Metallurgical Thermodynamics and Kinetics Lecture 8 PDF

Metallurgical Thermodynamics and Kinetics (MMC301) Dr. Manas Kumar Mondal Associate Professor Department of Metallurgical and Materials Engineering National Institute of Technology, Durgapur Disclaimer The study materials/presentations are solely meant for academic purposes and they can be reused, reproduced, modified, and distributed by others for academic purposes only with proper acknowledgements. SECOND LAW OF THERMODYNAMICS Lecture 8 References: 1. Stoichometry and thermodynamics of Metallurgical processes - Y K Rao. 2. Introduction to Metallurgical Thermodynamics – David R Gaskell. 5. Metallurgical Thermochemistry – O. Kubaschewski, E LL Evans and C B Alcock 3. Textbook of Materials and Metallurgical Thermodynamics – Ahindra Ghosh 4. Problems in Metallurgical Thermodynamics and Kinetics – G S Upadhyay and R K Dube. 5. Chemical Kinetics - Keith Laidler. Entropy change of the universe in an irreversible process Incorporation of an irreversible step into a Carnot cycle produces changes in the thermodynamic efficiency and the entropy of the universe. The conventional Carnot cycle consists of four reversible steps: Adiabatic compression (A), isothermal expansion (B), adiabatic expansion (C), and isothermal compression (D). We shall consider a modified cycle in which an irreversible step B’ is interposed between step B and C. The details of the modified cycle are shown in P-V diagram in Figure 1. There may be an increase in the entropy of the universe each time the working substance is taken through the modified cycle. For simplicity, 1 mole of ideal gas is considered as the working substance. Figure 1: Modified Carnot cycle with one Figure 2: (a) at the beginning of the irreversible free expansion (step B’) and irreversible step (B’) (b) at the completion of the irreversible free expansion (step B’) The additional process introduced here is an adiabatic free expansion this step is clearly irreversible. The practical details of how one might carry out step B in the context of the modified cycle are elaborated in Figure 2. Fig-2(a), at the completion of step B all the gas is on the right of this diagram that separates it from an evacuated chamber on the left. After the cylinder and piston are thoroughly insulated, suppose that an opportune moment the diaphragm is ruptured, where upon the gas rushes into the evacuated chamber. At the conclusion of step B, then this gas will occupy the entire space in the cylinder as shown in Figure 2(b). Because there is no change in the temperature of an ideal gas during an adiabatic free expansion at the completion of step B the thermodynamic state of the system is defined as (P3, V3 , T2). The remainder the modified cycle consists of two reversible adiabatic (A and C’) and two reversible isothermals ( B & D ). Step A : Heat absorbed QA = 0, Entropy change SA = 0 …………… (1) and work done WA = EA = CV(T1-T2) Step B : Heat absorbed QB = Q2 = RT2 ln(V3/V2) Entropy change SB = R ln(V3/V2) …………………(2) work done WB = Q2 = RT2 ln(V3/V2), Step B’ : Heat absorbed QB = 0, work done WB’ = 0 The entropy change can be calculated by assuming a reversible isothermal process between thermodynamic state 3 and 3′. Entropy change SB’ = R ln(V3’/V3) …………….. (3) Step C′ : Heat absorbed QC’ = 0, Entropy change SC’ = 0 …………… (4) and work done WC’ = EC’ = CV(T2 - T1) Step D : Heat absorbed QD = Q’1 = RT2 ln(V1/V’4 ) Entropy change SD = R ln(V1/V’4) …………………(5) work done WD = Q’1 = RT2 ln(V1/V’4), For the entire cyclic process, W  W i  W A  WB  WB  WC  WD  CV T1  T2   RT2 ln V3 V2   0  W T2  T1   RT1 ln V1 V4   RT2 lnV3 V2   RT1 ln V1 V4...........................................................(6). S S  S A  S B  S B   S C   S D  S s  0  R ln V3 V2   R ln V3 V3   0  R ln V1 V4  R ln V3 V2   R ln V1 V4   R ln V3 V3 .............................(7). The two reversible adiabatic steps, A and C′ ' 1 ' 1 TV  TV T1V1 1  T2V2 1 2 3 1 4  1 V   1 and V '  T1 T2   2   T1 T2   3  V '   V1   4  By considering this two reversible adiabatic steps, A and C’, we can show that    V2 V3 V3 V4   ....................8 V1 V  V2 V1 4     An substituting equation (8) into equation (7), we obtain S s  R ln V3   R ln V1   R ln V3    V    V3   V2   4    V V V   V V  R ln  3  1  3   R ln 1  3   V2 V  V3   V  V2   4   4  V V    V3 V4 S S  R ln  1  4   R ln 1  0........................(9) V4 V1   V2 V1 This result is entirely reasonable considering the fact that we are dealing with the entropy change in a cyclic in a process. By combining equations (8) and (6), we find  V3  V  Wnet  RT2 ln   RT1 ln 1   V2  V    4   V3  V   RT2 ln   RT1 ln 2   V2  V    3   V3  V V   RT2 ln   RT1 ln 2  3   V2   V3 V    3   V3   V3  V   RT2    RT1 ln   RT1 ln 3   V2   V2  V    3    V3  V     R ln  T2  T1   RT1 ln 3 ............................(9)  V2   V3     The efficiency of the modified cycle is given by  R lnV3 V2 T2  T1   RT1 lnV3 V3     Wnet Q2    RT2 ln V3 V2  T2  T1     RT1 ln(V3 V3 ) RT2 ln(V3 V2 ) T2   T1    1  RT1 lnV3 V3  RT2 ln V3 V2 .................................10  T2    Because V3 is greater than either V3 or V2 , the third term on the right hand side is a negative quantity. Hence, it follows that  is less than the Carnot efficiency. The diminution (decrease) in the efficiency of the modified cycle is directly linked to the presence of the irreversible step B’, the greater the departure of V3 from V3, the longer is decrease in efficiency. In order to determine the entropy change of the universe due to the modified cycle, we need to calculate the entropy change of the surroundings. The higher temperature reservoir “losses” an amount of thermal energy equal to Q2. The low-temperature reservoir “grains” an amount of heat equal to Q1’. Q2 V  Entropy change of high-temperature reservoir =    R ln 3 ..................11 T2  V2   V  Q Entropy change of the low- temperature reservoir =  1   R ln 1 ................................12  T1 V    4   V3  V  S l   R ln   R ln 1 ......................................13  V2  V    4  On combining equation (13) with equation (8), we find that  V3  V   V V  S l   R ln   R ln 2 ......................................14  as 3  4   V2  V    V2 V1   3   V  V V  S l   R ln 3   R ln 2  3   V2   V3   V3  V  V  V    R ln 3   R ln 3   R ln 3   V2   V2     V3  V    R ln 3   V3    Because V3’ is greater than V3, this quantity is positive the entropy change of the universe is given by V   S u  S l  R ln 3 .......................15  V3    It can be said that the presence of an irreversible step in an otherwise reversible cyclic process will reduce the efficiency of a heat engine operating on the cycle, furthermore, there will be an increase in the entropy of the universe each time the cycle is completed. Equation (9) and (15) can now be combined into a single expression Su  0 or dSu  0 ………….. (16) The clausius inequality The efficiency of a heat engine operating on a modified cyclic process containing an extra irreversible step is less than the Carnot-cycle efficiency.   RT1 lnV3 V3     T1     Q2  Q1    1     Q...........................17   T2  RT 2 ln V3 V  2  2      T Because V3’  V3  V2 it is clear that 1    Q  Q Q Q '  T  1 2 1 0 2  1  2  Q2 T2 T1 It is important to note that Q2 and Q1 are the heat exchanged reversibly by the system with the surroundings. Q2  Q2,r  heat absorbed reversibly by the system at T2  Q1  Q1,r  heat rejected reversibly by the system at T1 Q2,r Q1,r For the cyclic process containing are irreversible step   0   Qi ,r Ti  0...............(18) T2 T1 For the strict Carnot cycle consisting of four reversible step, Q i ,r Ti  0.............................19 Equations (18) and (19) can be combined to yield the following general expression: Q i ,r Ti  0.............................20 In other words, wherever a system is taken around a cycle consisting of several steps, the sum of the heat exchanged reversible Qi,r divided by the reservoir temperature, Ti for each step is less than equal to zero. This known as the Clausius in equality. Combined statement of first and second law The mathematical formulation of the first law of thermodynamics, in differential form, is as follows: Q  dE  W....................................1 In general, the work term W is composed of two parts:- Wm, the mechanical (or P-V-T) work and Wn, the non mechanical (i.e., magnetic, electrical etc) work, therefore Q  dE  Wm  Wn............................2  For a system for which only the mechanical work need to be considered, Wn = 0 and Q = dE + Wm ……… (3) The mechanical work done in a reversible process is Wm = P dV Equation (2) and (3) together yield Q  dE  PdV...........................4 The usefulness of the first law equation (4) is severely limited by the fact that is not a complete differential, that it is dependent on the both of the process. This short coming can be overcome by combining this relationship with the second law statement. According to the second law, for a reversible process spanning two equilibrium states, the entropy change , dS, is equal to the heat exchanged, Qr, divided by this temperature, T. Qr dS  Qr  TdS...............................5 T It is clear that for any reversible process involving only mechanical work, TdS  dE  PdV.............................6 For the general case in which several different terms of work are involved, TdS  dE  PdV  wn ,r.............................7  Equation (6) and (7) are formulations of the combined first and second laws for a P-V-T system and a generalized system, respectively. Some useful thermodynamic relationships (1) T and P are chosen as independent variables. TdS  dE  PdV dE  P   dS    dV T T  Considering E and V as function of T and P  E   E  dE    dP    dT............................................1 P  T  T P  V   V  dV    dP    dT............................................2  P T  T  P Substitution gives  E  1  E   P  V   V   dS    dP    dT     dP    dT  T   P  P  T  P  T  P  T  T P  1  E   V   1  E   V       P   dP     P   dT...........................3 T  P T  P T  T  T P   T P  Expressing S as a function of P and T  S   S  dS    dP    dT............................................4  P  T  T P Comparing the coefficient of dP and dT in equation (3) and (4)  S  1  E   V         P  ...........................................5  P T T  P T  T T   S  1  E   V         P  ...........................................6  T  P T  T  P  T  P  Differentiating equation (5) with respect to T at constant P, 2S 1  2E   2V   1   E   V      P    2     P  ..............................7  TP T  TP  T P   T   P  T  P T  Differentiating equation (6) with respect to P at constant T 2S 1  2E   2V   V      P     ..............................8 PT T  TP  T P   T P  The mixed second-order differentials S appearing on the left side of the equation (7) and (8) are equal. 1  2E   2V   1   E   V    1    E 2   2V   V      P      2     P       P      T  T P  T P   T   P  T  P  T  T   T P  T P   T P   1   E   V   1  V   0   2     P     T    P T  P T  T  T  P  1   E   V    1  V    0    2     P         T   P  T  P  T  T  T P   E  1  V    V      P        T  P  T   P  T   T P  E   V   V     T    P .................................9   P T   T P  P T  E   V   V     P   T  .................................10  P T  P T  T  P  S  1  E   V   From equation (5) and (10)       P   .........................................5 P  T T P  T  P T   S   V       ......................................11  P T  T  P From equation (6)  S  1  E   V         P    T  P T  T  P  T  P  1   E  PV  1  H       .............................12  T T  P T  T  P  V  1  H  As a result we have dS    dP    dT.....................................13  T  P T  T  P From the definition of CP , the heat capacity at constant pressure, we note that  1  H  CP      T  T  P T We can now define a quantity called the coefficient of thermal expansion, , as follows  1  V      .....................(14) V   T P Thus,  is the volume change per unit volume per degree change in temperature at constant pressure.  V     V  T P C  dS  VdP   P dT.......................................15  T   V  TdS  C P dT  T   dP........................................16  T P (2) T and V are chosen as independent variables TdS  dE  PdV.................................6  S   S  dS    dV    dT.......................................17   V T  T V Considering E and V as a function T and V  E   E   V   V  dE    dT    dV and dV    dT    dV  T V  V T  T V  V  T  1  E    1  E  dS      P dV     dT.................................18  T  V T   T  T V Because dV and dT are independent, comparison of the coefficient of these deferential yields.  S   1   E    S   1   E          ...................................................20         P ...................................................19  T V  T   T V   V T  T   V T  Differentiating equation (19) with respect to T at constant V, 2S 1   2 E  P   1  E           P ............................21 TV T  TV  T V  T 2  V T  Differentiating equation (20) with respect to V at constant T  2S 1 2E  ....................(22) VT T VT Setting equation (21) and (22) equal and cancelling the mixed second-order differentials,  1   E  P    1   E   1 2E 2   V       2     P . T T   T V  T V     T V T  T VT 1  P   1   E   0     2     P T  T V  T   V  T   E   P     P  T ............................................................23  V  T  T V  S   P  As a result    ..............................24  V T  T V  S   1  E  C   E   and        V As C V    .....................25  V T  T  T V T   T V   P  C  Put the value equation in (17) dS    dV   V dT..................................26   T V  T  The partial differential ⁄ can be expressed in terms of measurable quantities like the coefficient of thermal expansion and isothermal compressibility. Later its defined as follows 1  V    Isothermal Compressibility    ...............................27  V  P T  is defined on the change in volume per unit volume per unit change in pressure under isothermal conditions are dividing coefficient of thermal expansion  (equation 14) with isothermal compressibility  (equation (27))  1  V        V  T  P  1  V   V        V T T     P     P................................28   1  V   V        V  P T  P T Let us consider a constant volume process, for which equation (2) reduce to  V   V  dV    dPV    dTV  0...............................29  P T  T  P The subscript V indicates that the changes occur in a constant volume process. Rearranging term in equation (29).  V   V    dPV    dTV  P T  T  P  V    dPV  P   T  P    ....................................30 dTV  T V  V     P T  P   Comparing equation (28) and (30), we obtain   .................................31  T V  C   P  Substitute equation (31) into equation (27) dS   V dT    dV  T   T V C     V dT  dV..........................................32   T   T  TdS  CV dT  dV.........................................33  (3) P and V are chosen as independent variables Expressing S and E as functions of P and V, we can show that  S  1  E      .....................................34  P V T  P V  S  1  E  1       P.....................................35  V  P T  V  P T Adopting the same procedure as before, it can be shown that  S   C  T   C        V     V  ..........................36  P V  T  P V  T     S   C  T   C      V     V ..........................37   V  P  T  V  P  TV   T   T  TdS  CV   dP  C P   dV.................................38  P V  V  P Problem Calculate the coefficient of thermal expansion and isothermal compressibility for an (a) Ideal gas (b) a Van der Waals gas. By defining  1  V   1  V       and       V  T  P  V  P  T Solution (a) For 1 mole of an ideal gas, the equation of state is PV = RT  1  V        V  T  P  V  R 1 R  PV R 1    . T     T  P P V P  R PV T   R 1   VR T  V  RT    2  P T P Substitutions into the expression for  gives as  1  V       V   P T  1    RT       2  V   P  RT  PV R 1  R  T   VP 2  PV T  T 1   TP P  a  (b) For 1 mol of Van der Waals gas, the equation of state is  P  2 V  b   RT  V  Differentiating of the equation of state with respect V at constant P yields.  T      RTV 3 2a V  b  2   V  P RV 3 V  b  Again  V     1  V      RV 3 V  b    T  P  T    P RTV  2a V  b  3 2    T  V  P Substitute in equation of   1  V  1 RV 3 V  b          2   V  T  P V  RTV 3  2aV  b    RV 2 V  b     2   RTV 3  2 a V  b    P   To find the value of , we make use of equation     T V   P  R       V V V  b   V  b    R  RV 2 V   b  V  b  RTV 3  2 a V  b  2    V V  b  2 2 RTV 3  2a V  b  2 Difference between heat capacities CP and CV From first law consideration alone it can be shown that  E   V  C P  CV     P  ..............................................1  V T  T P  E   P  Again we derived    P  T   V  T  T V  P   V  C P  CV  T    ..............................................2  T V  T P  1  V  Again coefficient of thermal expansion       V  T  P  1  V  and isothermal compressibility       V  P  T  V  R     T  P P For 1mole of ideal gas PV = RT 1 R 1     V P T  V  RT    2  P  T P  1   RT  1         2   V   P  P  V   P      V and     T  P  T V   T 2V C P  CV  TV.    1 PV  1 1  C P  CV  TV .P     and   T2 T T P  PV  C P  CV   Rthe gas constant  T From the kinetic theory of the gases, we know that for a monatomic gas CV = 3/2 R and CP = 5/2 R with  = CP/CV = 1.667. The measured values of CP, CV and  for monatomic gases at room temperature are in good agreement with those values. For a diatomic gas, the values of CV, CP and  are 5/2 R, 7/2R and 1.40, respectively. Problem: Show that for an ideal gas the internal energy is solely function of temperature Solution: Because the internal energy E is a state function, it can be expressed as a function of state variables such as V and T.  E   E  E = f(V, T) and dE    dT    dV  T V  V T  E   P     P  T   V  T  T V  E   P     T  P  V  T  T V  P  R For 1 mole of an ideal gas, PV = RT. Differentiating at constant volume     T V V  E  TR  TR     P0  here  P   V T V  V   E    E   dE    dT  CV dT  as    CV E = f(T) for an ideal gas  T V   T V  Entropy of an ideal gas From useful thermodynamic correlation T and P are independent variables. Let us begin with equation according to the equation for 1 mole of an ideal gas  V  dT  V  TdS  C P dT  T   dP  dS  C P   dP.................................1  T P T  T P Let Sr, Pr and Tr denote the entropy, pressure and temperature of the substance in a reference state. Frequently Tr = 298 K and Pr = 1 atm defined the chosen reference state integration of equation yields. S T P  V  Sr dS  Tr P C d ln T  Pr  T  P dP T P  V   S  S r   CP d ln T     dP Pr T Tr  P P  V   dP.........................2 T  S  S r   CP d ln T    Pr T Tr  P Equation (2) is valid for any substance, including an ideal  V  R Equation for 1 mole of an ideal gas PV = RT, Integrating with constant pressure     T  P P T P  S  S r   C P d ln T   Rd ln P Tr Pr Because CP is independent of temperature for an ideal gas, T  P S  S r  CP ln   R ln .......................3  Tr   Pr  This gives the entropy as a function of temperature and pressure to calculate the entropy as a function of T and V. C    dS   V dT   dV  T    For an ideal gas,  = 1/T and  = 1/P, thus / = P/T = R/V [As PV = RT P/T = R/V] dV dS  CV d ln T  R  CV d ln T  R ln V V T  V  S  S r  CV ln   R ln ........................................4   Tr   Vr  Similarly, we can calculate the entropy as a function of P and V  T   T  TdS  CV   dP  C P   dV  P V  V  P CV  T  C  T  dS    dP  P   dV T  P V T  V  P C  C  V. dP  P dV T  TV C T CP  V  dP  dV T P 1  TV T  CV d ln p  C P d ln V P V  S  S r  CV ln   C P ln .................................6.97   Pr   Vr  Absolute value of entropy of a substance The absolute value for entropy, unlike that for enthalpy or internal energy can be determined. There are numerous stand order compilations, such as that by kubaschewski and Alcock, listing absolute entropy values for substances at a standard temperature of 298.15K (or 298⁰K, for short). The entropy change under isobaric condition the equation is expressed as C  dS  VdP   P dT  T  For isobaric dP = 0  CP   S 0  C dS   dT     P........................................1  T   T  P T CP  dS 0  dT T  CP  dT..............................................2 298 0 S 298  S 00    0 T  The degree sign denotes standard state at 1 atm. Here, S00 denotes the entropy of the substances at absolute zero. The quantity of the left side of equation (2) that is the difference between the entropies at 298⁰K and 0⁰K. It Can be determined by contracting a plot CP/T against T and then measuring the area under the curve. If we put S00= 0 CP dT..................................3 298 0 S 298  0 T At low temperature, specially near absolute zero, data on heat capacities are lacking for many substances. This lack of data is overcome by making extrapolations to lower temperature. In this regards, this following relationship for heat capacities CP has proved useful at low temperature. C P  aT 3  T  CV.........................................4  Where a and  are constants. Over this range of applicability of equation (4), the entropy ST0 at T can be obtained from equations (3) and (4). ST0   aT 3  T   dT    T  aT 3 S  0 T  T..............................................5 3 Over the range of validity of equation (4), because  is small, the entropy ST0 at TK tends to be less than CP at the same temperature. At low temperature, (but not T < 1K), can be neglected and T 3 CP S a 0 T ............................................6 3 3 With equation (6), the extrapolation down to 0⁰K is not satisfactory, because deviations occur from observations in the range T < 1K. The Helmholtz free energy and the Gibbs free energy Internal energy E and enthalpy ( or heat content) H are state functions, both of which have the dimensions of energy. We shall now define two additional state functions having the same dimensions. Helmholtz free energy, F: F = E  TS ……………………………… (1) Gibbs free energy, G: G = H  TS = E + PV TS …………………………………. (2) Both these function play on important part in thermodynamics. For an isothermal process, be it reversible or irreversible, the change in the Helmholtz free energy, F is given by F = E  T S …………………………… (3) Where , F = F2  F1, E= E2  E1 and S = S2  S1 , the subscripts 1 and 2 signify the initial and final states, respectively. Similarly the change in the Gibbs free energy of a system, in a isothermal process is G = H  TS ………………………….. (4) Here, G = G2  G1 , where G2 and G1 are the Gibbs free energies of the system in the initial and final states, respectively. From the first law, for any reversible or irreversible process, the total work done is given by W  Wm  Wn  Q  E..................................5 Where Wm is the mechanical work and the Wn is the nonmechanical (e.g., electrical) work. For any substance other than ideal gas, E is nonzero in an isothermal process. If E2 and E1 are the internal energies of the system in the initial and final states, then W  Q  E2  E1   Q  E.................................6 In equation (5) and (6), Q represents the amount of heat absorbed by the system from a single reservoir that is at a constant temperature of TK. Alternatively, the reservoir loses a quantity of heat equal to Q during the process. For the isothermal process, Entropy change of the system, SS = S2  S1 = S Energy change of the reservoir, Sl =  Q/T For any process Su  0 ……………. (7) Q Substitutions gives S u  0  S S  S l   0  S  0  T S  Q T Substitutions for Q from equation (6) WT  TS  E...............................8 The subscripts T signifies that the work done is for an isothermal process from equation (3) and (8) we have, WT   FT..............................................9 as F  E  TS   In this equation the equal sign holds for a reversible process and the inequality sign for irreversible process.  The decrease in the Helmholtz free energy in an isothermal process is greater than equal to the work done.  If the process is reversible, the work done is equal to the decrease in the Helmholtz free energy.  The work obtained in a reversible process is a maximum.  When the process is irreversible, the work obtained is less than the decrease in the Helmholtz free energy. When we consider an ideal gas undergoing an isothermal change, H = E = 0 and F = G =  TS For this particular case WT   G........................................10 Free-energy equations in differential form F  E  TS.........................11 On differentiating both sides of equation(11) dF  dF  TdS  SdT....................12 The combined statement of first and second laws is given by TdS  dE  PdV  Wn ,r............................12  Where Wn is the no mechanical work performed by the system, substituting for TdS in equation (12) dF  dE  dE  PdV  Wn ,r  SdT dF   PdV  SdT  Wn ,r........................................13 For a constant-volume, constant-temperature process, this reduces to dFV,T =  Wn,r or  FV,T = Wn,r ………….. (14) The non mechanical work preferred by the system at constant volume and constant temperature is equal to the decrease in the Helmholtz free energy. For a system in which no mechanical work is involved, the equation (13) simplifies to dF   PdV  SdT...........................................15 Differentiating equating (2) G  E  PV  TS dG  dE  PdV  VdP  TdS  SdT...................................16  Combining equation (12) and (16) dG  dE  PdV  VdP  dE  PdV  Wn ,r  SdT dG  VdP  SdT  Wn ,r............................17  For a constant temperature and constant-pressure, this equation becomes GP ,T  Wn ,r   GP ,T  Wn ,r.................................18  The decrease in the Gibbs free energy in a constant-pressure and constant-temperature process is equal to the amount of non mechanical work performed by the system. This is the maximum available work. Equations (14) and (18) apply for reversible processes. For irreversible process, these equation becomes Wn   FT ,V..........................................19  and Wn   GP ,T...........................................20  The maximum possible non mechanical work is obtained in a reversible process only. Free-energy change for an ideal gas (i) Isothermal process:- dF =  PdV For one mole of ideal gas, P = RT/V dF   RTd ln V  V1  ......................................21 V2 For the entire process F  F2  F1   V RTd ln V  F  RT ln 1  V2  On the other hand, the Gibbs free-energy change in isothermal process is given by RT dG  VdP , for 1 mole ideal gas V  P Substitution given dG  RT ln P P  G  G2  G1   RTd ln P  RT ln 2 .................................22 P2 P1  P1  P2 V1 For an ideal gas undergoing an isothermal process, P1V1  P2V2   P1 V2 V  F  G  RT ln 1   WT...........................23  V2  Where WT is the work done in the isothermal process. In an isothermal expression, V1 is smaller than V2, so that both F and G are negative in such a process. (II) Adiabatic process under reversible conditions: dF   PdV  SdT V2 T2 F    PdV  S  dT V1 T1 The first term on the right side is the work done in an adiabatic process. P1V1  P2V2  W   1 F  P2V2  P1V1   1  S T2  T1  Where  is the ratio of heat capacities W  P1V1  P2V2  /   1 for ideal gas P1V1 = RT1 and P2V2 = RT2  R  W    1  T1  T2   CV T1  T2  as R/CV =   1   F  CV T2  T1   S T2  T1   CV  S T2  T1 ...........................................24 The change in the Helmholtz free energy is directly proportional to the change in temperature. The Gibbs free- energy change for an ideal gas undergoing a reversible adiabatic process is given by dG  VdP  SdT For ideal gas PV  RT  PdV  VdP  RdT  VdP  RdT  PdV Substituting dG  RdT  PdV  SdT  dG   PdV  R  S dT G    PdV  R  S  dT V2 T2 Integrating it V1 T1 The first term right hand side is work done in an ideal adiabatic process. P1V1  P2V2   R  W    1   T1  T2   CV T1  T2     1  G  CV T2  T1   R  S T2  T1   T2  T1 CV  R  S  As C P  CV  R  T2  T1 C P  S .....................................25 Thermodynamic Potentials A closed P-V-T system with constant composition have two degrees of freedom. The change in the Helmholtz free energy and the Gibbs free energy in an infinitesimal process are given by dF   PdV  SdT................................26 dG  VdP  SdT................................27  So these we can add the equations giving the changes in the internal energy and enthalpy, using the first law and second law. dE  TdS  PdV...................................28 H EP dH  dE  PdV  VdP dH  dE  PdV  VdP  TdS  PdV  PdV  VdP  TdS  VdP.........................................29  Each of these equations has two terms on the right side corresponding to the two degrees of freedom of the system. Because E, F, G and H are functions of state, their differentials are exact. Each quantity can be expressed as a function of two state variables. Explicitly F  F V , T , G  G P, T , E  ES, V  and H  S, P  The exact differential dF is given by  F   F  dF    dV    dT.....................................30  V T  T V Combining equation (15) and (29) dF   PdV  SdT  F     P  V T  F      S....................................31  T V  G   G  Again, the exact differential dG is given by dG    dP    dT....................32  P T  T  P Comparing equation (32) and dG  VdP  SdT  G    V  P T  G      S.........................................33  T P  E   E  Again exact differential of dE dE    dS    dV.....................................34  S V  V  S  E   E  Comparing equation (29b) and (27) given    T     P..............................35  S V  V S  H   H  Again exact differential of H dH    dS    dP.....................................36  S P  P S Comparing equation (29c) with (28)  H    T  S  P  H     V.........................................37   P  S The equation F, G, E and F can be termed thermodynamic potentials, becomes as shown earlier, the partial derivative of these quantities yield the state properties such as P, S, V and T. EXTENSIVE AND INTENSIVE VARIABLES The various thermodynamic variables (in quantities) that are commonly used can be broadly divided into two categories, Extensive variables and Intensive variables. The extensive variables are those properties of the system that rang with its size or mass. The volume V, internal energy E, enthalpy H, entropy S, Helmholtz free energy, F and Gibbs free energy G of a substance of all dependent on its amount. For instance, a 10 g mass of solid copper at 0⁰C and 1 atm occupies a volume of 1.12 cm3 , whereas 1 kg of same substance has a volume 100 times as longer. The intensive variables are defined as those properties of the system that are independent of its amount. Temperature and pressure are two such variables, so are molar-specific quantities like molar volume V, molar internal energy E = E/n , molar enthalpy H = H/n , molar entropy S = S/n , molar Helmholtz free energy F = F/n, and, molar Gibbs free energy, G = G/n. For example, if the molar volume of solid copper at 0⁰C and 1atm is fixed at 63.54/8.9 or 7.14 cm3/g.mole, irrespective of the size of the lump of solid copper. It an intensive variable is known at are point in a system, it is known throughout. Knowledge of an extensive variable dependent's knowledge of the whole system. The Maxwell relations For closed P-V-T systems there are four valuable equations known as the Maxwell relation. These can be derived from the differential form of the thermodynamic potentials by equation (30) through (33).  F   F      P......................1a      S...................1b   V  T  T V  G   G     V......................2a      S...................2b   P T  T P  E   E     T......................3a      P...................3b   S V  V S  H   H  and    T.....................4a     V..................4b   S  P  P  S  E     P  V  T 2F  P  Differentiating with respect T at exact V   ...........................................34  TV  T V  F  and    S  T V 2F  S  Differentiating with respect V at exact T   ........................35 VT  V T  P   S  From equation (34) and (35)     .............................36 Maxwell relation.  T V  V T   2G   V  Differentiating equation (2a) with respect T at constant P     .............................2a.1  T P   T P   2G   S  Differentiating equation (2b) with respect P at constant T     .............................2b.1  TP   P  T   V   S From equation (2a.1) and (2b.1)  T    P .............................37 Maxwell relation.  P  T Differentiating equation (3a) with respect V at constant S   2 E   T      .............................3a.1  V S   V S  2E   P  Differentiating equation (3b) with respect S at constant V     .............................3a.2  VS   S V From equation (3a.1) and (3a.2)  T   P      .................................38 Maxwell relation.  V  S  S V Differentiating equation (4a) with respect P at constant S  2 H  T   .....................4a.1 PS  P  S   2 H   V  Differentiation equation (4b) with respect S at constant P     .............................4b.1  SP   S P  T   V  From equation (4a.1) and (4b.1)    ............................39 Maxwell relation.  P  S  S  P The cross product of the variables, either TS or PV, has the dimensions of energy. For systems with more than two independent variables, the number of Maxwell’s relation is correspondingly longer. Criteria of Equilibrium 0C ice  water 1 atm Mechanical (or pressure) equilibrium: When the pressure within a system is the same at all points, than the system is considered to be in a state of mechanical (or pressure) equilibrium. In such a system, pressure and velocity gradients are non-existent. Thermal equilibrium : A system in which the temperature is uniform throughout is said to be in thermal equilibrium. Chemical equilibrium : Chemical equilibrium is said to be exist when the species composing the system no linger a tendency to react. The absence of net reaction, when viewed in the light of reaction-rate theory, simply means that at chemical equilibrium the rate of forward reaction is exactly balanced by the rate of the reverse reaction, resulting is no net production or consumption of species. In a system that is in a state of chemical equilibrium, no net reaction can occur without external stimuli. State of complete equilibrium: When a system is in a state of complete (mechanical, thermal and chemical) equilibrium, it follows that all the parts or subsystems that compose it must also in equilibrium. Each subsystems examined separately must exhibit no pressure gradients, temperature differences, or net chemical reaction effects when the longer system itself in a state of complete equilibrium. The test for complete equilibrium consists in isolating the system and determining if any visible changes occur with time in its state properties. (1) A constant temperature and constant –pressure system: The system and the local surroundings, which previously labelled the universe, are also called the isolated system. At equilibrium, because a reversible condition exists, the entropy change for the isolated system is zero S i  0......................................1 Where the subscript i stand for the isolated system Figure: An isolated system consisting of the system and the local surrounding S i  S S  S l......................................(2) During the process, the system absorbs an amount of that equal to QS from the local surroundings. Heat lost by surroundings =  QS The exchange of this heat takes place at constant temperature, T. QS Sl  ..................................3 T Substitution into equation (2) yields QS  this is true only where no electrical  Si  S S ..................................4    T  or other type of mchanical work is being performed. For a constant-pressure process, QS = HS Where HS is the change in enthalpy for an isobaric, isothermal process, phase changes and chemical reactions included. TS S  H S  0 or GP ,T  0.......................................5 Where GP,T denotes the Gibbs free-energy change for the system along and does not refer to the local surroundings. For an infinitesimal process dGP,T = 0 ………………(6) or G is a minimum at equilibrium. Thus in equation (6) is a very simple criterion for equilibrium (or reversibility). Note: To use this criterion, it is sufficient if the pressure and temperature of the system at the start of a process are the same as at its completion. (2) A constant temperature and constant volume process: For a constant volume process Q = E S S T  E S 0 FV ,T  0................................7  For chemical transformation at constant temperature and fixed volume, the criterion for equilibrium is given by equation (7). Criteria of spontaneity (or irreversibility) The entropy change of the universe is positive in a spontaneous or irreversible process. Disadvantage: The entropy changes of the both the system and the surrounding must be computed to determine the spontaneity of the process. (1) A constant-temperature and constant-pressure process: For a spontaneous or irreversible change, entropy of the isolated system or the universe. S i  S u  0 or S S  S l  0............................8 During the process, heat is absorbed by the system from the local surroundings is QS. Heat lost by surroundings =  QS Figure: An isolated system consisting of the system and the local surrounding The entropy change of the surroundings Sl =  QS / T For an isobaric process, QS = HS H S S S  0    0....................................9  T GP ,T  0....................................10  For an infinitesimal process dGP,T < 0 ………………(11) dG P ,T  0....................................12  The equality sign is for equilibrium, and the inequality is for the spontaneous process. (2) A constant-temperature and constant-volume process: For a spontaneous change occurring at constant temperature and fixed volume FV ,T  0...............................13 The free –energy criteria are useful in chemical and physical thermodynamics. Problem: Show that 1 mole of a Van der Waals gas undergoing an isothermal process,  V  b   1 1  V2  b   1 1  1 1  F   RT ln  2   a    and G   RT ln    2 a   V V    bRT      V1  b   V2 V1   V1  b    2 1   V2  b  V1  b  Solution: dF   PdV  SdT For an isothermal change dF =  PdV The equation of state for 1 mole of Vander Waals gas  a   P  2 V  b   RT  V  a RT RT a P 2  P  2 V V b V b V  RT   a  dF    dV   2 dV V  b  V  V2 dV V2 dV F   RT   a V1 V  b V1 V 2  V2  b   1 1   RT ln    a     1V  b   V2 V1  dG  VdP  SdT For an isothermal change dG = VdP RT a P  2 V b V dV 2a dP   RT  dV V  b 2 V 3  RTVdV  2a  dG    2 dV V  b 2 V  V 1 b   V  b 2 V  b V  b 2  RTdV RTbdV  2a  dG     2 dV V  b V  b  V  2  V  b   1 1  1 1 G   RT ln  2   RTb     2 a     V1  b   V2  b  V1  b  V2 V1  The Gibbs-Helmholtz equation The Helmholtz free energy is a function of T and V.  F   F      P and    S  V T  T V F   F  From the definition of the Helmholtz free energy F  E  TS  E  T    F T  E  T V  T V Dividing both sides of this equation by  T2 ,   F  F 1  F  E   T       E  2    2  T T  T V T  T  T2   V 1  1      2 T T  T    F    T        E..........................................1   1     T   V This is one form of the Gibbs-Helmholtz equation. Gibbs-free energy function of T and P  G   G     V and    S  P  T  T P G  From the definition of the Gibbs free energy G  H  TS  H  T    T P  G   G T  H  T  P Dividing both sides of this equation by  T2 G 1  G  H       T 2 T  T  P T2   G    T   H       2  T  T    P   G    T        H.........................................2  This is the another form of the Gibbs-Helmholtz equation.   1     T   P Isothermal change of state at temperature T, such as a chemical reaction: A(P1,T) = B (P2, T) G1 and G2 denote the Gibbs free energies of the reactants and products, respectively. G is the Gibbs free-energy change for the isothermal process. G (P1, P2, T) = G2(P2, T)  G1(P1, T) Differentiation with respect to T at constant pressure yields G T  P1 , P2    G2    G1 ...................(3) T  P  T  P 2 1  G2   G      S 2 ,  1    S1  T  P2  T  P1 G T  P1 , P2   S 2  S1  S..................(4)  S  G  H ..................(5) T Combining equations (4) and (5) G T  P1 , P2  G  H   G  T G T T   P1 , P2  H  G T     1 T     P P 1, 2  H...............(6) G T   If P1 = P2 = P   1  T   H...............(6)  P This is the another form of the Gibbs-Helmholtz equation. From the Helmholtz free energy, the corresponding equation is F T  1T  V  E  F  T F T V  E.....(7) Problem: The Helmholtz free-energy change for 1 mole of a Van der Waals gas undergoing an isothermal process is given by F   RT V2  b  V1  b   a 1  1   V2 V1  Combine this with Gibbs-Helmholtz relationship to show that E   a 1  1   V2 V1  F   RT V2  b  V1  b   a 1  1  Solution:  V2 V1  Differentiating the for F gives with T at constant V  F      RV2  b  V1  b   T V As per Gibbs-Helmholtz relationship F  T F T V  E  RT V2  b  V1  b   a 1  1   RT V2  b  V1  b   E  V2 V1   E  a 1  1   V2 V1  The Gibbs free-energy changes in typical reversible processes (i) Fusion of a solid at its normal melting point, Tm: Iron melts at 1536C at 1 atm pressure. The fusion process absorbs 13810 J. This is a reversible process at constant T and P for which Hm0 = 13810 J/mole, and Sm0 = Hm0 /Tm = 7.634 J/mole.K For an isothermal process at a constant pressure of 1 atm G = H  TS For a fusion process G0m = 13810  7.634× 1809 = 0 (ii) Solidification of liquid at its normal solidification point: Liquid aluminium solidifies at 659C at 1 atm pressure. Accompanying the solidification process is a heat release of 10,460 J. Hsol0 =  10,460 J/mole, and Ssol0 = Hsol0 /Tsol =  10,460/932 =  11.2237 J/mole.K The Gibbs free-energy change for the solidification process is G0m = Hsol0  Tsol Ssol0 = 0 (iii) Vaporization of a liquid at its natural boiling point: Liquid magnesium vaporizes at 1105C = Tv at 1 atm pressure. During the vaporization process it absorbs a quantity of heat equal to 1,27,610 J/g.mole. Hv0 = 1,27,610 J/mole, Tv = 1378K and Sv0 = Hv0 /Tv= 1,27,610/1378 = 92.605 J/mole.K If vaporization occurs at constant pressure and temperature G0v = 1,27,610  1378× 92.605 = 0 (iv) Chemical reaction at equilibrium: Dissociation of calcium carbonate at 897C absorbs an amount of heat equal to 1,68,410 J. The entropy change for the process is 143.94 J/K. Hd0 = 1,68,410 J, Tv = 1170K and Sd0 = 143.94 J/K If isothermal dissociation occurs at 1 atm pressure at 897C. G0d = 1,68,410  1170× 143.94 = 0 The CaCO3-CaO-CO2 system is in a state of equilibrium at 897C and 1 atm pressure. The Gibbs free-energy changes in typical irreversible processes (i) Solidification of supercooled liquid silver: Normally, silver in the molten state solidifies on cooling at the normal solidification temperature of 960.5C. Sometimes it is possible to supercool ( i. e. by pass the normal solidification point) by rapid quenching. The liquid silver is supercooled at 940C. For solidification at 940C, the entropy change =  9.01 J/mole.K. The solidification temperature = 940 + 273 = 1213K. The heat released during the solidification of supercooled liquid silver is equal to 11,117 J/mole. Hs0 =  11,117 J/mole The Gibbs free energy change for any isothermal change at constant pressure G0 = H0  TS0 =  11,117 + 1213× 9.01 =  188 J/mole (ii) Irreversible chemical reaction: Calcium carbonate dissociates at 897C at 1 atm pressure. When CaCO3 and CaO are maintained in contact with CO2 at 1 atm pressure, no further dissociation of the CaCO3 occurs. That is the system is in state of equilibrium. The system CaCO3-CaO-CO2 is raised to and held at a temperature of 1050C, the pressure is held at 1 atm. The equilibrium CO2 pressure at this temperature is greater than 1 atm. By raising the temperature, while keeping the CO2 pressure constant at 1 atm, conditions are made conducive to dissociation. Hd0 = 168,410 J, Sd0 = 143.94 J/K At 1050C or 1323K G0d = 1,68,410  1323× 143.94 =  22,023 J < 0 The dissociation reaction is expected to produced spontaneously under these condition. The third law of thermodynamics  The third law of thermodynamics is concerned with the limiting behavior of the system in the vicinity of absolute zero.  The third law was first propounded by Nernst, who was influenced by the work of Richards.  The Gibbs free energy change, G for a reaction occurring isothermally is G = H -TS.  Richards found that for a given reaction, the values of G and H approach each other at low temperature.  Nernst postulated that only does G approach H in the vicinity of absolute zero but also the quantities (H/T)P and (G/T)P vanish as T approaches absolute zero. lim T 0G  H   0................(1)  The internal energy of a substance at absolute zero is not zero. lim  H  T 0   0................(2)  T  P lim  G  T 0   0................(3)  T  P Equations (1), (2) and (3) constitute the Nernst statement of third law. According to Kirchhoff’s law, equation (2) can be written as ?

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