Lecture Thermodynamics 9. Entropy form of the 1 st law. Let us start with the differential form of the 1 st law: du = d Q + d W

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1 Lecture hermodnamics 9 Entro form of the st law Let us start with the differential form of the st law: du = d Q + d W Consider a hdrostatic sstem. o know the required d Q and d W between two nearb states, we need to know the ath the sstem takes. his deends on the constraints we choose for the change (for eamle, constant ressure, adiabatic or isothermal). Let us consider a ver small change of a sstem. he change might involve a change in volume, a change of hase or a chemical reaction, for eamle. o sul heat to the sstem we use a single eternal reservoir at temerature (assumed to be constant during the ver small change). ) Reversible changes Start and finish on equilibrium states and in equilibrium on ath In this case d W =!d, where =ressure of sstem (also equal to the eternal ressure) d=volume change of sstem d Q = ds because the changes are reversible (= d Q rev ) herefore du = ds! d for reversible changes. 2) Irreversible changes (i) For changes that occur between two equilibrium states start and finish on equilibrium states he equation we derived for reversible changes above du = ds! d deends on variables that are all functions of state. So rovided that the initial and final states are equilibrium states, we can still use this equation even if in between (during the change) the sstem is not in equilibrium and the change is irreversible.

2 du = ds! d for irreversible changes if the start and end oints are equilibrium states his is an eamle of a small irreversible change in which the initial and final states are equilibrium states: iston 20 o C gas, gas 2 o C gas, +d Heat entering the gas, initiall all at 20 o C, suddenl increases the temerature of the lower art shown. After a short time, the temerature becomes uniform again at 2 o C. he ressure is constant so the volume increases slightl. he gas does work. he rocess is irreversible because of the finite temerature differences in the gas during the rocess (that includes thermal conduction in the gas). (ii) For general changes not between two equilibrium states In this case d W =!d, rovided that the volume of the sstem is changed in a mechanicall quasistatic wa so that the ressure of the sstem is well defined and equal to the eternal ressure. he change in entro of the sstem is ds. Let the change in entro of the reservoir be ds 0. Since this combined sstem is isolated, we know that ds + ds 0! 0 (from Lecture 8.) But ds 0 =!d Q. Here is the temerature of the reservoir during the small change. So ds! d Q 0. herefore d Q! ds. (We found the same result, ds! d Q, in Lecture 8, for a general change.) From the st law of thermodnamics, du = d W + d Q.! d Q = du d W = du + d ds 2

3 d Q du d W du d ds herefore du d ds 0 for irreversible changes sstem sstem temerature of reservoir he case when the change is not mechanicall quasistatic is treated later in this lecture. his is an eamle of a small change in a sstem in which the initial and final states are not equilibrium states of the sstem: 20 o C 60 o C 2 o C 59 o C In this eamle the sstem (enclosed b dashed lines) has two searate bodies in it, each one at a different temerature. he sstem as a whole is not in an equilibrium state. he ressure is the same in the bo so the sstem is in mechanical equilibrium. In this case the sstem is thermall isolated from the reservoir. In this eamle d W 0, but if we move the iston we can also do work. In this eamle of a small change, heat flows along the link joining the two bodies. Effect of different constraints: a) hermall isolated sstem Let us check that this agrees with the results we got last week: If the sstem is thermall isolated then d Q 0 du d W du d ds 0 ds 0 (as eected) Since the ossibilit of work is allowed we can continue to increase the entro indefinitel. Eamle: gas in thermall isolated bo W he gas kees on getting hotter, and so for a thermall isolated sstem the entro can increase without limit (if we kee doing work). b) Comletel isolated sstem If the sstem is comletel isolated then d Q 0, d W 0, du 0 3

4 ! ds 0! ds 0 (as eected) Note: when the entro has reached its maimum value for a given comletel isolated enclosure, it cannot increase an more. An eamle is the Joule eansion we treated in Lecture 8. In etreme nonequilibrium situations in which the constraints on the sstem are not clear it is not clear how to define the entro S. We will not be treating such nast roblems in this course. hermodnamic otentials (hdrostatic sstems with two degrees of freedom) So far, we have considered Internal energ U Enthal H=U+ Define: Helmholtz function (or free energ) F=U-S Gibbs function G=U-S+ For changes between equilibrium states, du = ds! d dh = du + d + d = ds + d df = du! ds! Sd =!Sd! d dg = df + d + d =!Sd + d his shows that each otential has natural variables: U=U(S, ), H=H(S, ), F=F(, ) and G=G(, ). Internal energ: For constant volume changes, du=ds! U = C = S Enthal: For an isobaric (constant ressure) change, dh=ds! H = C = S Helmholtz function: S =! F, =! F ' 4

5 We can also obtain U, H, and G from F: For eamle, U = F + S = F! F =! 2 ( ' F. ' F is an imortant thermodnamic function for rocesses at constant and when df=0. Gibbs function: =!G, S = (!G!! G is an imortant function for hase transitions at constant and when dg=0. It is an imortant quantit in chemical reactions (usuall in the atmoshere in which and can be aroimatel constant). Mawell relations Reminder of the mathematics of artial differentiation: Assume we have 3 variables related b F(,, z)=0, and eress as z=z(, ). dz =!z d +!z d. Define M =!z!!! N =!z! herefore dz = Md + Nd Also!M! ' =! 2 z!!!n! =! 2 z!! =! 2 z!! (see below)! M ' = N ' his is the condition for dz to be an eact differential. (See Lecture 3.) dz can alwas be integrated because z is a single-valued function of and. Proof of! 2 z!! =! 2 z!! assuming that z=z(, ): B 2 A Consider the small change from to 2 b the two different aths A and B: 5

6 At oint A, z A = z +!z +! (! 2 z 2! 2 At oint 2, z 2 = z A +!z A +! (! 2 z A 2! 2 Combining these two equations, we obtain z 2 = z +!z ( +!z +!! (! 2 z 2! 2 Similarl, going b route B, we get z 2 = z +!z ( +!z +!! (! 2 z 2! 2 (( ) z (( ) (( ) 2 +! 2 z 2! 2 (( ) 2 +! 2 z 2! 2 (( ) 2 +!!z ' (( !! (( ) 2 +!!z ' (( !! If z=z(, ) then these are equal, so!!z!! =!!z!! =! 2 z!! =! 2 z!!. Derivation of the Mawell relations: du = ds! d ' ) ( dh = ds + d ' ) ( df =!Sd! d S ' ) ( dg =!Sd + d S ' ) ( S S =! ' ) S( = ' ) S ( = ' ) ( =! ' ) ( Mawell Relations Note that the variables S and alwas occur in diagonal airs. If S and occur on the same side of the equations then we need a minus (-) sign. We have derived these Mawell relations for equilibrium states of a hdrostatic sstem. he are ver useful because the can eress difficult-to-measure things in terms of eas-tomeasure things!s! = (!! related to thermal eansion 6

7 If increases with then S decreases for an isothermal comression. Note that Mawell s relations al for an sstem with two variables, not just for an ideal gas. Eamles: ) Ideal gas: checking Mawell relations S = nc ln + nrln +constant caacit) (derived in the Lecture 8, where c is the molar heat = nc ln + nrln nr! nrln + constant = nc ln + nrln! nrln + constant = ( nc + nr)ln! nrln + constant! S = nc ln nrln + constant B the same method, we can also derive the following eression: S = nc ln + nc ln +constant Differentiate the equation S = nc ln + nrln +constant: ds = nc d + nr d = 0 if S is constant.! ' S = ) R c = ) nc Differentiating the equation S = nc ln + nc ln +constant, in the same wa:! S = nc herefore!! S = (!!S (in agreement with Mawell relation, no. ) In the same wa, b differentiating S = nc ln + nc ln + constant, we find! S ' = nc!s! = nc Also, differentiating S = nc ln! nrln + constant, we find!! S = nr nc = nc 7

8 herefore!!s =!! S (in agreement with Mawell relation, no. 2) 2) Ideal gas: derivation of equation of state Consider the following wa of defining an ideal gas: (i) (ii) =constant at constant (Bole s law)!u = 0! For changes between equilibrium states, du=ds-d!u!!! =!S! =!! = ( ( = 0 d = d! ln = ln + g( ) = ln + ln f ( ) = lnf ( ) where f and g are arbitrar functions of.! = f ( ) But from (i) = constant at constant. So = constant! = const. (the constant is in fact nr) 3) Calculating C -C for a general substance S=S(, )! ds = S ds =!S! d +!S d! d + S d '!S! =!S! +!S!!! 8

9 herefore, C! C = But!! (!! =!! (because!!!!!! = () From the definition of the volume thermal eansion coefficient! = comressibilit! = ' ), we find ( and the isothermal C! C = 2 his alies to an hdrostatic sstem, not just to an ideal gas. We know that β can be ositive or negative, but we alwas have κ >0. herefore C! C 0. An eamle of C! C = 0 is for water at 4 o C at which temerature β=0. Let us make a rough estimate for a metal of C! C. C! C = nr n densit). 2. Here /n=molar volume=m m /ρ (M m the molar mass and ρ the R C! C nr = M m 2 R For a tical metal at 300 K, with M m =0. kg, ρ =0 4 kgm -3, β=0-5 K -, and κ =0 - Pa - this gives C! C nr = ! ! 9

10 General conditions for thermodnamic equilibrium Interretation of the Helmholtz function: wh is it called the free energ? he idea of otential energ is often used in mechanics. In thermodnamics we do not have such a concet, but we can have a similar idea if a sstem is ket at a fied temerature. Consider a sstem at constant temerature consisting of several arts that ma not be in mechanical equilibrium. Here is a simle eamle of such a sstem: 2 atm atm d W reservoir at d Q OU In this eamle a gas is traed in a iston at atm ressure, and there is another enclosure inside the iston with gas at 2 atm ressure. A ta blocks this 2 atm enclosure. he whole sstem is in thermal equilibrium with a reservoir. During an irreversible rocess we might allow some gas to leak out of the 2 atm enclosure, for eamle. he work done on the sstem d W = du! d Q = du + d Q OU, where d Q OU is the heat entering the reservoir. For a general rocess d Q OU = ds 0, where the subscrit 0 refers to the reservoir, because the heat enters the reservoir that is also at the temerature. We know from the 2 nd law that ds 0 >-ds (since ds 0 +ds 0) so d W = du + ds 0! du ds he free energ is defined as U-S. Since we are considering a sstem and reservoir at the same temerature, then df=du-ds, so d W! df for a sstem at constant temerature in contact with a heat reservoir For a reversible change d W RE = df. We can think of reversible work as being stored as the free energ (Helmholtz function). So F las the role of a otential energ for sstems at constant temerature. For irreversible rocesses, the work done b the sstem!d W df. We can interret this as being caused b the resence of frictional forces that give rise to dissiation. Note that d W RE is directl related to F=U-S not to U. his is because there is a heat flow associated with a reversible change in volume at constant temerature at constant temerature (as we saw when we wrote d W = du! d Q above). 0

11 If we remember that S is a measure of disorder we can redict the direction of heat flow. If we comress a gas reversibl at constant, we decrease the disorder, so the entro decreases. o do this heat must therefore flow out (since d Q = ds ). Because of df = d W RE and du = d W RE! d Q OU the increase in U is less than the increase in F. (For an ideal gas, U does not change at all and all the energ is stored as heat transferred to the reservoir). We can get our energ back b allowing the gas to eand as the heat reenters the sstem. Mechanicall isolated sstem: If a sstem with several arts at a fied temerature is mechanicall isolated with fied volume ( d W = 0) but the arts are not in mechanical equilibrium, then df! 0. So during the irreversible aroach of this sstem to equilibrium, F decreases and reaches equilibrium when F is a minimum. his is similar to a mechanical sstem reaching the osition of minimum otential energ. Eamle of a mechanicall isolated sstem at constant temerature with arts not in mechanical equilibrium: 2 atm atm d Q OU, reservoir at Useful work and condition for general equilibrium: availabilit We have seen how a hot bod and a cold bod can be used to create work using a heat engine. We have also just seen how a bod at fied temerature can be used to create work. his section is a generalization of this. If we have a hot gas at 000 K and 5 atm ressure, what is the maimum useful work that can be etracted from it? he answer will deend on the surroundings. If the surroundings are at 000 K and 5 atm, we cannot do an work, obviousl. First consider a sstem made u of several arts at different temeratures that are not in equilibrium with the surroundings. We consider a small change in which we can use heat engines that use different arts of the sstem or the surroundings as hot and cold sources. he surroundings can be considered as a heat reservoir and a ressure reservoir at constant temerature 0 and constant ressure 0. Here is an eamle of such a sstem containing gases at different temeratures and ressures.

12 30 o C 4 atm 50 o C 2 atm ressure reservoir at 0 = atm d W Heat reservoir at 0 =00 o C d Q OU As before, for a small change, d W = du! d Q = du + d Q OU. Here d Q OU = 0 ds 0 because the heat enters the reservoir at temerature 0. herefore, d W = du + 0 ds 0! du 0 ds ds 0 +ds 0). (as before in the discussion about F, because Here, U and S are the total energ and entro of the sstem. d W is the total work done b the surroundings, either on the sstem directl or on heat engines. Write d W =! 0 d + d W U Here is the total volume of the sstem.! 0 d is the work done b the ressure reservoir and d W U is the other forms of work done on the sstem or on the engines. We call d W U the useful work inut. (he useful work outut, - d W U, is what we want to be a ositive quantit. If we can roduce some useful work outut, we can use it to drive something or lift something etc.) herefore, d W U! du 0 ds + 0 d, or d W U! da where A is the availabilit, defined as A = U! 0 S + 0. his result shows that the availabilit is a useful generalization of free energ (Helmholtz function). For a sstem that interacts with a single temerature 0 and ressure 0, we have d W U = da for a reversible change. An reversible work can be thought of as stored in the form of availabilit. In an reversible change the useful work that can be etracted, - d W U, is a maimum and equal to the decrease in A. For our eamle of a sstem that was homogeneous (all at one temerature and one ressure, such as our gas at 000 K and 5 atm, with, for eamle, surroundings at 300 K and atm), for a reversible change, for which du = ds! d, 2

13 !da =!du + 0 ds! 0 d = ( 0! )ds + (! 0 )d he first term ( 0! )ds is the useful work obtained from a reversible heat engine (e.g. a Carnot engine) that etracts heat reversibl from the sstem. his can be seen from the following diagram: dq = -ds dw OU = dq -dq 2 =-ds - 0 ds 0 =-ds+ 0 ds =( 0 )ds 0 dq 2 = 0 ds 0 he second term (! 0 )d reversibl. [= d! 0 d ] is the useful work out when the sstem eands total work done b sstem 0 d is the wasted work done in ushing back the atmoshere his eression shows that when = 0 and = 0 no further useful work can be etracted, as eected. Mechanicall isolated sstem: If a sstem with several arts is mechanicall isolated with fied volume ( d W = d W U = 0) but the arts are not in mechanical or thermal equilibrium, then da! 0. So during the irreversible aroach of this sstem to equilibrium, A decreases and reaches equilibrium when A is a minimum. his condition contains the revious result for F as a secial case. For a sstem held at fied volume in thermal equilibrium with the surroundings at constant temerature, da! 0 means da = du + 0 d! 0 ds = du! 0 ds = du! ds 0. his imlies that df! 0. We can also consider a new case for a sstem held at constant ressure and temerature. In that case da! 0 means da = du + 0 d! 0 ds = du + d! ds 0. his imlies that dg! 0. So Gibbs function G reaches a minimum value in equilibrium for such a sstem. (his lecture is artl based on the book b J. R. Waldram, he heor of hermodnamics, Cambridge Universit Press.) 3

1 Entropy 1. 3 Extensivity 4. 5 Convexity 5

1 Entropy 1. 3 Extensivity 4. 5 Convexity 5 Contents CONEX FUNCIONS AND HERMODYNAMIC POENIALS 1 Entroy 1 2 Energy Reresentation 2 3 Etensivity 4 4 Fundamental Equations 4 5 Conveity 5 6 Legendre transforms 6 7 Reservoirs and Legendre transforms