This follows from the Clausius inequality as a consequence of the second law of thermodynamics. Therefore. (for reversible process only) (22.

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Walter Rich
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1 Entropy Clausius inequality can be used to analyze the cyclic process in a quantitative manner. The second law became a law of wider applicability when Clausius introduced the property called entropy. By evaluating the entropy change, one can explain as to why spontaneous processes occur only in one direction. Figure 22.1 Consider a system in initial state 1. Let the system be taken from state 1 to state 2 along a reversible path 1-A-2, and then be restored to its initial state by following another reversible path 2BI (Figure 22.1). Then the two paths put together form a reversible cycle 1A2BI. Apply the Clausius inequality to this reversible cycle and obtain (22.1) (22.2) (22.3) 1

2 Since path 2B1 is reversible, the limits of the integral can be reversed. That is, has the same value whether the path followed is 1A2 or 1B2. It is possible to connect the states 1 and 2 by several reversible paths and see that has the same value irrespective of the path as long as the paths are reversible. Therefore is an exact differential of some function which we identify as entropy. Hence it can be said that there exists a function S, called entropy, the change in entropy is expressed as This follows from the Clausius inequality as a consequence of the second law of thermodynamics. Therefore Calculation of Entropy Change (22.4) (for reversible process only) (22.5) The following facts should be kept in mind while calculating the change in entropy for a process 1. for a reversible process 2. Entropy is a state function. The entropy change of a system is determined by its initial and final states only, irrespective of how the system has changed its state. 3. In analyzing irreversible processes, it is not necessary to make a direct analysis of the actual process. One can substitute the actual process by a reversible process connecting the final state to the initial state, and the entropy change for the imaginary reversible process can be evaluated. Entropy change for some elementary processes a. Absorption of energy by a constant temperature reservoir A certain amount of heat is added to a constant temperature reservoir. The actual process can be replaced by a reversible path in which an equivalent amount of energy is added to the reservoir. Then, the entropy change of the reservoir is given by b. Heating or Cooling of matter (22.6) 2

3 The heating can be carried out either at constant pressure or at constant volume. From the first law of thermodynamics for constant volume heating/cooling process for constant pressure heating/cooling process, for a constant pressure process (22.7), for a constant pressure process (22.8) Similarly,, for a constant volume process (22.9) ln, for a constant volume process (22.10) (c) Phase change at constant temperature and pressure Melting : [ solid to liquid] Evaporation: [ liquid to vapor] (d) Change of state for an ideal gas If an ideal gas undergoes a change of state from to (22.11) 3

4 (22.12) (22.13) ln (22.14) Again, (22.15) (22.16) (22.17) (22.18) 4

ln ln (22.19) For a constant temperature process is either ln ln. (e) Mixing of non-identical gases A rigid insulated container is divided into two compartments by a partition. Refer to Figure 22.2. One compartment contains Figure 22.

The partition is removed and the gases are allowed to mix. To evaluate the entropy change on mixing, let us device the following reversible process.

5 ln ln (22.19) For a constant temperature process is either ln ln. (e) Mixing of non-identical gases A rigid insulated container is divided into two compartments by a partition. Refer to Figure One compartment contains Figure 22.2 mol of an ideal gas A at pressure P and temperature T while the second compartment contains mol of an ideal gas B at the same pressure and temperature. The partition is removed and the gases are allowed to mix. To evaluate the entropy change on mixing, let us device the following reversible process. Let the gases A and B expand reversible and isothermally to their final partial pressures respectively. Then the gases are introduced through semi-permeable membranes into a container in which the partial pressure of gas A is and the partial pressure of gas B is. The reversible mixing process is, therefore, accomplished in the following two steps. Reversible isothermal expansion of gases to their respective partial pressures. Forcing the gases reversibility through semi permeable membranes. For gas A, and 5

6 ln (22.20) Similarly, for gas B, ln (22.21) Where, and Here and are the mole fractions of gases A and B respectively, in the final mixture. Hence, ln ln and (22.22) Therefore, the entropy change associated with the mixing of non-identical gases is In - R In (22.23) In In (22.24) In (22.25) Where is the molar entropy. Since, is always positive. 6

Principle of Entropy Increase Figure 23.1 Refer to Figure 23.1. Let a system change from state 1 to state 2 by a reversible process A and return to state 1 by another reversible process B.

7 Principle of Entropy Increase Figure 23.1 Refer to Figure Let a system change from state 1 to state 2 by a reversible process A and return to state 1 by another reversible process B. Then 1A2B1 is a reversible cycle. Therefore, the Clausius inequality gives: (23.1) If the system is restored to the initial state 1 from state 2 by an irreversible process C, then 1A2C1 is an irreversible cycle. Then the Clausius inequality gives: (23.2) Subtracting (23.2) from (23,1) (23.3) Since the process 2B1 is reversible (23.4) 7

8 (23.5) This can be generalized as (23.6) Where the equality sign holds good for a reversible process and the inequality sign holds good for an irreversible process. Now let us apply the above results to evaluate the entropy of the universe when a system interacts with the surroundings. Refer to Figure Figure 23.2 Let Temperature of the surroundings Temperature of the system and, (23.7) 8

9 If the energy exchange takes place, will be the energy transfer from the surroundings to the system. (23.8) (23.9) (23.10) [since ] (23.11) So, (23.12) If the system is isolated, that is, when there is no interaction between the system and the surroundings, then (23.13) Now let us redefine a system which includes our earlier system and its surroundings. Then for an isolated system (23.14) Through generalization we can write for an isolated system (23.15) Examine the statement critically. For a reversible process it is equal to zero. For an irreversible process it is greater than zero. 9

Similarly it is possible to write. The equality sign holds good if the process undergone is reversible; the inequality sign holds good if the process undergone is irreversible.

10 Similarly it is possible to write. The equality sign holds good if the process undergone is reversible; the inequality sign holds good if the process undergone is irreversible. Clausius summarized these as: Die Energie der Welt ist konstant. Die Entropie der Welt strebt einem Maximum. This means: The energy of the universe in constant (first law). The entropy of the universe tends towards a maximum (second law) Temperature Entropy Diagram Entropy change of a system is given by. During the reversible process, the energy transfer as heat to the system from the surroundings is given by (24.1) Figure

11 Refer to figure Here T and S are chosen as independent variables. The is the area under the curve. The first law of thermodynamics gives. Also for a reversible process, we can write, and (24.2) Therefore, (24.3) For a cyclic process, the above equation reduces to (24.4) For a cyclic process, the above equation reduces to For a cyclic process represents the net heat interaction which is equal to the net work done by the system. Hence the area enclosed by a cycle on a T S diagram represents the net work done by a system. For a reversible adiabatic process, we know that (24.5) (24.6) (24.7) Hence a reversible adiabatic process is also called an isentropic process. On a T S diagram, the Carnot cycle can be represented as shown in Fig The area under the curve 1-2 represents the energy absorbed as heat by the system during the isothermal process. The area under the curve 3-4 is the energy rejected as heat by the system. The shaded area represents the net work done by the system. We have already seen that the efficiency of a Carnot cycle operating between two thermal reservoirs at temperatures T 1 and T 2 is given by 11

12 (24.8) This was derived assuming the working fluid to be an ideal gas. The advantage of T S diagram can be realized by a presentation of the Carnot cycle on the T S diagram. Let the system change its entropy from to during the isothermal expansion process 1-2. Then, and, and, (24.9) (24.10) (24.11) This demonstrates the utility of T S diagram. Second law analysis of a control volume It was shown that the change in entropy of a system is given by (24.12) Where the equality sign holds good for the reversible processes and the inequality sign holds for all irreversible processes. This can be expressed as (24.13) 12

13 Where represents entropy generation in the system and it cannot take a negative value. for a reversible process. for an irreversible process. Consider a control volume through which material flows (Figure 24.2) Figure 24.2 Let us identify a system such that its mass remains constant during a given process. At time t, the system constitutes both the mass inside the control volume and the mass about the enter the control volume during a small interval dt (that is, the mass occupying the region A). At time (t+dt), the system is the mass inside the control volume as well as the mass in region B. During the time interval dt, the system undergoes a change in configuration and receives energy as heat from the surroundings. At time t, the entropy of the system (24.14) At time, the entropy of the system (24.15) Then, (24.16) 13

14 (24.17) (24.18) This can be rearranged as (24.19) In words: The rate of accumulation of entropy = Rate of inflow of entropy Rate of outflow of entropy + Rate of generation of entropy TdS Equations For a closed system contaning a pure compressible substance undergoing a reversible process. This is the famous Gibbsian equation. (24.22) or (per unit mass) (24.23) 14

15 Eliminate du by using the definition of enthalpy (24.24) Thus, (24.25) (24.26) Also, (24.27) Important: These equations relate the entropy change of a system to the changes in other properties :. Therefore, they are independent of the process. These relations can be used for reversible as well as irreversible processes. Example Problem-1 Consider steam is undergoing a phase transition from liquid to vapor at a constant temperature of 20 0 C. Determine the entropy change using the Gibbsian equations and compare the value to that read directly from the thermodynamic table. Corresponding to change from liquid to vapor 15

16 From table: It compares favourably with the tabulated value Entropy change of an incompressible substance For most liquids and all solids the density is not changed as pressure changes, that is,. Gibbsian equation states that,, for an incompressible substance is a function of temperature only. Therefore, Integrate to determine the entropy change during a process where is the averaged specific heat of the substance over the given temperature range. Specific heat for some common liquids and solids can be found in thermodynamic tables. Example Problem 2 An 1-kg metal bar initially at 1000 K is removed from an oven and quenched by immersing in a closed tank containing 20 kg of water initially at 300 K. Assume both substance are incompressible and c (water) = 4 (kj/kg K), c (metal) = 0.4 (kj/kg K). Neglect heat transfer between the tank and its surroundings. (a) Determine the final temperature of the metal bar, (b) entropy generation during the process. Solution 16

17 a. Energy balance from the first law: no heat transfer and no work done, both bar and water reach final temperature. b. No heat transfer with surroundings. The entropy balance of the system (generation) = In In In In ( kj / K ) The total entropy of the system increases, thus satisfies the second law. Example Problem 3 Steam enters an adiabatic turbine at 5 MPa and C and leaves at a pressure of 1.4 MPa. Determine the work output of the turbine per unit mass flowing through the turbine if we can assume the process is reversible and neglect all changes of KE and PE. Refer to Figure

18 Figure 24.3 State 1: F From table: State 2: at 1.4M Pa (from table) State 2 is superheated. From table, 18

IS A PARTICULAR PROCESS / REACTION FEASIBLE? TO WHAT EXTENT DOES THE PROCESS / REACTION PROCEED?

IS A PARTICULAR PROCESS / REACTION FEASIBLE? TO WHAT EXTENT DOES THE PROCESS / REACTION PROCEED? Limitations of First Law of Thermodynamics The first law of thermodynamics is a law of conservation of energy. It does not specify the direction of the process. All spontaneous processes processed in one