Unit II First law of thermodynamics, Internal energy, Heat capacity, Specific heat and - PowerPoint Presentation

1. Unit IIFirst law of thermodynamics, Internal energy, Heat capacity, Specific heat and latent heat, Enthalpy, Isothermal and adiabatic processes, State properties, Heat of reaction, Heat of formation, Standard heats, Heat of transition, Hess’s law, Kirchhoff’s law equation. Second law of thermodynamics, Entropy of irreversible processes, Auxiliary functions, combined statements of 1st and 2nd laws, Maxwell’s relations, Gibb’s-Helmholtz relations. Third law of thermodynamics, Clausius – Clapeyron equation, Temperature dependence of entropy, Statistical interpretation of entropy, Consequences of third law, Nernst heat theorem, Equilibrium constant, Van-Hoff equation, Concept of fugacity, activity and mole fraction

2. 1st Law of Thermodynamics (TD)Conservation of EnergyEffect of Heat on a System q = Ef – Ei = DEEffect of Work on a system Ef – Ei = DE = -W First Law of Thermodynamics (TD): DE = q-WW = work done by the system upon the surroundingFor infinitely small change in internal energy of the system, First law of TD: dE = dq – dW = dq – PdV – dW’ = dq – PdV (Assuming dW’ = 0)PdV = Work done against an external pressure; dW’ = all other form of work done (i.e. electrical, magnetic, gravitational etc.) Total Energy entering the System Total Energy leaving the System =Change in Total Energy of the System( EIN )( EOUT )( ΔE )_

3. Significance of 1st Law of TDIt is based on law of conservation of energyIt introduces the concept of internal energyIt separates heat interaction (Thermal) and work interaction (Mechanical)It treats internal energy as state property (i.e. independent of path for the change of state)

4. Quiz The internal energy of a system increases during some time interval. Which one of the following statements concerning this situation must be true?The increase in internal energy indicates that work was done on the system.The increase in internal energy indicates that heat was added to the system.The increase in internal energy indicates that work was done by the system.The increase in internal energy indicates that heat was removed from the system.The information given is insufficient to indicate the reason for the increase.

5. Quiz (Contd.)A gas is enclosed in a cylinder by a piston. The volume of the gas is then reduced to one half its original value by applying a force to the piston. Which one of the following statements concerning the internal energy of the gas is true?The internal energy of the gas will decrease.The internal energy of the gas will increase.The internal energy of the gas will neither increase nor decrease.The internal energy of the gas will equal the work done in moving the piston.The internal energy of the gas may increase, decrease, or remain the same depending on the amount of heat that is gained or lost by the gas.

6. Quiz (Contd.)Which one of the following statements is not consistent with the first law of thermodynamics?The internal energy of a finite system must be finite.An engine may be constructed such that the work done by the machine exceeds the energy input to the engine.An isolated system that is thermally insulated cannot do work on its surroundings nor can work be done on the system.The internal energy of a system decreases when it does work on its surroundings and there is no flow of heat.An engine may be constructed that gains energy while heat is transferred to it and work is done on it.

7. Basic TermsHeat Capacity: The amount of heat required to raise its temperature by 1oC. C = dq/dT (Joule/Kelvin or Joule /oC)Different materials require different amount of Energy for their temperatures to increase thought unit quantity ( i.e. 1 ºC) for identical mass. 1 kgFe20 – 30 ºC4.5 kJ1 kgH2O20 – 30 ºC41.8 kJHence, it is required to define a Property to compare the ENERGY STORAGE CAPACITY of different substances.This Property is known as SPECIFIC HEAT.

8. Basic Terms (Contd.)Atomic heat capacity is the heat capacity of 1 atom of an elementMolar heat capacity is the heat capacity of 1 mole of a substance. m = 1 kg∆T = 1 ºCSp. Heat = 5 kJ/kg ºC5 kJDefinition of Specific heat :The Energy required to raise the temperature of a unit mass of a substance by 1 degree.

9. Basic Terms (Contd.)dq is incomplete differential because it is not a state variable (i.e. it depends on the path followed)Heat Capacity at constant pressure (dq/dT)P : It is easy to determine heat capacity at constant pressure via experimentHeat Capacity at constant volume (dq/dT)V : Theoretically, it is easy to determine heat capacity at constant volume

10. Heat Capacity at Constant VolumedE=dq-PdV (from 1st law of TD) ------1Since, E = f(V,T) , ----2Combining Eq. 1 and 2,Or: At Constant Volume, dV = 0;Hence at Constant volume, ,

11. Heat Capacity at Constant PressuredE=dq-PdV (from 1st law of TD) ------1Since, E = f(P,T) , ----2Combining Eq. 1 and 2,Or: At Constant Pressure, dP = 0;Hence at Constant Pressure,

12. Relation Between CP and CVTherefore, Now dividing both sides by dT at constant pressure:Since, E=f(V,T)Therefore,

13. CP-CV for Ideal GasFor ideal gas: Therefore at constant pressure, For Ideal gas, energy is a function only of temperature and it does not change with volume at constant temperature, i.e. Therefore for an ideal gas at constant temperature: CP – CV = R

14. Effect of Temperature on Heat Capacity Where a,b,c are constants for a particular range of temperatureSalient Features:The term CT-2 is omitted because it is often too smallThis equation is used for solid metals and compounds at elevated temperatureFor liquid metals with insufficient data, Cp is assumed to be constant. CP (J/K)T(K)?

15. Importance of CP and CVWhich is more important to Metallurgist – CP or CV ?CP > CVConsidering: V = Constm = 1 kg∆T = 1 ºC3.12 kJCV = 3.12 kJ/kg.ºCCP = 5.19 kJ/kg.ºCP = Constm = 1 kg∆T = 1 ºC5.19 kJHe GasCP is always greater than CV; as the System is allowed to expand in case of Const. Pr. and the Energy for this expansion Work is also need to be supplied.

16. To relate heat and temperature change in solids and liquids, following equation is used:specific heatcapacityThe amount of a gas is conveniently expressed in moles, so following analogous equation is used:molar specificheat capacitySpecific Heat Capacities

17. For gases it is necessary to distinguish between the molar specific heatcapacities which apply to the conditions of constant pressure and constantvolume:constant pressurefor a monatomicideal gasSpecific Heat Capacities – Constant Pfirst law ofthermodynamics

18. constant volumefor a monatomicideal gasmonatomicideal gasany ideal gasSpecific Heat Capacities – Constant Vfirst law ofthermodynamics is the ratio of themolar specific heatcapacities at constantpressure and volume

19. Basic Terms (Contd.)Enthalpy (H): Greek word meaning worm – It is the total heat content of a systemE- Internal Energy, P – Pressure, V - Volume H is a state variable since E,P, and V are state variables. For a cyclic process, DH = 0First Law of TD (in terms of Enthalpy):dE = dq – dW By Definition: By differentiating: Therefore, at constant pressure (dP = 0): DH = qAt constant pressure, change in enthalpy of the system is equal to the heat absorbed by the system from surrounding or heat exchange between the system and the surrounding

20. Enthalpy (KJ/mole)Example 1: C(s) + O2(g)  CO2(g) + heat evolved (DH = -393 KJ/mol)DH = Hfinal - HInitialSince, DH < 0, Therefore, Hfinal < Hinitial  Exothermic reaction (i.e. Reaction that results in release of heat energy)Example 2: ZnO + C  Zn + CO – heat absorbed (DH = +350 KJ/mol)Since, DH > 0, Therefore, Hfinal > Hinitial  Endothermic reaction (i.e. Reaction that results in absorption of heat energy)Case I: Enthalpy of pure element is generally taken to be zero at reference temperature (298K) and 1 atm pressure provided it is in normal state. Example: H0298[C,graphite] = H0298[O2(g)] = H0298[Fe(s)] = H0298[Hg(l)] = 0H0630[Hg(g)] = 60.84KJ/molCase II: Enthalpy of a compound is taken to be its enthalpy change of formation.Example: C(s) + O2(g)  CO2(g) DH0298 = -393 KJ/mol H0298[CO2(g)] = DH0298[CO2(g)] = -393 KJ/mol

21. Enthalpy Change (KJ/mole)DH depends on: 1) Temperature, 2) Pressure, 3) Physical state of reactants and products, and 4) Amount of substances reactingSince,Therefore,On integratingCase I: Enthalpy change when substance is heated from T1 to T2 at constant pressureCase II: Enthalpy change when substance is heated from T1 to T2 at constant pressure with phase transformation from solid to liquid phase at TfWhere: CP(s) = Heat capacity of solidCP(l)=Heat capacity of liquidLf = Latent heat of fusionTf = Temperature of fusion

22. Enthalpy Change (KJ/mole) Due to Chemical ReactionsHeat of Reaction (DHr): Heat generated or absorbed due to chemical reaction3Fe2O3(s) + CO(g)  2Fe3O4(s) + CO2(g); DHr = -2.67 KJ/moleHeat of Formation of Compound (DHf): Enthalpy change when 1 mole of compound is formed from its constituent’s elements2Al(s) + 3/2O2(g)  Al2O3(s) ; DHf,Al2O3 = -1700 KJ/mole3. Heat of Combustion of a substance (DHc): Enthalpy changed when 1 mole of substance is completely burnt (Same as heat of formation for the compound)C(s) +O2(g)  CO2(g); DHc,298 = -393 KJ/moleMg(s) + 1/2O2(g)  MgO(s); DHc,298 = -605 KJ/mole4. Heat of Transformation or Latent Heat: Enthalpy change when 1 mole of the substance undergoes a specific physical change (melting, evaporation etc.). Latent heat of fusion is the heat supply for transforming solid to liquid at fusion temperature. Zn(s)  Zn(l) at 693K, Lf,Zn = 7.28 KJ/mole5. Heat of Solution: Enthalpy change when 1 mole of solute is dissolved in solvent. It depends on the concentration. Heat of solution increases with the decrease in concentration00

23. Hess LawHess Law states that for an isothermal process, the enthalpy change for a reaction is the same whether it takes place in one or multiples stages. i.e. Heat of reaction depends on the initial and final states of the processExample: Direct oxidation or the single stage process: i) CH4 + 2O2  CO2 + 2H2O; DHr,1 = -890 kJ/mole Indirect oxidation or three stages process: i) CH4  C + 2H2; DHr,2 = 77kJ/mole ii) C + O2  CO2; DHr,3 = -393kJ/mole iii) 2H2 + O2  2H2O; DHr,4 = 2(-287)kJ/mole CH4 + O2  CO2 + 2H2O DHr,1 = DHr,2 + DHr,3 + 2DHr,4 = 77+(-393) + (-574) = -890 kJ/mole ABCD

24. Salient Feature of Hess’s LawThe principle of algebraic addition, subtraction etc. of chemical equations is applicable to the energy changes or enthalpy changesChemical equations can be multiplied by suitable coefficients and then added or subtracted to get desired equation. The value of state variable (DE, DH etc.) should be manipulated accordinglyHess Law is applicable to isothermal process, i.e. constant temperatureThis law is of great help in TD calculation for the reactions that are difficult for experimental study.Hess Law is applicable to which process? TProcess State 1State 2Process 1State 2State 1TProcess Process 2

25. Reactions at Different Temperatures - Kirchhoff’s LawSince,Therefore, On integratingTherefore, Kirchhoff’s EquationAccording to Kirchhoff’s Law, heat of reaction is a function of temperature Kirchhoff’s Law states that if a system undergoes a change from one state to another state then both internal energy & heat occur would alter

26. 2nd Law of ThermodynamicsClausius Statement: it is impossible for a self acting machine working in a cyclic process without any external force, to transfer heat from a body at a lower temperature to a body at a higher temperature. It considers transformation of heat between two heat reservoirs.Kelvin – Plank Statement: it is impossible to construct an engine, which is operating in a cycle produces no other effect except to external heat from a single reservoir and do equivalent amount of work.Alternate Statements: 1) Heat absorbed at any one temperature can not completely transformed into workwithout leaving some change in the system or its surrounding.Examples: In electrical machines – The efficiency of the conversion of electrical to mechanical energy and vice-versa can be 90% max.In heat engine – The efficiency of the conversion of heat energy to work is in the range of 10%-40%. 2) Spontaneous processes are not thermodynamically reversible.Entropy distinguishes between reversible (i.e. equilibrium) and irreversible (i.e. non-equilibrium or spontaneous) process. 3) For a spontaneous (i.e. non-equilibrium) irreversible process, the entropy of an isolated system always increases.

27. Basic Definition: EntropyIn a reversible process, when a substance absorbs infinitesimal amount of heat (dq), in a reversible manner, at an absolute temperature (T), the entropy of the substance increases by dS: dS = change in entropy (J/K/Mole)Dqrev = heat absorbed by a system in a reversible processT=Temperature of the substance (K)Based on statistical concept, entropy may be defined as a measure of randomness or a disorder in the system – Boltzmann Entropy Equation:Using Stirling’s Approximation: Entropy is a thermodynamic property depending only on the state of the system Differential form: Integral form: It is for isothermal and reversible process It is a measure of disorder of the system, higher the disorder, larger is the entropy It is a measure of heat available for work

28. Entropy ChangeFor reversible process: For irreversible process: 2nd Law of TD: An irreversible process in an isolated system will occur with an increase in entropy1) Entropy change for a substance when it is heated from T1 to T2 Therefore,Since,

29. Entropy Change (Contd.)2) Entropy change with a change of state (Solid – Liquid)Example: Zn(s)  Zn (l)  Zn(g)DS = 10.5J/K/moleDS = 98.3J/K/mole3) Entropy change for a chemical reaction: When reversible reaction takes place: 1) At constant temperature: 2) At constant temperature and constant pressure: Entropy change for a chemical reaction: Where:

30. Combined Expression of 1st and 2nd Law of Thermodynamics1st Law of TD: 2nd Law of TD: By Differentiating:

31. Thermodynamic Equation of State I – DerivationSince, Therefore, Combining 1st and 2nd Law of TD:If z = f(x,y) ; then By comparison: and By differentiating: and Therefore,

32. Thermodynamic Equation of State I – DerivationBy comparison: and By differentiatingThermodynamic Equation of State, It correlates E,P,T, and V

33. Thermodynamic Equation of State II – DerivationSince Therefore,Combining 1st and 2nd Law of TD:If z = f(x,y) ; then By comparison: and By differentiating: and Therefore,

34. Thermodynamic Equation of State II – Derivation (Contd.)This equation is also known as Equation of State. It correlates H,P,T, and V

35. Free Energy and 3rd Law of ThermodynamicsHelmholtz Free Energy (Isothermal work) (A): Where TS is the bound energy and is dissipated as heat energy (i.e. not available for work)Differentiating A at constant T: From 1st law of TD: From 2nd law of TD: For mechanical work onlyTherefore, dW is maximum reversible work done by the systemHelmholtz Free Energy is applicable to Mechanical Thermodynamics

36. Free Energy and 3rd Law of Thermodynamics (Contd.)Gibbs Free Energy (G): By Differentiating at constant T and P: Reversible non-mechanical work = dW-PdV = -(dE-TdS) – PdV = -(dH-TdS) = -dG (i.e. decrease in free energy)Therefore, What is G and why it is important? The function G is a measure of stored non-mechanical energy available to the system for doing non-mechanical work. At constant P and T, G is a measure of the work obtainable from a reversible, isothermal and isobaric process and gives a direct indication of the possibility of chemical reaction. G serves as a criterion for equilibriumSince, H,T, and S are state variable, G is also a state variableHow to calculate G?At const. T and P

37. Free Energy and 3rd Law of Thermodynamics (Contd.)Free energy of a reaction at temperature T:If the reaction involves heating from temperature Ti to T, in a reversible process without the change of state

38. Free Energy as a Criteria of Equilibrium For a closed system undergoing a irreversible change, i.e. spontaneous, the free energy decreases because entropy increases. i.e. DG=DH-TDS;  DG <0, DH=0 (unchanged) For a closed system undergoing a reversible change, DG = 0, DS=0, DH=0If DG = 0, the system is in equilibrium if DG<0, the reaction tends to proceed spontaneously in forward direction if DG>0, the reaction tends to proceed spontaneously in the opposite direction

39. Partial Derivative of Free EnergyHelmholtz free energy (A): Combining 1st and 2nd law of TDGibbs free energy (G):  At constant T, dG=VdP and  At constant P, dG=-SdT and

40. Partial Derivative of Free Energy – Contd.By Definition: G=H-TSTherefore,  At constant P:Therefore, Dividing by T2 :

41. Maxwell’s RelationWe know, if z=f(x,y) dz=Mdx+NdyWhere, Therefore, Combining 1st and 2nd Law of TD:dE=TdS-PdVdH=TdS+VdPdA=-PdV-SdTdG=VdP-SdTWhere, P,V, T are state variable, E, S, H, A, and G are function of stateMaxwell’s RelationMaxwell’s relations are used to calculate thermodynamic parameters

42. 3rd Law of ThermodynamicsThe 3rd law of TD may be stated as the entropy of any homogeneous substance, which is in complete equilibrium, may be taken as zero as absolute zeroAt 0oK, DS = 0 Original Statement of Nernst heat theoremBy definition, for a reaction: dG=dH-TdS = dH-T(dDG/dT)PIf is a finite quantity, then at 0oK At 0oK, DG=DHBy differentiating with respect to T at constant P DG and DH are not 0 at 0oK but the slope of DG vs. T and DH vs. T curve have the same slope at 0oKSince (dDG/dT)P=-DS; At 0oK, , therefore, –DS=02) Since (dDH/dT)P=CP , At 0oK, , therefore, DCP = 0

43. 3rd Law of Thermodynamics (Contd.)Latest statement of Nernst heat theorem: For all reactions involving substances in condensed state, DS is zero at absolute zeroFor a chemical reaction: A+B  AB; AT 0oK, DS = 0Experimental verification of the 3rd Law of TDTReactionTemp.IIVIIIIIDS=DSI + DSII + DSIII + DSIV = 0or, DSIV = -(DSI + DSII + DSIII)According to 3rd Law of TD, for homogeneous substance in equilibrium, S0=0, but in practice, it is found that, S0 ≠ 0. This is because as temperature is lowered the more random arrangement at high temperature are frozen and the unique most order arrangement of the lower temperature is not actually obtained. Therefore, generally, there is a lack of equilibrium.

44. Fugacity and ActivityFugacity (f): It indicates the escaping tendency of the component or a substance. High gas pressure indicates high tendency of the gas molecules to escape outside the container. Therefore, high fugacity indicates a greater tendency of a component or a substance to escape, dissolve, intermix or react. Combining 1st and 2nd Law of TD: dG = VdP-SdTAt constant T, dT=0; Therefore, dG=VdPFor 1 mole of ideal gas, PV = RTTherefore, dG = (RT/P)dP = RT d(lnP)If gas is not ideal, P is replaced by f  dG = RT d(lnf)For ideal gas, f=PFor non-ideal gas, as P  0, f  P; Thus, for non-ideal gas, f=P only at low pressures Activity (a) : It is defined as the ratio of fugacity the substance in its actual state (f) to its fugacity in its standard state (fo). a = f/fo or, f=foadG = RTd(lnf) = dG = RT d(ln(foa)) = RT (d(lnfo)+ d(lna)) = RT d(lna) ; Since fo is constant.On integrating, Generalizing: Where, Gi = Partial free energy of ith species andGio = Partial free energy of ith species in standard state

45. Standard State and Equilibrium ConstantStandard State: It is the most stable state of the pure substance at 1 atm. pressure .Example: for Fe-C solid solution at 1700K, the standard state of iron is pure d-Fe; whereas the same solid solution at 1400K, the standard state is g-Fe. Activity of a solid or liquid in its standard state = 1 For ideal gas, at standard state  a=P=1 For non-ideal gas, at standard state  a=f=1Equilibrium Constant: For general reaction at constant temperature and pressure: bB+cC+…..=dD+eE+….At T, standard free energy change: Therefore, Where, Since at constant T and P, at equilibrium, (DG)T,P = 0; Therefore, Where, k=equilibrium constant=value of activity quotient at equilibriumEquilibrium constant can be calculated from activities of reactants and products at equilibrium

46. Importance of Equilibrium Constant and Lechatlier Principle If k is large (i.e. greater than 1), equilibrium lies well to the right since products predominate in the equilibrium mixture If k <1 , equilibrium lies to the left since reactants predominate Therefore, a reaction is feasible, if DG0 or k≥1Lechatelier Principle: If a system in equilibrium is subjected to external constraint (i.e. increase of T,P, V), the equilibrium of the system is shifted in such a way so as to counteract the external constraint.Example: C + CO2  2CO, DHo = +ve: Observation: Volume of product is greater than reactant and it is endothermic Case 1: Increase in T will favor forward reaction due to its endothermic nature Case 2: Increase in pressure will reduce the rate of forward reactionGas-Solid Reaction:MO(s) + CO(g)  M(s) + CO2(g) ….DG1Since, M and MO are pure solid in standard state, hence aM = aMO =1Therefore,

47. Inter-Relations Between Thermodynamics VariablesGibbs – Helmholtz Equation: It gives relation between DG, DH, and TCombining 1st and 2nd law of TD: At Const. PFor chemical reaction: Gibb’s-Helmholtz EquationGibb’s-Helmholtz Equation in standard stateDividing Gibb’s-Helmholtz equation by T2: Or,Or,Or,Since, Application of Gibb’s-Helmholtz Eq. To calculate DHo from DGo for a reaction To calculate DGo at T from DGo at T1

48. Inter-Relations Between Thermodynamics Variables – Contd.Van’t Hoff Equation: It gives relationship between k, DHo and T. From Gibb’s - Helmholtz EquationFrom earlier studyVan’t Hoff EquationVan’t Hoff Equation can be rewritten as: Since,

49. Inter-Relations Between Thermodynamics Variables – Contd.Integration of Van’t Hoff Equation and Sigma function: It gives relationship between k, Cp, and TVan’t Hoff Equation:By definition:Variation of CP with TFor standard state:On Integrating: Where, DHo is an integration constantBy integrating: Where, I is an integration constantWhere, Equilibrium constant of any reaction can be calculated from Cp at any T

50. Clausius – Clapeyron EquationFrom Maxwell’s relation:or, Since, Therefore, Where, DH=heat of transformation, DV = molar volumeClausius – Clapeyron EquationApplication of Clausius – Clapeyron Equation: To calculate the effect of pressure change on equilibrium transformation temperatureA. Liquid – Vapor Equilibrium:From Clausius – Clapeyron Equation:Since, Vvap>>Vliq, Therefore, Vvap-VliqVvapAssuming vapor behaves as an ideal gas, Case I: Integrating without taking limits of P and T (Graphical method)lnP1/TCase II: Integrating by limits of P and T (Numerical method)

51. Clausius – Clapeyron Equation – Contd.B. Solid – Vapor EquilibriumFrom Clausius – Clapeyron Equation:By integrating: Where, DHs = heat of sublimation and C is integration constantC. Solid – Liquid EquilibriumFrom Clausius – Clapeyron Equation:Or, Or, Assuming, DHf and DV do not change with T, by integrating, Therefore, C. Solid – Solid EquilibriumFrom Clausius – Clapeyron Equation:Where, DHf = heat of fusionOr, Assuming, DHtr and DV do not change with T, by integrating, Where, DHtr = heat of transformation and C = integration constant

52. Trouton’s RuleTrouton’s rule states that the ratio of latent heat of evaporation (DHv) to the normal boiling point temperature (Tb) is constant for all liquids, and is approximately equal to 21 cal/deg/moleThis rule is only an approximate one and is not followed by all liquid metals. However, it can be used to calculate DHv from boiling point of a metal.

53. Statistical Nature of EntropyIn Boltzmann's definition, entropy is a measure of the number of possible microscopic states (or microstates) of a system in thermodynamic equilibrium, consistent with its macroscopic thermodynamic properties (or macrostate)Where, KB is Boltzmann’s constant and  is the number of microstates consistent with the given macrostateExample: If N is the total number of sites and n is the number of atoms in a given system: Ln(N!)  NlnN-NTherefore, By Stirling FormulaOr,