Physical Chemistry 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 and0025

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 again0036

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 and0235

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 and0268

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 by0297

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 or0376

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 DP0419

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 temperature0607

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 going0734

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 it0768

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 change0809

that accompanies the melting + the temperature change from the melting temperature0832

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 vaporization0849

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 products0927

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 or0992

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 reactants1039

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 equations1223

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 this1400

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 DP2261

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 and2352

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 3732390

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 and2416

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 3732442

of the heat capacity of the liquid water which is 75.29/ T DT + 40,660 J/mol divided by the boiling temperature2466

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 change2578

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