CHEM Introduction to Thermodynamics Fall Entropy and the Second Law of Thermodynamics

Transcription

1 CHEM Introduction to Thermodynamics Fall 2003 Entropy and the Second Law of Thermodynamics Introduction It is a matter of everyday observation that things tend to change in a certain direction. For example, water flows down hill and an ice cube placed at room temperature melts. The scientific statement of this tendency is the Second Law of Thermodynamics. There are many different ways of stating this law; one of the simplest is: It is impossible to devise a process that has the sole effect of turning heat into work. This is known as Kelvin's statement of the second law. The physical reasoning behind this statement is as follows. Internal energy is present in all objects in the form of both random molecular motions and the potential energy of the interactions between the atoms and molecules. If Kelvin's statement were not true then we could extract this internal energy as heat and use it to lift the object (or to do work in some other form). Then, for example, we could make water flow up hill under its own power; something that is clearly impossible. So the Second Law as stated by Kelvin is actually a rather mild statement of what is possible in the real world. Although it is pretty easy to accept Kelvin's version of the Second Law as being reasonable, it is hard to see how to do anything with it. For example, consider the chemical reaction CO(g) + H 2 O(g) CO 2 (g) + H 2 (g). This reaction will proceed spontaneously in the forward direction under some conditions and in the reverse direction under other conditions. It turns out that the Second Law tells us the direction in which the reaction proceeds under any set of conditions. But how can we do this? That is, how can we apply a statement about heat and work to a chemical reaction? The answer involves the use of a state function called the entropy, S, which is defined so that in any reversible process it changes according to the formula. (1)

2 In this formula dq rev is an increment of heat transferred to the system. The subscript is just to remind us that the process must be reversible. Clausius used the entropy to restate the Second Law of Thermodynamics as: The entropy of an isolated system increases during any spontaneous process. The universe as a whole is an isolated system. So this form of the Second law can also be given as: The entropy of the universe increases during any spontaneous process. A spontaneous process means a process that can occur with no change other than what is part of the process. Thus, transferring energy as heat from an object at a high temperature to one at a low temperature is spontaneous. But transferring heat from low to high temperature (as in an air conditioner or refrigerator) requires some other change, such as the consumption of electricity. Clausius's statement of the Second Law is more abstract than Kelvin's statement. Its advantage is that the entropy is defined mathematically; thus, Clausius's statement can be used to apply the Second Law quantitatively to a variety of phenomena. These include the behavior of heat engines, melting of solids, freezing of liquids, evaporation of liquids and solids, condensation of gases, mixing of liquids, dissolution of gases and solids in liquids, and chemical reactions. As long Clausius's statement of the Second Law is true, we can make a vast array of predictions about how these phenomena behave. The fact that no violation of any of these predictions has ever been observed gives us great confidence that this Law is indeed universally valid. A key factor in using the entropy is the fact that it is a state function. This means that the change in the entropy of a system during any process is determined only by the states of the system at the beginning and end of the process. To show that this is so, we must prove that all reversible paths between the same two states give the same entropy change. We only need to consider reversible paths since the calculation of an entropy change requires a reversable path; entropy changes along any other paths are calculated by using a reversible path between the same initial and final states. As long as all reversible paths between two states give the same entropy change, then we can use any reversible path to calculate the entropy change between the two states. But if two reversible paths between the same states give different entropy changes, then we would no longer be able to unambiguously define the entropy change. If entropy is a state function, then we can compute entropy changes in irreversible processes (that is, real processes) as well as in reversible ones. All paths between two states produce the same change in all 2

3 state functions for the system. Thus, the entropy change for an irreversible process is given by the entropy change for any reversible path between the same two states of the system. The difference between reversible and irreversible processes is that they will leave the surroundings in different states. The sum of the entropy changes in the system and the surroundings is called the entropy change of the universe. For a reversible process, the change in the entropy of the universe is zero. For any spontaneous process, Clausius's statement of the Second law tells us that the entropy of the universe increases. Overview In the following we will first prove that the entropy is a state function and then show that Clausius's statement of the Second Law follows from Kelvin's statement of the Second Law. Our proofs will use the idea of a cyclic process. A cyclic process is one in which the system returns to its original state. As a result, in a cyclic process all state functions return to their original values. Therefore, if entropy is a state function then S = 0 for any cyclic process. If entropy is a state function, then all possible reversible paths must give the same change in entropy as computed using equation (1). There are many possible reversible paths by which we can change the state of a system from a given initial state to a given final state. Here are two examples of such paths. On path 1 we first reversibly change the temperature from T i to T f while keeping the volume constant and then reversibly change the volume from V i to V f while keeping the temperature constant. On path 2 we first reversibly change the volume from V i to V f while keeping the temperature constant and then reversibly change the temperature from T i to T f while keeping the volume constant. To show that entropy is a state function, we must show that any two reversible paths between the same states give the same change in entropy. Let S 1 and S 2 be the entropy changes on two different reversible paths from a particular initial state to a particular final state. Now consider a cyclic process in which we first follow path 1 and then return to the initial state by following the reverse of path 2; we can surely do this since the paths are reversible. The entropy change is S 1 during the first step and is - S 2 during the second step; therefore the entropy change for this process as a whole is S = ( S 1 - S 2 ). If we can show that S for the cyclic process as a whole is zero, then we have shown that S 1 = S 2 for any two reversible paths between the same two states. In other words, if we can prove that S=0 for any cyclic reversible process, then we will have proven that entropy is a state function. 3

4 We will go about doing this as follows. First, we will show that S = 0 for any cyclic reversible process of a system consisting of an ideal gas. Then we will show that if S is a state function for any system, then it is a state function for all systems. Thus, we will generalize our result for an ideal gas to everything. This will be done by considering a generalized cyclic, reversible process. We will then show that, if Kelvin's statement of the Second Law holds, the entropy of the universe must increase in any irreversible cyclic process. Thus, we will have obtained the Second Law in the form stated by Clausius. Entropy is a state function for ideal gases We will now prove that S = 0 for any cyclic reversible process involving only an ideal gas. From the First Law we have (2) Since we are dealing with an ideal gas, we can say that du = C V dt. Since the process is reversible, we have dw = -pdv. If we substitute these into equation (2) and use the ideal gas law to substitute for p we get. (3) Now we divide both sides by T and integrate from an initial state with temperature T i and volume V i to a final state with T f and V f. This gives us. (4) Since the process is reversible, the integral on the left is just the change in entropy for the ideal gas; call this S I. Evaluating the integrals on the right then yields. (5) For a cyclic process, T f = T i and V f = V i. Since ln(1) = 0, we have (6) for any cyclic reversible process involving only an ideal gas. Therefore, entropy is a state function for ideal gases. 4

5 Entropy is a state function for arbitrary systems Now let us consider the change in entropy for any arbitrary system that undergoes a cyclic, reversible process. We will refer to this arbitrary system as "System A". We require that the process be reversible since it is for reversible processes that we can calculate entropy changes. Since the process is otherwise arbitrary, it may exchange both heat and work with the surroundings, but these exchanges must be reversible. The changes within System A will only depend on what happens in System A and on the flows of heat and work across its boundaries. The changes within System A won't depend on how the surroundings provide the heat and work. We may therefore specify that the heat be provided to or removed from System A via a second, ideal system that we shall call "System I". We shall specify a particular set of properties for System I; since these do not affect what happens within System A, they do not restrict the arbitrary nature of the changes in System A. The properties that we specify for System I are: (1) it consists only of an ideal gas, (2) the changes that it undergoes are reversible, (3) it is always at the same temperature as System A (that is, T I = T A ), (4) all energy that is transferred to or from System A as heat is transferred from or to System I (that is, dq I = -dq A ). Conditions (1) and (2) will permit us to use the fact that we have shown entropy to be a state function for ideal gases. Condition (3) can be brought about by allowing System I to exchange work with the surroundings in order to adjust its temperature; this condition will ensure that the exchanges of heat between System A and System I are reversible. Condition (4) has already been justified in the preceding paragraph. Finally, we note that both systems are closed (no flow of matter in or out). We will carry out our changes in two stages. In Stage 1, System A will undergo an arbitrary cyclic process and System I will exchange heat with System A but not with any other part of its surroundings. In Stage 2, System A will undergo no change and System I will return to the state that it was in at the start of Stage 1. During Stage 2, System I will be allowed to exchange both heat and work with its surroundings. Since we have specified that the changes within the two systems are reversible and that the exchanges of heat and work are reversible, both stages represent reversible processes. We now analyze the changes in entropy and energy that occur during Stage 1. No heat has been transferred to the surroundings, so S surr = 0. The change in entropy of System A is given by equation (1). Since dq I = -dq A and T I = T A, this is just the negative of the change in entropy for System I; that is (7) 5

6 or, for Stage 1 as a whole,. (8) If System I has returned to its initial state, then S I = 0 (because we have already shown that entropy is a state function for ideal gases); therefore S A = 0. So if we can prove that System I has returned to its initial state, then we have proven that S = 0 for any cyclic reversible process of any arbitrary system. Thus we will have proven that entropy is a state function for any system. We shall now prove that System I has returned to its initial state. The temperature of System I has returned to its initial value since T I = T A and T A has returned to its initial value. The number of moles in System I have not changed, since it is a closed system. So all that is left is to show that the volume of System I has returned to its original value. To see what has happened to the volume of System I, we must consider the work done on the system. Since System A has undergone a cyclic change, we have U A = 0 so, from the First Law, we have. (9) Since System I is an ideal gas with T I = 0, we have U I = 0 and. (10) Adding equations (9) and (10) gives (11) but, since all heat transfers into System A have come from System I, we have. (12) Therefore, equation (11) reduces to. (13) In other words, no net heat or work is exchanged by the two systems with their surroundings during Stage 1. We now show that if the change in System A was reversible (as we have required), then V f = V i for System I during Stage 1. We do this by showing that if V f V i then we could use Stage 2 to violate Kelvin's statement of the Second Law. Suppose first that V f < V i. Then during Stage 2 we allow System I to expand isothermally until its volume is equal to V i, since System I is an ideal gas, we have U I = 0 6

7 and w I = -q I during Stage 2. System I is now back in its original state. During Stage 2 the system does work on the surroundings so the heat and work transferred to the surroundings is. (14) But, since there was no net heat or work during Stage 1, equation (14) gives the net heat and work for the two stages together. In other words, we have turned heat entirely into work in the surroundings. Since both systems have returned to their original states, there is no other change. 1 So this violates Kelvin's statement of the Second Law and we can not have V f < V i. Now suppose that V f > V i. Then, since Stage 1 was reversible, we could carry out Stage 1 in the opposite direction and have V f < V i. But, as we have just shown, this would be impossible. So we can not have V f > V i if our process was reversible. Therefore the only allowable result is V f = V i. Thus, at the end of Stage 1, System I has returned to its original state and, since entropy is a state function for the ideal gas in System I, we have S I = 0. Then by equation (8) we have (15) for any cyclic reversible process of any arbitrary system. Since we use reversible processes to define the entropy change, entropy is a state function for any arbitrary system. The entropy of the universe can only increase The final thing that we must show is that Clausius's statement of the Second Law follows from Kelvin's statement of the Second Law. To do this we must consider the change in entropy of the universe, that is, of the two systems plus the surroundings. In the reversible process that we have been working with, we have already seen that S I =0 and S A =0. We also must have S S =0 for the surroundings since we transferred no heat to or from the surroundings in Stage 1 and did nothing at all in Stage 2. Therefore, for any reversible process we have for the change in entropy of the universe:. (16) 1 We might ask "but what if something else has changed in the surroundings?" But, since System A is arbitrary we could have included in System A any part of the surroundings that changed other than by the removal of heat or the addition of work. So we can arrange things so that the only change in the surroundings is the conversion of heat into work. 7

8 Now let us consider what happens to S univ in an irreversible process. To do this, we relax the condition that the changes within System A be reversible. In other words, we allow the process in System A to be spontaneous. We still require that the changes in System I and the exchanges of heat and work be reversible. We can do this because System A is arbitrary; we can therefore include any irreversible processes within the boundaries of System A. Except for this change within System A, we carry out Stage 1 as before. System A undergoes an arbitrary, spontaneous, cyclic process while System I changes so that dq I = -dq A and T I = T A. Since the process is no longer reversible, we may have V f > V i at the end of stage 1 (we ruled this out for reversible processes since the reverse process would be impossible, V f < V i still violates Kelvin's statement of the Second Law while V f = V i is the reversible case). During Stage 2, we compress System I isothermally until its volume is equal to V i. System I is now back in its original state. During Stage 2 we have for the heat and work transferred to the surroundings. (17) In other words, we turned work entirely into heat in the surroundings. Kelvin's statement of the Second Law permits this. Since both System A and System I have returned to their original states, we have S I = 0 and S A = 0. Since no heat was transferred to or from the surroundings during Stage 1, any change in the entropy of the surroundings occurred during Stage 2. Since q S > 0 during this stage and T is always positive, we have S S > 0 from equation (1). Therefore the change in entropy of the universe is (18) for any spontaneous process. This is Clausius's statement of the Second Law of Thermodynamics. Therefore Clausius's statement of the Second Law follows from Kelvin's statement of the Second Law. Michael Mozurkewich, October