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Unformatted text preview: In the previous lecture we have introduced the Boltzmann formula for the entropy: S = kB ln Ω Examples of using this: Example 1. (see previous lecture). Irreversible gas expansion Example 2. We have also seen that hotter objects tend to have a higher entropy (other things being equal). For example, if the pressure of the system is fixed constant then S = c p ln T + constant (assuming that c p does not depend on temperature). How can we understand this from the point of view of the Boltzmann formula? Again, things are easier to understand if we think quantum mechanically. We have mentioned before that temperature is related to the average energy of molecules. If I have a gas at high temperature, this means that there is a lot of energy that’s available to be distributed among its molecules in many different ways (each corresponding to a different microscopic state). On the other hand, at low T the system must have low energy and there are only a handful of microscopic states, for which this can be achieved. So the higher the temperature, there more states the system can have, Ω is larger and the entropy higher, according to Bolzmann’s expression Using these arguments, you can also convince yourself that it’s not unreasonable that two objects will be in equilibrium when their temperatures are equal. E.g., if one object has T=0 and the other has a very high temperature, this is a very unlikely state (just like having no molecules in one half of a container). Two more notes on entropy and its statistical meaning: Note 1. Entropy is a measure of mess or randomness. A gas in one part of the vessel is more “ordered” than the same gas spread over the entire vessel. A stretched polymer chain is more ordered than one that is not stretched. If you keep your desk tidy (that is, every item is in a designated place, like the gas in a designated compartment of the vessel) then the entropy is low. If you deposit stuff randomly on your desk you make things more disordered and so you increase the entropy. You can reduce the entropy by doing work on the system (your desk). The entropy of your desk, however, is unlikely to decrease by itself. Note 2. A living organism is a state of matter that has a very low entropy (it is well organized). If we isolate it from the environment thermodynamics is telling us it’s bound to increase its entropy. That means die. A living organism needs energy and has to exchange heat with the surroundings to avoid increasing the entropy (the overall entropy of the world is still bound to increase). The free energy. The 2nd law states that the entropy of an isolated system increases in a spontaneous process. Eventually, the system will reach equilibrium, which will correspond to the maximum entropy. Equilibrium state of the system can be found from this maximum entropy principle. Unfortunately, very few systems of interest are isolated systems. More commonly, one deals with systems where (V,T) or (P,T) are controlled. For example, in chemistry, we’d often like to know how many moles of reactants and products we will have after a chemical reaction is over and the system has attained equilibrium. This reaction would normally be carried out in a room where both the temperature and the pressure are kept constant. We need a formulation of the 2nd law that would be easy to apply to constant (V,T) or constant (P,T) conditions. To this end, we start with the usual formulation of the 2nd law: dS > δ q / T or δ q − TdS < 0 in a constant volume process δ q = dU . At constant V and T the above equation can be rewritten as dU − TdS = d (U − TS ) ≡ dA < 0 The function of state A = U − TS is called the Helmholtz free energy of the system. In spontaneous processes in systems kept at constant V and T A always decreases. Equilibrium corresponds to the minimum of A. In a constant pressure process δ q = dH . At constant P and T we find dH − TdS = d ( H − TS ) ≡ dG < 0 The function of state G = H − TS = U + PV − TS is called the Gibbs free energy of the system. In spontaneous processes in systems kept at constant P and T G always decreases. Equilibrium corresponds to the minimum of G. To summarize: A=U
TS G=U+PV
TS dA ≤ 0 at constant V,T dG ≤ 0 at constant P,T Equilibrium = min A (at constant V,T) min G (at constant P, T) max S (for isolated systems) Example. liquids freeze and solids melt because A = U –TS (or G=H
TS) can be made small in 2 ways: by reducing U (H) or by increasing S. If T is small then U wants to be small. When molecules form a crystalline lattice, this minimizes U. However as T is increased, it becomes advantageous to increase S, that is, to increase the amount of disorder in the system. So the crystal melts – to be discussed in much more detail later Free energies and work. δ W + δ q = dU (1st law) δ q = dU − δ W ≤ TdS (2nd law)
For a process at a constant T: δ W ≥ d (U − TS ) = dA or, integrating this, A(2) – A(1) ≤ W A spontaneous process corresponds to A(2) < A(1). No work is needed to be done on the system as W=0 satisfies the above inequality. Consider the case where A(2)>A(1). The process will spontaneously take place from 2 to 1. However if we want to revert this process, we can achieve this by exerting work on the system. The minimum amount of work that has to be done equals the change in the Helmholtz free energy: Wmin = A(2)
A(1). This equality holds for a reversible (i.e., equilibrium) process – in the case of an irreversible process the work is larger. Since −W = Wsys is the work done by a system itself, we can rewrite the above inequality as Wsys ≤ A(1) − A(2) In a process where the free energy decreases, the system can do positive work; the maximum work is achieved in an equilibrium process and is equal to the free energy change. In non
equilibrium (fast) processes part of the free energy is wasted so that the work is less than the free energy change. (E.g., cars are more efficient when driven not too fast and rapid acceleration is avoided). ...
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 Spring '11
 Makarov

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