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In this lecture, we're going to attempt to refine

our understanding of the second law of thermodynamics.

This lecture is a little longer, and

a little more complex than what's been before.

I think you'll find it worth it.

But you might have to really stick with

this, and maybe even watch it more than once.

Remember, our foundation here at the outset

is to try to understand spontaneous processes.

And our goal at the outset was to say, our

first try at the second law of thermodynamics, that entropy defined

as k logarithm W, increases in any spontaneous process.

Because the probability is constantly increasing in spontaneous processes.

But we know that this statement is now not correct, as

we discussed after we looked at absolute entropies in the previous lecture.

As examples, spontaneous freezing, for example,

of liquid water below zero degrees centigrade.

Or spontaneous condensation,

for example, of gaseous water below 100 degrees centigrade.

Can occur in ways in which delta S is

in fact, a negative number, rather than a positive number.

In both of those cases what we have omitted from our discussion is

the fact that there is energy exchanged between the molecules that we're studying.

In those cases water molecules, and the surrounding molecules.

In particular, when gas condenses

to liquid, it releases energy.

It releases that energy into the surroundings.

And that will change the entropy of the surrounding molecules.

Likewise, when liquid water freezes, it releases energy.

Often to whatever, for example the refrigeration unit,

that we are using to freeze those molecules.

That energy released to the surrounding

molecules will change their entropy as well.

So we have to take into account the entropy

effects, resulting from transfers of energy.

We must also consider the energy changes.

Now, our statement before, that the entropy drives the processes by

increasing probability would be true, provided

that there aren't any energy changes.

And sometimes, we can actually isolate our system from energy changes.

Keeping there from being any energy transfer out of our system, and

into surrounding molecules.

In those cases perhaps, our previous statement is going to work okay.

That the entropy defined here will increase in any spontaneous process.

Provided that we are in an isolated system in which there are no en, energy changes.

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As one specific case of that, we might regard the

universe as a system in which there are no energy changes.

Because the energy of the universe

is constant according to the first law of thermo, thermodynamics.

Therefore, since the entropy increases in any isolated system, and the universe is

isolated, then S increases constantly for the entire universe.

We could then write the second law of thermodynamics is

that delta S of the universe, is always greater than 0.

For any

spontaneous process that actually occurs.

That actually turns out to be a general

statement, but it's not a very satisfying statement.

Because it requires us to calculate the entropy of,

essentially, an infinite body, as far as we're concerned.

And there's no way that we can calculate that in a simple way.

But it turns out there are ways that we can calculate it using thermodynamics.

We're going to develop those means in this lecture.

To do so, we need to understand the

consequences of the heat transfer that we discovered were

at the source of our problem a little while

ago, when talking about spontaneous condensation, or spontaneous freezing.

And we'll do that in a couple of different ways.

One is to notice that if we take an object which is hot, say at temperature T2.

And put it up against an object which is colder, at say

temperature T1. So T2 is greater thaen T1.

We know what will happen unless we thermally isolate these two.

Energy will transfer from the hotter body, to the cooler body in the form of heat.

So q is the heat, which is being transferred.

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And that transfer will occur spontaneously.

Until we wind up

with a situation in which we have exactly the same temperature for both of these.

And T3 will be somewhere between the temperatures of the two original bodies.

The cold body will heat up to T3.

The hot body will cool down to T3.

And it will happen because heat has transferred across.

At this point we will have achieved equilibrium.

But that means that heat spontaneously flows from a hot body to a colder body.

If entropy is in fact an appropriate measure here, then one of the things we

will also notice, is that energy increases when

a body is heated to a higher temperature.

We observed that when we studied the the, the

absolute entropies of materials such as the entropy of oxygen,

when we elevated it's temperature.

Looking at the data in the previous lecture.

Both of these tell us that temperature has

something to do with energy, with enthropy changes.

That in order to transfer energy from one place to

another spontaneously, it must be true that temperature is a factor.

It's not just the heat of course transferred.

Because the amount

of heat transferred from the first body to the second body, is exactly equal.

That is, the amount of energy lost by the first

body, equals the amount of energy gained by the second body.

So it cannot be that spontaneous processes

occur just because heat has been transferred.

Rather, it must be because of the temperature differences between the two.

That gives rise to the following equation.

That the change in entropy,

when a system is heated, within an amount of energy q.

Or loses heat by an amount of energy q.

The entropy change, is q divided by the temperature.

Notice what that means for the case we described up above.

The entropy change for the body on the left, delta S2, is minus q over T2.

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We can go back and look at our tables for entropy.

And we can pull out that the entropy of gaseous

water at 25 degree centigrade, is 188 joules per mole kelvin.

And the entropy for liquid water, at that

same temperature, is 69.9 joules per mole kelvin.

Just by taking the difference between these two, in

other words taking the entropy of the liquid and subtracting

the entropy of the gas, we wind up with an

entropy change which is negative 118 joules per mole kelvin.

Don't like those units, I'll fix that later.

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Clearly, the the entropy change is negative

here because we've gone from gas to liquid.

In the process though, of course, we've also released some energy.

To understand why we're releasing energy, think about the reverse process.

To transfer liquid into gas, we clearly have to elevate

the energy of the molecules, to overcome the intermolecular attractions.

So, liquid going to gas absorbs energy. Therefore, gas going to liquid

releases energy. The amount of energy released is 44 joules

per mole kelvin. If that's the case, then we can use those

data to calculate what the entropy change is for the surroundings.

Based upon the fact that the entropy change of the, or the energy change is 44.

The entropy change for the surroundings is negative delta H over T.

Let's stop for a second and remember why

it is negative.

The surroundings are absorbing an amount of energy 44 kilo joules per mole.

The change in energy of the surroundings is

the negative of the energy change of the system.

Therefore the energy change of the surroundings is positive 44 kilo joules

per mole divided by 298 kelvin, gives us an entropy change for the surroundings of

147 joules per mole kelvin.

Notice that is a larger positive change, than the

negative change of the entropy of the individual liquid molecules.

In fact, if we take the difference between

the two, and calculate the entropy of the universe.

The entropy of the universe is the difference between

those two numbers, and it is greater than zero.

Consistent with our previous idea about

the entropy of the universe always increasing.

The entropy of the universe increases during

the spontaneous process, provided that we take

into account both, the entropy change of the system, as we have done here.

And the entropy change of the surroundings as we have done here.

Adding those two together, delta S of the universe is delta S of the system.

In this case the water molecules, and the surroundings,

everything else.

And we've observed experimentally, that for

the spontaneous process of water condensing to,

I'm sorry a gaseous water condensing into liquid water at 25 degrees centigrade.

The entropy of the universe does in fact, increase.

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When we do that, we wind up with delta H of the system

minus T delta S of the system is a number less than zero.

Remember, if we multiply an inequality by minus 1 it reverses

the sign of the inequality, that gives rise to this equation.

And now we notice that we don't really need to any longer put that system

subscript on there because clearly we're no longer

going to be trying to calculate the properties

of the surroundings.

The means we now have a new statement of the second

law of thermodynamics here that describes a spontaneous process taking into account.

Both the energy changes and their effects on the surroundings

as well as the entropy changes in the individual system.

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We can apply this to the particular case of the condensation of water as follows.

Here is the delta S of the process calculated before.

Here is the delta H of the process calculated before.

If we plug those numbers into the equation from the previous slide.

We wind up with a negative number

exactly consistent with our second law of thermodynamics.

We thus just, we have thus managed to generalize our statement of law,

second law of thermodynamics, so that we no

longer have to restrict ourselves to an isolated system.

There's one last sort of tantalizing bit about this that's going

to lead to the next concept development study and it's this.

What if delta H minus T delta S were

to be zero, what would happen in that circumstance?

To ask that question, let's instead ask

the question, at what point are these materials

actually in equilibrium with one another rather than having a

spontaneous process take place with all of the materials at one atmosphere pressure?

Well, we know that the boiling point of water is 100

degrees centigrade or 373 degrees kelvin. For one

atmosphere of pressure. If that's the case, then rather than

calculate the delta H minus T delta S at 298, where we

know that the condensation is spontaneous, what if we calculated at 373.

100 degrees centigrade.

It turns out if we take these numbers here, just

take these values, plug them into this expression, you get zero.

That suggests that delta H minus T delta S is

not less than 0, nor is it greater than 0.

Meaning that the forward process is no more

likely than the reverse process to be spontaneous.

Meaning that the forward and reverse processes

must be in equilibrium with each other.

That means, that we have generated a condition, not just for spontaneity, when

we see a negative number, but for equilibrium when we see that zero.

That's going to lead us to the definition of

something we'll call free energy, and we'll use

free energy to help us understand the

conditions under which delta H minus T delta

S is equal to 0 and the

conditions therefore, under which we are at equilibrium.