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

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Del curso dictado por University of Minnesota

Statistical Molecular Thermodynamics

146 calificaciones

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

De la lección

Module 5

This module is the most extensive in the course, so you may want to set aside a little extra time this week to address all of the material. We will encounter the First Law of Thermodynamics and discuss the nature of internal energy, heat, and work. Especially, we will focus on internal energy as a state function and heat and work as path functions. We will examine how gases can do (or have done on them) pressure-volume (PV) work and how the nature of gas expansion (or compression) affects that work as well as possible heat transfer between the gas and its surroundings. We will examine the molecular level details of pressure that permit its derivation from the partition function. Finally, we will consider another state function, enthalpy, its associated constant pressure heat capacity, and their utilities in the context of making predictions of standard thermochemistries of reaction or phase change. 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

We're now going to look at characteristic ideal gas expansion paths.

So we've got a road map and it's time to follow some roads.

And I'll remind you that we're going to consider three paths, and a reversible

isothermal expansion that'll take us from conditions p1, v1, t1 to p2, v2 t1.

So, the final temperature is the initial temperature.

We're going to consider the isothermal expansion, that's where the temperature

will never change. We're going to consider an adiabatic

expansion followed by heating at constant volume, and we're going to consider a

reversible constant pressure expansion, followed by cooling at constant volume.

Delta u's going to be the same for all paths because it's a state function.

It only depends on the start and end point.

But we're going to explore how q and w heat and work differ.

So let me start with the reversible isothermal expansion.

And I'll remind you that the energy of an ideal gas only depends on temperature.

And so if you remember say our monotomic ideal gas, the internal energy is 3

halves rt. So if the initial temperature and the

final temperature are the same, delta u is 0.

And as a result, it must be the case, since delta u is equal to q plus w, that

negative q equals w. Or, I could've written q equals negative

w, but it's going to be more convenient to write minus q equals w.

And so, I'll emphasize that we're on path a, by using this a subscript, but we have

the relationship minus q and I'll also emphasize that the reverse ability.

Minus q reversible a is equal to w reversible a.

Because the process is reversible and we've already done this as a, a prior

study of the reversible path. The external pressure which is going to

dictate how much work is being done is given by the gas, the ideal gas law

itself. That is its going to be equal to rt over

v and I'll keep the one here to emphasize its t one the same temperature as the

initial and final states. And when I solve the integral in order to

get the work it is minus rt log v2 over v1.

And so to try to make more clear the relationship between the work and the

heat. Let's just remember I'm expanding.

My v2 is greater than my v1. So the gas is doing work on the

surroundings. And, sure enough, v2 greater than v1

means that the log is a positive number. I'm multiplying a bunch of positive

numbers by the minus sign so work is negative, as it should be, for doing work

on the surroundings. And q must be positive, right, because

there is this minus sign preceding it as well.

So, q positive means heat is transferred to the system.

And so if you like what's happening here is, to keep the temperature constant as

the gas is doing work, heat is being added, right, and the total amount of

these two is equivalent. Work and heat are equivalent to one

another, so I'm putting as much heat in as work is being done.

Now let's consider path b. Path b is a reversible adiabatic

expansion. So what does adiabatic mean?

Adiabatic means no energy transferred as heat.

So if you like, I perfectly insulate the system from the surroundings.

Heat is not allowed into the system from the surroundings.

That is q is equal to w. There is no, no sorry not w, zero.

q is equal to zero. There is no heat transfer.

As a result, delta u is equal to w, du is equal to little delta w.

Now, for an ideal gas, again, u only depends on t.

And so if I want to know what's been the change in u, well, let me recall for you

that the constant volume heat capacity is the partial derivative of u with respect

to t. And for an ideal gas u only depends on T

so I don't have to write a partial, I can just write a normal derivative.

If I now move dt over on the other side, I get that du is equal to constant volume

heat capacity dt. So if I want to know the work and I know

it's equal to the change in the internal energy Then I just want to integrate my

state function from t1 to t2. So remember I'm going to t2 that's what

two defines here. That's the final temperature in this

adiabatic expansion. That's going to be equal to the integral

from t1 to t2 of cv which is itself a function of t dt.

I don't know what that number is, but if I did know the form of the, the heat

capacity as a function of temperature, I could carry out that integral.

For now, I'll just keep that around as, it's a number, all integrals are numbers.

I'm not sure what the number is in digits, but it's a number.

Alright and that derives again I'll just emphasize by replacing this du with cvtdt

and the reason I want to do that replacement is I know how the temperature

is varying. I want to integrate from one temperature

to another. So now lets do path c.

Like path c which will get us back toward the final place we want to get to, this

is our other road to get to the same destination.

I'm going to reversibly heat at constant volume.

So I'm just going to put heat into the system in order to increase the pressure

but it's in a, it's in a fixed volume. So, some container that is rigid.

Well, given delta v is equal to 0, work must be 0 because work is minus p

external dv. And there is no dv here, dv is 0 so the

work is 0. So we have only heat, and so delta u

along path c is going to be heat plus work, but work is 0, so it's just the

heat. Well, once again, I'll, I'll play the

game of figuring out, delta u is the integral of the heat capacity at constant

volume dt. The difference here is that my limits

have changed compared to the last slide. I'm now going from temperature two to

temperature one, not from temperature one to temperature two.

So, that's worth emphasizing, that it's t2 to t1.

So gain, that integral is a number. I'm not sure if it's 23.9 or 14.1, but

anyway, it's a number. What I'm interested in is the sum of

these two. That's the road I took to get to the

final destination. And so, what was the reversible heat for

the sum of path b plus c, well remember B was 80 adiabatic, for that the heat was

zero. And along C it's the integral from t2 to

t1 of the heat capacity dt. What about the work?

Well that's the work along b plus the work along c.

The work along c was constant volume so its zero.

What was it along the first path? It was integral t1 to t2 cv.

So what's the energy change? Well its energy change for b plus energy

change for c that's the integral from t1 to t2 of the capacity plus the integral

from t2 to t1. But, when you reverse the limits on a

definite integral, you just change the sine of the integral.

So, the sum of these two is 0, as it must be.

Energy's a state function, and it only depends on t for an ideal gas, and since

the destination temperature and the The point of departure temperature are the

same, the energy change is zero. So, as it must be for a state function,

delta u equals 0 along this path. But clearly along paths b and c, we got a

different answer then what we did along path a, for the work and for the heat.

Alright, we've done two, let's embark on our last pv journey.

And so, we're going to consider the sum of paths d and e.

And we're getting good at this, so I'm just going to do them both at the same

time, basically. So, path d is an expansion at constant

pressure, so that's like a constant external pressure.

That's an easy one to evaluate. So the work along path d, is minus p

external, it's p1, is the name of it in this case.

v2 minus v1, and along path e, the volume does not change.

If delta v is 0, then the work is 0. So, this is the total work for paths d

and e, minus p1 times v2 minus v1, Meanwhile, what's the change?

I'm going to actually do the internal energy first, before doing the heat.

You'll see why in a second. What is the change?

Well, just as we derived in the last case, you'll get an integral of a heat

capacity from one temperature to another, and then, you'll reverse those

temperatures, those'll cancel. We already knew they were going to

cancel, we know delta u for this process is zero, because the initial and final

temperatures are the same. And so given that delta u is equal to q

plus w, q must be negative of whatever w was.

It it p1 times v2 minus v1, we just removed the negative sign.

So, this is this is an illustration of something that's generally true.

That it's typically easier to get heat as the difference between internal energy

and work; than any other particular way. Because we sort of know how to compute

work, and we've got some ideas about how to compute energy in cases like ideal

gases. Heat can be trickier although certain

limits are easy. Adiabatic for instance means heat is

zero. That's an easy way to get head but let me

compare these paths then and make clear the difference between the exact

differential and the inexact differential, the state function versus

the path function. So, when we followed the purple curve,

when we did the reversible isothermal expansion.

Here were the works, the work there's just one work.

Here was the work, here was the heat. Alright, they differ from one another by

sign because they must sum to zero. And they involve the logarithm of the

volumes that are involved. When we followed the blue path, the

adiabatic expansion, followed by the constant volume heating, here were the

work, here was the work, and the heat, again.

And they differ in sign, the exact number we don't know, but it involves some

integral over the heat capacity. And then finally the red path involves

the pressure times the volume change along the initial constant pressure

expansion. And those three quantities red, purple,

and blue, clearly are different from one another.

Of course, delta u, the state function is independent of the path and it's equal to

0 for an ideal gas. But work and heat differ.

Alright. Well, we have the traversal of those

various paths behind us at this point. And hopefully you've gained some facility

for working with work and heat and energy and we'll be doing more of that because

it's kind of a critical aspect of Internalizing the ideas that underlie

first law concepts. I want to spend the next lecture taking

an additional look at adiabatic processes.

All right. So introduced for the first time here but

of some interest in general where we're controlling heat flow.

And in particular preventing heat flow in a system.

Prior to getting to that lecture though, let's do one more demonstration.

And in particular I want to explore pressure volume relationships.

And see what happens with temperatures when we do in fact compress gases,

perhaps suddenly with external pressure to small volumes.

So it's an interesting demo I hope you enjoy it, then we'll get back to

adiabatic processes. This demonstration explores another

variation on the theme of the relationships between pressure, volume,

and temperature for a gas. In this case, I have a tube with a piston

at the top, and I'm going to push down on the piston suddenly and quickly.

Thereby compressing the gas in the tube to a much smaller volume, and a higher

pressure. As my compression will be sudden, there

will be no time for the heat generated from my work on the gas to be dissipated

to the outside surroundings. Instead, the temperature of the gas will

rise from absorbing that heat energy Now we could measure the temperature of the

gas with a thermocouple of some sort. But it's a bit more fun to demonstrate

the temperature rise by placing some substance that can be ignited into the

tube. In this case, there is a small wisp of

cotton at the bottom. If the temperature becomes high enough

cotton should ignite and smolder. Well, actually that's a perfectly okay

demonstration. But I've taken the liberty of making it

all a bit more exciting by using gun cotton instead of plain cotton.

Thus, the cellulose that is cotton has been soaked in nitric and sulfuric acid

to generate nitrous cellulouse, or gun cotton.

And the nitration causes it to burn much more quickly once ignited by providing a

local source of oxygen that's not simply the rather dilute atmospheric gas.

Should we give it a try? [SOUND] Wow.

That gun cotton flared and disappeared instantly on compression of the air in

the tube. Let's see that one more time.

[SOUND] Okay. That's a good illustration of how pv work

can be transformed into heat. We'll be doing many problems to compute

such conversion as we proceed in the course.

So that you'll see how to predict such changes quantitatively.

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