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Section 3 Entropy and Classical Thermodynamics

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Section 3 Entropy and Classical Thermodynamics Section 3 Entropy and Classical Thermodynamics 3.1 Entropy in thermodynamics and statistical mechanics 3.1.1 The Second Law of Thermodynamics There are various statements of the second law of thermodynamics. These must obviously be logically equivalent. In the spirit of our approach we shall adopt the following statement: • There exists an extensive function of state called entropy, such that in any process the entropy of an isolated system increases or remains constant, but cannot decrease. The thermodynamic definition of entropy is as follows. If an infinitesimal amount of heat dQ is added reversibly to a system at temperature T (where T is the absolute temperature of the system), then the entropy of the system increases by dS = dQ / T. We shall see that this definition (and note that in thermodynamics only entropy changes are defined) is consistent with the statistical mechanics definition of entropy S = k ln Ω (here the absolute value of the entropy is defined). We have seen already that the law of increasing entropy is a natural consequence of the equation S = k ln Ω. In some thermodynamics texts (the classic is the book by H.B.Callen) this law is given the status of a postulate or axiom. It may be summarised by S ≥ 0. 3.1.2 Restatement of the First Law Let us now ask what form the first law takes, given this connection between entropy and heat. For an infinitesimal change dE = d⁄ Q + d⁄ W. this is true always Taking a fluid (p − V variables) as our thermodynamic system and considering a reversible process we have: d⁄ W =−pdV for a reversible process and also d⁄ Q = TdS. for a reversible process Thus dE = TdS − pdV. for a reversible process But all the variables in this last expression are state variables. In a process connecting two states dE = Efinal − Einitial etc. This does not depend on any details of the process. So we conclude (Finn p86) that for all processes, reversible or irreversible dE = TdS − pdV. always This form of the first law is very important. You note that all the differentials are exact (i.e. they are differentials of a well defined function). You see that T = ∂ E / ∂ S|V in agreement with the statistical definition of temperature. PH261 – BPC/JS – 1997 Page 3.1 3.1.3 Microscopic interpretation of the first law Returning to the microscopic view, the internal energy of a system is given by E = ∑ pjεj j where pj is the probability that the system is in the εj energy state. So E can change because of changes in εj, the energies of the possible quantum states of the system, or changes in pj, the probability distribution for the system. In an infinitesimal change we can write dE = ∑ εjdpj + ∑ pjdεj. j j thermal mechanical The first term is identified with TdS, the second with −pdV. Entropy changes are associated with changes in the occupancies of states, while the mechanical term (work) relates to the energy change of the states. Let us establish this equivalence. a) Work term Suppose the energies εj, depend on volume (as they do for particles in a box – a gas). Then ∂ε dε = j dV. j ∂ V We suppose the derivative to be taken at constant pj i.e constant S. Then ∂ε p ε = p j V ∑ jd j ∑ j ∂ | d j j V S and since the derivative is taken at constant pj then the derivative can be taken outside of the sum: ∂ p ε = p ε V ∑ jd j ∂ | (∑ j j) d j V S j ∂ E = | dV ∂ V S =−pdV. Thus we have identified the ∑j pjdεj term of dE as the work done on the system. b) Heat term The fact that TdS =∑j εjdpj takes a bit longer to prove (see also Guenault p 25, although our approach differs in detail from his; we have already shown β = 1/kT) We must make connection with entropy, so we start for the expression for S for the Gibbs ensemble Sensemble = k N ln N − ∑ nj ln nj . j This is the entropy for the ensemble of N equivalent systems. What we are really interested in is the (mean) entropy of a single system — which of course should be independent of N. We express the first N as the sum ∑j nj, so that PH261 – BPC/JS – 1997 Page 3.2 Sensemble = k ∑ nj ln N − ∑ nj ln nj j j nj =−k ∑ nj ln ( ) . j N Then the (mean) entropy of a single system is one Nth of this: n n S =−k ∑ ( j ) ln ( j ) j N N th but here we can identify nj / Np as j – the probability that a system will be in the j state. So we have the useful formula for the entropy of a non-isolated system S =−k ∑ pj ln pj. j We shall now use this expression to make the connection between ∑j εjdpj and heat. The differential of S is dS =−k ∑ (dpj ln pj + dpj) . j Now the second dpj sums to zero since the total probability is constant. Thus we have dS =−k ∑ dpj ln pj j and for ln pj we shall use the Boltzmann factor −ε e j/kT p = j Z so that ε ln p =−( j + Z) , j kT giving ε = j + . dS k ∑ ( Z) dpj j kT The second term in the sum is zero since Z is a constant and dpj sums to zero as above. So only the first term contributes and we obtain 1 dS = ∑ εj dpj , T j or ∑ εj dpj = TdS j which completes the demonstration. ______________________________________________________________ End of lecture 11 PH261 – BPC/JS – 1997 Page 3.3 3.1.4 Entropy changes in irreversible processes We have seen that dS = d⁄ Q / T. And we stated that this applies in a reversible change. In a separate handout we gave some examples of calculating entropy changes. What happens to the entropy in an irreversible process? To highlight matters let us take an isolated system for which d⁄ Q = 0. Consider a process undergone by this system on removing a constraint. The microscopic understanding of the second law tells us the entropy will increase, as the system goes to the new (more probable) equilibrium macrostate with a larger number of microstates. Since for an isolated system d⁄ Q = 0d, the expression S = d⁄ Q / T certainly does not apply. Indeed, in this case TdS > d⁄ Q. This is generally true for irreversible processes. Consider a specific example of an irreversible process: the infinitesimal adiabatic free expansion of a gas (by the removal of a partition in a chamber). The partition is the constraint. Replacing the partition will certainly not get us back to the initial state. What will? In this case d⁄ Q = 0dand ⁄ W = 0 so then dE = 0 by the first law. Since the expansion is infinitesimal p is essentially constant, and pdV > 0. But we know that dE = TdS − pdV for all processes. Therefore in the free expansion, for which dE = 0, if pdV > 0, we must correspondingly have TdS > 0. So for this irreversible process we find TdS > d⁄ Q −pdV < d⁄ W. So although no heat has flowed into the system, the entropy has increased. This is a consequence of the irreversible nature of the process. To summarise: TdS > d⁄ Q in an irreversible process. 3.2 Alternative statements of the Second Law 3.2.1 Some statements of the Second Law Now when you read chapter 4 of Finn you will find that the second law is introduced through a discussion of cyclic processes and heat engines. The statement of the law takes a rather different form from the one we have given. That is how the subject developed historically. But we are approaching things from a somewhat different angle. We started from statistical mechanics and the principle of increasing entropy follows easily from that treatment. We’ll look at these cyclic processes in a bit. Here are various statements of the second law: PH261 – BPC/JS – 1997 Page 3.4 1a. Heat cannot pass spontaneously from a lower to a higher temperature, while the constraints on the system and the state of the rest of the world are left unchanged. 1b. It is impossible to construct a device that, operating in a cycle, produces no effect other than the transfer of heat from a colder to a hotter body. (Clausius) 2. It is impossible to constuct a device that, operating in a cycle, produces no effect other than the extraction of heat from a body and the performance of an equivalent amount of work. (Kelvin- Planck). These various statements can be shown to follow from each other (Finn p57) and from the law of increasing entropy. 3.2.2 Demonstration that the law of increasing entropy implies statement 1a Let us show that S ≥ 0 ⇒ statement 1a. (The argument here is identical to that in Section 2.3.3) Consider two systems A and B separated by a fixed diathermal wall. Thus the volumes of each system are constant, so no work can be done. Since the composite system is thermally isolated S ≥ 0. If the two systems are not in thermal equilibrium with each other (i.e. they are at different temperatures) then heat will flow between them. Consider the exchange of an infinitesimal amount of heat dQ. fixed diathermal wall ↑ d⁄ Q → B A heat flow The law of entropy increase tells us that dS = dSA + dSB ≥ 0 ∂ ∂ = SA + SB | dEA | dEB ∂ EA V ∂ EB V 1 1 = dEA + dEB TA TB since, in general, 1/T = ∂ S / ∂ E|V.
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