Physical Chemistry Entropy Changes for an Ideal Gas

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

Today, we are going to discuss entropy changes for an ideal gas.0005

The equations that we developed so far, as we said earlier they apply to any every situation solid, liquid, gas, real gas and ideal gas.0010

For all practical purposes we do not really need to take those equations and somehow fiddle around with them0019

in order to apply them to the case of an ideal gas because they apply as is.0026

However, it is nice to do this because most of the time when we are dealing with gases we are doing really specialized work.0031

We always going to treat gas as an ideal gas so it is nice to have some equations specifically for ideal gases0037

that you can just turn to when you know you are dealing with an ideal gas.0044

You do not have to if you do not want to.0049

Again, all the other equations are absolutely valid.0050

If you want to do the derivations for yourself, it is always nice mathematically.0053

Personally, it gives you a nice sense of accomplishment.0058

I think it would be nice to go ahead and actually discuss it specifically for an ideal gas because it does tend to be the system that we talk about the most.0061

Let us jump right on in.0069

What is nice about the equation for ideal gas is that they tend to be really simple.0076

Let us go ahead and start off first of all we will do a little bit of a mathematical derivation, very quick one though.0081

In general, you remember that we had for energy with respect to temperature and volume,0086

the differential change in energy was equal to CV=CV DT + Du DV under constant T DV.0096

But because we are dealing with an ideal gas, what was nice about it, you also remember hopefully.0111

For an ideal gas, this term the DU DV T that is equal to 0, that was Joule’s law.0117

For an ideal gas, this change in energy with respect to a change in volume under constant temperature actually equal 0.0124

This was Joule’s law.0133

When you change the volume of an ideal gas, you do not change the energy of the system.0138

Basically, what you have when this term goes to 0, what you are left with is just that.0144

We have DU = CV DT.0148

I will go ahead and put that aside for a second.0152

Let us deal with fundamental theorem of thermodynamics.0155

The fundamental theorem of thermodynamics, it says the following.0158

It said that DS = 1/ T DU + P/ T DV.0168

We had DU under conditions of an ideal gas so we will just go ahead and put this into here.0177

What we end up with is DS = 1/ T CV DT + P / T DV.0186

I will go ahead and write this as DS = CV/ T DT + P / T DV.0199

This particular differential, it looks familiar.0210

It does not really look altogether that different than what we just did with the other questions, it is the same equation.0212

Notice, this is a function of temperature and volume but notice we have pressure here.0219

The ideal gas law, again, we are dealing with an ideal gas so it make things simpler.0225

We have an ideal gas we have a relationship between pressure, volume, and temperature.0229

Because this is temperature and volume every single variable in here of this side has to be expressed in terms of temperature and volume.0235

As pressure, we have to express it in terms of temperature and volume.0243

We can do that it is right here P = nRT / V.0246

I take this and I put in there.0253

Here is what I get so DS = CV/ T + nRT / V / T DV.0256

The T cancels the T and I'm left with the following equation DS = CV/ T + nR/ V DV.0278

When we consider entropy as a function of,0293

I’m sorry I forgot my differential here DT.0297

When we consider entropy as a function of temperature and volume for an ideal gas this is what0299

the general equation takes the form as it is CD / T DT that part is the same + nR/ V DV.0306

In other words the number of moles × the gas constant divided by the volume of the system.0314

This is the differential coefficient for the volume component of the entropy.0318

Let us go ahead and call this equation 1 and for an ideal gas this is the equation that you want to memorize if you want to.0326

It is pretty easily derived so it is not a problem but there it is.0333

We recall the basic total differential expression for S = the function of T and V.0339

What you have is DS = DS DT constant V DT + DS DV under constant T DV.0361

I will call this one equation 2.0380

When we compare equations 1 and 2, the DT DV DT DV that means this and this and this and this are identified with each other.0381

We already know the DS DT under constant V is just a constant volume capacity divided by temperature.0391

Nothing is new there.0397

However, the change in entropy with respect to volume under conditions of constant temperature for ideal gas is equal to nR/ V.0399

That is really nice.0408

What we have is DV T =nR / V.0409

Notice, this is always positive.0423

Therefore, because it is always positive that means an increase in the volume implies an increase in the entropy of the system.0432

In other words, if I have a smaller volume, if I have a particular ideal gas and if I make the volume bigger now the gas has more room to bounce around in.0446

It becomes more chaotic, it becomes more disorder.0455

It is entropy rises.0459

T is positive S has to be positive.0461

This is positive because this is positive.0465

An increase in the volume implies an increase in entropy, we knew that already but it is always nice to have this corroboration,0470

that is what is nice about this.0476

For a finite change in state, we just integrate the differential expression.0479

Integration of equation 1 which I have to go ahead and write again, it is not a problem.0496

DS = CV/ T DT + nR/ V DV.0506

When I integrate that, I integrate this, I integrate this, I integrate this, and I get the following.0514

I get δ S = the integral from temperature 1 to temperature 2 of CV/ T DT + the integral from volume 1 to volume 2 of nR/ V DV.0523

If the constant volume heat capacity happens to be constant over the temperature range of T1 and T2.0548

In general, we tend to use constant values for heat capacities.0556

Heat capacity is actually a function of temperature so as the temperature of the system rises, if heat capacity changes0562

but in general over a reasonably good temperature range, it tends to be pretty constant.0567

If CV is constant which in most cases it will be, constant over the particular temperature range, we can pull this out of the integral.0574

nR our constant so we could pull those out of the integral then we get, when we pull these out of the integral.0598

I will go ahead and write that down, it is not a problem.0616

We get DS = CV/ the integral from T1 to T2 of DT/ T + nR × the integral from V1 to V2 of DV/ V.0619

These are very simple integrations that we can do and we end up with the following equations.0634

CV × log of T2/ T1 + nR × log of V2/ V1.0639

This is the equation that we want, the final version for a finite change from state 1 to state 2.0651

The change in entropy of the system, if I change its temperature and I change its volume or either one, it is a whole lot of constant,0659

I will just knockout, I do not need it.0666

It is going to be the constant volume heat capacity × the log of T2/ T1 + n × R × nat log of T2/ T1.0668

For an ideal gas this will give me the entropy change, nice and simple.0676

For a temperature change and volume change or just a temperature change or just the volume change, that is all, very straightforward.0682

This was temperature and volume, let us deal with temperature and pressure like we always do.0691

For the δ S of an ideal gas, as a function of temperature and pressure here is what we have got.0700

In general, we will go ahead and do the derivation.0716

We have DH= CP DT + DH DP under constant T DP this was the general expression for the change in enthalpy of the system.0722

The enthalpy is just a different way of looking at energy that includes the pressure, volume, work, that was the definition of enthalpy.0737

For ideal gas though, this term is 0.0747

I would actually write it down.0755

For an ideal gas, this term DH DP T = 0.0760

Therefore, we just have DH = CP DT.0771

Another form of the fundamental thermodynamics that deals with enthalpy instead of just energy is the following DS = 1/ T DH - D/ T DP.0780

We have seen this before, this is just the fundamental equation of thermodynamics except now in terms of energy DU0801

we have used enthalpy because under conditions of constant pressure.0806

That is all that is going on here.0810

We have DH = this, so we go ahead and put this there and we end up with DS = CP/ T DT - V/ T DP.0812

Notice, this differential expression is in terms of temperature and pressure which means that every single variable0832

on the right has to be expressed in terms of temperature and pressure.0839

This is volume not a problem, we are dealing with an ideal gas PV= nRT.0842

I will just go and ahead express volume in terms of temperature and pressure nRT / P.0847

I take this and I put in there I get the following.0853

I get the DS = CP/ T DT - nRT/ P / T DP.0859

The T cancel and I'm left with DS = CP / T DT –nR / P.0879

Let us correct here R / P DP this is the other equation for the differential change in entropy with respect to temperature and pressure.0891

I will go ahead and call this equation 3.0906

Recall the basic total differential expression for S as a function of T and P it is DS = DS DT P DT + DS DP T DP.0911

T and P these 2 I call this equation 4.0955

If I compare these 2 that means I have identified this with this which we know already.0964

In this case, I’m identifying this for the case of an ideal gas with this right here.0971

What I have got is a change in entropy with respect to a change in pressure under conditions of constant temperature for an ideal gas = -nR/ P.0976

In other words, if I have an ideal gas and if I hold a temperature constant and I change the pressure, the entropy change is this right here.0994

Notice that it is negative.1004

What this says that if I increase the pressure of the system and keep the temperature constant, the entropy has to go down,1005

that is what is interesting here which says under isothermal conditions, you know what isothermal means, it means T constant.1013

If I increase the pressure of the system that implies that the entropy of the system has to go down.1046

That is what this says.1056

This change is negative, this is the entropy change per unit change in pressure.1059

If I change the pressure that entropy change is negative which means it goes down.1066

Rising pressure of the system means a decrease in entropy of the system.1072

Notice this is isothermal that is what makes this important.1077

I will get to that in just a minute, I will say actually a couple of things.1083

Let us first talk about a finite change.1087

For a finite change, we just integrate the DS expression.1091

We have the integral of DS = the integral of CP/ TD T from T1 to T2 - the integral from P1 to P2 and nR/ P DP.1100

Again, if the constant pressure heat capacity is constant / the particular temperature range,1125

this of course is constant we can pull those out of the integral.1131

When we pull these out from under the integral sign we end up with the following.1135

Let me actually right them out.1139

I get CP × the integral from T1 to T2 of DT/ T - nR × the integral from P1 to P2 of DP/ P.1141

I get the change in entropy of the system = CP × LN of T2 / T1 - nR × LN of P2 / P1 this is the entropy change for an ideal gas1155

whenever I change the temperature and the pressure or either one.1181

Let us go back and talk about what I was going to talk about up here regarding the negative.1191

Notice that the change in entropy with a change in pressure under constant temperature is actually negative.1197

It is -nR/ P so once again an increase in the pressure of the system implies a decrease in the entropy of the system.1211

Here is why it makes sense, you probably think yourself and say if I increase the pressure of the system1227

does it mean that the volume is actually going to get bigger?1232

If the volume gets bigger does not the entropy increase?1237

Yes, but we are holding the temperature constant, an ideal gas PV = nRT.1239

If temperature is constant that means this n, this R, this T, they are all constants.1248

That means that the right side of the equation is constant.1252

If I increase the pressure of the system the volume of the system have to go down in order to retain the fundamental equality.1255

When a volume of the system goes down, we already know that when volume goes up entropy goes up1262

which means that when volume goes down entropy goes down which makes sense physically.1267

If the volume increases now there is less room for the molecules to bounce around, there is less disorder in the system, the entropy goes down.1271

Pressure actually alternately comes down to a volume.1281

Let us go ahead and finish this off with a little bit of example.1291

Again, let us finish with a theoretical discussion, we would be doing a lot of example problem.1295

Example problems are going to come in a couple of lessons and there are going to be several of them.1303

We are going to do in bulk, all at once.1308

We have a chance to reuse these equations over and over again in the context of problems.1310

Do not worry there are plenty of examples to come, just like there were for energy.1315

2 mol of oxygen gas we will go ahead and treat it as an ideal gas is taken from 25°C.1326

I’m going to do this one in red, I think.1332

It is taken from 25°C and 1 atm pressure to 100°C and 5 atm pressure.1334

Here you are changing the temperature and we are changing the pressure.1341

Calculate δ S for this transformation, the molar constant pressure heat capacity = 7R / 2.1344

This is nice and simple.1353

Since we are dealing with constant pressure and it is an ideal gas, I can go ahead and just fall back on the equation.1354

If I do not know it, I can derive it, it is not a problem.1363

I have CP × LN of T2 / T1 -n × R × LN of P2 / P1.1366

Just plug my values in δ S.1380

The biggest problem you run into is arithmetic.1383

We have 7/2 of 2 mol, the molar CP is 7 R/ 2, molar CP that is CP/n = 7/2 R.1387

The CP = 7/2 Rn that is what it means, molar means divided by the number of moles.1406

For actual heat capacity you have just move the n over so you have 7/2 Rn.1413

We have 7/2 × 8.314 J/ mol-°K × n which is we have 2 mol of the oxygen gas.1418

mol and mol cancel × the log 25° to 100° so 100°C is 373°K, we have to use Kelvin temperature divided by 25 which is 298 - the number of moles which is 2 × R which is 8.314.1434

I’m going to keep writing the units, it is not a problem 0.314 × log of 5 atm/ 1 atm.1459

Hopefully, I have done the arithmetic correctly and I hope you will go and check this out.1472

I know I tend to make arithmetic mistakes but this is what is important, this right here, not the actual arithmetic.1476

You end up with - 13.7 J/-°K.1482

In this particular case, notice the change in entropy is negative.1492

The affect of temperature on the entropy is going to be positive.1497

The effect of pressure on the entropy is going to be negative.1501

In this particular case, the temperature from 25 to 100°C is overcome by the increasing pressure from 1 atm to 5 atm.1503

In this particular case, the effect of pressure, the effect of P exceeds the effect of T which is why it ended up with a negative entropy change.1516

There you go.1537

Thank you so much for joining us here at www.educator.com.1539

We will see you next time, bye.1541