From 1st law of thermodynamics, dq = du + PdV At constant volume, ΔQ v =ΔU Again H=U+PV so, ΔH = ΔU + PΔV + VΔP At constant pressure, ΔH = ΔQ p

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1 Chapter 2 HERMODYNAMICS 2.1 hermochemistry All chemical reactions and few physical changes are associated with energy change. In most cases this energy is heat energy but there are cases where this may be light or electricity also. hermochemistry deals with energy change associated with chemical changes and their application. A thermochemical equation is a chemical equation where not only reactants and products but also their complete states (solid, liquid or gas) have to be mentioned.( since the heat quantities depend on the physical state of matter). C(s) + O 2 (g)co 2 (g) + Q kcal at o C and atm. H 2 (g) +1/2 O 2 (g) H 2 O(l) kj at 25 o C. his reaction normally takes place in open vessel so that is constant of 1 atm. he reaction in which heat escapes from the system to the surroundings is termed as exothermic. While if heat is absorbed by the system from surroundings is endothermic Heat of reaction: According to old convention heat of reaction is the amount of heat evolved or absorbed during reaction. According to modern convention heat of reaction is the difference in total energy of products and reactants. From 1st law of thermodynamics, dq du + d At constant volume, ΔQ v ΔU Again HU+ so, ΔH ΔU + Δ + Δ At constant pressure, ΔH ΔQ p his means for reactions involving gaseous substances, energy change i.e. heat of reaction in any chemical reaction is related to ΔH provided the reaction is performed at constant pressure and it will be ΔU the reaction is performed at constant volume. hus heat of reaction is denoted as ΔH or ΔU But in most of the cases reactions are carried out in open vessels where pressure is constant that s why mostly heat of reaction is denoted as ΔH. It will be denoted as ΔU when we carry out a reaction in closed container where volume remains constant. 37

2 38 CHAER 2. HERMODYNAMICS Considering enthalpy as total energy heat of reaction at constant, At constant volume, For exothermic reaction ΔH -ve and for endothermic reaction ΔH +ve. C(s) + O2(g) CO2(g) ΔH - 94 kcal N2(g) + O2(g) 2NO(g) ΔH + 42 kcal ΔH value should be reported by mentioning and Relation between ΔH and ΔU: Let us consider a reaction involving gaseous substance r p he corresponding state change r(,,) p(,, ) We know, H p U p + and H r U r So, ΔH H p -H r (U p -U r )+ ( -) Assuming the gases behave ideally. ΔH ΔU + ( npr ΔH ΔU + ΔnR - nrr ) Δn is the difference in number of moles of gaseous products and reactants. When, Δn 0, ΔH ΔU Δn+ve, ΔH> ΔU Δn-ve, ΔH< ΔU Kirchoff s equation Heat of reaction depends on temperature. How it will vary with temperature is expressed by equation, called kirchoff s equation. Let us consider a reaction AB If this reaction is carried out at constant pressure, the heat of reaction of this reaction at any temperature. ΔH H B H A (2.1) o find the dependence of this quantity on temperature let us differentiate it with respect to temperature. dδh d dh B d But by definition at constant pressure, ( dh d ) C p Hence, dh A d dδh d C B C A (2.3) dδh d ΔC (2.4) Let at temperature o and heat of reaction for this reaction ΔH 0 and ΔH respectively.hen integrating equation 3.73 we get, (2.2) 1 0 d(δh) ΔH 1 ΔH ΔC d (2.5) 1 0 ΔC d (2.6)

3 2.1. HERMOCHEMISRY 39 If the range of temperature covered in equation 2.5 is short, the heat capacities of all substances involved may be considered constant. hen, ΔH 1 ΔH 0 ΔC ( 1 0 ) (2.7) If the temperature interval is very large, the heat capacities must be taken as function of temperature. For many substances this function has the form, C p a + b + c 2 + d 3 + where a, b, c, d, are constant for specified material. ΔH 1 ΔH 0 + ΔH 1 ΔH 0 + ΔH 1 ΔH ΔC d (a B + b B + c B 2 + ) (a A + b A + c A 2 + )d (Δa +Δb +Δc 2 + )d ΔH 1 ΔH 0 +Δa( 1 0 )+ Δb 2 ( )+ Δc 3 ( )+ (2.8) Similarly, ΔU 1 ΔU 0 +ΔC ( 1 0 ) (2.9) and, ΔU 1 ΔU 0 +Δa( 1 0 )+ Δb 2 ( )+ Δc 3 ( )+ (2.10) Equation (2.7), (2.8),2.9) and 3.8) are called Kirchoff s equation Heat of formation (ΔH f ) Heat of formation is the energy change when 1 mole of any substance is produced from its constituent element. he formation reaction of a compound has one mole of the compound and nothing else on the product side; only elements in their stable states of aggregation appear on the reactant side. e.g. H 2 (g)+ 1 2 O 2(g) H 2 O(l) Heat of formation; ΔH f(h2 O) H H 2O H 1 2 O2 H H 2 Standard heat of formation (ΔHf 0 ) is the heat of formation when all the constituent element of a reaction are in their standard state which is assigned as the state with pressure 1 atm and temperature 298 o K.he superscript o indicates standard state. It is however not possible to determine the absolute value of the standard enthalpy (H 0 ) of a substance. In order to measure enthalpy change it is arbitrarily assumed that standard enthalpy of elements to be ZERO. hus, ΔH f(h2 O) (H0 f ) here are two laws of thermochemistry. i) Lavoisier and Laplace law: his law states that the heat change for the forward reaction is exactly identical in magnitude but opposite in sign to the heat change for backward reaction. ii)hess s law(law of constant heat summation): It states that the heat change of a reaction is independent of the number of steps. Let for the reaction; hus by this law x y + z A+2B C ; ΔH x and f A+B I ; ΔH y I+BC; ΔH z

4 40 CHAER 2. HERMODYNAMICS Heat of combustion It is the change in enthalpy when 1 gm mole of a substance is completely burnt in presence of oxygen at a specified pressure and temperature Heat of Solution Heat of solution is the enthalpy change when a specified amount of solute is dissolved in specified amount of solvent Heat of dilution It is the heat change when some moles of solvent is added to an aqueous solution Heat of neutralization It is the change in enthalpy when 1 gm equivalent of an acid reacts with 1 gm equivalent of a base in a dilute solution. 2.2 Joule-homson effect he term throttling means sudden lowering of pressure.joule and homson carried out experiment on throttling of gases.hey found that at room temperature all gases except H 2 and He are cooled by throttling; H 2 and He are warmed up. Later experiment revealed that during throttling a gas may be cooled, heated or may retain its temperature unchanged. What will happen that depends on experimental temperature.if the the experimental temperature is equal to certain value, known as inversion temperature ( i ) then nothing will happen and if > i gas will be heated and if < i gas wil be cooled.all gases, except H 2 and He have i value much above room temperature where as i for H 2 and He are much below 0 o C, so throttling at room temperature produces heating effect. As for example in bi-cycle tube the gas is kept under very high pressure, compare to its outside pressure.if there creates any leakage the gas will come out and if we touch that gas generally we feel cold. In bi-cycle tube the gas is kept under very high pressure, compare to its outside pressure.if there creates any leakage the gas will come out and if we touch that gas generally we feel cold.if we fill the same tube with H 2 or He gas we will see the gas is heated up.during this process the gas is allowed to pass from very high to low pressure region through a small orifice (pore) and the gas experiences very low pressure suddenly and the gas expands very fast in such a way that there is is no time to exchange its energy with its surrounding, to make the process adiabatic. Figure 2.1: J expansion Let a volume 1 of a gas at a constant pressure 1 be forced by a piston through a porous barrier or plug from left to right. As it passes through the gas, pushes a piston at a constant pressure

5 2.2. JOULE-HOMSON EFFEC 41 2, ultimately occupying a volume 2.If the pistons, cylinder and plug are poor conductor of heat, the process can essentially be adiabatic. he work done by the piston on the left 1 1 Work done by the gas 2 2 heoretically, 1 1 may be greater, less or equal to 2 2, thus temperature of the gas may increase, decrease or remain unchanged. From the 1st law of thermodynamics, dq du + W 0 (U 2 U 1 )+( ) U U H 2 H 1 ΔH 0 his means that in the adiabatic expansion of a gas through fine orifice the heat content remains unaltered i.e. J- expansion is an isoenthalpic process. Joule-homson Coefficient(μ J ): he rate of change of temperature with pressure when enthalpy remains constant is called Joule- homson coefficient. ( ) δ μ J (2.11) H Its value can be evaluated as follows: We know for closed system, In an isoenthalpic change dh0 H f(, ) ( ) δh dh d + ( ) δh d δ Now,HU+ So, ( ) δh So, ( ) δh In J- expansion gas is real. So, ( δu δ ( ) δh So, ( ) δh d ( ) d d H μ J 1 C ) ( ) δu ( ) δu δ a 2 ( ) δh d δ C d 1 ( ) δh C ( ) δh ( a 2 ) ( δ ( ) δ() + δ δ() + ) ( ) δ() + (2.12) (2.13)

6 42 CHAER 2. HERMODYNAMICS ( ) o find δ(), for vander Waals gas we write for one mole gas, ( + a )( b) 2 R neglecting ab 2 which is very small, we get o evaluate ( ) δ ( ) δ() + a b R a ( ) δ 2 b 0 ( ) δ() a ( ) δ 2 + b (2.14) ( ) δh ( a ) ( ) δ hus, 2 + a ( ) δ 2 + b 2a ( ) δ 2 + b (2.15) we again write vander Waal s equation as So, equation 2.15 becomes herefore, ( ) + δ 1 2a 3 ( + a 2 ) R b ( a ) 2 δ ( ) δ ( ) δ ( ) δ ( ) δh Since 2a <<R ( R b ) R ( b) ( ) 2 δ 1 2a R 3 ( b) 2 1 2a [1 R 3 3 2a( b) ] 2 1 2a [1 R 3 2a ] (2.16) b + 2a 1 2 2a [1 R 3 b + 1 R 2a 2a b + 2a R b 2a R μ J 1 ( b 2a ) C R ( ) 1 2a R b C 2a ] (2.17) (2.18)

7 2.2. JOULE-HOMSON EFFEC 43 Conclusion: (i) At inversion temperature i, μ J 0 ( ) 1 2a so, C R b 0 i 2a Rb (ii) At high temperature 2a R <b, hence μ J is negative. (iii) At low temperature 2a R >b, hence μ J is positive. For H 2 and He a is very small, and so at ordinary temperature 2a R effect in J- expansion. (2.19) <b, so there is heating

8 44 CHAER 2. HERMODYNAMICS

9 Chapter 3 2nd law of thermodynamics 3.1 Drawbacks of 1st law From the 1st law of thermodynamics we learn that the different forms of energy are interconvertible and when one form of energy disappears an equivalent amount of another kind must appear. But it does not indicate whether a transformation of energy would at all occur or not and if it occurs, to what extent, in which direction the transformation would take place. o ascertain the direction of a chemical of physical process we need some more information or knowledge of some other criteria beyond 1st law.he principle which would determine the direction as well as extent of change is provided by 2nd law of thermodynamics.experience teaches us that water flows down the hill nd law of thermodynamics Clausious Statement: It is impossible for any self acting machine, unaided by any external agency, to convey heat from a body of lower temperature to higher temperature region. lanck-kelvin Statement: Heat can not be converted completely work in a cyclic process Identical nature of the two statement he two statement are very different to look at. but they are basically identical. Let us consider one heat engine (E) and one refrigerator (R) working between two heat reservoir at temperatures 1 (HR) and 2 (L R). Let us assume that the engine follows lanck-kelvin statement i.e. heat Q taken from HR is not completely converted into work; heat Q Q W is rejected to L R. Otherwise, that the refrigerator does not obey Clausius statement i.e. heat Q Q W taken by the refrigerator from L R is conveyed to HR without doing any work as shown in following figure(3.1). In the whole cycle, let us consider what happens. Heat lost by HR, Q Q Q (Q W ) W Work done in the whole cycle, W 0 W Heat gain by LR, Q Q 45 0

10 46 CHAER 3. 2ND LAW OF HERMODYNAMICS Figure 3.1: -K right hus in the whole cycle the HR losses heat W which is completely completely converted into work. Consequently, it is found that -K statement is not obeyed if the Clausius statement is not obeyed, although we assumed that -K statement is obeyed. Figure 3.2: Clausius right Now let us assume that the refrigerator follows Clausius statement i.e. it absorbs heat Q from LR, work W is done on it and it transfers Q + W heat to HR. And let the engine disobey -K statement i.e it absorbs heat Q from HR and all of it is converted into work W as shown in figure (3.2). In the whole cycle Heat lost by LR, Q Work done in the whole cycle, W W 0 Heat gain by HR, Q + W Q Q since Q W Hence heat, by itself, passes from low to high temperature. hus Clausius statement is not obeyed if -K statement is not obeyed although we assumed that Clausius statement is obeyed. herefore the two statement are identical.

11 3.1. DRAWBACKS OF 1S LAW Carnot s cycle French engineer Sadi Carnot explained how and to what extent work is obtainable from heat. Heat engine is a device which can convert heat to work. o estimate the work that can be obtained from heat during its passage from a higher to a lower temperature, the external agency (engine) must come back to its initial state so as to exclude any work involved in its own change i.e. the engine must operate in cyclic manner. o obtain maximum work in a cyclic process every steps should be carried out in a reversible fashion. In forward path engine does work and in backward path work is done on it to take it to the source again. Backward path should be such that net work obtained is always maximum. Efficiency (η) of an engine is the fraction of heat converted to work.if Q heat is absorbed, out of which W amount is converted to work then, η W Q. Figure 3.3: Carnot s cycle Carnot consider an imaginary cyclic process to find out an expression for highest efficiency of an engine. he typical Carnot cycle consists of four successive steps of operations using one gm-mole of an ideal gas as the working substance, enclosed in a cylinder fitted with frictionless and weightless piston. here are two heat reservoirs, one at high temperature ( 1 ), named as source and another at low temperature ( 2 ), named as sink. Now a reversible cycle consisting of four steps is carried out. Let initial state of the gas in the engine is ( 1, 1, 1 ) Step-I: he gas is placed over the source ( 1 ), external pressure is now lowered infinitesimally small amount to ( d ); volume expands to ( + d ). Some work is done by gas but temperature is not dropped as equivalent of heat has been absorbed from source. he process is continued till the volume becomes ( 2 ) and pressure becomes ( 2 ). his is an isothermal reversible expansion represented by AB portion of the figure (3.3). Heat absorbed by the gas dq Q 1 (say).

12 48 CHAER 3. 2ND LAW OF HERMODYNAMICS From 1st law of thermodynamics, for reversible process dq du + W AB dq du + int d For isothermal process, du 0 then, Q 1 0+ int d 2 Q 1 R 1 1 d Q 1 W AB R 1 ln 2 1. (3.1) Step-II: he engine is now covered by thermally insulated enclosure (adiabatic wall). he gas is allowed to expand further from volume 2 to volume 3 adiabatically and reversibly until the temperature falls down to that of sink 2 and pressure drops from 2 to 3. Heat absorbed by the gas, dq 0 dq du + W BC For adiabatic process, dq 0 so. W BC C d (3.2) 1 his is represented by BC portion of the diagram. Step-III: he adiabatic wall is now removed and system is isothermally reversibly compressed to such volume 4 2 Heat given out by the gas (to the surrounding) Q 2 Work done on the gas, W CD R 2 3 d W CD R 2 ln 4 3 (3.3) Step-I: he system is again covered by adiabatic wall and it is compressed adiabatically reversibly till the cycle is completed (DA portion of the cycle). dq du + W DA For adiabatic process, dq 0 so. W DA C d (3.4) 2 he total work produced in the cycle is the sum of the individual quantities. W cy W AB + W BC + W CD + W DA R 1 ln C d + R 2 ln C d R 1 ln R 2 ln 4 3 (3.5)

13 3.1. DRAWBACKS OF 1S LAW 49 So, and From BC adiabatic we have, 1 γ γ 1 3 From DA adiabatic we have, 1 γ γ 1 4 W cy R 1 ln R 2 ln 4 3 hus, (3.6) ( 1 2 )Rln 2 1 (3.7) η W cy Q cy ( 1 2 )Rln 2 1 R 1 ln (3.8) 1 his relation express the maximum amount of work obtainable from the heat flowing from 1 to 2. From the equation of efficiency 3.8 of a carnot engine it can be concluded that (a) efficiency depends only on the temperature of the heat source and sink not on the nature of working substance(gas). (b)he efficiency of carnot s engine will be unity if 1 or 2 0K. As infinite temperature or 0 K can not be attained, efficiency of carnot s engine can t be ONE, even under reversible condition. his means heat can not be converted into work completely, without rejecting some part of it to the sink. (c) If the source and sink are at the same temperature the efficiency of carnot s engine will become zero. his implies that conversion of heat into work is impossible without having source and sink at different temperatures. NB:In practice, the low temperature reservoir of a heat engine is often the atmosphere so that economic cost of producing work in the surrounding is mainly that for supplying Q Carnot heorem It states that no engine working between two temperatures can be more effcient than a Carnot s engine working between the same temperatures. Alternative statement: All reversible engines working between the same two temperatures have the same efficiency, whatever be the nature of the substance Entropy (i)dq rev is not a perfect differential but dqrev is a perfect differential : Suppose one mole of ideal gas undergoes a reversible expansion from volume 1 to volume 2 when the temperature changes from 1 to 2. hen from the 1st law of thermodynamics II I dq du + d dq II I 2 du + II 1 C d + d I 2 1 R d (3.9)

14 50 CHAER 3. 2ND LAW OF HERMODYNAMICS he integral on the right hand side 2 1 R d can t be evaluated unless we know the relation between and. If is constant, the integral will have one value and if varies during the process the integral will have different values. hus the magnitude of II dq depends on the way of expansion. I Now dividing by we get II I dq 2 1 C d R d C ln R ln 2 1 (3.10) hat means dq for a given change from state (I) to (II) can be evaluated without any reference is independent of the way the change has been is supposed to be state function. to the path of the transformation. Hence dq done. So dq (ii) In a cyclic process dq rev 0 From Carnot s cycle we know η Q 1 Q 2 Q Where Q 1 and Q 2 are heat absorbed and rejected from source at 1 and sink at 2 respectively. Now Q 1 Q 2 Q Q 2 Q Q 2 Q Hence for carnot s cycle which has four steps Q 2 2 Q 1 1 (3.11) dqrev Q Q Any arbitrary cycle can be regarded as being made up of a number of carnot s cycle. Since for a single cycle dq rev 0 so for any cycle dq rev 0. So dq rev is a state function and hence perfect differential. It is an extensive property. It is named as entropy (S) which is mathematically defined as S ds dq rev o understand the physical significance of the entropy let us calculate the entropy change during the following processes.

15 3.1. DRAWBACKS OF 1S LAW Heating (or cooling) of any substances (solid or liquid) of mass m: Let a solid (or liquid) of mass m and specific heat c is heated from 1 to 2. Its entropy change S2 S 1 ds ΔS II I 2 1 dq rev mcd mc ln 2 1 (3.12) During heating 2 > 1 i.e. ΔS is positive and during cooling it is negative isothermal expansion of an ideal gas: Let one mole of an ideal gas expands isothermally from ( 1, 1 )to( 2, 2 ).Its change in entropy S2 S 1 ds ΔS II I II I dq rev C d + d For isothermal process d 0 and R For one mole ideal gas.so, ΔS 2 1 R d R ln 2 1 R ln 1 2 (3.13) During expansion 2 > 1 so ΔS +ve i.e. entropy increases during expansion.

16 52 CHAER 3. 2ND LAW OF HERMODYNAMICS Non isothermal expansion of an ideal gas Let one mole of an ideal gas expands non-isothermally from ( 1, 1, 1 )to( 2, 2, 2 ).Here S2 II dq rev ds S 1 I II C d + d ΔS I 2 d C + R 1 2 C ln R ln 2 C ln R ln 1 1 R 2 2 R 1 1 C ln R ln d C ln R ln R ln 1 2 C ln 2 + R ln 1 (3.14) 1 2 o find out the value of absolute entropy let us integrate the following equation without putting any limit. We get dq ds C d + d d C + R d so, S C ln + R ln + constant (3.15) S C ln + R ln R + constant C ln + R ln R ln + R ln R + constant C ln R ln + S 0 (3.16) Where S 0 is new constant. Equation 3.15 and 3.16 are the expression for absolute entropy. However we can not use them because of our ignorance regarding the value integration constant. In fact classical thermodynamics can t measure absolute entropy. However at present with the help of statistical thermodynamics and third law we can calculate absolute entropy Entropy change during transition of state ransition at normal transition temperature is a reversible process. So heat change (absorbed or rejected) during transition (latent heat) is ΔQ rev. Hence entropy change during transition ΔS ΔQ rev ±L t (3.17)

17 3.1. DRAWBACKS OF 1S LAW 53 Where L t is the latent heat of transition per mole and is the transition temperature.during melting and vaporization heat is absorbed and hence entropy of the system increases whereas during solidification (from liquid to solid) and condensation (vapor to liquid) entropy of the system decreases Entropy change on mixing of gases wo or more gases when brought in contact would immediately diffuse into one another and mix up irreversibly. Such a spontaneous irreversible would lead to increase in entropy. Let n A and n B moles of two gases A and B be mixed at a constant temperature and under constant pressure. Before mixing their entropies are S A n A (C A ln R ln + SA) 0 S B n B (C B ln R ln + SB) 0 After mixing the total pressure remaining constant( ), the partial pressures are p A and p B ; such that p A + p B.he entropies of the two components in the mixture are he entropy change due to mixing S A n A (C A ln R ln p A + SA) 0 S B n B (C B ln R ln p B + SB) 0 ΔS m S Am + S Bm S A S B n A (C A ln R ln p A + SA)+n 0 B (C B ln R ln p B + SB) 0 n A (C A ln R ln + SA) 0 n B (C B ln R ln + SB) 0 n A R ln p A + n B R ln p B n A R ln p A n BR ln p B (3.18) Let n (n A + n B ) moles, so their mole-fractions are X A na n and X B nb n ; again, X A pa Now and X B pb ΔS m n A R ln X A + n B R ln X B nr( n A n ln X A + n B n ln X B) nr(x A ln X A + X B ln X B ) In general for mixing of any number of gases ΔS m nr X i ln X i (3.19) herefore ΔS m /mole X i R ln X i (3.20) +ve

18 54 CHAER 3. 2ND LAW OF HERMODYNAMICS Helmholtz free energy(f) Mathematically, Helmholtz free energy is expressed as : FU-S o understand its physical significance let us differentiate it. df du ds Sd From 1st law, dq rev du + W max ds du + W max So, df W max Sd df W max (3.21) So,Helmholtz free energy is the function decrease of which measures maximum amount of work which is obtainable from the system isothermally and reversibly.it is a state function and extensive property Gibbs free energy(g) If S is the entropy of a system at 0 K and H is its enthalpy then, Gibbs free energy is mathematically expressed as : GH-S On differentiation we get, dg du + d + d ds Sd W max + d + d Sd dg, W max d (3.22) W net Where d is the mechanical work involved in the system itself during the transformation and W max denotes the maximum work output. Hence (W max d) is the amount of work received for any external use, exclusive of the mechanical work in the change of the system itself. Hence, Gibbs free energy is the property of the system whose decrease is the measure of the maximum external work available during the transformation of system reversibly at constant pressure and temperature.

19 3.1. DRAWBACKS OF 1S LAW Some formulae: From 1st law of thermodynamics, for mechanical work, dq rev du + d ds du + d du ds d (3.23) H U + dh du + d + d dh ds + d (3.24) F U S df du ds Sd using eq. [3.23] df d Sd (3.25) G H S dg dh ds Sd using eq. [3.24] dg d Sd (3.26) ( ) df Now from eq and 3.26 we get respectively : d ( ) dg and d S (3.27) S (3.28) So, entropy (S) is the decrease in work function (Gibbs or Helmholtz free energy) per degree rise in temperature at constant volume (or constant pressure) Condition of spontaneity of any change of a system Many spontaneous changes are accompanied with a decrease in enhalpy. But this is not universally true. Reaction with increase in enthalpy may also happen spontaneously. We have seen that process which lead to a net increase in entropy would spontaneously occur. When the system is in equilibrium, for any virtual change, ΔS net 0 For any spontaneous irreversible process,δs +ΔS > 0 For any reversible process (equilibrium)δs +ΔS 0 Where ΔS and ΔS are the entropy changes of the system and surroundings respectively.hus in the computation of entropy change, the changes in entropy of both system as well as surroundings have to be considered and which is not feasible. herefore, there must be some other function we need to consider, and the function is Gibbs free energy change of which (ΔG) will be more convenient to find out possible direction of the transformation of a system. From 1st law of thermodynamics, For a reversible occurrence, ΔQ rev ΔU + W rev (3.29) For a irreversible occurrence, ΔQ irrev ΔU + W irrev (3.30)

20 56 CHAER 3. 2ND LAW OF HERMODYNAMICS For a process taking place from a given initial to a given final state, ΔU would be the same. But we know W rev >W irrev ; hence ΔQ rev > ΔQ irrrev. herefore for an infinitesimal change δq irrev δq rrev < 0. By definition G H S U + S dg du + d + d ds Sd dq + d ds Sd Work is supposed to be purely mechanical. Chemical reactions are carried usually at a fixed temperature and pressure, therefore For a spontaneous process hence irreversible dg, dq ds ΔG, ΔQ irrev ΔS ΔQ irrev ΔQ rev but, ΔQ rev > ΔQ irrrev so, ΔG, < 0 (3.31) For reversible process (equilibrium), ΔG, ΔQ rev ΔQ rev 0 (3.32) i.e. in a spontaneous change ΔG must be negative. o find out any process to be spontaneous or not at specified and, we have to calculate only ΔG of the system only. If ΔG is NEGAIE, the change shall spontaneously occur and if OSIIE the process will not occur spontaneously but its reverse may be.if the ΔG is ZERO, the initial and final states would exist at equilibrium with each other Gibbs-Helmholtz equation Let G 1,H 1 and S 1 be the Gibbs potential, enthalpy and entropy of a system at temperature initially.it undergoes isothermal change to G 2,H 2 and S 2, then G 1 H 1 S 1 and, G 2 H 2 S 2 So, G 2 G 1 (H 2 H 1 ) (S 2 S 1 ) { } δg2 δg1 ΔG ΔH δ δ ΔH + δ δ (G 2 G 1 ) So, ΔG ΔH + δ δ (ΔG) (3.33) Similarly it can be shown that, ΔF ΔU + δ δ (ΔF ) (3.34) Equation 3.33 and 3.34 are known as Gibbs-Helmholtz equation. Show that ( δ G ) δ 2

21 3.1. DRAWBACKS OF 1S LAW 57 δ δ ( ) G 1 δ δ (G) G 2 S G 2 H 2 (3.35) Clausius-Clapeyron equation ransition of state (e.g. melting, boiling and sublimation) takes place at a particular temperature at a given pressure. ariation of transition temperature with pressure is given by an equation in known as Clapeyron equation. o derive this equation let us consider an equilibrium between two phase I and II of the same substance at transition temperature at pressure. I II We know at equilibrium ΔG, 0 At equilibrium if little amount of I converts to II an equal amount of II will convert into I. If dg 1 and dg 2 are corresponding free energy change then, dg 1 dg 2 1 d S 1 d 2 d S 2 d Where ( 1, 2 ) are molar volume and (S 1,S 2 )are molar entropies of phases I and II respectively. ( 2 1 )d (S 2 S 1 )d d S 2 S 1 d 2 1 ΔS 2 1 L t (3.36) ( 2 1 ) Where ΔS is molar entropy change during transition of state and L t is latent heat of transition per mole. he equation 3.36 is called Clapeyron equation. Clausius simplified the Clapeyron equation specially for vaporization.in the case of vaporization Clapeyron equation 3.36 can be written as d L v d ( g l ) where L v is latent heat per gm-mole Since volume of gas, g >>volume of liquid, l, then d d L v. g Assuming the vapor to behave like a ideal gas, for 1 mole g R d d d L v. R so, L v d (3.37) R 2

22 58 CHAER 3. 2ND LAW OF HERMODYNAMICS If we further assume that the latent heat practically remains constant over the temperature range from 1 to 2, where 1 and 2 are boiling point at pressure 1 to 2 respectively, then the integration of the eq gives 2 1 d 2 1 ln 2 L v 1 R his eqn is the integrated form of Clausius-Clapeyron eqn routon s rule From eqn for the vaporization of a liquid L v R 2 d [ ] 1 (3.38) ln Lv R + C 1 for 1 gm-mole. If we consider the liquids at their boiling temperature ( b ), at the 1 atmospheric pressure ( 1), then L v + C 1 ln1 R b L v R b C 1 L v b RC 1 Constant. (3.39) Where L v is the molar latent heat of vaporization. routon had made this important generalization from experimental observation and stated that the ratio of the molar heat of vaporization to the boiling temperature is constant for simple or non-associated liquids and the constant is approximately 21. Now we know ΔS ΔQ Lv for vaporization so in otherwords the molar entropy change on vaporization at the boiling temperature is constant Standard molar free energy We know dg d Sd At constant temperature, dg d G So, dg G 0 G G 0 Where G 0 is the standard free energy of the system at 1 atm and same temperature.if the system is one of n moles of a perfect gas at temperature, we can write, G G 0 nr + 1 d G G 0 + nr ln (3.40) 1 1 d d G n + R ln n μ μ 0 + R ln (3.41) G0

23 3.1. DRAWBACKS OF 1S LAW 59 Usually free energy of a gm-mole of a substace is expressed by the symbol μ Free energy change in mixing of gases At a constant temperature under a pressure, the molar free energy of a pure ideal gas i is μ i μ 0 i + R ln (3.42) Under the same total pressure if this gas be present mixed with other gases and its partial pressure be p i then its molar free energy μ i will be μ i μ 0 i + R ln p i μ 0 i + R ln x i (where mol-fraction x i pi ) μ 0 i + R ln + R ln x i μ i + R ln x i (3.43) he indicates the pure state of the substance. Suppose at a constant temperature and under the same pressure; n 1,n 2,n 3, moles of different gases are mixed when their initial molar free energies are μ 1,μ 2,μ 3,.After mixing, let the molar free energies be μ 1,μ 2,μ 3, etc. Since free energy is an extensive property, hence G before mixing n 1 μ 1 + n 2 μ 2 + n 2 μ 2 + G after mixing n 1 μ 2 + n 2 μ 1 + n 3 μ 3 + so, ΔG mixing G after mixing G before mixing n 1 (μ 1 μ 1)+n 2 (μ 2 μ 2)+n 3 (μ 3 μ 3)+ n 1 R ln x 1 + n 2 R ln x 2 + n 2 R ln x 2 + [ n1 nr n ln x 1 + n 2 n ln x 2 + n ] 3 n ln x 3 + where n n 1 + n 2 + n 3 + and x 1,x 2,x 3, are the mol fractions of the different components. ΔG mixing nr (x 1 ln x 1 + x 2 ln x 2 + x 3 ln x 3 + ) (3.44) nr x i ln x i (3.45) his means the free energy decreases in mixing of ideal gases. When only two ideal gases are mixed, ΔG nr [x 1 ln x 1 + x 2 ln x 2 ] nr [x 1 ln x 1 +(1 x 1 ) ln(1 x 1 )] (3.46) When ΔG is plotted against x 1 from this relation, a curve as in 3.4 with a minimum is obtained. he free energy decrease is maximum at the minimum point. So, at that minimum d(δg) dx 1 0, solving, it is found at the minimum point,x hus,free energy Figure 3.4: Free energy change on mixing decrease will be largest when equimolar quantities of the two components are mixed.

24 60 CHAER 3. 2ND LAW OF HERMODYNAMICS Maxwell s relation dq du + d du ds d ( ) δu So, δ S [( ) ] ( ) δ δu δ S ( ) δu Again, [( ) ] ( ) δ δu δ δ δ S S Since U is perfect differential, [( ) ] δ δu δ ( ) So, S δ [( ) ] δu δ S ( ) δ δ S (3.47) (3.48) (3.49) H U + dh du + d + d dh ds + d ( ) δh So, S [( ) ] ( ) δ δh δ S ( ) δh Again, [( ) ] ( ) δ δh δ S S Since H is perfect differential, [( ) ] δ δh δ [( ) ] δh S S δ δ So, S (3.50) (3.51) (3.52)

25 3.1. DRAWBACKS OF 1S LAW 61 F U S df du ds Sd df d Sd ( ) δf So, δ [( ) ] ( ) δ δf δ δ δ ( ) δf Again, S δ [( ) ] ( ) δ δf δ δ δ Since F is perfect differential, [( ) ] δ δf δ [( ) ] δf δ δ δ δ So, δ δ (3.53) (3.54) (3.55) G H S dg dh ds Sd dg d Sd ( ) δg So, [( ) ] ( ) δ δg δ δ δ ( ) δg Again, S δ [( ) ] ( ) δ δg δ Since G is perfect differential, [( ) ] δ δg δ [( ) ] δg δ δ δ δ (3.56) (3.57) (3.58) Equation 3.49,3.52,3.55 and 3.58 are known as Maxwell s relation Open system thermodynamics So far we have discussed, the system was assumed to be closed. So the free energy (G) ofa given system as a state function depends upon the parameters (,,). Since these three are interrelated we may write G f(, ). But if we take an open system in which quantities of the components may also vary. the free energy will depend upon the amounts os the components. Suppose we have a system containing n 1 moles of A 1, n 2 moles of A 2,n 3 moles of A 3 etc. then its free energy G f(,, n 1,n 2,n 3, ) (3.59)

26 62 CHAER 3. 2ND LAW OF HERMODYNAMICS If all the variables undergo change, the total differential will be given by dg ( ) δg d +,n 1,n 2, ( ) δg d + δ,n 1,n 2, ( ) δg δn 1 dn 1 +,,n 2,n 3, ( ) δg δn 2 dn 2 +,,n 1,n 3, We know, dg d Sd ( ) δg so,,n ( ) i δg S δ,n ( ) ( ) i δg δg dg d Sd + dn 1 + dn 2 + (3.60) δn 1,,n 2,n 3, δn 2,,n 1,n 3, ( δg he term δn 2 means that, and all other mole numbers except n 2 have been kept ),,n 1,n 3 constant during differentiation.it is called partial molal free energy of that component(a 2 ) he change in free energy per mole addition of a component keeping other components fixed at a fixed temperature is called partial molal free energy. he partial molal free energy expressed as ( δg δn i ),,n j for the i th component is called its chemical potential and is denoted by the sumbol μ i.hus, chemical potential is the rate of increase in Gibbs free energy per mole of the component (i) added or it is the free energy increase of the system when one mole of this component is added to an infinitely large amount of the system so as to keep the composition practically unaltered. Now we can write the equation 3.60 as dg d Sd + μ 1 dn 1 + μ 2 dn 2 + dg, μ 1 dn 1 + μ 2 dn 2 + (3.61) Now imagine at constant and all the constituents are increased in the same proportion (Δx) i.e. A 1 is incresed by an amount n 1 Δx, A 2 is incresed by an amount n 2 Δx moles,,a i is incresed by an amount n i Δx moles. Hence dn 1 n 1 Δx, dn 2 n 2 Δx, dn i n i Δx Since G is an extensive property, this will also increase by an amount GΔx i.e. Differentiating we get dg GΔx GΔx μ 1 n 1 Δx + μ 2 n 2 Δx + G n 1 μ 1 + n 2 μ 2 + (3.62) n i μ i dg (μ 1 dn 1 + μ 2 dn 2 + μ 3 dn 3 + )+(n 1 dμ 1 + n 2 dμ 2 + n 3 dμ 3 + ) (3.63) Substracting eqn 3.61 from 3.63 n 1 dμ 1 + n 2 dμ 2 + n 3 dμ (3.64) he eqn 3.61,3.62 and 3.64 are known as Gibbs-Duhem relations. ***** One of the important aoolication of Gibbs-Duhem equation is prediction of condition of equilibrium.for simplicity we consider the phase of a component I and II at equlibrium, having

27 3.1. DRAWBACKS OF 1S LAW 63 molal free energies ḠI and ḠII.Now at equlibrium if an infinitesimal amount (dn) gm-moles of a component passes from phase I to II. hen from Gibbs-Duhem relation we get μ 1 dn 1 + μ 2 dn 2 dg At equilibrium, μ 1 (dn)+μ 2 ( dn) 0 μ 1 μ 2 Ḡ I ḠII (3.65) he chemical potential or partial molal free energy of a component in every phase must be the same under condition of equlibrium. hus in a closed vessel containing liquid water in eqm. with its vap the chemical potential of water in the liquid and in the vapor phase shall be same, though the masses may widely differ roblems Q:1:Show that ( ) δu δ We have a 2 for 1 mole real gas. For closed system, S f(,) ds d + d δ δ ds d + d δ δ ( ) d + C v d (3.66) δ Comparing 3.66 and 3.67 Again, U f(,) δu δu So, du d + d δ δ ( ) δu ds d d + C v d (3.67) δ ( ) ( ) δu d + d δ δ δu + δ δ From Maxwell relation, δ δ δu so, + δ δ d (3.68)

28 64 CHAER 3. 2ND LAW OF HERMODYNAMICS (i)for ideal gas ( R ) δ So from eqn 3.68, ( ) δu + δ ( ) δu δ R R (ii)for one mole real gas obeying vander Waal s equation + a R 2 b ( ) R δ b So from eqn 3.68, ( ) δu + δ ( ) δu δ 0 (3.69) R b + a Standard equation of state for any material a 2 (3.70) Let x, y, z be the three properties of system so that, x f(y, z); y f(z,x); z f(x, y) then δx δx dx dy + dz (3.71) δy z δz y δy δy dy dx + dz (3.72) δx z δz x δz δz dz dx + dy (3.73) δx y δy x utting the value of dx in the equation 3.72 herefore, dy ( δy δx dy + So, ) z [ (δx ) ( δy δx δy ) z dy + z ) ( δx δz δy δx δx z δz δy δx δx δz ( ) δy δx z ( δx δz ) z y ( ) δz δy ( ) δx δz dz + y dz y ) ( δy δz dz y y x ] + ( ) δy dz δz x dz (3.74) x ( ) δy dz δz x ( 1 ) δz δy x 1 (3.75)

29 3.1. DRAWBACKS OF 1S LAW 65 hus for the three variables,, ( ) δ δ δ δ 1 (3.76) his is a standard equation of state true for any material.