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Lecture Comments (4)

1 answer

Last reply by: Professor Hovasapian
Wed Nov 4, 2015 9:57 PM

Post by Manish Shinde on November 4, 2015

How can we show the dG = VdP - S dT for pure phases where the super bars refer to molar quantities. (Note:- the above equation has a bar)

1 answer

Last reply by: Professor Hovasapian
Wed Feb 4, 2015 4:16 PM

Post by matt kruk on February 4, 2015

hi professor i dont understand how you came up with the d(g/t)/dt = [T(dg/dt)p - G] /T^2 expression

Properties of the Helmholtz & Gibbs Energies

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
  • Properties of the Helmholtz & Gibbs Energies 0:10
    • Equating the Differential Coefficients
    • An Increase in T; a Decrease in A
    • An Increase in V; a Decrease in A
    • We Do the Same Thing for G
    • Increase in T; Decrease in G
    • Increase in P; Decrease in G
    • Gibbs Energy of a Pure Substance at a Constant Temperature from 1 atm to any Other Pressure.
    • If the Substance is a Liquid or a Solid, then Volume can be Treated as a Constant
    • For an Ideal Gas
    • Special Note
  • Temperature Dependence of Gibbs Energy 27:02
    • Temperature Dependence of Gibbs Energy #1
    • Temperature Dependence of Gibbs Energy #2
    • Temperature Dependence of Gibbs Energy #3
    • Temperature Dependence of Gibbs Energy #4

Transcription: Properties of the Helmholtz & Gibbs Energies

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

Today, we are going to talk about the properties of the Helmholtz and Gibbs energies.0004

Let us jump right on in.0009

For the Helmholtz energy, one of the fundamental equations that we have is the following.0013

We have DA = - S DT – P DV.0018

In this particular case, the temperature and the volume are the natural variables for the Helmholtz energy.0030

Essentially, what we are saying is that the Helmholtz energy is a function of temperature and volume.0043

We can go ahead and express the total differential this way.0052

DA is the general total differential, it is going to be DA DT we already know this, holding the other variable constant not pressure, 0058

this is volume, DT + the partial derivative with respect to the other variable DA DV holding that one constant DV.0069

This is a general total differential.0080

This is the one that we actually extracted from the fundamental equation of thermodynamics.0082

This is one of the 4 fundamental equations of thermodynamics.0087

If we equate this and this, and this and this, here is what we get.0090

Equating the differential coefficients, we get DA DT with respect to V = - S and DA DV holding T constant = -P.0100

Basically, what this says is that for every unit change in temperature, 0134

a change in the Helmholtz energy of that system is going to equal negative of the entropy.0140

What these negative signs tell us, this is a rate of change, the rate at which 0146

the Helmholtz energy changes as you change the temperature is going to be a negative of the entropy at that moment.0151

That is all this is saying.0161

If we make a change in the volume of the system, the Helmholtz energy of that system 0163

is going to equal the negative of the pressure of the system at that point.0169

We are holding V constant, here we are holding T constant.0175

Because of these negative signs, this is what it means.0180

Because of the negative signs these equations they say the following.0187

Number 1, an increase in the temperature implies a decrease in the Helmholtz energy.0206

That is what the negative sign means, if you increase this, this is dependent on this.0222

If this goes up, this negative sign means that this goes down.0229

An increase in temperature implies a decrease in the Helmholtz energy.0234

The higher the entropy, the higher this value is, the more quickly the Helmholtz energy decreases.0242

It is important when you see equations like this, you are equating some thing, some value.0251

You are equating it with a rate, this is a rate.0259

It is very easy to forget that we are talking about a rate of change.0261

An increase in temperature implies a decrease in A, the higher the entropy, in other words, if S itself is a bigger number, 0266

the higher the entropy of the substance the greater the rate of change at which it is changing when you change T.0282

Let us say I’m at a particular temperature and let us say that my system has an entropy of 50 units.0315

If I raise the temperature by a certain number of degrees, the Helmholtz energy of the system is going to decrease at a rate of 50 J/°K.0322

At that point, if my entropy was actually higher, if it was 100 then at that point it is going to decrease by 100 J for every °K.0345

That is what this means.0356

The rate, the higher S is, the fact that there is something going to drop the negative sign.0358

The second one says, exactly what you think based on what we did for the first one.0370

An increase in the volume implies a decrease in the Helmholtz energy.0375

Increase in V implies a decrease in A.0386

I’m just making sure that I'm keeping all of my variables clear here.0396

What this says, the higher the pressure at that moment, the higher the P the greater the rate of change.0406

In other words, the greater the rate of change or decrease.0425

If I’m at a particular temperature and if I'm at a particular pressure, 0436

if I change the temperature the Helmholtz energy is going to change by whatever the pressure happens to be.0439

It is going to be a rate of change.0446

If the pressure is higher, the rate of change is going to be bigger, the decrease is going to be bigger if I'm increasing the temperature 0447

and it is the other way around.0454

If I’m decreasing the temperature it is going to go the other correction, the Helmholtz energy is going to increase.0456

It is very important.0461

We are talking about, what we want to do here is we want to cover every single base possible.0464

As far as chemistry is concerned, we are going to concern ourselves with Gibbs energy,0471

we are not going to concern ourselves with the Helmholtz energy but what we need to know that it is there.0474

If for any reason we decide to run an experiment where we are actually holding temperature and0479

volume constant, then we have to use the Helmholtz energy and not that Gibbs energy,0485

because the Gibbs energy is for conditions of constant temperature and pressure.0489

Constant temperature and pressure is what 99% of all chemical reactions are run under so 0494

that is the standard laboratory conditions for chemistry, not for other things.0501

We would do the same thing for G.0509

We do the same thing for G, the Gibbs energy.0517

I think I would go back to blue.0523

Once again, we take a look at our fundamental equations for G so we have DG = - S DT.0526

This is going to be + V DP.0534

In the case of free energy, the Gibbs energy it is the temperature and pressure that are the natural variables.0537

The total differential DG is going to be DG DT constant P × DT + DG DP at constant T × DP.0545

When we equate the differential coefficients, this and this, we have the following relationships.0562

The rate of change of the Gibbs free energy per unit change in temperature holding pressure constant is – S, 0570

the same thing as we saw before the Helmholtz.0579

The other relationship is the rate of change of the free energy as you change the pressure of the system 0584

while you hold the temperature constant = V.0591

This is very interesting.0595

What this says, I’m going to write down in just a second.0596

I will go ahead and tell you.0598

As you change the temperature, if you increase the temperature, the free energy, the Gibbs energy of the system is going to decrease.0600

It is going to decrease as a rate, at a rate of whatever entropy that happens to be at that moment.0608

Here, if you have a system and if you increase the pressure because this is positive, 0615

if you increase the pressure you increase the free energy of the system and the rate at which you are increasing the free energy 0620

is going to equal whatever the volume of the system happens to be.0628

That is all these says, these equations are not mysterious, just take them for what they are.0631

Completely a phase value.0636

As you change the pressure, the free energy of the system is going to change the rate 0638

at which it changes is going to equal the volume at that particular moment.0644

That is all this is saying.0650

The first one, it says that an increase in T implies a decrease in G.0653

A decrease in G is –S.0667

Once again, the higher the entropy, in other words, the higher S is the higher the entropy S, the greater the rate of decrease.0671

I will just say the rate of change because it might be an increase and decrease in temperature.0689

The second one says, now this is positive so an increase in the pressure implies an increase in G.0698

The bigger the volume is at that particular moment, the greater the V or volume the, the greater the rate of change of G.0718

That is all that is going on here.0753

These are very important relationships and again what is important to know is that,0756

it is the 4 fundamental equations of thermodynamics.0765

They are not really 4 fundamental equations, there is only 1 fundamental equation of thermodynamics.0769

These other ones were derived simply based on the definitions of enthalpy, the Helmholtz energy, and Gibbs energy.0774

Those are the composite functions.0783

Remember what we did, the only real thermodynamics functions are temperature, energy, and entropy.0786

The composite thermodynamic functions, the ones that are accounting devices that take other things into consideration, 0793

that was the enthalpy, the Helmholtz energy, and the Gibbs energy.0800

Again, all of these equations they come from only one fundamental equation.0804

These are just different variations of it and it is a way of expressing some particular property of the system, some state property in this case.0809

G in terms of other properties of the system that is all we are doing.0816

At entropy volume, we are expressing G as a function of temperature and pressure but it is also going to be volume and entropy.0821

That is all we are doing, we are expressing one state property in terms of other state properties.0832

Things that are reasonably easily measurable.0837

Let us go ahead and take this particular equation right here and we can actually integrate this.0843

Let us go ahead and go to the next page.0850

We can integrate the DG/ DP at constant T = V to obtain an expression for the Gibbs energy of a pure substance at a constant temperature.0853

We took an expression for the Gibbs energy of a pure substance at constant temperature 0912

from standard pressure which is equal to 1 atm to any other pressure P. 0917

Let us go ahead and do that here.0933

We have our DG/ DP at constant T = V temperature is being held constant so essentially what this is, this is going to end up being just DG/ DP.0935

When we integrate this, let us go ahead and write this way.0949

DG = V DP again, we are holding the temperature constant.0954

For all practical purposes, the G is just a function of the pressure so T is held constant.0965

That is all it says, the function of two variables.0974

If we hold one constant it essentially just becomes a function of one variable.0977

In this case, the one variable is P so I can go ahead and use regular differential notation.0980

I can go ahead and integrate both of these expressions and end up getting the following.0985

I end up with δ G = the integral of V DP from P0 which is a 1 atm to any other value of P.0990

Well, δ G is just G - G0, the G0 on top is a standard free energy, free energy at 1 atm pressure.1003

Remember, this little degree sign it does not mean standard temperature and pressure, 1016

it only refers to standard pressure of 1 atm because temperature can change.1020

It is equal this P0/ P V DP and then when we end up moving this over, we end up with the following.1026

G a function of temperature and pressure is going to equal G0 which is just a function of temperature.1037

I will explain this in just a minute, 0 to P V DP.1046

If you are going to find free energy at a specific pressure, you are fixing the pressure, say we fix the pressure 1 atm that is what P0 is.1060

The 0 on top of the degree sign, that is standard conditions, standard condition is 1 atm.1071

If you fix the pressure, now the pressure is no longer variable.1076

The free energy becomes only a function of the temperature that is why we have G °of T.1080

However, over here when we are actually calculating what the new G is at a new temperature and at a new pressure, 1087

This G that you are looking for is a function of the temperature and pressure.1095

That is why this is G (TP) it is a function of temperature and pressure but this one is only a function of temperature because of that degree sign.1101

That degree sign says we are fixing the pressure at 1 atm, we are locking it in.1110

There is our standard, therefore, I no longer have to consider the pressure.1114

If you happen to know what the standard free energy is at 1 atm and 1121

if you want to know what the free energy is off the system at another pressure, this is the equation that you use.1125

Let us go ahead and say a little bit more.1138

If the substance is a liquid or a solid, because all of these equations that we are developing they actually apply across the board, 1140

to any substance, any system, and anytime.1156

These are universal, it does not apply just to gases or liquid or solids.1160

If the substance is a liquid or a solid then the volume can be treated as a constant.1165

We know already that for a liquid and a solid, unless you are applying a lot of pressure and by a lot of pressure I’m talking thousands of atmospheres, 1180

the volume of liquid or solid is not going to change very much.1191

In fact, it is going to be very little.1194

You remember K, the compressibility coefficient is very tiny on the order of 10⁻⁵ or 10⁻⁶, things like that.1196

For all practical purposes, the volume is constant so it actually ends up becoming independent of the pressure.1204

If the substance is a liquid or a solid, then V can be treated as a constant and we can pull that V out from under the integral sign.1212

You will end up with the following once you integrate it.1223

For a liquid or solid it is still going to be a function of temperature and pressure but you do not have to worry about the integration, 1226

It is going to be + the volume × just the δ P which we will write as P - P0, something like that.1232

Unless, P is really high like the difference between 1 and let us say 3000 atm, P is extremely high.1246

This term V × P - P0 is going to be very small which means we can usually ignore it.1263

Therefore, our G is actually just going to equal G(T).1287

Basically, what we are saying is if you are dealing with a liquid or solid this is the general equation for it right here.1293

Because V is essentially constant, you can pull it out from under the integral sign of the equation 1299

that we just had at the previous slide and you end up with just the integral of TP which is δ P.1303

That is what this is right here.1311

But unless, the pressure is really high this can pretty much drop out.1313

When this drops out, you are just left with the free energy of a system of a liquid or solid, 1320

it is pretty much just equal to free energy under standard conditions.1325

For liquids or solids, the free energy of the system is not going to change that much under pressure, that is all this is saying.1330

For an ideal gas, let us go ahead and write our expression again.1340

We had G as a function of T and P is going to equal G (T) + the integral of V DP from there to there.1350

For an ideal gas, V = nRT/ P, just rearranging the ideal gas law.1364

We go ahead and put this in here and we get the following.1377

We will get G (TP) = G standard function of T + nRT integral DP/ P when P is 0 to P.1382

We get our final equations.1401

This is for an ideal gas.1403

G0 at T + nRT LN (P)/ P0.1406

We said that P0 is 1 atm so this just becomes LN P.1415

We will go ahead and use this one as my basic equation.1421

That is fine, I will just go ahead and write it down. 1434

Since P is 0 = 1 atm we end up with G of TP = G0 T + nRT × log of the pressure,1436

if you are looking for the free energy of a particular substance, of an ideal gas.1460

I’m sorry this is an ideal gas under a certain temperature and pressure conditions, you find the standard free energy 1465

which is under 1 atm conditions and you just take the number of moles you are dealing with × R × T × log of the new pressure.1472

That is all that is going on here.1480

This is just an equation to find the free energy of an ideal gas at a particular temperature and a particular pressure.1486

Special note, the free energy is a very special thing.1497

If the function G (TP) is known, if you happen to know the functional form, this is free energy is some function of temperature and pressure.1507

If you happen to know what that function is, it is known, then all other thermodynamic functions can be derived from it.1523

Can be derived not just from it, all to say all the other thermodynamic functions can be derived 1545

using the relations that we have + the definitions of the composite functions.1560

We are actually could be doing one of these problems when we do the whole problem sets for free energy.1564

At a specific temperature, they should be obvious from the equation above but not a problem, we will go ahead and write it down anyway.1577

At a specific temperature, the G of an ideal gas, the free energy of an ideal gas is a function of pressure only.1588

Of course, if you hold one of these constant at a certain temperature, that temperature it is just going to be a function of P.1606

Let me go ahead and talk about the temperature dependence of the Gibbs energy.1618

We already have a relation for the temperature dependence of the Gibbs energy.1636

Depending on a particular problem that you are working on or a problem that you run across, 1643

we have developed different relations to express the relationship,1651

that is temperature dependence of Gibbs energy to what extent does Gibbs energy rely on temperature.1657

It is not just one equation, we actually have several equations.1663

We will go ahead and derive those now and they are actually quite simple.1666

Let us go ahead and start with the first one, the one that we already know.1670

We know that DG DT under constant pressure = -S.1673

We know this already, as the temperature changes the free energy of the system is going to change according to –S.1682

This is the relationship, this expresses the temperature dependents of G on T at constant pressure.1689

I’m going to call this equation number 1.1697

We know the G = H - TS that is the definition of G.1700

Let us go ahead and move this around a little bit.1709

G - H = - TS and let us go ahead and divide by T so we get G - H ÷ T = - S.1712

-S = G - H/ T - S = DG DT at constant P.1729

We are going to go ahead and put this into here and we are going to get our second equation.1735

DG DT at constant P = G - H/ T.1745

This is just another way of expressing so I will call this equation number 2.1754

This is just another way of expressing the temperature dependence of the Gibbs energy.1759

For every unit change in temperature the Gibbs energy is going to change according to what the G is 1763

at the time - the enthalpy that quantity ÷ the temperature of the system.1769

That is all we are doing, let us say I do not have the entropy but let us say I have the G at the time, let us say I know the temperature and enthalpy.1775

This just gives me another way of expressing the change in the Gibbs energy as a function of temperature.1783

Sometimes, in this case I’m going to say do not worry about Y.1792

Sometimes, we will want the derivative of the function G/ T with respect to T.1798

In this particular case, we get the following.1815

D (G/T DT) is equal to, we have a ratio.1820

When you take the derivative of a ratio you get this × the derivative of that - that × the derivative of this all over that squared.1831

1 - S let me go ahead put P here.1853

We will go ahead and substitute the equation.1860

We already have the equation DG, always variables floating around.1860

T at constant P = - S so this is our first equation, equation number 1.1869

We know the DG DT P = - S so we can go ahead and put that - S in for this DG DT P and we end up with the following.1876

The partial derivative of the function G/ T with respect to T under constant pressure is going to equal 1888

- TS - G/ T² that = - S/ T - G/ T² or it also = -ST + G ST + G/ T².1898

We know that G = H - TS so H = G + TS, G + TS which is ST + G that is actually equal to H.1929

I can go ahead and put that in there and I have my final form which it can end up being equation number 3.1947

Let me write it again.1956

The derivative of the function G/ T with respect to T constant P is going to equal - H/ T².1958

This is equation 3 and this is usually the equation let us call the Gibbs Helmholtz equation.1970

This is the form that is called the Gibbs Helmholtz equation.1978

Actually, any one of these equations I’m actually going to write one more you can consider as the Gibbs Helmholtz equation.1983

This is just variations of the same equation, pretty much this one.1988

We just took it and we are expressing the temperature dependence of the free energy in different ways depending on the particular problem we want.1991

We can go ahead and stick with this and at a particular problem, on a particular situation, 1999

we can derive all of these which is how it actually happened.2004

What they decided to do was they decided they were talking about free energy so when we present this material to you as students,2009

we are just going to present all of these variations of the equation for you.2015

That is all we are doing here.2020

This is called the Gibbs Helmholtz equation and this is usually the form that you see it in.2021

The derivative of G/ T with respect to T holding P constant = -H/ T².2027

If you want to know how the free energy ÷ temperature changes as you change the temperature, 2033

it is equal to the -enthalpy/ the temperature².2041

It is just equations.2045

A lot of times in thermodynamics, especially when it comes to things like free energy, when we are dealing with G,2051

I do not worry about that, I am actually going to explain what G is and what Helmholtz is in relation to the system.2059

I’m going to give a physical meaning to it, particularly when I start doing the problems.2065

A lot of this is just mathematical so just create it mathematically, sort of pull yourself away a little bit.2069

And if you do not completely understand what is happening physically, that is not a problem, that is natural.2075

Just go ahead and treat it mathematically.2082

Let us go ahead and give one more version of the Gibbs Helmholtz equation.2085

Let us go ahead, if I take the derivative with respect to T of 1/ T that is going to equal -1/ T².2091

We have D of 1/ T = -1/ T² DT.2110

I'm going to move the T² over so I have - T² D of 1/ T = DT.2121

I can do this, I can go ahead and create this thing D of G/ T with respect to D of 1/ T.2134

I can go ahead and rewrite this as DT = - T² D of 1/ T.2160

I have D of GV/ DT, I have this thing.2170

I have this, I'm going to go ahead and put this in the denominator so I end up with.2177

Let me go ahead write this whole thing out, things are confusing enough as it is.2186

We have the D of the G/ T DT = - H / T², that is the Gibbs Helmholtz equation, that is equation number 3.2193

I went ahead and I did a little bit more manipulation with this D of 1/ T.2204

I expressed it this way so DT = -T² D 1/ T.2209

I went ahead and converted to a partial derivative and now wherever I see the DT which is right here, I can go ahead and replace it with this.2214

I end up with the following.2222

I end up with D of G/ T - T² D of 1/ T = - H / T².2224

The - T² cancels and I end up with a partial derivative of G/ T with respect to 1/ T = H.2238

And this is going to be equation number 4 for the Helmholtz equation.2253

And again, these are all just 4 different variations of the same equation.2256

The temperature dependence of the free energy, when I change the temperature how does the free energy change?2262

This is just a different way of expressing it.2271

Personally, I do not like this form, it is confusing and you probably not going to run across it 2272

altogether that much except maybe in a couple of problems.2278

The most problems are really only going to require some mathematical manipulation.2281

The way we put those problems in these problem sets with the free energy section of your books 2286

simply to get you familiar with the mathematics of the equations.2291

You have noticed, in this free energy section when we dropped all of those fundamental equations 2295

at Maxwell relations there is a lot of mathematics going on.2302

One of the problem sets in thermodynamics books for the free energy section 2305

tend to involve mathematical manipulation a little more so than with energy and entropy.2311

That is all that is going on here, it is not altogether important that you totally grasp this.2318

Again, if there is one equation that you should probably remember more than any other it is the original one, the DG DT.2324

This one that was the equation number 1 that gave rise to the equation number 2 and equation number 3.2333

This one is the Gibbs Helmholtz equation.2341

They are all Gibbs Helmholtz equations, they are all just variations of this one.2344

In any case, I will go ahead and leave it that.2348

We will see some more of this when we actually do the problem sets and probably make more sense.2351

Thank you so much for joining us here at www.educator.com.2355

We will see you next time, bye.2357