aper-ii Chapter- S-equation, Maxwell s equation Let heorem: Condition o exact dierential: Where M Hence, z x dz dx and N Q. Derive Maxwell s equations z x z zx, z dx + dz Mdx + Nd z d Now 2 z x M N x x M x 2 z x N x d x i Helmholtz ree energ: Helmholtz ree energ or Helmholtz unction is deined as F U S For a pure substance undergoing an ininitesimal reversible process df du ds Sd As ds du + d, So, we can write df d Sd As F is exact dierentiable tpe z Mdx + Nd, so we get S 1
S his is Maxwell s 1st equation. his relation can be written as S From this relation we see that or an thermodnamic sstem, the increase o entrop per unit increase o volume at constant temperature is equal to the increase o pressure per unit increase o temperature in an isochoric constant process. iimaxwell s 2nd relation: Gibbs ree energ: Gibbs ree energ is deined as G U S + For an ininitesimal reversible process dg du ds Sd + d + d As ds du + d dg Sd + d As G is exact dierentiable, so we get S his is known as Maxwell s 2nd relation. his relation can be written as S From this relation we see that the decrease o entrop per unit increase o pressure at isothermal process is equal to the increase o volume per unit increase o temperature at isobaric constant process. iii Maxwell s 3rd relation: Internal energ: We have rom irst law o thermodnamics dq du + d 2
ds du + d du ds d As U is exact dierentiable, S S his is Maxwell s 3rd relation. his relation sas that the increase o temperature per unit increase in volume in a reversible adiabatic isentropic, S constant process is equal to the decrease o pressure per unit incease in entrop in an isochoric constant process. iv Maxwell s 4th relation: We have enthalp H U + dh du + d + d dh dq + d dh ds + d As H is an exact dierentiable, S S his is Maxwell s 4th relation. his relation sas that the increase o temperature per unit increase o pressure in a reversible adiabatic process isentropic, S constant is equal to the increase o volume per unit increase o entrop in an isobaric process. heorem II: I x,, z x z x z 3
roo: Let Let z x dx d x z x x, x utting the value o in eq1, we get Let dx x dx x dx x d + d + + dx d +, z x d + z x z x x 1 d + z d + z z x x, z x d + z d + x z Comparing 3 and 4, we get coeicient o d is x z x + d 1 dz 2 dz z x z x z x dz dz 3 dz 4 z and that o dz x x z z z x 1 x z 4
heorem: I x x, z and z z, x, then x z 1 z x z x roo: Let Let dx dz x x, z x z d + z z z, x d + x x z z x utting the value o dz in eq1, we get x x z dx d + d + z z x x x z x dx d + d + z z z x x x z x dx + d + z z From 3 we get coeicient o dx Coeicient o d z x x z 1 z x 1 x z x + z x z 5 z x z 0 x dz 1 dx 2 z dx x z x z x dx dx 3
x z x z x x z x z x z z z x x x 1 z z 1 Q. rove that roo: Let Let d d utting the value o d in eq1, we get d d d d + d + + From 3 we get coeicient o dx, d +, d + x 1 d + d + 6 d 1 d 2 d d + 1 d d 3
Coeicient o d 1 + 0 1 1 Q. Derive -ds equations First ds equation Let entrop S is a unction o and. i.e. S S, S S ds d + d 1 Multipling both sides b, we get S S ds d + d 1 S S ds d + d Q S ds d + d 7
As speciic heat at constant volume C Q ds C d + From Maxwell s irst equatin we get S So, we get rom 2 ds C d + his is known as irst ds equation. S d 2 d 3 Second ds equation Let entrop S is a unction o and. i.e. ds S Multipling both sides b, we get S ds S ds Q ds S S, As speciic heat at constant pressure C d + S d + d + d + ds C d + 8 Q d 1 S d 1 S S S d d d 2
From Maxwell s second equation we get So, we get rom 2 S ds C d his is known as second ds equation. d 3 hird ds equation Let entrop S is a unction o and. i.e. S S, S S ds d + d 1 Multipling both sides b, we get S S ds d + d 1 S S ds d + d Q S ds d + d Q S ds d + d ds C his is known as third ds equation. d + C d 2 Q. Caculate C C 2 9
We have irst ds equation and second ds equation From 1 and 2 we get C d ds C d + ds C d C C d d d 1 d 2 d C d + d C C d + Let temperature is a unction o and. i.e. d +, C C d d 3 d d + d 4 Comparing 3 and 4, we get coeicient o d C C C C C C 10
Coeicient o d, we also get Again we have From 5 and 6 we get Case-1: Here 2 C C C C C C +ve alwas and 1 1 2 C C +ve e. Hence 5 6 C > C Case-2: When 0 i.e. at absolute temperature C C 0 Case-3: I C C 0. e.g. Water at 4 0 C, densit maximum. C C 0 11
C C Case-4: For perect gas R d + d Rd When constant, When constant, So we get R R d + d 0 2 C C R 2 C C Case-5: or rove that C C 9 α2 K and volume compressibilit C C R2 C C R. We have volume expansion coeicient γ 1 γ K 1 12
K 1 K So we get 2 C C C C γ 2 2 1 K C C γ2 K In terms o linear expansion coeicient α γ, we get 3 C C 9 α2 K Q. rove that S C C C. We have irst ds equation ds C d + For isentropic process S constant, ds 0 Again we have 0 C d + C d C C C d 1 d d S 13 1 2
Dividing 1 b 2 we get C C C C C S C S Q. rove that S C C C. We have second ds equation ds C p d For isentropic process S constant, ds 0 0 C d d C d d C Again we have Dividing 1 b 2 we get C C C C C S d 1 S C C C 14 S 1 2
Q. State and explain Joule s law For an ideal gas, the internal energ does not change with volume and pressure when temperature remains constant. i.e. U and U 0 Hence internal energ U does not change unless temperature changes. 0 U U onl or an ideal gas. his is known as Joule s law. So, Explaination: he properties o a substance are related b ds du + d Let the internal energ U is a unction o and. i.e. ds 1 du + d 1 U U, U U du d + d 2 Substituting the value o du in 1, we get Again, we consider ds 1 ds 1 U U U d + ds S d + 1 U S S, d + 15 S d + d + d 3 d 4
Comparing 3 and 4, we get coeicient o d 1 U S coeicient o d 1 U S + Now dierentiating 5 w.r.t. when constant, 1 2 U 2 S Dierentiating 6 w.r.t. when constant, From 7 and 8 we get For all ideal gas 1 2 U + 1 U 2 S + 2 1 2 U 1 2 U + 1 U + 2 1 U + 1 2 utting this value in 9 we get U + R R R U + 5 6 7 8 9 16
U + U 0 So, internal energ U does not change when volume changes at constant temperature. Similarl, i we take U U, then we get For all ideal gas utting this value in 9 we get U U + R R R 10 + U + U 0 So the internal energ does not change when changes at constant temperature. Hence, U U hase transition: here are three tpes o states o a substance. e.g. i Solid 17
ii Liquid iii Gas Each o these states is called a phase. A phase is deined as a sstem which is homogeneous having deined boundaries. hase transition is a process in which a thermodnamical sstem changes rom one state to another with dierent phsical properties: Solid Liquid gas In most cases, phase transition takes place b appropriate changes in temperature and / or pressure. All phase transitions are essentiall due to the interpla o attractive orces between the particles o the sstem and their thermal motion and trap them in bound states, thermal energ leads to random and ree motion. Melting, vaporisation and sublimation are the most amiliar phase transitions. First order phase transition: I in transitions, the temperature and pressure remain constant, but the entrop and volume change. his is known as irst order phase transition. We have Gibb s thermodnamical potential G U S + dg du ds Sd + d + d dg d Sd 1 For the irst order phase transition d 0 and d 0 So we get rom 1 dg 0 Hence G constant 18
So we see that in the irst order phase transition Gibb s thermodnamical potential or Gibb s energ is constant. Let G G, then we get dg G d + Comparing 1 and 2 we get G and G S So, we get the properties o the irst order phase transition i Gibb s unction is constant at isothermal and isobaric process. i.g. G i G G d 2 ii here are change in entrop S i S and volume i. Hence, in irst order phase transition, Gi Hence, in irst order phase transition, S i S G Gi i G Q. rove that in the irst order phase transition o a one component sstem, the speciic Gibb s potential have the same value. 19
We consider a sstem o liquid with its vapour. Let m l be the mass o the liquid m v be the mass o the vapour g l be the speciic Gibb s energ o liquid g v be the speciic Gibb s energ o vapour So or the whole sstem G m l g l + m v g v dg dm l g l + dm v g v 1 We have Gibb s thermodnamical potential G U S + dg du ds Sd + d + d dg d Sd 1 For the irst order phase transition d 0 and d 0 So we get rom 1 dg 0 2 Comparing 1 and 2, we get 0 dm l g l + dm v g v 3 Again or a closed sstem we have dm l + dm v 0 dm l dm v 4 So, we get rom 3 and 4 g l g v Hence, the speciic Gibb s potentials must have the same value. Q. Derive Clausius-Claperon equation. 20
Wa 1: Let us consider the irst order phase transition o one mole o a substance rom phase i to phase. he Gibb s unction remains constant during a reversible isothermal-isobaric process. i.e. in the irst order phase transition at and and or a phase -change at + d and + d From 1 and 2 we get g i g 1 g i + dg i g + dg 2 dg i dg s i d + v i d s d + v d v v i d s s i d d d s s i v v i d d s s i v v i d d L v v i where L s s i latent heat per mole. Eq.3 is known as Clausius-Claperon equation. 3 Wa 2: Let us consider the irst order phase transition o one mole o a substance rom phase i to phase. Using the irst ds equation ds C d + d 1 Let is a unction o onl. Hence So, we get ds C d + 21 d d d d d
d ds du + d d Integrating both sides at constant, we get For reversible process s s i ds U U i du + s i s U i U + U U i d v d d v i d v i v d and latent heat per mole L s s i So, we get d L v v i d d d L v v i his is Clausius-Claperon equation. his is known as irst latent heat equation. So, we get the properties o the irst order phase transition i Gibb s unction is constant at isothermal and isobaric process. i.g. G i G ii here are change in entrop S i S and volume i. Hence, in irst order phase transition, Gi Hence, in irst order phase transition, S i S G i 22
Gi G Q. Calculation o boiling point o water inside the cooker We have Clausius-Claperon equation or irst latent heat equation as d L v v i d d L d v v i Integrating both sides we get p i d p i Here is the boiling point o water. L v v i i d L v v i ln i ln p i v v i i L p i v v i exp i L p i v v i i exp L Q. Derive second latent heat equation. Let s i and s be the entropies o unit mass o the substance in the initial and inal phase respectivel. During the phase transition, the temperature remains constant. So we get Dierentiating w.r.t. we get L s s i 1 dl d L ds 2 d ds i d 23
dl d L ds d ds i 1 d Now, we consider the transition rom liquid to vapour phase. So, the speciic heat o saturated vapour and that o liquid ds d c v ds d c l So we get rom 1 dl d L + c v c l 2 his is known as Clausius equation and is also known as second latent heat equation. It gives the variation o latent heat equation. Note: o compute the variation o latent heat o usion, we consider S s, s s ds d + d ds s d s + d d sat sat ds s d s + d d sat Using Maxwell s equation, we get c sat c So at saturated condition or vapour or liquid c v c v c l c l sat v d d sat vv d d sat vl d d sat 1 2 24
With the help o 1 and 2 we get rom second latent heat equation dl d L + c v vv dl d L d + c v c l d dl d L + c v c l d c l + d sat vv sat L vv v v v l vl vl vl d d sat Q. Derive Ehrenest s equation or second order phase transition. In second order phase transition, the entrop and volume do not change. First we consider the entrop S be constant at the temperature and pressure. So, we get the properties o the second order phase transition i Gibb s unction is also constant i.g. G i G ii here is no change in entrop S i S and volume i. Hence, in second order phase transition, S i S Gi G Hence, in second order phase transition, Gi i G Let S i and S be the entropies o the substance in the initial and inal phase respectivel. So we get at and S S i 25
ds ds i ds ds i Apping second ds equation, we get C p d d C p d i d i 1 We have coeicient o expansion at constant pressure α 1 α So, we get rom 1 C p d α d C p d α i d v C p i C p α α i i For Constant volume, we get at and i d d i 1 Let d, d + d K d + α d d So we get rom 1 K d + α d K i d + α i d 26
α α i K v K i his is known as Ehrenest s equation or second order phase transition. Q. rove that there is a discontinuit or change o ree energf, enthalph and internal energu during the irst order phase transition. We know that in the irst order phase transition temperature and pressure remain constant. i Internal energu: We have rom irst law o thermodnamics Q S U + we consider the entrop S at the temperature and pressure Let S i and S be the entropies o the substance in the initial and inal phase respectivel. So we get or initial state S i U i + i 1 that or inal state S U + 2 From 1 and 2 we get S S i U U i + i So, we get the properties o the irst order phase transition i Gibb s unction is constant at isothermal and isobaric process. i.g. G i G ii here are change in entrop S i S and volume i. Hence, in irst order phase transition, Gi S i S 27 G
Hence, in irst order phase transition, Gi Again, latent heat L S S i. i G U S S i i Gi G Gi G U 0 As in irst order phase transition, Gi Hence, in irst order phase transition, Gi G G U L 0 3 Hence change o internal energ U 0. Hence there is a change in internal energ during second order phase transition. ii Free energf : From the deinition o ree energ, we have F U S For initial state F i U i S i For inal state F U S So the change o ree energ F F F i 28
Hence, in irst order phase transition, F U U i S S i Gi Hence, in irst order phase transition, S i S G Gi i G U L 0 3 Hence change o internal energ U 0. Hence there is a change in internal energ during second order phase transition. F U L F L L F 0 It is equal to the work done b the sstem during isobaric -isothermal process. iii EnthalpH: From the deinition o enthalp, we have H U + For initial state H i U i + i For inal state H U + So the change o enthalp H H H i 29
Hence, in irst order phase transition, U U i + i Gi Hence, in irst order phase transition, S i S G Gi i G U L 0 3 F 0 So H U + H L + H L 0 It is equal to the latent heat absorbed during the irst order phase transition. Q. rove that there is no change o ree energf, enthalph and internal energu during the second order phase transition. We know that in the second order phase transition temperature S and pressure remain constant. i.e. S S i and i. In second order phase transition, the entrop and volume do not change. First we consider the entrop S be constant at the temperature and pressure. So, we get the properties o the second order phase transition i Gibb s unction is also constant i.g. G i G 30
ii here is no change in entrop S i S and volume i. Hence, in second order phase transition, S i S Gi Hence, in second order phase transition, G Gi i G i Internal energu: We have rom irst law o thermodnamics Q S U + we consider the entrop S at the temperature and pressure Let S i and S be the entropies o the substance in the initial and inal phase respectivel. So we get or initial state S i U i + i 1 that or inal state S U + 2 From 1 and 2 we get S S i U U i + i L U + here latent heat L S S i 0 and 0 U 0 3 U U i 31
Hence change o internal energ U 0. Hence there is no change in internal energ during second order phase transition. ii Free energf : From the deinition o ree energ, we have F U S For initial state F i U i S i For inal state F U S So the change o ree energ F F F i U U i S S i F U L F L L F 0 F F i here is no change o ree energ during second order phase transition. iii EnthalpH From the deinition o enthalp, we have H U + For initial state H i U i + i For inal state H U + 32
So the change o enthalp H H H i U U i + i H U + H L + H L 0 H H i here is no change o enthalp during second order phase transition. Q. Show that or an ideal gas µ µ 0 +R ln 0, where µ 0 is the chemical potential at pressure 0 We have the relation G U + dg du + d + d dg du + Rd + d 1 As Constant, d 0 and also change o internal energ du 0 So dierentiating 1 w.r.t. at constant temperature, we get For a pure substance, we get Integrating both sides, we get G dg d G G 0 dg 0 d G G 0 + 0 d 33
For an n moles o an ideal gas nr. So we get G G 0 + nr 0 d G G 0 + nr ln 0 For 1 mole G n G 0 + R ln n 0 µ µ 0 + R ln 0 Where µ G n Gibb s energ per mole or chemical potential. Q. For a two phase sstem in equilibrium, is a unction o onl so that S Show that E S C 2 irrespectie o the tpe o transition, E S being the adiabatic elasticit. Let the entrop ds S S S, d + For an adiabatic process ds 0 and we get S d S S d + d 0 S S S d S d S 34
Using Maxwell s relation, we get S Multipling both sides b, we get We have adiabatic elasticit So we get S C S S E S S S C E S C 2 E S Q. Deine the riple point he triple point reers to a particular temperature and pressure at which the three phases i.e. solid, liquid and vapour o a substance coexist. I we draw a curve showing change in melting point with pressure o a substance, we get a usion curve. Similarl, we will get vaporisation and sublimation curve. hese three curves intersect at a point, called the triple point, at which all the three phases coexist. 35
From these curve we see that a solid cannot alwas be transormed into the liquid state b heating. I a bod, below the pressure is heated, it will not melt, but will directl pass into the vapour state skipping the liquid phase-it will sublimate instead o melting. Melting transition to liquid state is possible onl heating above a pressure. We have Clausius-Claperon equation Case-1: For usionsolid liquid d d d d For most pure substances v l > v s So, we get L v v i L u v l v s d d +ve But or water ice water For this case v i > v w, so we get Fusion curve o water: d d L iw v w v i d d e 36
Case-2: or vaporizationliquid gas d L vap d v g v l For most pure substances v g > v l, So, we get aporisation curve: d d +ve Case-3: sublimationsolid gas d d For most pure substances v g > v s So, we get Sublimation curve: L sg v g v s d d +ve Compact orm o the curve: 37
For an substance these curves are given bellow. Let t m melting point t b boiling point We know i or most substance, t m increases i pressure increases. ii t b increases as pressure increases. So we get the ollowing curve. Q. rove the exitance o triple point rom second law o thermodnamics. We consider ice and water in equilibrium with their vapour at the pressure 1 over ice and 2 over water at a constant temperature. i Let δm gm ice is converted into its vapour. So heat absorbed Q 1 δml sv 38
Here L sv is the latent heat o sublimation. Work done W 1 δm 1 1 2 Next the vapour is allowed to expand rom 1 to 2 at constant. Work done b the vapour or gas 2 Q 2 W 2 δm d 1 2 2 Q 2 W 2 δm d Q 2 W 2 δm 1 1 3 he vapour is then condensed into water. Heat rejected Work done on the vapour 4 Heat rejected b water to orm ice 2 2 1 1 Q 3 δml vl W 3 δm 2 2 2 1 2 1 d d Q 4 δml ls his is a reversible process. So the second law o thermodnamics can be applied to it. From second law o thermodnamics we get ds Σ dq 0 As temperature is constant, ΣdQ 0 Q 1 + Q 2 + Q 3 + Q 4 0 δm L sv + 2 2 1 1 2 1 d + L vl + L ls 0 L sv + 2 2 1 1 2 1 d + L vl + L ls 0 1 39
otal work done during the ccle is zero. i.e. W 1 + W 2 + W 3 0 1 1 1 1 + 2 2 2 1 d + 2 2 0 2 1 d 0 1 2 Hence, 1 2. So, we get rom 1 L sv + L vl + L ls 0 L sv L vl L ls L sv L lv + L sl Hence, latent heta o sublimation Latent heat o vaporisation+latent heat o usion. 40