, calculate the standard entropy for aluminum at 498°K.*1855

*Let us take this + some integral which is actually going to be the integral from part 1 so it is really easy.*1869

*Let us go ahead and do part A, taken from 25°C constant pressure heat capacity.*1877

*I have my basic equation of DS = CP/ T DT - VA DP this is going to go to 0 because we are dealing with constant pressure here.*1903

*I have δ S = it is going to be the integral of CP/ T DT as we go from temperature 1 to temperature 2.*1919

*And because it is not constant, it is going to be 298 to 498, I think that is correct.*1932

*The constant pressure heat capacity is this thing right here so I have 20.7 + 12.4 × 10⁻³ T/ T DT.*1940

*What I end up solving this integral which I want to do right now.*1964

*I'm hoping that you can actually take this into separate out, it is going to be 20.7/ T + the integral this thing.*1970

*The T and T cancel so I have two integral.*1976

*And when you solve this, you are going to end up with δ S= 13.11 J/ mol-°K.*1979

*And because it is 1 mol it could be 13.1 J/°K.*1995

*This is just 1 mol, I will just go ahead and leave this mol here.*1998

*That is the δ S for the particular process.*2001

*Part B says, the S at another temperature = the S at a given temperature that I do know +*2005

*the integral from that temperature to the next temperature of the CP/ T DT.*2016

*S of 498= 298, the 298 they tell me is equal to 28.4.*2025

*This integral right here that is exactly what which we calculated, it is the 13.11.*2034

*It equals 41.51 J/mol-°K.*2041

*Let us see example 4, we have carbon tetrachloride and has the boiling point of 77°C and the heat of vaporization of 32.5 kJ/ mol.*2062

*Calculate the entropy change of vaporization for CCL4.*2071

*Nice and easy problem.*2076

*We know that the δ S of vaporization is just equal to the δ H of vaporization divided by the boiling temperature.*2077

*We have 32.5 kJ/mol ÷ 77°C, 350°K.*2094

*I get the δ S of vaporization equal to, I'm going to go ahead and change this to J and ends up being 93 J/mol-°K.*2106

*For every mol of carbon tetrachloride and I take from a liquid to the gas phase, I increased its entropy by 93 J/°K.*2125

*Let us go ahead and see what we can do with this one.*2138

*Part A says, what is the δ S for the transformation of 1 mol of water from 0°C to 100°C at constant pressure?*2148

*The molar constant pressure heat capacity is 75.29 J/mol-°K.*2158

*This is pretty straightforward.*2163

*The second part says the δ H of fusion is 6.01 kJ/mol, the δ H of vaporization is 40.66 kJ/mol,*2166

*the constant pressure heat capacity of steam of water vapor is 37.5 J/mol-°K.*2175

*This is liquid and this is gas, we want to calculate δ S for the following transformation.*2182

*I take ice which is at 0°C and 1 atm pressure and I take it and convert it to steam which is now at 125°C still at 1 atm pressure.*2191

*What I'm doing is I’m taking the solid and I'm converting it into a liquid.*2202

*I'm raising the temperature of the liquid from 0°C to 100°C, to 100°C liquid.*2208

*At 100°C I’m going to take the liquid to a gas and now that gas I’m going to take from 800°C, I'm going to take the gas to 125°C.*2220

*The change in entropy for the system is going to be the change in entropy for each of these.*2232

*The melting, the rise in temperature of a liquid, the vaporization of a liquid and the rise in temperature of the gas.*2238

*I'm just going to add all the δ S so let us go ahead and do that.*2250

*This is going to be constant pressure.*2254

*Under conditions of constant pressure I have the following.*2257

*I have my basic equation and I just write it over and over again because it is nice to keep in my mind, equal CP/ T DT - VA DP*2261

*this goes to 0 because P is constant.*2274

*Therefore, I got DS = CP/ T DT which implies that δ S is going to equal the integral from T1 to T2 of CP/ T DT.*2280

*When I do that, I end up with the following.*2301

*When I do this integration, I end up with DS = CP LN T2 / T1 which = 75.29 J/mol-°K × 1 mol × log of 373 divided by 273.*2303

*I end up with δ S of 23.50 J/°K.*2333

*Just taking water at constant pressure, the liquid water from 0°C to 100°C.*2339

*I have increased the entropy by 23.5 J/°K.*2346

*Part B, part B is going to look like this, the δ S of the total is going to equal the δ S of the melting of the ice + the δ S of the liquid water and*2352

*I take it from 0 to 100 + the δ S of the vaporization of that liquid at 100 + the δ S, as I take that gas from 100 to 125.*2368

*All these 4 δ S, I have to calculate them.*2385

*δ S T = the δ H of fusion divided by its melting temperature + the integral from 273 to 373*2390

*that is this one that is the melting of the CP of liquid phase / T DT.*2405

*That is that one, the δ S of vaporization is the δ H of vaporization divided by the boiling temperature and*2416

*this is going to be the integral from 373 to 398 of the constant pressure heat capacity of the gas or T DT.*2423

*Let us go ahead and put the values in.*2438

*δ S total = 6010 convert it to J/ mol divided by 273°K, that is the melting temperature of the ice + the integral 273 to 373*2442

*of the heat capacity of the liquid water which is 75.29/ T DT + 40,660 J/mol divided by the boiling temperature*2466

*which is 373°K + my final integral which is going to be the integral from 373 to 398= 37.50/ T DT.*2486

*I end up with δ S total = 22.01 + let us go ahead and I will not do these integrals.*2505

*Let us go ahead, this is going to be + 75.29 × log of 373/ 273 + 109 + 37.50 × log of 398/ 373.*2522

*I get δ S total = 22.01 + 23.50 + 109 + 2.43, I get the δ S total of 156.94 J/°K.*2549

*There we go, per 1 mole.*2572

*In order to take solid ice from 0 all the way to water vapor which is 125 the entropy change*2578

*for all of that I had to account for the entropy change of the melting of the ice.*2585

*I had to account for the entropy change in raising the temperature of the now melted ice from 0 to 100.*2590

*I have to account for the entropy change in going from liquid to gas + now the entropy change in going from now gas from 100 to 125.*2598

*That is my final answer.*2611

*Thank you so much for joining us here at www.educator.com.*2616

*We will see you next time, bye.*2618

3 answers

Last reply by: Professor Hovasapian

Mon Sep 21, 2015 1:01 AM

Post by rafael delaflor on October 15, 2014

Why does this video stop running at 27:30?