Caratheodory1 showed that dqrev/T is an exact differential by a more formal mathematical procedure. His argument is sketched briefly in Appendix D. It begins with the fact that two reversible adiabats cannot cross. We now show that this is a fact. We have already seen an example of it in the previous chapter when we derived a formula for the reversible adiabat for an ideal gas with a constant heat capacity.
Equation (3.4-21b) is
T T1
V1
V nR/CV
(3.2-7) This equation represents a family of functions ofT as a function ofV, one for each initial state:
T T(V) (3.2-8)
No two curves in this family can intersect.
Constantin Caratheodory, 1872–1950, was a Greek-German mathematician who made many contributions in mathematics in addition to his work in thermodynamics.
To show that two reversible adiabats cannot cross for other systems we assume the opposite of what we want to prove and then show that this assumption leads to a contradiction with fact and therefore must be false. Assume that there are two different reversible adiabats in the state space of a closed simple system and that the curves coin- cide at state number 1, as depicted in Figure 3.6. We choose a state on each reversible adiabat, labeled state number 2 and state number 3 such that the reversible process leading from state 2 to state 3 hasq >0. Now consider a reversible cyclic process 1→2→3→1. Since steps 1 and 3 are adiabatic,
qcycleq2>0 (3.2-9)
Since∆U0 in any cyclic process,
wsurr −wcycleqcycleq2 (3.2-10)
Heat transferred to a system undergoing a cyclic process has been converted completely to work done on the surroundings, violating the second law of thermodynamics. The source of this violation is the assumption that two reversible adiabats can cross. There- fore, only one reversible adiabat passes through any given state point. The rest of Caratheodory’s argument is summarized in Appendix D.
V
T
Two proposed adiabats that cross 2
3
1
Figure 3.6 Two Reversible Adiabats That Cross (Assumption to Be Proved False).
V 1
T
V
T
(a) (b)
1
3
2
1
2
3
Figure 3.7 Reversible and Irreversible Adiabats. (a) Impossible case. An irreversible adiabatic process cannot lead to the low- temperature side of the reversible adiabat. (b) Possible case. The irreversible adiabatic process can lead to the high-temperature side of the reversible adiabat.
Figure 3.7 shows schematically two possibilities for an irreversible adiabatic process of a closed simple system in which the initial state (state 1) and the final state (state 2) are equilibrium states. During the process, the state of the system is not an equilibrium state and cannot be represented by a point in theV–T plane. The broken curve in the figure indicates that the state point leaves theV–T plane and then returns to theV–T plane at the end of the process.
We now show that an irreversible adiabatic process must lead to a higher temperature than the reversible adiabatic process starting at the same initial state (state 1). The solid curve in Figure 3.7a represents the reversible adiabat passing through state 1. We first assume that state 2 lies below this curve (an assertion that we want to disprove). Let state 3 be the state on the reversible adiabat that has the same volume as state 2. After the irreversible step 1 has occurred, we carry out a reversible constant-volume step from state 2 to state 3 (step 2). For a constant-volume process,
q2
c
dq
c
CVdT (3.2-12)
It is an experimental fact that the heat capacity of any system is positive. Therefore, q2>0, since the temperature of state 2 is lower than that of state 3. After step 2, we carry out a reversible adiabatic step from state 3 to state 1 (step 3). Step 1 and step 3 are both adiabatic, so that
qcycleq2>0 (3.2-13)
SinceUis a state function,
∆Ucycle0 (3.2-14)
The work done on the surroundings in the cycle is
wsurr −wcycle −∆Ucycle+qcycle −∆Ucycle+q2q2 (3.2-15) Heat transferred to the system in a cyclic process has been completely turned into work done on the surroundings, which is a violation of the second law. An irreversible
adiabatic process cannot lead to a state that is lower in temperature than the reversible adiabat.
If state 2 is above the reversible adiabatic curve as in Figure 3.7b, we carry out a constant-volume reversible step (step 2) from state 2 to state 3, and an adiabatic reversible step from state 3 to state 1. This time, because state 2 is at a higher temperature than state 3 and because the capacity of the system must be positive,
qcycleq2<0 (3.2-16)
so that
wsurr −wcycle −∆Ucycle+qcycle −∆Ucycle+q2q2<0 (3.2-17) In this case, heat transferred to the surroundings has been turned completely into work done on the system. This does not violate the second law of thermodynamics because the surroundings do not undergo a cyclic process. The final temperature for an irreversible adiabatic process cannot be lower than for a reversible adiabatic process with the same final volume, but it can be higher.
Now consider the entropy change for the irreversible adiabatic process that was depicted in Figure 3.7b. BecauseSis a state function,
∆Scycle∆S1+∆S2+∆S30 (3.2-18) Because step 3 is reversible and adiabatic,∆S30, and
∆S1 −∆S2 (3.2-19)
Because step 2 is reversible, we can integrate Eq. (3.2-1) for this step:
∆S2 T3
T2
dqrev
T
T3
T2
CV
T dT <0 (3.2-20) The inequality comes from the fact that the temperature and the heat capacity are both positive and the fact that the temperature of state 2 must be greater than that of state 3. Because∆S1 −∆S2,∆S1must be positive. Therefore,
∆Sirrev ∆S1>0 (irreversible adiabatic process) (3.2-21)
Combining Eqs. (3.2-11) and (3.2-21),
∆S≥0 (any adiabatic process) (3.2-22) where the equality holds for reversible processes.For any adiabatic process, the entropy of the system cannot decrease. This is the most important consequence of the second law of thermodynamics. Because we define the universe to include all objects that interact with each other, the universe can undergo only adiabatic processes.In any reversible process, the entropy of the universe remains constant. In any irreversible process, the entropy of the universe increases.