So ItS Some randomness of distribution. entropP

GIBBS AND HELMHOLTZ FUNCTIONS function as Gibbs function (G) and Helmholtz (A) thermodynamic quantities it is the Gibbs function G is otherwise called Gibbs free energy and called Helmholtz net work function. Helmholtz Function A is otherwise function free energy and it is the maximum work

The Gibbs free energy G - Net work function

Definition G H-TS Gibbs ree energy (also called enelgy funetion) G is detined. thetmodynamie a n which e. 1G property, decrease gives thee e wotk that can be obtained frem a system during a change

G W PAV Relationship between Gibd's tree energy change and enthalpy chans. From detinition

Since HT and S are state properties, G Is also a state property For change at constant temperature change in Gis given as Fora

dG dH -T dS dG IS an exact differential Gibbs free energy is useful to study ch. at constant pressure and temperature For a ftinite change 1G= H TAS Relation between helmholtz free energy and Gibbs tree energy we knew

H=NE PAV

1G E PAV-T 1S

But E TIS = 4

So AG= 14 PIV

work To show that the in decrease Gibb's tree energy gives the net work other than be work work of expansion done by a system) ree We knew IG JA PAV stud But A-W NG-W PAV Difs or-AG W-P1G= W The Helmholtz free energy A Maximum work function Definition : A=E- TS

Helmoholtz tree also caled work A energy function) s defined as

a a decrease in which i thermodynamie property, e,A gives he maximum work that can be otained trom a system duning a qive change A 91

Toshow AA W.

From definition A E TS

Since E T and S are state properties A is also a state property he change in A for a change in the system at constant temperature is given below dA dE TdS Since A is a state property, dA Is an exact different:al For a finite change A E- TAS Where A Is the change in the function A E Is the correspondi ng change in internal energy and 1S Is the change in the entropy of the system For a reversible change at constant T the heat absorped is

From definition Q,T=AS Substituting this in the above equation we

get AA E Q But from First law of thermodynamics E = W W E e A =-W rev A W

done is the max1mum We know in a reversible process the werk -JA work thust can work So, the decrease in A, e, gives the maximum So the Heimholts be obtained from a system during a given change A, useful to work function Heimhoitz free energy is free energy is called a and volume study change at constant temperature and Gibb's free Differences between Helmholtz free energy energy

Helmholtz Free ener9y Gibb's Free energy

A E- TS Definition that ceig be Maximum work Usefu wers Indicatess content of a system octaned (Networ

alue A W 92

Two show how TAS determines the randomness of a reactinn We know AG= AH - TAS action

or TAS AH AG

=Enthalpy useful work (Heat absorbed) unavailable enerqy

Thus TaS is a measure of which unavailable energy, i.e., the ene. could not be converted into work energ, If TAS > AH; then AG negative

Such a process will be All lead spontaneous. spontaneous to an increase in the processe disorderliness or randomness of the So if TAS is > AH system randomness will increase. if < randomness will decrease. Similarly TAS AH then

Thus TAS determines the randomness of a reaction. TAS us an idea of the also gives unavailable energy. AA AND aG AS EQUILIBRIUM AND SPONTANEITYCRITERIA FOR THERMODYNAMIC General condition for equilibrium and spontaneity / condition for a reversible and Criteria Helmholts free spontaneous process in terms of energy (or) Condition for under Constant Equilibrium and spontaneity termperature and constant volume Consider the general conditions for equilibrium and spontaneity - dE-Pon dV - du + T dS 20 At Constant volume PopdV = 0 du

- o-(dE - du dV TdS) >0

At constant = temperature dA dE - T dS

So - dA - du 2 0

No work other than work of expansion takes in reactions and place the chemical ordinary physical changes. For these changes du=0 and o-dA 20 dA 0 Condition for equilibrium dA 0 Condition for spontaneity dA < 0 93

Criteria/ condition for a reversible and spontaneous procesS In terms of Gibb's free energy (or) Condition for equilibrium and Spontaneity under constant Temperature and Pressure.

Consider the general condition for equlibrium and spontaneity dE-Pan dV - du + T dS >0

At constant pressure P Pop Therefore

-dE- PdV du + Tds 20

(dE PdV) - du+ TdS20;-dH- du+ TdS 20

- At constant temperature dH - TdS = dG; -dG du>20 reactions takes in chemical No work other than work of expansion place For these changes and ordinary physical changes, dG20: dGS0

T dG0 Condition for equilibrium at constant and P and P dG <0 Condition for spontaneity at constant.T constant temperature carried out at chemical reactions are Many free energy of the system reactions the Gibbs In these will not and pressure. the free energy attains equilibrium, a reaction decreases. When

decrease.

Conclusion tell whether a change law is unable to earlier that first have seen not. That was why We spontaneously or reaction can take place law is able to state or a how the second of let us see second law. Now needed a a reaction. we spontaneity of or the predict the feasibility

accordit is reaction own or on its Feasibility ofa spontaneously take place /f a reaction can reaction said to be feasible of a the feasibility method in predict feasible gives a will be Second law AS> 0 the reaction AG <0 or if to second law, if We know According spontaneous. reactions are exothermic Normaliy G IH TAS n Pvoetle feattions H is So hegative 1 will be in te invided 1S ie poeitive So nermally exothermic reactins, 1 hanentis bwe fat an bvothetmic et reactions are spontana. i h owee exetherimic teaction where AH IS anenus ale le negativa hegative is will be negative Ih such a case - 1S) 1G-H TAs Now ill be negative only if- H> T1S peelle iT1S>- H then iG i thie imeane that the reaction e cannot take place Thus ws ha ales sene cattot determine the feasibiity of a reaction ah ehdothetimie teartion is positive 11S beeohmee hedative 1S is negative than

e - H-1 1S ositive hegativel= positivePositive Positive hat enine stuch a teaction cannot take place t atn ehdothettmie procese to S eitive take place nust be highly so that-1S is greater than Hwhich is positive ie- 1S >H hett il wil ecome hegativve tth tottee fot We maturat changes khow tree that the enetdv of a system decreases during ptaheois A natura lhange definite change in Iree ih the energy for a ahge system at constant natural eietelhh temperature is given by the following 13 - T1S When C ie hegative the change is eld t spontaneous or natural The depends AH and 1S If H is e megative and 1S hegative is elated to the is positive 1G take ener gv ef the system For natural t rlare the etnetdy of the changes systetm should decrease hisutd netease is and his observed in entropy excthermic If teee in a reactions rettain eharge then the enthalpy itst entropy should inctease to ramp eneate the increase in larger enthalpy This is etdti te tom observed in artime thesp We can conchide ie t th at the drivina haas ap fhp fart eaee entlalp a/1d inre ase in 95 CHANGE ENTROPY ADVANTÁGES OVER is AND equilibrium (AA AG) towards THEIR physico-chemical systerm movement of the energy The the other entropy and one involving towao by two factors, ontrolled processes proceed we know, spontaneous of entropy the system. the value f At equilibrium, with increase of entropy process equilibrium of a spontaneous the other hand the passage of the maximum. On of energy, a decrease s accompanied by equilibrium is is minimurm the system towards the energy of Hence at equilibrium. may or sportaneity system. thermochemical equilibrium for a Thus the criteria follows be given as (SVT20 (i) (OE) v1 () conditions of studied under are rarely at chemical reactions carried out Dut reactions are Usually or constant energy. temperature entropy constant constant and at (atmospheric pressure) volume (in constant pressure out at constant are carried Sometimes they necessary thermostat). it becomes (in a temperature. So and at constant or AG. bomb calorimeter) in terms of AA or spontaneity the criteria for equilibrium to specify as follows: They are given (AT0:(G)p0 more that are over AS is they of AA and AG Thus the advantage concrete problems. some the of practicaly, in study convenient will be whether the process will indicate of AG or AA Thus The sign is spontaneous. 2 the reaction is negative or not, If it spontaneous with respect to AS can not say the we in deciding more important and AS. AS is Between AG reaction: than AS is spontaneity advantageous AS, ofa is more seen that AG is we have as AG itself Though a reaction spontanity of in deciding the more important dependant on 1S AG AH- TJS and AH negative and If,AS is positive on AH 1S. so see G, depends sufficiently large We is negative and But if 1S not will be negative. the reaction is than 1G positive. Then 1G will become be > -H, than must that then JS TiS (endothermic reactions) is positive may be feasible if AH so that G numerical value should have large positive and 96

important G in deternh that 1S is more than. negative Thus we see fU the spontaneity of a reaction. from a practical advantageous pain At more Thus 1G and A are of from a theoritical point vie more important view where as iS is T gu VARIATION OF G ANDA WITH P,V AND is given by the followi free energy Change in Helmholtz wing fundamental equation dA -SdT PdV (11)

= 0 Ai constant volume PdV GA -S

Tv free Therefore the energy is is always positive. Entropy of a system of a gas is very high Since the entropy to temperature. related an increase in inversely drastically with the decreases tree energy of gas becomes equation (l1) constant temperature, temperature At A -P

decreases with volume. The free energy The following fundamental is related to T and P. Gibbs free energy in T and i) change with change P relates Gibb's free energy equation (12) dG=-SdT+VdP

dP = 0 At constant pressure -S

temperature =0 At constant dT G V

related to temperature But at 1G is inversely constant pressure At related to pressure 1G is directly constant temperature 97

dG-SdT Vd P Further

temperature dT = 0; At constant dG VdP

ideal gas PV =nRT But for an nRT dP

- So dG nRT V = and P P 62 P2 dP Integrating G1 dG nRTp P

G,-G, nRTIn P2 or P P2 or AG 2.303 nRT log.

AG 2.303 nRT log

RELATIONS EQUATIONS OF STATE MAXWELL'S THERMODYNAMIC dE + w thermodynamics q= From the First law of to work of expansion the work is restricted is reversible and If the process dE + PdV alone W =PdV; Dividing throughtout byyT dEPdV But rev dS by definition rev T T dEPdV dE PdV dS or So dS T T

dE = TdS - PdV (13) dE PdV TdS or It law of thermodynamics. first and second This equation incorporates The definitions of thermodynamics fundamental equation is known as of H, A and G are H A E - TS G E+PV-TS H E PV; 98

VdP Equations 9 10 22) are On differentiation dH dE PdV rio as Marwes eatos rom(13) and (1(14) we can dern dA dE- TdS - SdT dE PdT VdP. TdS-SdT dG (13) these equations becom. After substituting dE from equation dH TdS VdP and (11) dertye - we can dA SdT PdV Crom13) dG-SdT VdP (12 dE TdS- Pdv Let us take equation (13) At const. V dE=TdS = T we derive or and (12) can (15) From (14

At const. S dE =-PdV (16) we can derve (11 ) and (12) From S Differentiating (15) with respect to V at constant E (17) known as ma (23) to (28) are aiso (S) OV) Eguatons

Differentiat1ng (16) with respect to S at constant V Solved University Proble- to convet one E the heat requried Calculate

(aV) 0s) (into its vapour) (18) cas steam at Cis540 heat of 100 (Latent 100C =54 and we have steam at Comparing (17) (18) Latent heat of Solution water Molecular weight of 540 x (19) 1 moie 9720 cals Imole 540x 18 = one ma to convert Heat required Following the same mathematical procedure (20). (21) and (22) Answer vapour 9720 cal. can be derived from (14). (11) and (12) respectively at 25 C and of nitrogen intially 2 100g 1 atm pressure of 20) constant against a gas (C,for temperature the OP the finai of

21) volume Calculation final Solution of is adiabatic change for an and v P.V (22) 1atm 10atms (given) P, 99

to Equations (19) (22) are known as om (13) and (14) we Maxwells can derive reiations. Similariy E OH (23) OS OS From (13) and (11) we can derive

(24) doStivdp »1 From (14) and (12) we can derive d64 ds.vd d (25) OP

From (11) and (12) we can derive OH aG (26) OT

maxwells relations also known as to are Equations (23) (26)

Solved University Problems C of water at 100 one mole to convert the heat requried 1 Calculate -Mar 89. (into its vapour) 540 cals / gm) 100 C is at cals/ gm of steam = 540 (Latent heat at100 C heat of stam Latent water Solution: of

Molecular weight into its = 540 x 100"C mole mole 1 cals / water at 9720 mole of x 18 one 540 convert to required adiabatically Heat expands Answer: atm and and 10 volume vapour= 9720cal. 25°C the final initially at Calculate deg/mole) nitrogen atrn. cais / of of 1 is 6.96 - Nov. 79 2100g pressure 2 for the gas constant P gas (C, between against a the of relation temperature

w the