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

Loading...

From the course by University of Minnesota

Statistical Molecular Thermodynamics

118 ratings

University of Minnesota

118 ratings

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

From the lesson

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]

Coursera provides universal access to the world’s best education, partnering with top universities and organizations to offer courses online.