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Entopy

Chapter 20

Entropy and the Second Law of Thermodynamics

Key contents: The arrow of time Entropy and the 2 nd law of thermodynamics Efficiency of engines Macrostates and microstates Probability and entropy

Entropy change of an irreversible process can be found with a reversible one connecting the initial and final states. free expansion an isothermal process

20.3 Change in Entropy: Entropy is a State Function Suppose that an ideal gas is taken through a reversible process, with the gas in an equilibrium state at the end of each step. For each small step, the energy transferred as heat to or from the gas is dQ, the work done by the gas is dW, and the change in internal energy is dEint. We have: Since the process is reversible, dW = p dV and dEint = nCV dT. Therefore, Using ideal gas law, we obtain: Integrating, Finally, The change in entropy  S between the initial and final states of an ideal gas depends only on properties of the initial and final states;  S does not depend on how the gas changes between the two states.

Example, Change of Entropy, Free Expansion of Gas: Suppose 1.0 mol of nitrogen gas is confined to the left side of the container of Fig. 20-1a. You open the stopcock, and the volume of the gas doubles. What is the entropy change of the gas for this irreversible process? Treat the gas as ideal. Calculations: From Table 19-4, the energy Q added as heat to the gas as it expands isothermally at temperature T from an initial volume Vi to a final volume Vf is Here n is the number of moles of gas present_._ The entropy change for this reversible process in which the temperature is held constant is

20.4 The Second Law of Thermodynamics If a process occurs in a closed system, the entropy of the system increases for irreversible processes and remains constant for reversible processes. It never decreases. Here the greater-than sign applies to irreversible processes and the equals sign to reversible processes. This relation applies only to closed systems. The reversible processes as dictated in a P-V diagram, however, can have any signs of entropy change since they describe only part of a closed system, which includes the environment.

20.5 Entropy in the Real World: Perfect Engines To have a ‘prefect’ engine, i.e., all the absorbed heat transferred to work, we require Q L=0. With the engine entropy change being zero, and the environment entropy change being the total entropy change for such an engine to work will be negative, violating the 2 nd law. The 2 nd law of thermodynamics can be stated as: No perfect engine! (The Kelvin-Planck statement)  S env   | QH | T H  0

20.6 Entropy in the Real World: Perfect Refrigerators The entropy change for the cold reservoir is -| Q|/TL, and that for the warm reservoir is +| Q|/TH. Thus, the net entropy change for the entire system is: TH >TL , and the right side of this equation is negative and thus the net change in entropy per cycle for the closed system refrigerator reservoirs is also negative. This violates the second law of thermodynamics, and therefore a perfect refrigerator does not exist. The 2 nd law of thermodynamics can be stated as: No perfect refrigerators! (The Clausius statement)

20.5 Entropy in the Real World: Carnot Engine Heat: Entropy Changes: Efficiency: The reverse of a Carnot engine is an ideal refrigerator, also called a Carnot refrigerator, whose efficiency, the coefficient of performance is

20.7 The Efficiencies of Real Engines Fig. 20-16 ( a) Engine X drives a Carnot refrigerator. ( b) If, as claimed, engine X is more efficient than a Carnot engine, then the combination shown in ( a) is equivalent to the perfect refrigerator shown here. This violates the second law of thermodynamics, so we conclude that engine X cannot be more efficient than a Carnot engine. Suppose there is an engine X, which has an efficiency  X that is greater than  C, the Carnot efficiency_._ When the engine X is coupled to a Carnot refrigerator, the work it requires per cycle may be made equal to that provided by engine X. Thus, no (external) work is performed on or by the combination engine +refrigerator, which we take as our system. We have the assumption , where the primed notation refers to the engine X. Therefore, which finally leads to : > 0 This shows that the net effect of engine X and the Carnot refrigerator working in combination is to transfer energy Q as heat from a low-temperature reservoir to a high-temperature reservoir without the requirement of work. This is a perfect refrigerator, whose existence is a violation of the second law of thermodynamics. Carnot’s theorem is proved!

Example, Impossible Engine:

20.8 A Statistical View of Entropy Macro-states and Micro-states Example: a two-phase-block case with 6 identical but distinguishable particles A certain set of numbers of particles in each phase block corresponds to a certain macrostate. {ab, cdef}and{ab, cedf} are the same macro and microstates. {ab, cdef}and{ac, bdef} are the same macro but different microstates. {abc,def}and{def,abc} are the same macro but different microstates. {ab, cdef}and {abcd,ef} are different macrostates. The thermodynamic probability of a macrostate is the corresponding number of microstates:

Every microstate is assumed to be equally probable.

Example, Microstates and multiplicity:

The arrow of time points to the direction of a state of more microstates, i.e., a more probable state. Sometimes it is called a more disordered state. Entropy is an extendable (i.e. additive) state variable. Thermodynamic probability is multiplicative. It’s likely that One may show in fact that the constant a is equal to the Boltzmann constant k , i.e., 20.8 A Statistical View of Entropy: Probability and Entropy Stirling’s approximation is often used for ln N! when N is large : ln N! ≈ N(ln N) - N (Stirling’s approximation). S  a ln W