!W "!#U + T#S, where. !U = U f " U i and!s = S f " S i. ! W. The maximum amount of work is therefore given by =!"U + T"S. !W max

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1 1 Appendix 6: The maximum wk theem (Hiroshi Matsuoka) 1. Question and answer: the maximum amount of wk done by a system Suppose we are given two equilibrium states of a macroscopic system. Consider all the processes, reversible and irreversible, starting with one of these two states and ending with the other. In addition, let s also assume that during each of these processes, the system exchanges heat with a heat reservoir at temperature T and does wk on the outside. The question now is what is the maximum amount of wk that can be done by the system going through one of these processes. It would be nice if we can answer this question in terms of the values of some state variables at the initial and the final state. In fact, the wk! W done by the system satisfies where!w "!#U + T#S,!U = U f " U i and!s = S f " S i.! W ( U, S) ( U + "U, S + "S) Q The maximum amount of wk is therefe given by!w max =!"U + T"S. Actually, this is not mysterious because this equation is nothing but the first law equation f a quasi-static process:!u = Q " W max,

2 2 where Q is the heat flowing into the system through a quasi-static isothermal process at the temperature T so that Q = T!S. This means that the maximum amount of wk is achieved by a reversible (i.e., quasi-static) process connecting the initial and final states. Keep in mind that the system is assumed to exchange heat with the reservoir at T and that this assumption puts a constraint on the quasistatic process that results in the maximum amount of wk. In fact, such a process is uniquely determined f a particular choice f the initial and the final states. Me specifically, we must first follow the adiabat going through the initial state until the temperature of the system reaches the temperature T of the reservoir and then go through the quasi-static isothermal process at the temperature T to reach the adiabat that goes through the final state and finally follow this adiabat to arrive at the final state. 2. The maximum wk by a process that starts and ends at T: the Helmholtz free energy F a process that starts and ends with equilibrium states at the temperature of the reservoir T, we find!w "!#( U! TS) =!#F, where F is the Helmholtz free energy of the system. The maximum amount of wk in this type of processes is therefe given by! W max =!"F. 3. Derivation of the theem As with all the results in thermodynamics, this theem is a direct consequence of the first and the second law. Using the first law, we get!u = Q + W!W =!"U + Q. The total entropy difference f the combined system of the system of our interest and the heat reservoir is just the sum of the entropy difference f the system and the entropy difference f the heat reservoir:

3 3!S total =!S +!S reservoir =!S " Q T, where!s reservoir = "Q T. Since the combined system does not exchange heat with other systems, we can apply the second law and the total entropy difference f the combined system either increases stays constant:!s total " 0 Q! T"S, where the equality holds when the process is reversible.!w =!"U + Q, we get Combining this inequality and!w =!"U + Q #!"U + T"S. 4. Example: the efficiency of a heat engine We can apply this theem to a heat engine by regarding a hotter heat reservoir at temperature and the engine receiving an input heat Q in from the reservoir as a combined system. This combined system does wk!w on the outside and discards heat!q ex to a colder heat reservoir at temperature T c. We consider one cycle of the engine f which the initial and final equilibrium states are identical.

4 4 Since the combined system loses energy through the wk output!w > 0 and the exhaust heat!q ex, the internal energy difference after one cycle is then given by!u total = Q ex + W = "Q in, where we have used!u enginel = Q in + Q ex + W = 0. After one cycle, the entropy of the engine does not change and therefe the entropy difference of the combined system is just the entropy difference of the hotter heat reservoir:!s = " Q in. By applying the maximum wk theem to the combined system, we get (!Q!W "!#U total + T c #S = Q in + T in ) c = 1! T ' c ) ( Q, in from which we can derive an inequality f the efficiency f the engine:! " #W Q in 1# T c "! Carnot, where! Carnot is the efficiency f a reversible engine based on a quasi-static Carnot cycle wking between the same pair of the heat reservoirs. 5. Example: a monatomic ideal gas. Suppose n mole of monatomic low-density gas is initially in a container of volume V and at temperature T. Through a process, the gas is brought to a final equilibrium state with V! and T!. During the process, the gas exchanges heat with a reservoir at temperature T r. We assume that the molar heat capacity at constant volume is constant and c v = ( 3 2)R. We will assume that T r is not equal to T n T! so that to make the heat exchange between the gas and the reservoir be reversible, we must somehow bring the temperature of the gas to T r by either adiabatically compressing expanding the gas. What is the maximum wk!w max that can be done by the gas? Using the maximum wk theem, we find! W "!#U + T r #S,

5 5 which leads to! W max =!"U + T r "S. The difference in the internal energy is given by!u = 3 2 nr ( T " # T) while the entropy difference is given by )#!S = nrln+ * + T " ( T ' # V " V, (. '-.. The maximum wk W max is then given by!w max =! 3 2 nr T "! T )# ( * + T ' ( ) + nrt ln + T " r # V ", (. V '-.. We can appreciate the power of the maximum wk theem by noting that without this theem, we must calculate the wk done by the gas in a series of reversible quasi-static processes connecting the initial and the final equilibrium state. F example, if V! > V and T > T! > T r, then we first expand the gas adiabatically so that its temperature becomes T r, then expand it isothermally allowing it to absb heat from the reservoir at T r and finally compress it adiabatically so that its temperature reaches T! gas in each of these three processes. We must then calculate the wk done by the

6 6 F the adiabatic process from T to T r, Q = 0, and using the first law, we find the wk done by the gas as! W 1 =!"U =! 3 2 nr ( T r! T). F the adiabatic process from T r to T!, Q = 0, and using the first law, we find the wk done by the gas as!w 3 =!"U =! 3 2 nr ( T #! T r). F the isothermal process at T r, we need to know the initial and the final volume, V! V!!. In the first adiabatic process, the volume increases from V to V! and therefe and TV 2/3 = T r V!! 2 / 3 V! = " T # T ' r V. In the last adiabatic process, the volume decreases from V!! to V! and therefe T! V! 2 / 3 = T r V!! " T! V!! = ' # T r 2/3 V!. The wk done by the gas during the isothermal process is then given by!w 2 = V " " # PdV = # nrt r V dv = nrt ln V "" ' * r ) = nrt V " ( r ln, V "" +, V "" V "" T "' ) T ( V "' - ) / V (./. The total amount of wk done by the gas is then given by!w max =!W 1! W 2! W 3 =! 3 2 nr T "! T )# ( * + T ' ( ) + nrt ln + T " r # V ", (. V '-.

7 7 as we have found by applying the maximum wk theem. With the maximum wk theem, all we need to know is the difference in the internal energy and the entropy difference between the initial and final equilibrium states: we do not need to know what happens to the system during the process how the system reaches the final state starting from the initial state.

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