(1)Fundamental Definition of Equilibrium
The second law of thermodynamics in equation form is,
where the = sign holds when the process is reversible and the > holds when it is irreversible. Our first application of the second law will be to provide a thermodynamic definition of equilibrium. We indicated very early in the course that at equilibrium none of the variables was changing in time. (Of course, a system in a steady state has to be excluded from this definition.) Another possible definition of equilibrium was that all the variables have the values they would have at time equals infinity. Neither of these "definitions" provides an equation we can use to discuss equilibrium systems carefully.
The second law provides us with a definition of equilibrium that we can use to derive equilibrium properties of thermodynamic systems. Consider a closed isolated system. In a closed isolated system dq = 0 and dV = 0 which implies that dU = 0. (Actually, we should really say that dw for all forms of work are zero, but for the time being we will only consider pV work.)
Under the conditions of constant U and V the second law, Equation 1, becomes
This may look simple, but it is a very profound statement. It says that in a closed isolated system (a system which us not being disturbed from the outside) any spontaneous change must increase the entropy.(2)
In a closed, isolated system the entropy seeks a maximum.
If you plot the entropy of a closed isolated system against some system variable any spontaneous change in the system must take the system to higher entropy. If the system is not in equilibrium then
but if the system is at equilibrium any spontaneous change in the system must leave the entropy unchanged,
Equation 2 is the origin of the somewhat arrogant statement that you may have heard, "The entropy of the universe is increasing." If you regard the universe as a closed isolated system then the statement is probably true, even though it is difficult to take such broad statements concerning the nature of the universe seriously.
Combined First and Second Laws
The first thing we must do is incorporate our new-found equation for the second law into what we already know. Going back to the first law, with pV work only, recall that we can write
If we restrict our attention to reversible processes this becomes,(3)
but we know from the second law that(4)
Merging Equations 4 and 5 we arrive at what is called the combined first and second laws,(5)
(Later on, when we want to include work other than pV work, we will add it in to Equation 6,(6)
but for now we will just stick with pV work.)(7)
Equation 6 implies that the natural variables of internal energy, U, are S and V. In previous calculations we have regarded U and a function of T and V or of T and p, but nature - in the form of the first and second laws of thermodynamics - gives us U as a function of S and V.
Divide Equation 6 by dT and hold V constant:
Equation 8b provides us with a way to calculate entropy changes for a certain class of processes, namely processes at constant volume. Set up Equation 8b for integration,(8a, b)
and calculate the entropy change for a constant volume process as,(9)
Continuing to incorporate the second law into our set of thermodynamic tools, recall that(10)
(Note that equation 11d implies that the natural variables of H are S and p.)(11a, b, c, d)
We can use Equation 12b to calculate entropy changes for processes at constant pressure. Set up Equation 12b for integration,(12a, b)
and integrate,(13)
What about processes at constant T? We can calculate the entropy change for a process at constant temperature by rearranging Equation 5 for integration,(14)
The finite entropy change is,(15)
If we now restrict consideration to processes at constant T Equation 16 becomes,(16)
(17)
Example Calculation
Calculate the entropy change in heating 1.00 mol of Al from 300 K to 500 K at constant pressure. The constant pressure heat capacity of Al is given to good approximation by,
The calculation is
Another Example - An irreversible Process
Often we must calculate entropy changes for irreversibly processes. We don't know how to calculate entropy changes for irreversible processes, but it doesn't matter. Entropy is a state function so ΔS is independent of path. All we have to do is imagine a reversible path which will effect the same change and calculate the entropy change for the reversibly path.
Suppose we start with a 100. g block of Cu at 500 K and a 100. g block of Cu at 300 K. Bring the two blocks into thermal contact and heat will flow from the hotter block to the cooler block until they reach the same temperature. This is definitely an irreversible process. However, let's imagine that we can reversibly cool the hot Cu block down to the final temperature and reversibly warm the cold Cu block up to the final temperature. We can calculate the entropy change for both of these processes from Equation (14). The total entropy change is just the sum of the two individual entropy changes. Note that we expect the entropy for this process to be positive because the process is spontaneous and the two Cu-block system can be regarded as isolated.
We need the following data:
First we must find the final temperature. We do this by recognizing
that the heat lost by one Cu block is gained by the other Cu block. This
is called "heat balance."
This is easily solved to obtain Tfin = 400 K. Then
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