Thermodynamics II

NEED FOR THE SECOND LAW OF THERMODYNAMICS - STATEMENT - CARNOT CYCLE - REFRIGERATOR CONCEPT - CONCEPT OF ENTROPY - FREE ENERGY FUNCTIONS - GIBB'S HELMHOLTZ EQUATIONS - MAXEWELL'S RELATIONS - THERMODYNAMICS EQUATION OF STATE - CRITERIA OF SPONTANITY - CHEMICAL POTENTIAL - GIBB'S DUHEM EQUATION

1. Dr.P.GOVINDARAJ   Associate Professor & Head , Department of Chemistry SAIVA BHANU KSHATRIYA COLLEGE ARUPPUKOTTAI - 626101 Virudhunagar District, Tamil Nadu, India THERMODYNAMICS II

2. THERMODYNAMICS Need for second law : To determine the direction in which energy can be transferred To predict whether a given process can occur spontaneously To convert maximum fraction of heat into work in a given process Second law helps

3. It is a process in which a system can change from one state to another state rapidly (at a definite rate) when the difference between external pressure (opposing pressure) and internal pressure is maximum (pressure of the gas) Spontaneous reaction : If P ext > > P int THERMODYNAMICS

4. Water flows downhill spontaneously Heat flows spontaneously from the hot end to the cold end A gas expands spontaneously from a region of high pressure to the region of low pressure Examples: THERMODYNAMICS

5. It is impossible to convert heat into work without compensation Statement of II law of thermodynamics : Statement I: It is impossible to transfer heat from a cooler body to a hotter body without external agency Statement II : The energy of the universe remains constant and the entropy of the universe tends towards a maximum Statement III : It is impossible to use a cyclic process to transfer heat from a reservoir (source) and to convert it into work without transferring a certain amount of heat from a hotter to a colder part of the body Statement IV : THERMODYNAMICS

6. It is the process in which a system can undergo a series of change and return to original state i.e., for cyclic process the initial and final state are same so , ∆E = 0 and q = ∆E + W q = 0 + W q = W Cyclic process: For cyclic process THERMODYNAMICS

7. “Carnot cycle is a reversible cycle process involved in four on a system containing one mole of an ideal gas in order to demonstrate the maximum convertibility of heat into work” Carnot cycle: Graphical representation of C arnot cycle Definition: q 2 q 1 THERMODYNAMICS

8. The gas expand reversibly and isothermally from state A (P 1 V 1 ) to the state B (P 2 V 2 ) by absorbing q 2 amount of heat from the source at constant temperature T 2 According to first law of thermodynamics q 2 = ∆E + W 1 Since ∆E =0 for isothermal expansion q 2 = W 1 Step 1: Isothermal expansion W 1 = q 2 = RT 2 ln ----------(1)   THERMODYNAMICS

9. Then the gas expand reversibly and adiabatically from state B (P 2 V 2 ) to the state C (P 3 V 3 ) by decreasing the temperature from T 2 to T 1 . Let W 2 be the work done by the system. According to first law of thermodynamics q 2 = ∆E + W 1 Since q =0 for adiabatic expansion 0 = ∆ E + W 2 W 2 = - ∆ E W 2 = - C v ∆T W 2 = - C v (T 1 – T 2 ) W 2 = C v (T 2 – T 1 ) ---------(2) Step 2: Adiabatic expansion THERMODYNAMICS

10. Then the gas undergo isothermal and reversible compression from state C (P 3 V 3 ) to the state D (P 4 V 4 ) by given up q 1 amount of heat to the sink at constant temperature T 1 .Let W 3 be the work done on the system and the sign of W 3 is negative. According to first law of thermodynamics q 1 = ∆E + (- W 3 ) Since ∆E =0 for isothermal process q 1 = -W 3 = RT 1 ln   Step 3: Isothermal compression W 3 = q 1 = RT 1 ln ----------(3)   THERMODYNAMICS

11. Finally, the gas arrived to original state A (P 1 V 1 ) from the state D (P 4 V 4 ) by adiabatic reversible compression by increasing the temperature from T 1 to T 2 . Let W 4 be the work done on the system and its sign is negative . According to first law of thermodynamics q = ∆E + (-W 1 ) Since q =0 for adiabatic process 0 = ∆ E – W 4 W 2 = ∆E W 2 = C v ∆T = C v (T 2 – T 1 ) W 2 = C v (T 2 – T 1 ) ---------(4) Step 4: Adiabatic compression THERMODYNAMICS

12. The net heat absorbed is (q) q = q 2 – q 1 q = RT 2 ln - RT 1 ln Since = q = R (T 2 – T 1 ) ln ----------(5) The net work done is (W) W = (W 1 + W 2 ) – (W 3 +W 4 ) W = RT 2 ln + C v (T 2 – T 1 ) – RT 1 ln – C v (T 2 – T 1 )   Calculation THERMODYNAMICS

13. W = RT 2 ln + C v (T 2 – T 1 ) –RT 1 ln – C v (T 2 – T 1 ) Since = W = R (T 2 – T 1 ) ln ---------(6) We know that q 2 = RT 2 ln ---------(7) = = W = q 2   THERMODYNAMICS

14. Since < 1, it shows that W < q 2 , i.e., a part of the heat absorbed by the system at the higher temperature T 2 is transferred into work and the rest of the heat (q 1 ) is given out by the system to the surrounding (sink at lower temperature)   W = q 2   THERMODYNAMICS

15. Carnot theorem : Engine 1 Engine 2 Engine 3 Gas 1 Gas 2 Gas 3 Source T 2 Source T 2 Source T 2 Sink T 1 Sink T 1 Sink T 1 Efficiency of engine 1 =   Efficiency of engine 2 =   Efficiency of engine 3 =   “Carnot theorem states that every perfect engine working reversibly between the same temperature limits has the same efficiency whatever be the working substance THERMODYNAMICS

16. Thermodynamics efficiency for Carnot engine : “It is defined as the fraction of heat absorbed by the C arnot engine used to do work” Mathematically  = =  = ---------(1) Substitute the value for q 1 and q 2 in equation (1) we get  =  =   THERMODYNAMICS

17.  = =  = So,  = = From the above equation, it is found that The efficiency of an actual engine is always less than unit When T 1 = 0 , the efficiency become unity i.e., the engine is 100% efficiency engine, but the absolute zero (T 1 = 0) cannot be arrived in actual practise . That means 100 % efficiency machine is an ideal machine   THERMODYNAMICS

18. The reverse of C arnot cycle is called refrigerator cycle and is diagrammatically shown Refrigerator cycle : THERMODYNAMICS

19. The net effect of the cycle ABCD is q 1 = RT 1 ln of heat will be removed from the sink which is at a lower temperature T 1 K q 2 = RT 1 ln of heat will be given to the source at high temperature T 2 K Net work W = R (T 2 – T 1 ) ln ln = (q 2 – q 1 ) The co-efficient of performance of refrigerator is  = =  = [ Since =  = =   THERMODYNAMICS

20. Derivation of the concept of entropy: For a C arnot cycle, the efficiency is  = = ---------(1) 1- = 1- = = ---------(2) q 2 ------ is the heat absorbed and its sign is (+) ve q 1 ------ is the heat given out and its sign is (-) ve   THERMODYNAMICS

21. Now equation (2) becomes = + = 0 ------------(3) i.e., for a single C arnot cycle + = 0 Like wise for a reversible cycle ABA containing a series of C arnot cycle equation (3) becomes + + + ….. = 0  = 0 -------------(4)   THERMODYNAMICS

22. When the changes are infinitesimally, equation (4) becomes  = 0 -------------(5) Since the cycle process is carried out in 2 steps Step I = A to B Step II = B to A Equation (5) can be written as  = = 0 = 0 =   THERMODYNAMICS

23. From this it was found that the quantity is independent of the path and depends upon the initial and final states of the system. So that is a state function called entropy change (∆S) = S B – S A ---------(6) where S B – entropy of the state B S A – entropy of the state A   THERMODYNAMICS

24. Relationship between T & V for adiabatic reversible expansion : We know that, for adiabatic reversible expansion W = - ∆E ∆E = -W ----------(1) Let P is the external pressure ∆V is the increase in volume, then Work done by the system is W = P ∆V -----------(2) Substitute equation (2) in (1) we get ∆ E = - P∆V THERMODYNAMICS

25. Substitute (2) in (1) we get ∆ E = - P∆ V ----------(3) If ∆ T is the change in temperature then ∆E = C v ∆T ----------(4) Equating (3) & (4) C v ∆T = - P ∆V For infinitesimally small change, C v dT = - PdV C v dT = - (For 1 mole of the gas, PV = RT) = - ----------(5)   THERMODYNAMICS

26. Integrate the equation (5) with in the limits C v = - R C v ln = - R ln V C v ln = - R ln C v ln = R ln ln = ln Since C P – C V = R , we get ln = ln   THERMODYNAMICS

27. Since =  , we get ln = (  - 1) ln ln = ln (  - 1) = (  - 1) (or) T 2 V 2 (  - 1) = T 1 V 1 (  - 1)   This is the relationship between T & V for adiabatic reversible expansion THERMODYNAMICS

28. Relationship between T & P for adiabatic reversible expansion : We know that = (  - 1 ) ----------(1) P 1 V 1 = RT 1 ---------(2) P 2 V 2 = RT 2 ---------(3) → = ---------(4) = = x ---------(5)   THERMODYNAMICS

29. Substitute equation (5) in (1), we get = x (  - 1) = (  - 1 ) x (  - 1) (  - 1 ) = (  - 1) (1+  - 1) = (  - 1)  = (  - 1)  = (1- ) (or) 1   THERMODYNAMICS

30. Variation of entropy with temperature and volume : We know that for an infinitesimally small change for one mole of an ideal gas ds = --------(1) According to the first law of thermodynamics = dE + PdV --------(2) We know that C v = dE = C v dT ---------(3)   THERMODYNAMICS

31. Substitute equation (3) in equation (2) = C v dT + PdV --------(4) For one mole of an ideal gas PV = RT P = --------(5) Substitute equation (5) in equation (4) = C v dT + dV = C v dT + RT --------(6)   THERMODYNAMICS

32. Divide equation (6) by T = + = + R ----------(7) Since = dS , equation (7) becomes dS = + R Integrating the above equation with in the limits T 1 and T 2 , V 1 and V 2 = + R   THERMODYNAMICS

33. = + R S 2 – S 1 = ln + R ln ∆S = ln + R ln   Variation of entropy with temperature and pressure : We know that P 1 V 1 = R T 1 P 2 V 2 = R T 2 Dividing these equation we get = --------- (1) ∆S = ln + R ln ---------(2)   THERMODYNAMICS

34. Substitute (1) in (2) we get ∆S = ln + R ln ∆ S = ln + R ln + R ln ∆S = ln + R ln + R ln ∆S = ln + R ln - R ln ∆S = + R) ln - R ln ∆S = ln - R ln   THERMODYNAMICS

35. So for ‘n’ moles of the ideal gas ∆ S = ln + n R ln and ∆S = ln - n R ln   THERMODYNAMICS

36. Free energy function: There are two free energy function 1. Helmholtz free energy (A) 2. Gibbs free energy (G) Helmholtz free energy (A) Definition : It is defined as the isothermally available internal energy of the system. M athematically defined as A = E - TS THERMODYNAMICS

37. Variation of Helmholtz free energy with temperature and volume : We know that A = E – TS --------(1) On differentiating the equation (1) dA = dE – TdS – SdT --------(2) Put dE = dq – PdV in equation (2) we get dA = dq – PdV – TdS – SdT --------(3) According to entropy concept dS = dq = T dS --------(4)   THERMODYNAMICS

38. Substitute (4) in (3) we get dA = - PdV – SdT -------(5) This equation gives the dependence of G of a system on T & V When the process is carried out at constant volume i.e., dV = 0 equation (5) becomes dA = - S dT v = - S i.e., The rate of change of Helmholtz free energy with temperature is the measure of Entropy in isochoric processes   THERMODYNAMICS

39. Gibbs free energy : Definition: It is defined as the isothermally available enthalpy of the system Mathematically, G = H – TS Variation of Gibbs free energy with Temperature & Pressure: We know that G = H – TS ---------(1) Differentiating the equation (1) we get dG = dH – TdS – SdT ---------(2) THERMODYNAMICS

40. We know that H = E + PV dH = dE + P dV + V dP ---------(3) and also dq = dE + PdV ---------(4) Since dS = , equation (4) becomes TdS = dE + PdV ---------(5) Substitute equation (5) & (3) in (2) we get dG = dE + P dV + V dP – dE – PdV – SdT dG = V dP – SdT This equation gives the dependence of G of a system on T & P   THERMODYNAMICS

41. Gibbs – Helmholtz equation : Consider an isothermal process A B Let G A is the Gibbs free energy for A (initial state) G B is the Gibbs free energy for B (Final state ) ∆G = G B – G A ---------(1) We know that G = H – TS So that G A = H A – TS A G B = H B – TS B At constant T & P ---------(2) THERMODYNAMICS

42. Substitute equation (2) in (1) ∆ G = H B – TS B – H A + TS A ∆ G = H B – H A – T (S B – S A ) ---------(3) Since ∆H = H B – H A , ∆S = S B – S A , equation (3) becomes ∆ G = ∆ H – T ∆ S ---------( 4) Let small rise in temperature from T to T + dT on the initial state and final state make small rise in Gibbs free energy from G A to G A + dG A and G B to G B + dG B for initial and final state A A’ B B’ G A , T G A + dG A , T + dT G B , T G B + dG B , T + dT Infinitesimally small change Infinitesimally small change THERMODYNAMICS

43. We know that dG = VdP – SdT At constant pressure for all processes dP = 0 dG = - S dT So dG A = - S A dT ---------(5) dG B = - S B dT ---------(6) (5) – (4) gives dG B - dG A = - S B dT + S A dT d(G B – G A ) = - (S B - S A ) dT d(∆G) = - ∆ SdT P = - ∆ S ---------(7)   THERMODYNAMICS

44. Substitute equation (7) in (4) we get ∆G = ∆ H + T P --------(8) Similarly ∆A = ∆E + T V --------(9) Equation (8) & (9) are known as Gibbs – Helmholtz equation   THERMODYNAMICS

45. Variation of internal energy with S & V : We know that dq = dE + dW ----------(1) Since dW = PdV , equation (1) becomes dq = dE + PdV ----------(2) Since dS = , dq = TdS , equation (2) becomes TdS = dE + PdV dE = TdS – PdV --------- (3)   THERMODYNAMICS

46. Variation of enthalpy with S & P : We know that H = E + PV -------------(4) Differentiating on both side, we get dH = dE + PdV + VdP ------------- (5) Substitute (3) in (5) dH = TdS – PdV + PdV + VdP dH = TdS + VdP -------------(6) THERMODYNAMICS

47. We know that dE = TdS - PdV -----------(1) dH = TdS + VdP -----------(2) dA = - SdT - PdV -----------(3) dG = - SdT + VdP -----------(4) Maxwell relationship from equation (1) At constant volume, dV = 0, equation (1) becomes dE = T dS V = T ------------(5)   Maxwell’s relationship: THERMODYNAMICS

48. At constant entropy , dS = 0, equation (1) becomes dE = -P dV S = -P ------------(6) Differentiating equation (5) with respect to volume at constant entropy , we get = S ------------(7) Differentiating equation (6) with respect to entropy at constant volume, we get = - V ------------(8)   THERMODYNAMICS

49. From equation (7) and (8) it is found that S = - V ------------(9) Maxwell’s relationship from equation (2) dH = TdS + VdP ------------(2) At constant entropy, dS = 0, equation (2) becomes dH = V dP S = V -----------(10)   THERMODYNAMICS

50. At constant pressure , dP = 0, equation (10) becomes dH = T dS P = T ------------(11) Differentiating equation (10) with respect to entropy at constant pressure , we get = P ------------(12) Differentiating equation (11) with respect to pressure at constant entropy , we get = V ------------(13)   THERMODYNAMICS

51. From equation (12) and (13) it is found that P = S ------------(14) Similarly we have arrived from equation (3) and (4) as T = v ------------(15) T = - P ------------(16) Equation (9), (14), (15) & (16) are Maxwell’s relationships   THERMODYNAMICS

52. THERMODYNAMICS EQUATIONS OF STATE Thermodynamic equation relating state variables which describe the state of matter under a given set of physical conditions, such as pressure, volume, temperature (or) internal energy (or ) enthalpy. Derivation of thermodynamic equations of state: we know that dE = TdS – PdV ------------(1) Divide on both side with dV , we get = T - P ------------(2)   THERMODYNAMICS

53. At constant temperature, equation (2) becomes T = T T - P ------------(3) We know that one of the Maxwell’s relationship is T = v ------------( 4 ) Substitute equation (4) in (3) we get T = T V - P ------------(5) Equation (5) is the First Thermodynamic equation of state   THERMODYNAMICS

54. we know that dH = TdS + VdP ------------(6) Divide on both side with dP , we get = T + V ------------(7) At constant temperature, equation (7) becomes T = T T + V ------------(8) We know that one of the Maxwell’s relationship is T = - P ------------( 9 )   THERMODYNAMICS

55. Substitute equation (9) in (8) we get T = - T P + V ------------(10) Equation (10) is the Second Thermodynamic equation of state   THERMODYNAMICS

56. Problems: Using the thermodynamic equation of state, Show that T = 0 for an ideal gas T = 0 for an ideal gas T = for real gas T = b for real gas obeying the P(V-b) = RT equation   THERMODYNAMICS

57. Criteria for spontaneity in term of entropy: For a reversible small change dS = TdS = dq rev Tds = dE + PdV --------(1) At constant E & V TdS = 0 dS = 0 ( ∂ S) E,V = 0   THERMODYNAMICS

58. For a irreversible small change, absorbed dq irr is less than dq rev but the entropy change is same ( dS ) i.e ., dq irr < dq rev otherwise dq irr < TdS dE + PdV < TdS TdS > dE + PdV At constant E & V TdS > dS > ( ∂ S) E,V > 0 for spontaneous (or) irreversible process (∂S) E,V = 0 for reversible process THERMODYNAMICS

59. Criteria for spontaneity in term of Gibb’s free energy: We know that G = H – TS dG = dH – TdS – SdT -----------(1) Since H = E + PV therefore dH = dE + PdV + VdP -----------(2) Substitute (2) in (1) dG = dE + PdV + VdP – TdS – SdT -----------( 3 ) Since for reversible process, TdS = dE + PdV , equation (3) becomes dG = TdS + VdP – TdS – SdT dG = VdP – SdT THERMODYNAMICS

60. At constant T & P dG = 0 (∂G) T,P = 0 for reversible process Since for spontaneous process , TdS > dE + PdV , equation (3) becomes dG < VdP – SdT At constant T & P ( ∂G) T,P < 0 for spontaneous process THERMODYNAMICS

61. Criteria for spontaneity in term of Helmholts’s free energy: We know that A = E – TS dA = dE – TdS – SdT At constant temperature, dT = 0 dA = dE – TdS -----------(1 ) Since for reversible process, TdS = dE + PdV , equation (1) becomes dA = dE – dE – PdV dA = – PdV THERMODYNAMICS

62. At constant Volume dA = 0 (∂A) T,V = 0 for reversible process Since for spontaneous process , TdS > dE + PdV TdS = dE + PdV + x ---------(3) Substitute (3) in (1), we get, dA = dE – dE – PdV – x dA = – PdV – x dA < – PdV At constant Volume (∂A) T,V < 0 for spontaneous process THERMODYNAMICS

63. Chemical potential: For a closed system , G = f (T, P) But, for open system , G = f (T, P, n) Where n be the number of moles of gas For open system containing various gases, G = f (T, P, n 1 , n 2 , n 3 ,…. n j ) -------(1) Where n 1 , n 2 , n 3 ,…. n j be the number of moles of the gases of 1, 2, 3, .. and n 1 + n 2 + n 3 +…. + n j = Total number of moles (N) THERMODYNAMICS

64. Differentiating equation (1) dG = P,N dT + T,N dP + T,P, n 2 , n 3 ,…. nj dn 1 + T,P, n 1 , n 3 ,…. nj dn 2 + … + T,P, n 1 , n 2 ,….nj-1 dn j Where the quantity T,P, n 1 , n 2 ,….nj-1 is the chemical potential ( μ j ) T,P, n 1 , n 2 ,….nj-1 = = μ j Definition: Chemical potential ( μ ) of a given substance is defined as change in free energy of the system on the addition of one mole of that substance at a constant temperature and pressure. i.e., T,P = μ   THERMODYNAMICS

65. THERMODYNAMICS Gibbs- Duhem equation : For a open system containing various gases 1, 2, 3, …. i and j of moles n 1 , n 2 , n 3 , ….. n i and n j . G = f (T, P, n 1 , n 2 , n 3 ,…. n i , n j ) dG = P,N dT + T,N dP + T,P, n 2 , n 3 ,…. ni,nj dn 1 + T,P, n 1 , n 3 ,…. ni,nj dn 2 +…. + T,P, n 1 , n 2 ,…. nj + T,P, n 1 , n 2 ,…. ni dn j dG = P,N dT + T,N dP + μ 1 dn 1 + μ 2 dn 2 + …. + μ i dn i + μ j dn j At constant T & P ( dG ) T,P = μ 1 dn 1 + μ 2 dn 2 + …. + μ i dn i + μ j dn j -----(1)  

66. Integrating the equation (1), we get G = μ 1 n 1 + μ 2 n 2 + …. + μ i n i + μ j n j Differentiating the above equation, we get ( dG ) T,P = μ 1 dn 1 + n 1 d μ 1 + μ 2 dn 2 + n 2 d μ 2 + …. + μ i dn i + n i d μ i + μ j dn j + n j d μ j ( dG ) T,P = ( μ 1 dn 1 + μ 2 dn 2 + …. + μ i dn i + μ j dn j ) + ( n 1 d μ 1 + n 2 d μ 2 + …. + n i d μ i + n j d μ j ) --------(2) Substitute (1) in (2) ( dG ) T,P = ( dG ) T,P + n 1 d μ 1 + n 2 d μ 2 + …. + n i d μ i + n j d μ j n 1 d μ 1 + n 2 d μ 2 + …. + n i d μ i + n j d μ j = 0  n j d μ j = 0 This relationship is called as Gibbs- Duhem equation THERMODYNAMICS

67. Variation of chemical potential with temperature We know that, the chemical potential of i th gas of a system is μ i = T,P, n 1 , n 2 ,…. nj T,P, n 1 , n 2 ,…. nj = μ i Differentiating with respect to temperature P,N = P,N ---------(1) We know that, dG = VdP – SdT At constant pressure and number of moles dG = – S dT = – S P,N = – S   THERMODYNAMICS

68. Differentiating with respect to n i P , n 1 , n 2 ,…. nj = P , n 1 , n 2 ,…. nj = - -------(2) Where is the partial molar entropy of the component i Equating the equation (1) & (2) we get P,N = - i.e., change of chemical potential of i th gas with temperature is the negative of its partial molar entropy   THERMODYNAMICS

69. Variation of chemical potential with pressure We know that, the chemical potential of i th gas of a system is μ i = T,P, n 1 , n 2 ,…. nj T,P, n 1 , n 2 ,…. nj = μ i Differentiating with respect to pressure T ,N = T ,N ---------(1) We know that, dG = VdP – SdT At constant temperature and number of moles dG = VdP = V T ,N = V   THERMODYNAMICS

70. Differentiating with respect to n i T , n 1 , n 2 ,…. nj = T, n 1 , n 2 ,…. nj = -------(2) Where is the partial molar volume of the component i Equating the equation (1) & (2) we get T ,N = i.e., change of chemical potential of i th gas with pressure is the partial molar volume   THERMODYNAMICS

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