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

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From the course by University of Minnesota

Statistical Molecular Thermodynamics

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

From the lesson

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

It's time to introduce a new state function, and that state function is

Â enthalpy. So, let me consider constant pressure

Â conditions. Up till now we've been working with

Â ensembles, nvt ensembles that have often involved constant volume.

Â But in practice you know, it's pretty convenient to do chemistry at constant

Â pressure. We have open vessels on a desk, for

Â instance. And, let's consider the reversible

Â process where work is restricted to pressure volume work.

Â Alright? And so, first law says delta u is q plus

Â w. That's q minus integral v1 to v2 pdv.

Â At constant volume we have the no change in volume v1 equals v2.

Â Delta u is equal to the heat transfer at constant volume.

Â Right? In that case heat would become a state

Â function. That is actually a measurement that can

Â be done. It's an experiment called bomb

Â calorimetry, so you put some sort of reacting system perhaps into a so-called

Â bomb. That is a, a vessel of volume that is

Â extremely rigid and you screw the top down tight and you watch the heat change

Â and so you also know the internal energy change.

Â And as I said, it's more convenient to work at constant pressure sometimes, so

Â when that's the case if it is indeed constant pressure, well, then again I'll

Â rearrange the first law. The heat at constant pressure is going to

Â be delta u plus p external. So plus because I moved it around to the

Â other side; work is minus p external. Integral from v1 to v2, dv.

Â It's a constant pressure, so I can pull it out from the integral.

Â So, I get that the heat at constant pressure is delta u plus p delta v.

Â So, qp, unlike qv is not equal to delta u.

Â There's also this p delta v term. So, at constant volume, qv is a state

Â function, it's equal to delta u. At constant pressure, qp is also a state

Â function. Delta u plus p delta v.

Â In fact, we're going to give it a name. We're going to call it the enthalpy.

Â So, more generally speaking, enthalpy, indicated by h, is equal to u plus p

Â times v. And the differential form is going to be

Â dh equals du plus, and when I take the differential of a product, I have to

Â employ the chain rule, pdv plus vdp. But, if I am at constant pressure, that

Â last term, dp, is zero. And I would get delta h is equal to delta

Â u, plus p delta v. And that confirms then that enthalpy is

Â equal to the heated constant pressure and also that it's a state function.

Â Right? Because U is a state function.

Â And p is being held constant, it's just some value and v is certainly a state

Â function, you specify what the volume is, the volume doesn't depend on how you got

Â to the volume it is the volume. And so, h has the same role at constant

Â pressure that u has at constant volume. That is enthalpy has the role at constant

Â pressure that internal energy has at constant volume.

Â So, let's look at a comparison between the 2.

Â And let's consider a specific system and we'll actually work with some numbers

Â here. let's take ice, water, solid water.

Â Melts at 273 kelvin at 1 atmosphere of pressure.

Â And, I'll just tell you that if you did the measurement, the heat required to

Â melt that ice, the heat of fusion, if you will, 6.01 kilo-joules per mole.

Â So, delta h molar, delta h is equal to the heat transfer constant pressure 6.01

Â kilo-joules per mole. That's now, the 273 kelvin molar volumes

Â for solid and liquid water are something you can measure, and I've shown them

Â here. So for the solid, it's 0.0196 liters per

Â mole. And for the liquid, it's 0.0180 liters

Â per mole. So I, I have a little note here, rare and

Â important. I guess this is something that every,

Â every living breathing organism should appreciate, and that is, that the volume

Â occupied by solid ice at, at the freezing point is s little bit larger than the

Â volume occupied by liquid water. That is the density of the ice is a bit

Â lower than that of water. Ice floats, now that's something everyone

Â knows, you encounter that in everyday life at a pretty early age.

Â Ice floats, imagine what would happen if ice did not float?

Â Imagine if every lake froze from the bottom up.

Â That would be very bad for the fish. And so, you know, some of the

Â responsibility for life on earth as we know it is, that ice covers bodies of

Â water allowing the life in the body of water to persist through the winter

Â through the Ice Age, perhaps. This property that the solid is less

Â dense than the liquid is very, very rare. Water is one of the very few substances

Â for which this is true, and it's terrifically important that it is true.

Â Well, in any case, what is the change in u, the internal energy for this process?

Â Well, let me turn around the definition of enthalpy as a means to determine that.

Â We're going to ask, what's the difference between internal energy and enthalpy?

Â So, delta u is going to be equal to delta h minus p delta v, at constant pressure.

Â So, I'll just plug in the numbers. Here is delta h, here is the pressure,

Â one atmosphere. Here is the difference in the volumes.

Â I will express the d-d-d-d-d-, I will express this in common units, that's what

Â I want to do here. So I've got atmospheres, I've got liters

Â per mole, I've got 1.6 times 10 to the minus third liter atmosphere per mold.

Â That's this difference. I'm going to turn atmospheres into joules

Â by dividing by liter atmospheres in order to get at a final quantity of well not

Â much changed. Notice that this is 0.008 divided by 0.08

Â so that's about 10 to the minus 1 multiplying something times about 10 to

Â the minus 3 so I get 10 to the minus 4 joules, that's what 10 to the minus 7

Â kilojoules. And so what I get is delta u is just

Â about equal to delta h to within part per million.

Â That is the internal energy change is the same as the enthalpy change.

Â Why is that? Well, because delta v is really, really

Â small. So, p delta v is small, and so, there's

Â not much difference between the internal energy and the enthalpy.

Â On the other hand, let's consider a different process, let's warm our water

Â up. It's no longer melting, now it's boiling.

Â So we're at 373 Kelvin, still one atmosphere.

Â The heat required to boil water is 40.7 kilojoules per mole at its boiling point

Â at one atmosphere pressure. So delta h, then, is equal to qp.

Â It's 40.7 kilojoules per mole. Let's look at the different molar

Â volumes. Now there's quite a difference because

Â the gas occupies much more volume than the liquid.

Â So, the gas occupies 30.6 liters per mole, the liquid, as we saw in the last

Â slide, 0.018. I'll again ask the question, what's the

Â internal energy? It is the enthalpy minus p delta v.

Â I'll again plug in the numbers. Here's the pressure and atmospheres, here

Â are the the molar volumes and liters per mole.

Â I again do a transformation to turn atmosphere liters per mole into joules,

Â and I end up with 37.6 kilojoules per mole.

Â Alright. So, the internal energy now differs from

Â the enthalpy by 3.1 kilojoules per mole. And why is that?

Â Well, because I have to do pv work to expand the liquid to the gas.

Â So, one way to think about this is, this delta u value, the internal energy change

Â It's most of delta h. And that's associated with the energy

Â required to tear the molecules apart, basically.

Â There they are in the liquid phase, happily interacting with one another,

Â hydrogen bonding to their heart's content, and you force them to become a

Â gas. You, you ripped them from their, from

Â their compatriots. That was a large part of the energy.

Â But in addition you force them to expand into a much larger volume, at constant

Â pressure, and that is work. And that is why it is more enthalpy

Â change than internal energy change, because of that extra pv work.

Â All right, so that's two examples. Let's pause for a moment.

Â I'll let you do one example for yourselves, and then we'll move on.

Â Great, hopefully now you're getting a handle on internal energy and enthalpy,

Â so let's continue and take a closer look at an old friend, heat capacity.

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