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Entropy as a Function of Temperature & Pressure

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  • Intro 0:00
  • Entropy as a Function of Temperature & Pressure 0:17
    • Entropy as a Function of Temperature & Pressure
    • Rewrite the Total Differential
    • Temperature-dependence
    • Pressure-dependence
    • Differentiate with Respect to Pressure & Holding Temperature Constant
    • Differentiate with Respect to Temperature & Holding Pressure Constant
  • Pressure-Dependence of Entropy for Liquids & Solids 18:45
    • Pressure-Dependence of Entropy for Liquids & Solids
  • Example I: ∆S of Transformation 26:20

Transcription: Entropy as a Function of Temperature & Pressure

Hello and welcome to and welcome back to Physical Chemistry.0000

In the previous lesson, we discussed entropy as a function of temperature and volume but in this lesson we are going to do the same thing0004

and we are going to discuss entropy as a function of temperature and pressure.0010

Let us get started.0015

Let us take entropy as a function of temperature and pressure and let us write our total differential expression0021

which is going to be DS = DS DT at constant pressure + DS DP at constant temperature.0034

This is going to be × DT and this is going to be × DP.0050

Let us recall the fundamental equation.0054

The fundamental equation is DS = 1/ T DU + P/ T DV.0057

Let us recall the definition of enthalpy is H = U + PV.0070

Therefore, I will go ahead and move this around and it becomes U = H – PV.0081

Therefore, the differential DU = DH, we have this U = H – PV.0091

I’m going to differentiate this expression so I get DU = DH or when I differentiate this is the product rule.0102

This is going to be – P DV – V DP.0111

I'm going to put this expression for DU into the fundamental equation.0120

I’m going to put this for that and end up with the following.0128

I'm going to end up with DS = 1/ T × DH – P DV – V DP + P/ T DV.0135

I’m going to multiply it out DS = DH/ T - P/ T DV - V/ T DP + P / T DV – P/ T DV + P / T DV these terms cancel and I'm left with the DS = DH/ T - V/ T DP.0155

This is just another version of the fundamental equation.0191

I just manipulated with respect to enthalpy and energy.0195

U was here and it also shows up in the definition of enthalpy so this is just a version of0200

the fundamental equation of thermodynamics involving enthalpy instead of energy.0204

This is a fundamental equation of thermodynamics.0210

This is just the enthalpy version of it, that is all.0215

There we go, in other words it expresses the change in entropy in terms of the changing of enthalpy and the change in pressure.0223

From our discussion of energy we had the following, it was the constant pressure version of the DU.0231

Let me go ahead and do this in red, we had DH is equal to CP DT + DH DP under constant T DP.0238

Let me go back to black.0261

I have this DH is this one so I'm going to end up putting this expression or I see DH in this expression and we get the following.0266

We get DS = 1/ T × CP DT + DH DP constant T DP - V/ T DP.0278

I get the DS = CP/ T DT + DH DP T DP - V/ T DP.0302

DP and DP I'm going to combine some terms here so when I do that I end up with the following.0322

I end up with DS = CP/ T DT + 1/ T × DH DP T - V DP.0333

Let us go ahead and rewrite the differential.0356

Rewrite the general total differential expression of S = S(T and P).0364

I get DS = DS DT at constant P DT + DS DP at constant T DP.0377

Notice both of these equations, I will call this one equation 1 and I will call this one equation 2.0394

Both of these equations express entropy as a function of temperature and pressure DS temperature pressure,0400

that means that this is identifiable with that and this is identifiable that.0410

We are doing the same exact thing we did last time except now in terms of temperature and pressure.0421

We have the following, we have DS DT under constant P = CP/ T and we have DS DP sub T = 1/ T × DH DP T – V.0428

Once again, CP/ T is always positive.0465

If I hold the pressure constant and raise the temperature, this is positive and this is positive that means if this is positive,0479

if I hold the pressure constant and raise the temperature I raise the entropy.0487

We already know it.0491

At constant P, I will just say a rise in temperature implies a rise in the entropy.0499

The temperature dependence on the entropy again is very simple, it is just the appropriate heat capacity0503

in this case, constant pressure heat capacity divided by the temperature.0518

Pressure dependence not so simple.0524

Again, we have to form our mixed partials.0526

Temperature dependence is simple, that is just this.0533

Pressure dependence not so simple.0544

For pressure dependence, we are going to do the same thing.0550

We are going to form DDP of DS DT and we are going to form DDT of DS DP.0555

These are mixed partials when I take the derivative of this, the derivative of this we set equal to each other we are going to find the relationship.0574

Let us go ahead and go back to black for our purposes here.0590

For DS DT under constant pressure equal to CP/ T we are going to differentiate with respect to P and we are going to hold T constant.0596

When I do that, I get the following.0617

Differentiate with respect to P so this is going to be DDP of DS DT of P and it is going to equal 1/ T × the derivative of the CP0623

with respect to P which = the constant pressure heat capacity so DDP, by definition at constant pressure heat capacity was just DH DT.0645

Therefore, D² S DP DT, DP DT = D² H DP DT.0663

1/ T, sorry I forgot the 1/ T right there.0681

1/ T DP DT.0685

We have that one.0688

For the other one, for our DS/ DP under constant T which we set = 1/ T × DH DP T - V0689

this time we are going to differentiate with respect to T and we are going to hold P constant.0709

This end up looking like also this equation, we are going to differentiate this with respect to T holding P constant.0725

Here we differentiate with respect to P holding P constant or we are going to set it equal to each other to see what we get.0734

We get D DT of DS DP T = this × the derivative of that + that × the derivative of this.0739

We get 1/ T , D DT DH DP sub T - DV DT constant pressure -1 / T² × DH DP sub T – V.0753

We get D² S/ DT DP = 1/ T × D² H DT DP.0787

I will put this together - 1/ T DV DT P- 1 / T² × DH/ DP constant T – V.0811

It is just mathematical manipulation.0831

I will do this in red, D² S DT DP= D² S DP DT.0836

I'm going to set this equal to this and when I do that I get the following.0844

I get 1/ T D² H DP DT = 1/ T D² H DT DP -1/ T × DV DT sub P -1/ T² × DH DP sub T – V.0853

D² H DP DT D² H DT DP, this cancel because mixed partial for enthalpy are the same.0891

I'm going to move one of those over and I end up with the following.0898

1/ T² DH DP sub T - V = - 1/ T × DV DT/ P.0903

I'm going to cancel one of the T's so I going to end up with 1/ T × DH DP sub T - V = DV DT sub P.0923

This thing right here, this thing happens to equal DS DP at constant T.0945

We have DS DP at constant T, one of the differential coefficients that we wanted, we want to know the pressure dependence of entropy0959

is actually equal to something that is easily measurable is equal to the relative is equal to0970

the change in the derivative of volume with respect temperature DV DT at constant pressure.0977

This is what we were looking for, the differential coefficient or expressing it in terms of something that is easily measurable.0987

How much does the volume change when I change the temperature at constant pressure?0995

Recall that A, the coefficient of thermal expansion = 1/ V DV DT P that is this thing right here.1000

Therefore, V A = DV DT P.1018

Therefore, wherever I see DV DT sub P I'm going to put that right there.1027

Our final relation is pressure dependence of the entropy at constant temperature = - V/ A.1034

This is the other differential coefficient, this is what we wanted.1045

This equation right here is the pressure dependents of the entropy at constant temperature, in terms of very easily measurable quantities the volume1051

and the A the coefficient of thermal expansion or final equation.1060

The one that we want to memorize is the following.1065

With respect to temperature and pressure, instead of temperature and volume, let me go back to black.1068

I have DS = CP/ T DT – V A DP under conditions when I’m changing the temperature and the pressure1075

and I want to know what the entropy change is, this is the equation that I use.1092

If pressure is held constant it goes to 0.1096

If temperature is held constant this goes to 0.1099

Let us talk about the pressure dependence of the entropy of liquid and solids.1107

The pressure dependents of entropy for liquids and solids.1128

For liquids and solids, it is going to be the same thing as it was for volume.1145

For liquids and solids, δ S with respect to pressure is very small.1152

In effect, it is so small that you can pretty much ignore it.1169

It is very small so we usually ignore it, in other words we are dealing with a system which happens to be liquid or solid1173

and we change the pressure on it from let us say 1 atm to 10 atm, the entropy change1182

that is contributed by the pressure change is just too small to be to be significant.1187

Just ignore it.1194

What that means is that this thing goes to 0.1197

We just take the 0 and we just use this DS=CP/ T DT.1200

In this particular case, let us see how small, let us actually get an example of this.1207

Let us see how small.1212

We have at constant temperature.1220

I actually want to start with it again and do it the way we are supposed to do it.1224

DS = CP/ T DT - V A DP.1229

This time we want to see how much the entropy changes when I hold the temperature constant and when I change the pressure.1240

I just want a sense of a how a thing is held constant.1250

If I change the pressure how much does the entropy change?1253

I want to see how small it is so we are going to be holding temperature constant so this is going to go to 0.1255

This is the equation that I used, therefore, DS = -V A DP.1261

I can integrate this so when I integrate this and that, this becomes δ S - V A that is the constant.1276

It just becomes - V A × the integral of DP and I end up with δ S = - V A DP.1286

If I want to know the change in entropy is when I change the pressure on a liquid or a solid, I take the volume1298

and I multiply by its coefficient of thermal expansion then multiply by the change in pressure.1303

This is the equation that I'm going to take a look at.1309

For solids, A is approximately equal to 10⁻⁴.1314

It is on the order of 10⁻⁴.1329

For liquids, A is on the order of 10⁻³/°K.1332

Let us go ahead and examine what happens when we look at a particular amount of liquid.1341

We are going to take a look at 100 g of H2O when the pressure goes from 1 atm to 2 atm.1346

If I change the pressure from 1 to 2 atm, I double the pressure what is the entropy change going to be in that 100 g of water?1369

Let us see what we got, we have 100 g of water so let us go ahead and find out how many moles that is.1378

That is going to equal 5.56 moles so our molar volume we have 100 g of water which is 100 ml/ 5.56 moles that is going to give me 18 ml/mol.1387

This is the molar volume or it is going to be 0.018 L/ mol.1413

Basically, 1 mol of liquid water occupies a volume of 0.018 L and we are going to need that.1423

Δ S = our volume negative volume × A × the change in pressure where δ S = -0.018 L/ mol.1433

We said that for a liquid, the A is an order of 10⁻³/°K.1452

Our δ P is going to be final - initial so it was going to be 2 atm -1 atm.1464

When I do this calculation, I get the δ S = -1.8 × 10⁻⁵ L/ atm /mol °K.1474

Let us go ahead and convert this to Joules.1493

This is going to be 8.314 J/ L/ atm.1496

L/ atm and L/ atm cancels L/ atm and I'm left with an amount of -0.0018 J /mol-°K.1502

Let me write that over here δ S= -0.0018 J /mol-°K.1515

I double the pressure on something on 100 g of water and entropy only changes by 0.0018 J.1525

This is very small.1535

For liquid and solids, unless the change in pressure is massive and I do mean massive,1537

you can pretty much ignore the second term on the equation DS = CP/ T DT – V A DP.1543

The pressure dependence of the entropy for a liquid or a solid it does not even matter.1555

It is the temperature that is going to have the deepest affect, the largest effect for a particular liquid or solid.1561

I just want you to see how small it is.1569

Let us go ahead with an example.1574

1 mol of solid gas is raised from 25°C to 125°C at constant pressure.1581

Its constant pressure molar heat capacity is 23.7 + 0.0052 T so this is temperature dependent.1591

As the temperature changes that heat capacity changes so it is not constant anymore.1598

What is the δ S for this transformation?1603

Molar heat capacity is this.1607

Let us start with our general equation which is this.1610

We always want to start with our general and let the problem tell us what the constraints are so we can pair down the general equation if we need to.1614

We are going to have CP/ T DT - V A DP.1623

What is constant pressure?1632

Constant pressure is if pressure does not change that means DP 0 so this term goes away.1633

We are left with DS = CP/ T DT.1638

The next thing we want to do is we want to integrate it because we want the finite change.1647

The δ S is going to equal the integral from temperature 1 to temperature 2 of the CP/ T DT.1653

Let us go ahead, this is molar heat capacity.1675

Remember molar heat capacity is this, it is CP/ n so that is equal to 23.7 + 0.0052 T.1680

We have to multiply by n so the actual the heat capacity itself is going to be 23.7 n + 0.0052 n × T.1692

We are dealing with 1 mol this n is just 1.1705

For all practical purposes I can either include it just put 1 in the computation or I can just know that I’m dealing with 1.1708

I’m dealing with that heat capacity.1714

It was very important in these problems.1717

If they are talking about molar heat capacity, it is CP/ n.1719

If they just say heat capacity it is already included in the n.1723

Be very careful about that.1727

This is going to equal, we are taking it from 25° to 125° that is going to be 298 to 398 and this is going to be 23.7 + 0.0052 T DT.1731

The δ S is going to equal 23.7 × the integral of 298 to 398.1752

This is over T because CP/ T so this is going to be DT/ T + the second integral which is this one.1765

This was going to be 0.0052 T/ T, the T's are going to cancel so the 0.0052 was going to come out of the integral 298 to 398 of just DT.1781

The δ S is going to equal this one was going to be 23.7 × the nat log of 398/ 298 + 0.0052 × 398 -298 just δ T.1797

When I do this calculation, I end up with 6.86 + 0.52 I get δ S = 7.38 J/ mol-°K because we are dealing with 1 mol.1818

It is that simple, you start off with your general equation, you take a look at what the constraint is, in this particular case pressure is constant so it goes to 0.1841

We are left with this part of the equation, we integrate it, we put in our values,1849

we check to see that we are dealing with molar heat capacity and see how many moles.1856

As I take 1 mol of solid gold and I raise it from 25° to 125°, I'm changing the entropy, I'm raising the entropy 7.38 J /mol-°K.1862

Thank you so much for joining us here at

We will see you next time, bye.1878