Thermodynamics i

Statement - Cp and Cv relation - Applications of the first law

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

2. THERMODYNAMICS Zeroth law of thermodynamics When two objects are in thermal equilibrium with the third object, then there is thermal equilibrium between the two objects itself

3. THERMODYNAMICS Nature of Work and Heat: The changes of a system from one state to another is accompanied by change in energy . The change in energy may appear in the form of heat, work, light, etc., The mathematical relation that relate the mechanical work done (W) and heat produced (H) is W  H (or) W = JH ------(1) Where J is Joule mechanical equivalent of heat When H = 1 Calorie then the equation (1) becomes W = J i.e., J is the amount of mechanical work required to produce one calorie of heat energy and i ts value was calculated as 4.184 Joules

4. THERMODYNAMICS Internal energy (E ) The energy associated with a system (matter) by means of its molecular constitution and the motion of its molecules is called internal energy. It is a state function property First law of thermodynamics Energy can neither be created nor destroyed but one form of energy can converted to another form of energy It is impossible to construct a perpetual motion machine. Perpetual machine is a machine which can produce energy without expenditure of energy 3. The total mass and energy of an isolated system remains unchanged (or) constant

5. THERMODYNAMICS Mathematical derivation of first law of thermodynamics When a system absorbs ‘q’ amount of heat energy and changes from one state (A) to another state (B)

6. Heat absorbed by the system is used by two ways Increasing the internal energy of the system i.e., ∆E = E B – E A Doing work ‘w’ by the system so that the piston can move from A (initial state) to B (final state) i.e., Heat absorbed by the system = Increase in internal energy + Work done by the system (Disappear) (Appear) q = ∆E + W ------ (1) THERMODYNAMICS

7. THERMODYNAMICS Since w = F x displacement w = P x area x displacement w = P x volume change w = P∆V Equation (1) becomes q = ∆E + P∆V -------(2) For small changes dq = dE + Pdv --------(3) equation (1), (2) and (3) are the mathematical statements for first law of thermodynamics

8. THERMODYNAMICS Enthalpy (H) It is defined as the total energy stored in the system Mathematically, H = E + PV i.e., Enthalpy of a system is obtained by adding the internal energy and the product of pressure and volume When a system changes from state A to state B at constant pressure by absorbing ‘q’ amount of heat energy

9. THERMODYNAMICS W = P∆V = P (V 2 – V 1 ) According to First law of thermodynamics q = ∆E + W -------(1) q = ∆E + P (V 2 – V 1 ) q = E 2 – E 1 + PV 2 – PV 1 q = (E 2 + PV 2 ) – (E 1 + PV 1 ) q = H 2 – H 1 q = ∆H -------(2)

10. THERMODYNAMICS i.e., The change in enthalpy of a system is the amount of heat absorbed at constant pressure So equation (1) can become ∆H = ∆E + W ∆H = ∆E + P∆V -------(3) Note , Absolute value for E & H cannot be determined but change of internal energy (∆E) and Enthalpy(∆H) are measured quantities

11. THERMODYNAMICS It is defined as the amount of heat energy required to raise the temperature of the system by 1 c Mathematically C = For small changes C = --------(1) There are two types of heat capacities 1. Heat capacity at constant volume (C V ) 2. Heat capacity at constant pressure (C P )   Heat capacity of a system

12. THERMODYNAMICS Heat capacity at Constant V olume (C V ) It is defined as the amount of heat energy required to raise the temperature of the system by 1 C at constant volume

13. THERMODYNAMICS According to first law of thermodynamics q = ∆E + P∆V q = ∆E + 0 (since ∆V = 0 ) q = ∆E For small changes dq = dE Equation (1) becomes C = = = v   i.e., C v is the increase in internal energy of the system on raising the temperature by 1 C and C v for one mole gas is called molar heat capacity at constant volume i.e., C v = v  

14. THERMODYNAMICS Heat capacity at C onstant Pressure(C P ) It is defined as the amount of heat energy required to raise the temperature of the system by 1 C at constant pressure

15. According to first law of thermodynamics q = ∆E + P∆V = ∆H For small changes dq = dE + PdV = dH Equation (1) becomes C = = = = P   i.e., C P is the increase in enthalpy of the system on raising the temperature by 1 C and C p for one mole gases is called molar heat capacity at constant pressure i.e., C p = p   THERMODYNAMICS Heat capacity at C onstant Pressure(C P )

16. THERMODYNAMICS Relation between C P & C v C v is the amount of heat energy required to raise the temperature from T 1 to T 1 +1 C by increasing the internal energy (∆E = E 2 – E 1 ) for one of gases i.e., C v = ∆E C V C P

17. C P is the amount of heat energy required to raise the temperature from T 1 to T 1 +1 C by increasing the internal energy (∆E) and performing work (P∆V) for one of gases i.e., C P = ∆E + P∆V THERMODYNAMICS So, C P is greater than C v by P∆V i.e., C P > C v by P∆V and C P - C v = P∆V --------(1) For one mole of an ideal gas PV = RT --------(2) When the temperature is raised by 1 C ie ., from T →T +1 , so that the volume of the system is V+∆V , then equation (2) becomes P(V+∆V) = R(T +1 ) --------(3)

18. THERMODYNAMICS Equation (3) – Equation (2), becomes P∆V = R ------- (4) Subtracting Equation (4) in Equation (1) we get C P – C V = R q = ∆E + W q = ∆E + P∆V H = E + PV ∆H = ∆E + P∆ V ∆H = ∆E + W C v = v C p = p C P – C V = R   Note :

19. THERMODYNAMICS Process: Process is an operation by which the system can change from one state to another state Types of Process : Reversible process Irreversible process Isothermal reversible process Adiabatic reversible process Isochoric process Isobaric process Isothermal irreversible process Adiabatic irreversible process

20. THERMODYNAMICS Reversible process: Reversible process is the process in which the system can change from one state to another state in a series of small steps by successive small decrements( dP ) or increments( dP ) in external pressure that makes successive small increments ( dV ) or decrements ( dV ) in volume

21. THERMODYNAMICS Irreversible process: Irreversible process is the process in which the system can change from one state to another state suddenly in a single step when the pressure difference between the system and the surrounding is very high

22. THERMODYNAMICS Compare reversible and irreversible process: REVERSIBLE PROCESS IRREVERSIBLE PROCESS Slow process Sudden process Pressure difference is infinitesimally small Pressure difference is high Involved in many small steps to complete Involved in single step to complete Maximum work is obtained Maximum work is not obtained Equilibrium existed between two successive small steps No existence of equilibrium between initial and final stage

23. THERMODYNAMICS Isothermal reversible process: Isothermal reversible process is the process in which the system can change from one state to another state in a series of small steps by successive small decrements or increments in external pressure ( dP ) that makes successive small increments or decrements in volume ( dV ) at constant temperature Heat conducting material (so temperature remains constant during the process)

24. THERMODYNAMICS Change in internal energy and enthalpy for isothermal reversible process: The temperature of the isothermal reversible process is constant. Since the internal energy(E) depends upon the temperature of the system, internal energy (E) also constant i.e., Change in internal energy (∆E) = 0 We know that H = E + PV ∆ H = ∆ (E + PV) (For 1 mole of gas ) ∆ H = ∆ (E + RT) ∆ H = ∆E + R ∆T --------(1) Since, ∆E = 0 & ∆T = 0 at constant temperature equation (1) becomes ∆ H = 0 + R (0) ∆ H = 0

25. THERMODYNAMICS Heat energy (q) for the isothermal process: According to first law of thermodynamics ∆E = q – w -----(1) Since ∆E = 0 for an isothermal process, equation (1) becomes 0 = q – w q = w -----(2) Equation (2) shows that the work(w) is done at the expense of the heat absorbed(q)

26. THERMODYNAMICS Work done in isothermal reversible expansion of an ideal gas: Let dw be the small amount of work done by the system in each step of isothermal reversible expansion and it is given as, dw = (P – dP ) dV dw = PdV – dPdV --------(1) Since dPdV is very small and this value is neglected, then the equation (1) becomes dw = PdV --------(2) The total work (w) is obtained by integrating equation (2) with in the limits V 1 & V 2 i.e., --------(3) Since for an ideal gas PV = RT and P = , then equation (3) becomes  

27. = = ] --------(4) Since for an ideal gas P 1 V 1 = P 2 V 2 at constant temperature = So equation (4) becomes   THERMODYNAMICS V 2 V 1

28. THERMODYNAMICS Adiabatic reversible process: Adiabatic reversible process is the process in which the system can change from one state to another state in a series of small steps by successive small decrements or increments in external pressure ( dP ) that makes successive small increments or decrements in volume ( dV ) at various temperature i.e., no heat is enter or leave the system Heat insulating material (so temperature changes during the process)

29. THERMODYNAMICS Since no heat is enter into the system q = 0 According to first law of thermodynamics , q = ∆E + W 0 = ∆E + W W = - ∆E i.e., work is done at the expense (decrease) of the internal energy of an ideal gas

30. THERMODYNAMICS ∆E, ∆H and work done for adiabatic reversible process for Enthalpy and work done We know that C v = v   C v =   For finite change, equation (1) becomes = C v ∆ -----------(2)   Since for adiabatic reversible process W = - ∆E, equation (2) becomes = C v ------------(1)   = - C v ∆ -----------(3)  

31. For finite change, equation (4) becomes = C P ∆ ------------(5)   THERMODYNAMICS From equation (2), (3) & (5) it is clear that the value for ∆E , W and ∆H depends upon the v alue of ∆T i.e., by knowing the initial temperature (T 1 ) and final temperature (T 2 ) for the adiabatic reversible process, the value for ∆E , W and ∆ H are calculated easily = C P -----------(4)   C P =   C p = P   And also we know that

32. THERMODYNAMICS Comparison of workdone in isothermal & adiabatic reversible expansion Consider the isothermal & adiabatic expansion of ideal gas from initial volume V 1 and pressure P 1 to a common final volume V t If P iso and P adia are the final pressure then

33. THERMODYNAMICS For isothermal expansion P 1 V 1 = P iso V t ---------(1) For adiabatic expansion P 1 V 1 γ = P adia ( V t ) γ ---------(3) Where γ =   From equation (1) = ---------(2)   From equation (3) γ = ---------(4)  

34. THERMODYNAMICS Since for expansion V t > V 1 and for all gases γ >1 and hence γ >   &   and also P adia < P iso Since P adia < P iso the workdone in isothermal expansion ( W iso = P iso ∆ V t ) shown by spotted area ABCD is greater than the workdone in adiabatic expansion ( W adia = P adia ∆V 1 ) shown by spotted area AECD

35. THERMODYNAMICS Application of first law of thermodynamics Joule Thomson effect Hess’s law of heat summation Kirchoff’s equation Bond enthalpy

36. “The phenomenon of change of temperature (cooling effect) produced when a gas is m ade to expand adiabatically from a region of high pressure to a region of very very low (vacuum) pressure is called Joule Thomson effect” THERMODYNAMICS Joule Thomson effect

37. THERMODYNAMICS While adiabatic expansion a part of kinetic energy of the gaseous molecules used to overcome the vanderwaal’s forces of attraction existed between the molecules. So that the kinetic energy decreased and the temperature of the gaseous molecule decreased. So cooling effect takes place. Reason: Applications of Joule Thomson effect:

38. THERMODYNAMICS Let us consider a volume V 1 of the gas enclosed between the piston A and the porous plug G a t a pressure P 1 , is forced slowly through the porous plug by moving the piston A inwards a nd is allowed to expand to a volume V 2 at a lower pressure P 2 by moving the B outwards, as shown Derivation of Joule – Thomson coefficient :

39. THERMODYNAMICS Work done on the system at the piston A (W 1 ) = - P 1 V 1 Work done by the system at the piston B (W 2 ) = P 2 V 2 Net work done by the system W = W 1 + W 2 W = P 2 V 2 – P 1 V 1 ------------(1) Since q=0 for adiabatic expansion and the first law (q =∆E + W) becomes 0 = ∆E + W W = -∆E -----------(2) i.e., the work is performed at the expense of internal energy and the internal energy of the system decreased from E 1 to E 2 . Substitute equation(1) in equation (2) , we get P 2 V 2 – P 1 V 1 = -∆E P 2 V 2 – P 1 V 1 = - (E 2 - E 1 ) P 2 V 2 – P 1 V 1 = E 1 - E 2 E 2 + P 2 V 2 = E 1 + P 1 V 1 ------------(3)

40. THERMODYNAMICS Since E + PV = H, equation (3) becomes H 2 = H 1 H 2 - H 1 = 0 ∆H = 0 and dH = 0 ----------(4) i.e., The adiabatic expansion of a gas occurs at constant enthalpy Since H = f (P,T) dH = T dP + P dT -----------(5) Since P = C P equation (5) becomes = T dP + C P dT ------------(6)  

41. THERMODYNAMICS Rearranging the equation (6) we get C P dT = - T dP = - T x H = - T x -------(7) where H = Joule Thomson coefficient and its symbol μ J.T Definition: Joule Thomson coefficient is defined as the decrease of temperature with pressure at constant enthalpy  

42. THERMODYNAMICS For a finite change equation (7) becomes = - T x = - T x This equation says that the decrease in temperature (∆T) is directly proportional to the d ifference of pressure (∆P)  

43. THERMODYNAMICS We know that The mathematical statement for μ J.T is For μ J.T is μ J.T = H = - T --------(1) Since H = E + PV equation (1) becomes μ J.T = - T μ J.T = - T T μ J.T = - x T T μ J.T = - T T T   Joule Thomson Coefficient ( μ J.T ) for an ideal gas

44. THERMODYNAMICS When an ideal gas adiabatically expands into vacuum W = 0 and q = 0 So from first law of thermodynamics q = ∆E + W , we get ∆E = 0 i.e., the change of internal energy with volume at constant temperature = 0 ----------(3) Equation (2) becomes, μ J.T = - T ----------(4)  

45. THERMODYNAMICS Since PV = constant for ideal gases at constant temperature T = 0 -------(5) Equation (4) becomes μ J.T = - i.e., Joule Thomson coefficient for an ideal gas is zero   Joule Thomson Co-efficient for real gases : μ J.T for real gases can be obtained used Vander Waals equation PV = RT - + bp - -------(1)  

46. THERMODYNAMICS Since both a and b are small, the term ab /v 2 can be neglected then equation (1) becomes V = - + b ---------(2) Since PV = RT, then the equation (2) becomes V = - + b ---------(3) Differentiating with respect to temperature at constant pressure P = ---------(4) Substituting the value of V from equation (3) and P from equation (4) in the following thermodynamic equation of state T = V - T P we get  

47. THERMODYNAMICS T = - + b - T T = - + b - T = b - ---------(5) we know that , μ J.T = - T ----------(6) Substitute equation (5) in equation (6) we get μ J.T = - b - μ J.T (real) = - b  

48. THERMODYNAMICS Inversion temperature: We know that Joule Thomson coefficient for real gas is When = b , equation (1) becomes μ J.T (real) = (0 - 0) μ J.T (real) = 0 The temperature at which μ J.T (real) = 0 is called inversion temperature and its value is obtained from the equation = b as T i = Below the inversion temperature μ J.T (real) is positive i.e., cooling effect occurs Above the inversion temperature μ J.T (real) is negative i.e ., heating effect occurs At the inversion temperature μ J.T (real) is zero i.e ., neither cooling nor heating effect occurs   μ J.T (real) = - b ------(1)  

49. THERMODYNAMICS Hess’s Law of constant Heat summation Hess’s Law : “The enthalpy change of a given chemical reaction is the same whether the process takes place in one or several steps” Explanation: Suppose a substance A can be changed to Z in two ways First way: Involved in one way A → Z + Q 1 Where Q 1 ---- amount of heat evolved i.e., equal to change in enthalpy ∆H

50. THERMODYNAMICS Second way: Involved in three steps A → B + q 1 B → C + q 2 C → Z + q 3 B C Z A q 1 q 2 q 3 Q 1 INITIAL STATE FINAL STATE Where q 1 ---- amount of heat evolved in first step ( ∆ H 1 ) q 2 ---- amount of heat evolved in second step (∆ H 2 ) q 3 ---- amount of heat evolved in third step (∆ H 3 )

51. THERMODYNAMICS The total evolution of heat in second way is q 1 + q 2 + q 3 = Q 2 (or) ∆ H 1 + ∆ H 2 + ∆ H 3 = ∆ H’ According to Hess’s law : Q 1 = Q 2 o r Q 1 = q 1 + q 2 + q 3 o r ∆ H = ∆ H’ o r ∆H = ∆H 1 + ∆H 2 + ∆H 3

52. THERMODYNAMICS Example : The burning of carbon into carbon dioxide involved in 2 ways First way: C + O 2 → CO 2 ∆H = -393 KJ Second way: C + ½ O 2 → CO ∆ H 1 = -110.5 KJ ii) CO + ½ O 2 → CO 2 ∆H 2 = -283.2 KJ ∆ H = ∆ H 1 + ∆ H 2 -393 KJ = -110.5 KJ – 283.2 KJ -393 KJ = -393.7 KJ According to Hess’s law

53. THERMODYNAMICS Kirchhoff’s equation: The variation of ∆H with temperature at constant pressure of a given reaction is r epresented by Kirchhoff’s equation Derivation : Consider a reaction at constant pressure A → B The change of enthalpy ∆H = H B – H A -------(1) Differentiate equation (1) with respect to temperature at constant pressure = P -------(2)  

54. THERMODYNAMICS Since P = (C P ) A then equation (2) becomes P = P = --------(3) = dT = 2 - 1 = (T 2 – T 1 ) --------(4) Equation (3) and (4) Kirchhoff's equations where & 2 are the heats of reaction at constant pressure at temperature T 1 & T 2  

55. THERMODYNAMICS Bond enthalpy : Bond enthalpy is defined as the amount of energy required to dissociate one mole of bonds present in a compound in the gaseous state For example H 2(g) → 2H ∆H = 433 KJ / mol That is 433 KJ of energy is required to break the H – H bond in one mole (Avogadro number) of Hydrogen gas molecule

56. THERMODYNAMICS Application of the first law of thermodynamics to real gases : Work done for the isothermal expansion of real gases : We know that dw = Pdv ------(1) For n moles, the vanderwaal’s equation for real gas is P + (v – nb ) = nRT P + = P = - -------(2)  

57. THERMODYNAMICS Substitute equation (2) in (1), we get dw = - dv --------(3) Let the system containing real gas expand isothermally from volume v 1 to v 2 , the work done for the process is obtained by integrating the equation (3) with limit v 1 and v 2 = dv w = dv w =  

58. THERMODYNAMICS w = w = nRT - w = nRT [ ln ( v 2 - nb ) – ln (v 1 - nb )] – an 2 w = nRT ln + an 2  

59. THERMODYNAMICS Change in internal energy for isothermal expansion of real gases : In Vanderwaal’s equation for real gases, the factor is representing the internal pressure a nd also T is also corresponding to internal pressure i.e., T = dE = dv --------(1) Integrating the equation (1) with in the limit = dv E 2 – E 1 =  

60. THERMODYNAMICS E 2 – E 1 = ∆E = ∆ E = an 2 ∆ E = an 2 ∆E = - an 2 This is the Change in internal energy for isothermal expansion of real gases  

61. THERMODYNAMICS Heat absorbed for isothermal expansion of real gases : From the first law of thermodynamics q = ∆E + w ---------(1) We know that ∆E = - an 2 ---------(2) w = nRT ln + an 2 ---------(3) Substituting (2) & (3) in (1) we get q = - an 2 + nRT ln + an 2 q = nRT ln  

62. THERMODYNAMICS Work done for adiabatic expansion of real gases : From the first law of thermodynamics q = ∆E + w ---------(1) For adiabatic process, q = 0 so the equation (1) becomes 0 = ∆E + w ∆E = -w ---------(2) Since E is a state function of T and V i.e., E = f (T,V) dE = V dT + T dV ---------(3)  

63. THERMODYNAMICS For a vanderwaal’s equation, P + (v – nb ) = nRT V = n C v -------(2) T = --------(3) Substitute (2) and (3) in (1) we get dE = n C v dT + dV Integrating the equation within the limit = n C v + E 2 – E 1 = n C v (T 2 – T 1 ) - an 2 ∆E = n C v ∆T - an 2  

64. THERMODYNAMICS Limitations of first law of thermodynamics: First law of thermodynamics does not tell The direction of flow of heat Whether a process occur spontaneously i.e., whether it is feasible Heat energy cannot be completely converted into an equivalent of work without producing some changes elsewhere

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