This introductory physical chemistry course examines the connections between molecular properties and the behavior of macroscopic chemical systems.

Loading...

Do curso por University of Minnesota

Termodinâmica Estatística: Dinâmica Molecular

165 classificações

No Coursera, você encontrará as melhores aulas no mundo. Estas são algumas das nossas recomendações personalizadas para você

This introductory physical chemistry course examines the connections between molecular properties and the behavior of macroscopic chemical systems.

Na lição

Module 7

This module is relatively light, so if you've fallen a bit behind, you will possibly have the opportunity to catch up again. We examine the concept of the standard entropy made possible by the Third Law of Thermodynamics. The measurement of Third Law entropies from constant pressure heat capacities is explained and is compared for gases to values computed directly from molecular partition functions. The additivity of standard entropies is exploited to compute entropic changes for general chemical changes. Homework problems will provide you the opportunity to demonstrate mastery in the application of the above concepts.

- Dr. Christopher J. CramerDistinguished McKnight and University Teaching Professor of Chemistry and Chemical Physics

Chemistry

Welcome to Week 7 of Statistical Molecular Thermodynamics.

Today, I want to talk about entropy in the context of other thermodynamic

functions. And, in particular, I want to start with

what we have available now, from the first and second law of thermodynamics.

So the first law says that dU is equal del w reversible, plus del q reversible.

And we know, for an ideal gas at least, the reversible work is minus PdV, and the

reversible heat is TdS by the definition of dS.

So if I make those replacements, I can also write dU is equal to TdS minus PdV.

But let me consider the total differential of U with respect to T and

V. And so if I express U as a function of T

and V, I'll get then that dU is equal to the partial derivative with respect to T

holding V constant times dT, plus the partial derivative with respect to V,

holding T constant, times dV. And actually we've already seen one of

these partial derivatives before, dU, partial U partial T is defined to be the

constant volume heat capacity. Since we're holding volume constant.

So that's CV, and it may be a function of T.

And so, if I now equate these two expressions for dU.

So I have the one I derived from the first and second law.

And I have this other one, which is just expressing the total differential of

view. And I'll put that up on the next slide.

I have the TdS, minus PdV Is equal to. I've made the substitution of the heat

capacity. DT, plus partial U partial V, dV.

And I'm going to rearrange that in order to place entropy entirely on its own on

the left hand side. And so you see I just put PdV over on the

other side. That's actually just two terms

multiplying dV, so I've grouped them together.

And I've divided both sides by T, so there's a 1 over T appearing in front of

the terms on the right hand side. And remember, for an ideal gas that the

internal energy depends only on the temperature.

So partial U partial V, if the temperature is held constant, then the

energy, the, the energy is constant and, hence, the partial derivative is zero.

And so I can simplify that and say that well, I'll simplify that in a moment.

Let me also consider the total differential of S with respect to T and

V. And so, again, that's just partial S,

partial T, holding V, constant dT. It's partial S, partial V, holding T,

constant dV. So here, I have two terms multiplying

each, multiplying dT, and different terms multiplying dV.

Since these are both expressions of dS, these factors multiplying the

differentials must be equal to one another.

That is, partial S, partial T, holding volume constant is equal to the heat

capacity, the constant volume heat capacity, over T, and partial S, partial

V is 1 over T times P plus partial U, partial V, which would be zero for an

ideal gas, but it doesn't have to be an ideal gas.

This is a more general expression. So we'll come back in a moment and look

at why this is useful for actually measuring entropy and assigning values to

entropy. But first, let me pause for a moment and

let you try some differential manipulation of your own, and then we'll

come back and look at enthalpy, as opposed to internal energy with this kind

of discussion. All right.

Well, here's the the way I would have derived that last problem.

If you want to know dh, we replace H with its definition, U plus PV.

So as I take the differential operator through there, I'll get dU.

Plus, and I need to use the chain rule for this product.

So I get a VdP term, and a PdV term. But I know that dU is TdS minus PdV.

So I swap that in for the first term. Now I've got a minus PdV, I've got a plus

PdV. Those will cancel out, and I'm left

behind TdS plus VdP. And once more, I can play the game of

looking at the total differential and equating common factors, multiplying

differentials. So, if I take the total differential of

the enthalpy with respect to temperature and pressure, we've seen this before.

I won't read every character. But its partial derivative with respect

to T, and with respect to P, multiplying their respective differentials.

Again, this first term is one we've seen before, that's the constant pressure heat

capacity. And, if I equate the two expressions and

solve for dS, I end up with, here's the equating of the two expressions.

And when I rearrange, I get something similar looking to the internal energy

case, except we're now involving pressure instead of volume.

So those, those symbols have been [INAUDIBLE] swapped to some extent and

enthalpy instead of internal energy. Constant pressure heat capacity instead

of constant volume heat capacity. Once more remember that, for an ideal

gas, there would be no pressure dependence of the enthalpy.

It depends only on the temperature. And we're holding temperature constant.

But it's a more general expression, so we'll keep it there.

But it's just, it's good to always be thinking about these things.

And recalling some of the past work that we've done.

And why ideal gases are so pleasant to work with.

Again, I'll take the total differential, with respect to T and P, and that allows

me to equate this term with this term. They both multiply dT in an expression

for dS, this term with this term. So I'll just write that out then, that

the partial derivative of S with respect to T at constant pressure is the heat

capacity at constant pressure over T, and then this is the pressure dependence of

entropy. And, the reason that this is useful

[INAUDIBLE] so far it looks like it's just a lot of differential calculus, and

it may have a certain abstract beauty to it.

But why would it be useful to you in the laboratory, potentially?

Well let's look at that. If you want to know what, what really is

this left hand side. It's the change in entropy associated

with a change in temperature. And that's a very practical quantity you

might like to know if I increase the temperature of my substance by 50 degrees

for example 50 kelvin. by how much did the entropy change.

And what this tells us is that we should be able, in fact, to determine that

change by integrating the heat capacity, with respect to that temperature change.

Or, if you like, let me write it out more, more generally.

So, if I want to know the entropy at temperature 2, relative to the entropy at

temperature 1, ST2 minus ST1. It's the integral from T1 to T2 of the

heat capacity, which may depend on the temperature.

So we'll keep it inside the integral, dT over T.

And indeed if we start from 0 Kelvin as T1, that actually gives us a way to

express the absolute entropy at a given temperature.

It's equal to whatever the temperature, sorry, excuse me, whatever the entropy is

at absolute zero, plus the integral from zero to that target temperature, T2.

Again, CPT dT over T. So what this says is that we can

calculate the entropy of a substance at any temperature, T2, if we know the

entropy at zero kelvin and the constant pressure heat capacity.

And you may recall, when we talked about enthalpy, we talked about the measurement

of the heat capacity. That's just how much heat does it take to

raise the temperature 1 degree. And so that's a readily accessible

quantity. So with heat capacities in hand, we can

go and measure entropies. So that's actually something we're going

to take a look at in, in more detail and the first step of doing that will be to,

in fact, explore and express the third law of thermodynamics.

Now, before we do that I think it is time maybe to slot in another demonstration.

And in particular, a demonstration that illustrates the importance of entropy as

one goes to higher temperatures. So I'll let you watch that, and then,

we'll return to look at the third law of thermodynamics.

[SOUND] Have you ever tried to polish silver in order to remove tarnish?

It can be a lot of work with a physical polish, rubbing every square centimeter

of surface. Some of you might know, that you can also

immerse silver, in a hot solution sodium bicarbonate, that is, baking soda, that

also includes immersed aluminum foil. That's an example of an electrochemical

process that can be described in fascinating detail using thermodynamics.

But that is beyond the scope of this course, unfortunately.

However, there is another approach that one might employ.

And one that we already have the thermodynamic tools necessary to

understand. The tarnish on silver is silver oxide.

That is, there is a small layer on the surface of the silver where oxygen atoms

have bonded to the surface to form silver oxide, which instead of being a lustrous

metallic color, is dark and opaque. Tarnish on silver can also be silver

sulfide but we'll ignore that complication here.

Oxidation of silver in air at room temperature is driven by enthalpy.

It is exothermic to react oxygen from the atmosphere with metallic silver to

generate silver oxide. However, the process clearly is

unfavorable from an entropic standpoint because free molecules of oxygen are

removed from the gas phase and their atoms tied into the solid.

Thus, at sufficiently high temperatures, we should be able to reverse the

spontaneity of the reaction. And use entropy to overcome enthalpy, and

drive silver oxide to become silver and molecular oxygen.

Thanks to the favorable and tropic release of molecular oxygen.

Let's see if that works. I have here a fabulously tarnished piece

of silver. Pretty ugly isn't it?

But now I'm going to light this torch and try heating it up.

[SOUND] Notice, as it's warming, getting hotter, I see oxide disappear, and whiter

metallic silver reasserts itself. Pretty, isn't it?

Because this silver is rough and not polished to a uniform layer, we don't see

the luster that we could otherwise achieve for metallic silver.

But certainly, the loss of tarnish is evident.

Unfortunately, if I let it cool down in air, the reaction returns to its normal

room temperature direction and I end up with tarnished silver again.

To avoid that I'd have to put the hot silver into an inert atmosphere, that is

one lacking in oxygen, and let it cool. Of course, it will still tarnish at room

temperature, too. But the reaction is slower at lower

temperature, which is why people don't have to polish silver every day.

So, if you have a torch and a box full of argon, now you know a simple way to

polish silver. Or, you might want to try that

electrochemical trick instead. [SOUND] [BLANK_AUDIO]

O Coursera proporciona acesso universal à melhor educação do mundo fazendo parcerias com as melhores universidades e organizações para oferecer cursos on-line.