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Entropy Example Problems III

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: Isothermal Expansion 0:09
    • Example I: Calculate W
    • Example I: Calculate ∆U
    • Example I: Calculate Q
    • Example I: Calculate ∆H
    • Example I: Calculate ∆S
  • Example II: Adiabatic and Reversible Expansion 6:10
    • Example II: Calculate Q
    • Example II: Basic Equation for the Reversible Adiabatic Expansion of an Ideal Gas
    • Example II: Finding Volume
    • Example II: Finding Temperature
    • Example II: Calculate ∆U
    • Example II: Calculate W
    • Example II: Calculate ∆H
    • Example II: Calculate ∆S
  • Example III: Calculate the Entropy of Water Vapor 25:20
  • Example IV: Calculate the Molar ∆S for the Transformation 34:32
  • Example V 44:19
    • Part A: Calculate the Standard Entropy of Liquid Lead at 525°C
    • Part B: Calculate ∆H for the Transformation of Solid Lead from 25°C to Liquid Lead at 525°C

Transcription: Entropy Example Problems III

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

Today, we are going to continue our entropy example problems, let us jump right on in.0004

Our first example for today says that 1 mol of an ideal gas with a constant volume heat capacity of 3/2 Rn is in the initial following state 350°K and 1.5 atm.0011

This is just a continuation of the example problems from the previous section.0025

We have the same initial state, we are just subjecting this initial state to various transformations or transformations and the various circumstances.0030

This time the gas expands isothermally, we have a nice isothermal expansion against 0 external pressure.0038

This time it is going to be a free expansion.0046

A free expansion is when there is nothing on the other end keeping it from expanding.0048

0 external pressure of the gas is 0.75 atm, calculate Q, W, δ U, δ H, δ S and again we want compare the δ S with Q/ T.0054

0 external pressure free expansion until the pressure goes from 1.5 atm to 0.75 atm.0066

Let me go ahead and do this in red since the problems are concerned.0074

Let us go ahead and start with a definition of work.0079

The definition of work is DW = P external × DV.0083

The P external is 0 external pressure so this is 0.0090

Therefore, DW =0 which means that our work for this process = 0.0095

In a free expansion no work is done because we are not pushing against anything.0100

There is no force resisting our push, our expansion.0104

This is an ideal gas which makes our job a little bit easier.0110

We are dealing with something that is an isothermal process.0115

Isothermal implies that δ U=0.0119

The whole bunch of things might actually end up being 0 here.0123

δ U = Q - W so Q = δ U + W.0129

Q = 0 + 0 so it looks like the heat in this is also going to be 0.0133

Let us take a look at δ H so DH = let us go ahead and use our equation just so we can actually practice writing it down.0148

CP DT + DH DP DP, this is an isothermal process so this is 0.0159

This is an ideal gas so this is 0 so we have δ H = 0.0170

It looks like everything is going to be 0 here.0177

Let us see what happens with δ S so DS = CV.0179

It is going to be CV we are using pressure here so this is going to be CP/ T DT - nR/ P DP.0194

This is an isothermal process so the change in temperature is 0 so this term goes to 0.0215

Therefore, we have just DS = - nR/ P DP and then when we integrate this differential equation 0221

we are going to get the equation δ S = -n × R × log of P2/ P1 the nat log.0233

Let us go ahead and put some numbers in so we get - there is 1 mol that is going to be 8.314 nat log that we have 0250

0.75 is the second temperature and 1.50 is the initial temperature.0260

We get δ S = 5.76 J/°K.0267

We end up with 5.76, the initial state the 350°K and 1.5 atm is the final state 350°K, 0.75 atm.0278

You remember the previous lesson, the last two examples, these were the initial and 0290

final states because entropy is a state variable, the state function, state property.0297

The path does not matter how you do it.0304

It does not matter what you do, a reversible expansion against a particular pressure.0307

A constant pressure expansion or a free expansion, the entropy ends up being the same because the initial final states are the same.0313

Let us go ahead and do our comparison.0323

We have got δ S = 5. 0327

We are slowing down in writing the numbers out properly 5.76 J/°K and then we have Q/ T is going to equal 0 J/ 350°K = 0 J/°K.0336

In this particular case because that Q was 0, our Q/ T = 0 and they do not match.0358

Let us go ahead and take a look at the next example.0370

1 mol of an ideal gas with this particular constant volume heat capacity is initially in the following state.0372

Again, same initial state this time the gas expands adiabatically and reversibly.0380

It is no longer isothermal so now are expanding it adiabatically and we are expanding it reversibly until the pressure of the gas is 0.75 atm.0385

That part stays the same.0395

Calculate all of these variables and compare δ S with Q/ T.0397

This time we are doing it adiabatically and reversibly.0403

Let us see here, let us see what this looks like if you remember this is the P and this is the V.0409

We have this which is an isotherm and we have this which is an adiabat.0419

We might have an initial state here, state 2.0427

If we follow the isotherm keeping the temperature constant, it is going to be this.0433

If we follow the adiabat, in other words an adiabatic condition is where there is no heat allowed to transfer anywhere.0438

You are going to end up with a huge temperature drop and reversibly just means we are actually following that path.0446

We are not going this way to that state.0455

We have an isotherm and we have an adiabat.0459

Adiabat that just means that Q = 0, DQ = 0.0463

Adiabatic we automatically know what Q is.0474

Adiabatic implies that Q = 0 so we have taken care of Q.0479

If we look back at reversible adiabatic expansion of an ideal gas, we find the following.0487

Let us go ahead and start with the definition.0494

This is going to be a little bit of a review of adiabatic processes.0497

We have DU = DQ - DW this is the basic equation.0502

Since Q was 0, adiabatic DQ is 0.0509

What you end up with is DU = - DW.0513

For an ideal gas DU = CV DT and the definition of work DU is just P external × the change in volume.0519

What you end up with is CV DT = - P external × DV so this is the equation.0543

They also said that it is going reversibly and we know what reversible means.0563

Reversible means that the external pressure is actually equal to the external pressure on the system = the pressure in the system.0568

They are always in equilibrium, that is what reversible means.0578

That means we are following this path closely in terms of stair stepping.0581

Therefore, we can replace P external with P and what we get is CV DT =- P DV.0587

This is an ideal gas, P = nRT/ V and we did this derivation back when we discussed adiabatic processes but I thought it would be nice to do it again.0601

We have CV DT =- nRT/ V DV.0617

Let us go ahead and divide both sides by this variable T and we would end up with CV/ T DT = -nR/ V DV.0628

When we go ahead and integrate this equation, we are going to end up with the following.0644

We are going to end up with CV × the nat log of T2/ T1 = - nR × the nat log of V2/ V1.0652

This is our fundamental equation for an adiabatic reversible expansion or a compression.0669

This is the relationship between the constant volume heat capacity are the volume and the temperature for an ideal gas.0678

This is the basic equation.0689

This was the equation that we derived when we discuss this, when we talked about the energies.0694

This is the basic equation for the reversible adiabatic expansion of an ideal gas.0700

Let us go ahead and continue a little bit more.0721

Let me write the equation here on this page so we have it.0722

We have CV × nat log of T2/ T1 is going to equal - n × R × the nat log of V2/ V1.0726

We are going to use this thing.0746

Do you remember the ratio of the constant pressure heat capacity divided by the constant volume heat capacity we called it γ.0749

We are going to go ahead and use this.0757

Let me make this V a little more clear and we are going to do a little bit more mathematical manipulation on this using PV = nRT to derive some equations here.0758

I’m not going to go ahead and go through the mathematical manipulation.0771

If you want you can check your books, it is usually in every single thermodynamics book or you can go back to the previous lessons.0774

I think that I did it there.0782

We got the following relations, we got T1 V1 ⁺γ = T2 V2 ⁺γ.0783

I will go ahead and call this equation 1.0795

It expresses a relationship between temperature and volume.0797

This was also a relationship between temperature and pressure, it is T1 ⁺γ P1¹ - γ = T2 ⁺γ × P2¹ - γ.0801

I will call this equation 2 and there is one for relationship between pressure and volume.0815

We have P1 V1, I’m sorry this is γ -1.0822

The temperature and volume is γ -1, P1 V1 γ = P2 V2 γ.0829

I will go ahead and call this equation 3.0837

Using this CP/ CV calling it γ and then using the ideal gas PV = nRT, I can actually manipulate this fundamental equation 0840

to derive these three relationships temperature volume, temperature pressure, and pressure volume, based on the heat capacity.0851

This is for an adiabatic process.0858

We are going to start, manipulating this.0862

The first thing I'm going to do is I’m going to start with equation number 3.0864

It just happens to be the one that I picked, so let us just jump right on in.0869

Let us go ahead and do this.0875

First of all, let us go ahead and see what γ is, so CP/ CV = γ so CP =5/2 Rn and CV is going to be 3/2 Rn.0877

γ is going to equal 5/3.0897

Let us go ahead and list what it is that we actually have.0905

We know our first pressure is 1.5 atm, we have our first volume which we can get from nR T1/ P1.0907

Our first volume = nR T1/ P1.0933

I’m just trying to get as many of the variables as possible.0944

I have the first pressure and I have the second pressure.0947

I can calculate the first volume but I want to get the second volume.0950

I have the first temperature I want to get the second temperature.0954

In other words, for this thing I have temperature 1 which was 350°K.0960

We are going to be able to find what the temperature 2 is.0965

I want to be able to find this and that so I have some variables to work with this that is why I'm doing this.0968

I have 1 mol, in this case R, we are using the ideal gas law so I have to use 0.08206.0976

Be very careful with that.0985

Temperature is going to be 350°K and the initial pressure is going to be 1.5 atm.0988

V1 is actually equal to 19.15 L.0997

I know what P2 is, P2 is 0.75 atm.1005

I want to know what volume 2 is so I have P1 V1, I have P2, I want to know what volume 2 is.1013

I’m going to go ahead and go to this equation right here and I know what γ is 5/3 so I get the following.1021

Pressure 1 is 1.5 atm and volume 1 is 19.15 L, 5/3 = 0.75 atm and I'm looking for volume 2 is going to be 5/3.1029

When I solve for volume 2, I get volume 2 = 29.03 L is my volume 2.1056

Let us see what I can do with that.1068

Since I have volume 2, now I want to find temperature 2.1072

Let us find temperature 2.1080

Temperature 2 is just nR T2.1084

Again, PV = nRT therefore T2 = P2 V2/ n × R.1096

The pressure 2 = 0.75 atm, volume 2 = 29.03 L, n is 1, and R is 0.08206.1114

If you use the ideal gas make sure you use the proper form of R and we end up with temperature 2 equal to 265.3°K.1130

This is cooler, we started at 350 and the temperature 1 =350°K.1143

Temperature 2 = 265.3, the gas is expanding.1154

It is expanding adiabatically, we want it to be cooler so our number matches, it is in the proper direction.1159

Cooler which makes sense, an adiabatic expansion gives you the largest temperature drop.1168

We can go ahead and actually work out the problem.1179

We just needed to find what these other values were.1182

We needed to find what V2 was, we need to find what T2 was, so we can use these in our equations.1185

That is what all of this led to.1192

Let us go ahead and start doing this.1194

DU = CV DT which means that δ U = CV × δ T which means 3/2 Rn δ T.1198

Therefore, δ U = 3/2 × 8.314 × 1 and δ T that is going to be the final temperature which is going to be 265.3 -350 because this is the final - the initial.1220

You end up with δ U = -1056 J.1242

In the process of expanding adiabatically and reversibly, the system ends up losing 1056 J of energy.1252

This is an adiabatic process so we said that δ U = - work = - δ U.1260

Let me go ahead and keep writing this – δ U, - - so 1056 J.1279

The gas is expanding therefore, it is doing work on the surroundings.1287

Work is positive from the surroundings point of view.1292

It is expanding so the expanding gas does 1056 J of work on the surroundings.1295

Let us go ahead and take a look at DH, DH = CP DT.1303

Therefore, δ H = CP × δ T this is what we want, we want δ T.1311

We wanted to find that first temperature so we can actually do the problems that we needed to do because DU is CV DT.1318

CV δ T this is CP δ T, that is why we went through the initial process.1329

Remember, we were only given the initial temperature, the initial and final pressures.1334

I use those relationships in order to find the second volume and the second temperature so that I can do these problems.1339

That = 5/2 Rn δ T and again CP if the 3/2 Rn is the constant volume capacity, this is an ideal gas at constant pressure heat capacity of 5/2 Rn.1348

This is just the 3/2 + 1.1364

Based on this relation, from ideal gas the difference between the CP and the CV= R.1373

If I’m given this I can find this.1379

Therefore, we have δ H = 5/2 0.314 × 1 × 265.3 -350.1383

Therefore, our δ H = - 1760 J.1400

Let us go ahead and do δ S.1419

What shall we do?1423

δ S for an ideal gas, I’m not going to go through the entire equation, I’m going to go through the final version of the equation.1426

It is equal to the constant pressure heat capacity × LN of T2/ T1 -nR × LN of P2/ P1.1432

That = 5/2 × 8.314 × 1 × the nat log of 265.3/ 350 -1 × 8.314 × the nat log of 0.75/ 1.5.1449

When I actually do that I end up with δ S= 0.1480

If I do Q/ T this is actually equal to Q reversible/ T.1487

Q reversible is adiabatic 0/ the temperature which is 350 so we end up with 0.1496

You see they match.1503

Reversible processes the entropy δ S and Q reversible/ T they will match.1507

Let us see what else have we got.1518

This one, the entropy of water is 69.95 J/°K at 25°C.1522

This is the entropy not that change in entropy, this is S not δ S, 25°C and 1 atm, 1 atm so it is standard.1531

This is S standard 298 for water is 69.95.1546

In other words, the general sense of disorder of liquid water and 25°C 1 atm pressure is 69.95 J/°K.1558

Given the following, calculate the entropy of water vapor at 180°C and 0.75 atm.1569

Let us see what we have got.1580

They give us the constant pressure heat capacity of the liquid water which is 75.29 J/ mol °K.1582

They give us the constant pressure heat capacity of the gas which is 33.58 J/°K.1589

They give us the heat of vaporization of water 40.656 kJ have to be injected into 1 mol of water that much energy has to be put into water to convert it from liquid to gas.1595

That is what this means.1607

Assume ideal behavior for the water vapor accounts, we can assume that it is an ideal gas.1609

Let us see what we can do here.1615

The basic equation that you are going to work with here is the following.1617

Two things that are happening, you are going for 25°C where water is a liquid.1621

You are going to a 180°C where water is a gas.1629

You have to calculate the entropy of the phase change, the entropy of the temperature rise, and there is one other thing going on here.1633

You are going from 1 atm to .75 atm so you have to account for the entropy change for the pressure change.1642

Here is what it looks like.1649

The standard entropy at any temperature is going to equal the standard entropy at the temperature that you know which in this case is 298 + 1653

the entropy in going from this initial temperature all the way to the boiling point for water of the liquid state of water/ T DT.1671

This from the third law.1684

The third law says that if you want to calculate the entropy from temperature 1 to temperature 2 1688

it is going to be the constant pressure heat capacity of the particular phase divided by the temperature DT.1694

The entropy that we start off with a given temperature now we are going to calculate, 1703

we are going to add to that the entropy in going from, in this particular case 25 C to 100°C.1709

The entropy change that accompanies the vaporization of the water at the boiling temperature, 1719

the conversion of going from liquid to water vapor, we have more temperature rise.1724

The temperature boiling up to the particular temperature that we want and this time, since it is going to be water vapor it is going to be this.1733

That takes care of the temperature part, now we have to adjust for the pressure part - this is an ideal gas nR × the nat log of pressure 2/ pressure 1.1743

I hope that makes sense.1760

Remember that the CP/ T DT - nR/ P DP this accounts for the pressure change in entropy, the rise in entropy.1762

This accounts for the change in temperature.1780

All I have done is I have accounted for the entropy that I start off with.1782

The entropy of the temperature rise going from 25 to 100.1787

The change in entropy in going from liquid to the water vapor.1791

The change in entropy in going from 100°C to 180°C.1795

I account for the entropy decrease in the pressure difference.1799

In this particular case, because the pressure is going down 1 atm to 0.75 atm, that pressure decline is going to actually accompany volume increase.1804

Our volume increase is going to be actually you can end up getting a positive number here.1814

It is going to be slightly more entropy than usual.1818

Let us go ahead and put the numbers in.1821

This is going to be, we said 180°C.1824

It is going to be 453°K.1830

Our S and now it can no longer standard pressure 1 atm.1837

It is going to be 0.7518 atm and 453°K that is going to equal, the entropy of the temperature that we do know so the S at 298 = 69.95 + 1845

the integral from 298 to 373 of the constant pressure heat capacity 75.29.1866

This is going to be 75.29/ T DT + 40656 J divided by the boiling temperature which is 373°K + this 1.1878

This is going to be 373 to 453, this time we are going to use the 33.58 which is the constant pressure heat capacity for water vapor, 33.58/ T DT.1900

We are going to subtract so it was going to be 1 mol and this could be 8.314 and this is going to be the nat log of 0.75 atm 1917

which is P2 or the initial which is 1 atm.1936

This is the integral that we solve.1942

Let us go ahead and write out a little bit more.1945

S at 0.75 atm and 453 = 69.95 + 75.29 × the nat log of 373/ 298 + when I do the 4656 divided by the 373 I get 109.1950

I'm going to add to that + 33.58 × the nat log of 453/ 373.1983

It is going to be -2.39 and I get the entropy of water vapor at 180°C which is 453°K and 0.75 atm is going to equal 205 J/ mol/°K.1996

Here we go, just a basic application of the mathematical portion of the third law of thermodynamics.2028

There is nothing strange going on here.2036

You start with a particular entropy that you know and you account for the any temperature change.2038

If there is a phase change you account for the entropy change for the phase change.2044

After the change of phase there is still the temperature increase then you account for that temperature increase.2048

If there is a pressure increase or decrease, you account for that pressure increase or decrease.2054

That is all that you are doing.2059

You are accounting for every single change that takes place in the system.2060

Let us see what we have got here.2068

Silicon dioxide experiences the following transformation.2076

We go from 25°C and 1 atm pressure, we are going to raise the temperature to 225°C and we are going to raise the pressure to 1500 atm.2080

There are two things going on, change in temperature and a change in pressure.2090

Given the following data, calculate the molar δ S for the transformation.2093

The constant pressure heat capacity for the solid is this, notice it is not a constant, it is going to be a function of T.2099

The density of silicon is 2.648 g/ cm³ and the coefficient of thermal expansion is 3.530 × 10⁻⁵.2107

Sorry I forgot the unit here, for thermal expansion this is going to be per °K.2117

Let me do it in black, my apologies.2124

It is important that we get these.2128

Coefficient of the thermal expansion this is α.2132

α = 3.530 × 10⁻⁵ /°K.2137

It represents the percentage change in the volume of something when you heated up, relative to how much you started off with.2142

Let us go ahead and see what we can do.2154

We are changing temperature and we are changing pressure, let me go back to red here.2157

We are going to use this equation, is our fundamental equation so the DS = CP/ T DT – V A DP.2164

We are dealing with a solid here, this is a general system.2181

We are no longer dealing with an ideal gas.2183

For the most general equation, the entropy change when you change temperature and pressure this is it right here.2185

There is nothing strange going on here.2194

δ S when we integrate this so we get δ S = the integral from temperature 1 to temperature 2 of CP/ T DT - V × A × the integral from pressure 1 to pressure 2 of DP.2199

We have our final equation of DS = this stays the same, we cannot pull the CP out because CP is now no longer constant.2229

It is going to be the integral of CP/ T DT from T1 to T2 , this is just DP so - V A and δ P.2241

This is the equation that we are going to use.2256

It looks like we are going to need the sub V.2259

We are going to need the molar volume.2264

We need to know what the volume is of this particular thing.2266

They wanted us to calculate the molar δ S, they gave us the density, they did not give us the molar volume.2271

The first thing we have to do is we have to calculate V.2277

Let me go ahead and do that in blue.2280

We have silicon dioxide and that is going to be 60.09 g/ mol, let us start off with that way.2287

We have 60.09 g/ mol and we have 1 cm³ is 2.648 g.2308

2.648 g that takes care of that so that = 22.69 cm³/ mol so that is the molar volume.2323

This is V but it is in cm³.2339

I want to express it in dm³ which is the same as the L.2342

That is going to equal 22.698 × 10⁻³ cm³.2348

I hope you guys are okay with converting from cm to dm to m, things like that.2355

Dm/ mol which is the same as 22.69 so dm³ is a L.2360

22.69 × 10⁻³ L/ mol this is what I wanted, this is my V right here.2370

In 1 mol of this stuff because we are calculating molar, the volume is 22.69 × 10⁻³ L.2380

1 mol of silicon dioxide has this volume, now I can go ahead and do my problem.2387

Let us go ahead and rewrite what I need.2396

I need to go back to red.2401

There we go so δ S = the integral from T1 to T2 of CP/ T DT – V A δ P.2406

I got my δ S = the integral from 298 to 498 and now I write 46.94 + 34.31 × 10⁻³ × T -11.30 × 10⁵ T⁻² 2421

that is the constant pressure heat capacity of silica/ T that is the integral part and it is going to be –volume.2455

The volume is 22.69 × 10⁻³ L/ mol and A which is the coefficient of thermal expansion they give us at 3.530 × 10⁻⁵/°K.2467

The change in pressure δ P is going to be 1500 – 1.2494

I’ m going to keep multiplying this by, notice this is going to end up being in L atm.2501

I need to put this into J.2512

I’m going to multiply it by 8.314 J = .08206 L atm.2515

It is important if you are going to be working in L atm, you want to convert this to J.2531

This is the conversion factor, it is just the ratio of the 2 R.2537

8.314 J = .08206 L atm they are both units of energy.2542

When I do that, atm cancels atm, L cancels L , you end up with J/ mol °K.2548

Everything works out right.2563

When you do all this, let us actually calculate this.2564

The integral comes out to be 26.88 J / mol °K and this ends up being 0.12 J/ mol °K.2567

Our final δ S is 26.76 J/ mol °K.2590

Clearly the change in temperature accounts for 26.88 J/ mol °K change in entropy.2600

The change in the pressure from 1 to 1500 atm, 1500 atm is massive, it is huge.2607

1500 atm, all that pressure change.2614

This is solid and a solid is not going to contract all that much under that much pressure, even that much pressure.2618

Its only difference is 0.12.2627

Clearly, the change in entropy of the system especially of the solid or liquid, under a variation of pressure is virtually negligible.2630

You are not losing anything by just ignoring this 0.12 but we want you to see it because we want to be able to solve the problem.2640

This is how you solve the problem, your basic equation.2646

Let us see what else have we got here.2657

The standard entropy of lead at 25°C is 64.8 J/ mol °K.2663

Standard means 1 atm it does not necessarily mean 25°C.2671

In general chemistry, when you see this little degree sign on top, it generally means 25°C and 1 atm.2679

The degree sign ° really just means the temperature.2686

We specify the temperature because in these problems now we are a little bit more sophisticated than we were in General Chemistry.2689

We can calculate S, δ H, δ S at different temperatures.2696

This degree sign ° represents the temperature, that is what standard means.2702

Standard pressure is 1 atm.2708

The constant pressure heat capacity is this, you notice it is not at constant, it is a function of the temperature.2712

The constant heat capacity to liquid at 32.51 again it is not a constant.2720

The melting temperature of lead is 327.4°C, the δ H of fusion of melting is 4.770 kJ/ mol.2728

They want us to calculate the standard entropy of liquid lead at 525°C.2738

Here it is going to be solid lead at 25, what is the entropy of liquid lead and calculate the δ H 2744

for this transformation of solid lead from 25°C to liquid lead at 525°C.2749

This is going to be exactly like the problems that we did a little bit earlier when we calculated the entropy change for water, 2758

from liquid water to water vapor.2763

We are just going from solid lead to liquid lead.2765

Let us see what we can do.2768

We have got 25 and 525.2770

Let us see what we have.2774

We have S at 298 = 64.80 J/°K and we want S.2780

Still standard so there is no change in pressure, all we are doing is actually changing the temperature, it is actually melting the lead.2799

It is going to be at 823 this is what we want.2805

Let us see what we have got.2812

We are going to write S of 823 = S of 298 + the integral from 298 to the melting temperature of the solid heat capacity/ T DT.2814

The entropy at a given temperature + the new entropy at the new temperature which is melting and we are going to account 2839

for the entropy of the phase change δ H of fusion/ the temperature of melting, boiling from going to solid to liquid.2848

This is the melting temperature.2857

We are going to heat that liquid some more.2860

We are going to go from the melting temperature all the way to this 823, except this time we are going to use the constant pressure heat capacity of liquid.2862

That is the integral that we want.2875

Let us go ahead and put in our numbers so that S 823 = 64.80 + the integral from 298 to the temperature of melting 2877

which is 300 and something, they said would actually ends up being 600.4.2891

It is going to be the 22.13 + 0.01172 T + 0.96 × 10⁵ T⁻² / T DT.2900

We are going to add to that the 4770 J divided by the melting temperature which was 4850.4 °K, that is going to be the transition.2924

We are going to add to that the entropy change in going from 600.4 to 823.2937

This is going to be the 32.51 -0.00301 T/ T DT.2947

I can go ahead and let you work out the arithmetic here.2963

I’m not going to give you the answer, this is the answer right here.2967

You can work out the integral and whatever number you get that is the answer.2970

I just decided that I did not feel working this particular thing out, this is what is important.2974

You are just taking the entropy at any other temperature is the entropy at some temperature that you know.2983

For all practical purposes, the entropy of the temperature you know comes from the table of thermodynamic data in the back of your books.2990

Just like in the back of your General Chemistry books, there was the enthalpy, the free energy, and the entropy.2997

Those are standard third law entropy at 25°C and 1 atm pressure, that is going to be your starting point.3004

If you want to calculate the entropy of any other temperature at any other pressure, you have to account for the change.3016

25°C this is the entropy, you add to it the entropy that you get from the temperature rise from 298 to 600.3024

This is the entropy rising going from the solid to liquid state and this is going to be entropy change in going from 600.4 to 823°K.3033

That is all you are doing and it is going to be the same thing for δ H.3045

You remember when we did that earlier for energy, we are going to be doing it again in just a moment.3049

If we want the standard entropy at still higher temperatures, at temperatures above boiling,3061

let us say we took this solid lead to liquid lead, that is why we boiled off that lead to turn it into lead vapor 3086

and we raise the temperatures too, then we just add more terms.3092

Temperatures above boiling then we just add the appropriate terms.3096

In this particular case, they would add the δ H of vaporization divided by the temperature of boiling, that is the entropy in going from liquid to vapor.3113

We add to that if we are going to still raise it above the boiling temperature, temperature of boiling to whatever final temperature and 3124

this time we are going to use the constant pressure heat capacity of the gas / T DT.3131

That is all nice and simple.3138

Let us do part B, this time it asks for the δ H.3145

I’m going to go ahead and write δ H = CP DT + DH DP T DP.3157

The pressure is constant in this particular case so this is 0.3166

δ H is just equal to the integral from T1 to T2 of CP DT.3171

If there are any phase changes we simply add the δ H for the phase change, same as before.3181

Our general equation becomes the δ H at any temperature T that I want is going to equal T0 to T melting of the CP solid DT.3189

In this particular case, lead going from 298 to its melting temperature + the δ H of fusion and we are talking about δ H here not δ S.3207

Once I have actually melted it, I raise the temperature some more, it is going to be the melting temperatures 3220

to whatever final temperature that I want.3227

This time it is going to be the constant pressure heat capacity of the liquid DT.3230

We have δ H at 823 = 298 to 600.4 of the 22.13 + 0.01172 T + 0.96 × 10⁵ T⁻² DT.3239

I'm going to add to that the 4770 J and I'm going to add to that the change in enthalpy for going from the melting temperature of 823.3274

This time I’m going to use the heat capacity of liquid lead which is 32.51 -0.00301 T DT.3291

I'm going to let you work out whenever this number is.3305

This is the answer, the rest is just arithmetic and integration and stuff like that.3312

If I needed to know the δ H at higher temperature still, let us say I want to take solid lead turn it into liquid lead like I did here and 3320

then take the liquid lead and turn it into a gaseous lead and still raise the temperature, I would add the two following terms.3329

I have to account for the δ H that comes from the vaporization and then if I raise the temperature beyond that, 3336

I would go from the boiling temperature to whatever temperature I wanted and this time I would use the constant pressure heat capacity of the gas.3345

It is solid to liquid, transition from solid to liquid temperature change during the liquid phase, if I needed to,3366

I would include the δ H of the transition from liquid to gas and then any other temperature change during the gas.3385

This is just normal heat.3392

This is the solid phase, the liquid phase, the gas phase, these are constant temperature processes.3399

These are the phase changes so these are accounted for by the δ H.3406

This is a vaporization, this is the δ H of fusion.3411

Here I have to do them in terms of integration.3416

That is all that is happening here.3420

Thank you so much for joining us here at www.educator.com.3424

We will see you next time, bye.3426