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8.044 Lecture Notes

Chapter 3: Thermodynamics, rst pass

Lecturer: McGreevy

Reading:CJ Adkins,Equilibrium Thermodynamics, Chapters 1, 2, 3.1-3.7.

Skip Adkinsx3.5.4.

R Baierlein,Thermal Physics, Chapter 1.

The four laws of thermodynamics [according to Prof. Yosi Avron]:

0th Law: You have to play the game.

1st Law: You can't win.

2nd Law: You can't even break even.

3rd Law: And you can't quit.

To counter your impatience to get to statistical mechanics, let me oer the following [from

Prof. David Tong]:

...the weakness of thermodynamics is also its strength. Because the theory is ignorant of the underlying nature of matter, it is limited in what it can tell us. But this means that the results we deduce from thermodynamics are not restricted to any specic system. They will apply equally well in any circumstance, from biological systems and quantum gravity. And you cant say that about a lot of theories! 3-1

3.1 Basic terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3-3

3.2 0th Law of Thermodynamics and denition of temperature . . . . . . . . . .

3-7

3.3 Dierential changes of state and exact dierentials . . . . . . . . . . . . . . .

3- 13

3.3.1 Appendix: Reciprocity theorem, derivation . . . . . . . . . . . . . . .

3-19

3.4 Work in quasistatic processes . . . . . . . . . . . . . . . . . . . . . . . . . .

3- 22

3.5 First law of thermodynamics; internal energy and heat . . . . . . . . . . . .

3-28

3.6 Heat Capacities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3-31

3.7CVandCPfor a hydrostatic system; enthalpy; heat reservoirs . . . . . . . .3-32

3.8CVandCPfor ideal gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3-35

3-2

3.1 Basic terminology

We will often refer to system(s) and surroundings.Systems can be separated by boundaries of \walls" that can be either

adiabatic: No heat can ow between them. Adiabatic walls keep the systems isolated.system 1 T

1system 2

T

2... or ...

diathermal: Heat can ow. (Particles cannot.) This means that after a little while T

1=T2=T.system 1

Tsystem 2

T3-3 The macroscopic state of each system is described by thermodynamic varaibles.

These can be

extensive: proportional toN, the number of particles ... or ... intensive: not proportional toN. Local in character. The number of particlesNis extensive. If I divide the room in half, (roughly) half the atoms will end up in each half. Similarly, energy is extensive. On the other hand, the temperature of each half of the room will be unchanged; temperature is intensive. I claim that no other dependence onNwill arise in the thermodynamic (largeN) limit, at least in 8.044. Examples:Type of systemExtensive VariableIntensive Variable

Gas or liquid in a cylinderVvolumePpressure2-dimensional lmAareasurface tensionlament or elastic rodLlengthftensionelectrically charged objectqchargeEvoltage, EMFdielectric material~

Ppolarization~

Eelectric eldparamagnetic material~

Mmagnetization~

Hmagnetic eldanysystemSentropyTtemperatureThe product of each pair has units of energy.

PV;A;fL;Eq;~E~P;~M~H

are all measures ofwork. Such pairs of thermodynamic state variables and their associated generalized force are calledconjugate variables. TandSare also called conjugate variables, althoughTSis related to heat, not work. 3-4

Thermodynamic, or Thermal, Equilibrium

When the surroundings of a system change, the system itself will change. After a time, no further macroscopically-detectable changes take place. Transients have died down. We say the system is inequilibrium. The equilibrium state of a system is completely specied by a certain number (this number is called the number of (thermodynamic) degrees of freedom) of independent variables. In equilibriumhese specify the other variables viaequations of state. Classic example: ideal gas, with a xed number of particlesN.

Equation of state (EoS):PV=NkBT.

So, we can takeP;VORP;TORV;Tas the two independent varaibles. The third is the determined by the EoS. An EoS is a functional relationship among thermodynamic variables that is satisedin equilibrium. Another example: Curie's Law for a paramagnet.M=CCHT .Mis magnetization,His an applied magnetic eld,Tis temperature.CCis a constant. Valid forH=Tnot too large. These examples above have two thermodynamic degrees of freedom (dofs). It's easy to generalize:e.g.consider a magnetic gas, where the atoms have spins. Then we needP;V andM;H.

We could also add more variables

(like say the compressibilityof the gas), but these would satisfy the same number of additional relations (in this case, the new relation would be=V1 @V@p T In general there may be other variables of the system which are independent, but which we're not going to vary. In that case we can ignore them. 3-5

Thermodynamic Reversibility

A series of changes undergone by a system isreversiblei its direction can be reversed by an innitesimal change in the condition. (e.g.: yanking the piston out of the cylinder is not reversible. Moving it out very slowly may be.)

Reversibility requires:

1.Quasistatic processes: (slowness) All changes must be made so slowly that every

state through which the system passes may be considered an equilibrium state.

In particular:PV=NkTalways satised

no shock waves no energy dissipated due to friction: frictional forces/v.

Elost to friction/vdistancev!0!0:

These are limits of real processes, performed arbitrarily slowly. Drawing lines on a PV diagram only makes sense for such quasistatic processes.

2.No hysteresisEven if all the stu is done arbitrarily slowly, there are situations where

what seemed like a closed loop in parameter space is not closed.e.g.bar of iron. Even if we quasistatically vary the applied eld, the state of a ferro-

magnet depends on its history. e.g.rod under tension, when it deforms. Having said this, in 8.044 we will only discuss systems for which quasistatic = reversible. That is, we will have no truck with hysteresis here. That's for 8.08. [End of Lecture 7.] 3-6

3.2 0th Law of Thermodynamics and denition of temperature

Dene: A is "in equilibrium with" B if when we stick them next to each other, separated by a diathermal wall, nothing happens { no macroscopic change occurs. Zeroth law: "in equilibrium with" is a transitive property.

That is: If systemAis in equilibrium with systemB

and systemBis in equilibrium with systemC then systemAis in equilibrium with systemC.

Once more, in symbols:AB|{z}

in equilibriumandBC|{z} in equilibrium=)AC|{z} in equilibrium Now, think ofBas a thermometer! This fact makes thermometers possible. 3-7

Observation: many macroscopic states ofCcan be in equilibrium with a given state ofA.On the right, I've drawn the locus of states ofCthat areallin equilibrium with the same

state ofA. This is an isotherm. Similarly, if I x the state ofC, there will be a locus (a curve in this case) of states of systemAwhich are in equilibrium with it. The property that these states have in common is calledtemperature. We can use the 0th Law to dene an \empirical temperature" called by Adkins. Idea: the temperature of a system is some function of the other state variables that deter- mines whether or not the system is in equilibrium with other systems. Two systems are in equilibrium if they have the same temperature.

Note the order of the logic here:

1. de nitionof equilbrium 2. statem entof 0th La w 3. de nitionof isotherm 4. de nitionof temp erature 3-8

A pedantic implementation of this denition

Consider two systemsA;Bwhich are initially isolated:P

A;VAisolated fromP

B;VBP A;VA;PB;VBare all unspecied here. We're not assuming that these are ideal gases, but as always that's a good example to keep in mind. Now bring them in contact, and adjustVB(move a piston) until they are in equilibrium.P A;VAP B;VB(If you like, it is some additional input from our experience of the world that it is possible to do this.) Demanding equilibrium imposes one relation among the four state variables: e:g: V

Bjequilibrium with A=f1(PA;VA;PB) (1)

Now repeat forCandB,i.e.bring a second systemCin contact withB.P B;VBP C;VCHold xed the parameters ofBand adjustVCto putB+Cin equilibrium. Again this gives one relation between the thermodynamic variables (this time ofB+C), which I choose to express as: e:g: V

Bjequilibrium with C=f2(PC;VC;PB)

Claim:f1(PA;VA;PB) =f2(PC;VC;PB) (2)

by the 0th Law { if we got dierent values forVBin these two cases, \equilibrium with" wouldn't be transitive. This equation (2) is one equation for 5 variables. Solve (2) forVA: V

A=g(PA;PC;VC;PB) (3)

Now repeat the story forA+C(adjustVAto put it in equilibrium withC): V

A=f3(PA;PC;VC) (4)

Eqn. (4) determinesVAin terms ofPA;PC;VC.

But now combine the information in (3) and (4): if we plug the expression in (4) forVAinto (3) we discover that we can ndgwithout knowing anything aboutPB{gis independent of P B. This is only possible if the equation (2) relatingf1andf2is independent ofPB.1This allows us to rewrite equation (2) in a way that makes no reference toB: f

1(PA;VA) =f2(PC;VC)1

Note that in factf1;2themselves can depend onpBin such a way that the dependence cancels. This is what happense.g.for the ideal gas, wherePAVA=NkBT, so in this casef1(PA;VA;PB) =PAVAP Band f

2(PC;VC;PB) =PCVCP

B. Equating these, we learn thatPAVA=PCVC, each side of which is independent of

systemB(and is proportional to the temperature). Thanks to Sabrina Gonzalez Pasterski for correcting an

error on this point. 3-9 We've found a function of state variables of systemCwhich is equal to a function of the state variables of systemAwhen they are in equilibrium. For each of the systems in equilibrium, there is a function of its state variables that takes on the same value: =f1(PA;VA) is EoS for A (gives shape of isotherm) =f2(PB;VB) is EoS for C (gives shape of isotherm) Note that we could have calledP;VhereX;Yfor some generic state variables. Details about gases are not relevant at all. For any of the systems, we can now map out isotherms with varying .We can also determine that

3>2>1by watching thedirectionof heat

ow when we bring systems on the respective isotherms into contact.

1>2if heat goes from 1 to 2.

Q: SupposeA;Bhave the same temperature. How do we know that the direction of heat ow is transitive {i.e.that it is the same betweenA+Cand betweenB+C? A: We know because the states ofA;Bdeneoneisotherm ofC, by the 0th Law. To choose units of temperature, we need to make some choices. 3-10

Thermodynamic temperature

... is a concept dened by Atkins, roughly with units. It is measured in Kelvin (dened below). Based on the fact (we'll explain this later) that all gasesin the dilute limitsatisfy

PV=NkBT :

The dilute limit isP!0;V! 1, withPVxed. (Note that the curves I've been drawing are the hyperbolae in (P;V) specied by constantTin this ideal gas law.) So we can useanygas to dene a temperature scale, via

T= const limP!0;V!1PV :

This is is nice in that it doesn't depend on which gas we use.

The choice of constant is a convention:

T(in Kelvin) = 273:16limP!0;V!1(PVof the gas whoseTwe're dening)lim

P!0;V!1(PV)TP

where (PV)TPis thePVof the gas in question when it is in equilibrium with water at its

Triple Point.

Denition of Triple Point: liquid water, solid water (ice), and gaseous water (vapor) are all in equilibrium with each other at T

TP273:16K= 0:01CandPH2O

TP= 611Pa= 0:006atm:

This is a useful convention because the triple-point of water is reproducible. Then:

T(water boils, at 1 atm) = 373:15K= 100C

T(water freezes, at 1 atm) = 273:15K= 0C

(The number 273.16 was chosen so that we get 100 between freezing and boiling of water.) Although it's reproducible, this scale is not so practical. We can now use this dilute gas thermometer to calibrate other, more practical, thermometers. Any physically observable quantity which varies with temperature can be used. You can imagine therefore that lots of things have been tried. e.g. expansion of liquids, resistance (temperature dependence of resistance of metals and semiconductors), thermocouples... Dierent systems are useful for dierent temperature ranges and dierent sensitivities (there is generally a tradeo between how sensitive a thermometer is and how large a range it can measure). Combining all these together is something called the "International Practical Temperature

Scale" which you can read more about in Adkins.

3-11 Then we can use these thermometers to map out isotherms of other systems. They are not always hyperbolae. e.g. a paramagnet.T= constHM Mapping isothermsempirical determination of equation of state. 3-12

3.3 Dierential changes of state and exact dierentials

Imagine an equation of state written as

V=f(P;T)(think of V=NkTP

(5) An innitesimal change from one state of equilibrium to another state of equilibrium involves tiny changes of of each of the state variables:dV;dT;dP. Since both states of equilibrium satisfy (5), we must have dV=@f@P T dP+@f@T P dT We can avoid the waste of an extra symbol by writing the EoS asV=V(P;T) and therefore dV=@V@P T dP+@V@T P dT I remind you that when we have several independent variables, we need to be careful in specifying what we mean by a derivative { we have to specify what is held xed; this is indicated by the subscript: @x@y z means the derivative ofxwith respect toywithzheld xed. The coecients above have names, because they can be measured. T 1V @V@P T is called theisothermal compressibility 1V @V@T P is called thevolume expansivity: Both are determined from the two independent variables (whichever two we pick) by the equation of state.

Similarly,

P=P(T;V) =)dP=@P@V

T dV+@P@T V dT

T=T(P;V) =)dT=@T@V

P dV+@T@P V dP These partial derivative quantities (likeTand) of the form @something@something else a third thing 3-13 are calledresponse functions. How does the system'ssomethingrespond when we make a perturbation of itssomething elseunder conditions where itsa third thingis held xed? This kind of thing is often what gets measured in experiments. We've written six of them on the board. They are not all independent. You can show (see Adkins and recitation and subsection 3.3.1) that if three variablesx;y;z are related by one equation of the form x=x(y;z) ory=y(x;z) orz=z(x;y) then: @x@y z =1 @y@x z(`Reciprocity Theorem') and the two other equations related by permuting the varaibles. Note that the same quantity is held xed on both sides. @x@y z @y@z x @z@x y =1These identities (just math, no physics) reduce six response functions to two independent ones. We'll use these over and over. 3-14

Exact dierentials.

A third useful identity is@

z@x@y z =@y@x@z yThis identity says that it doesn't matter which path you take from the point1to the point

2. The end result,x(2), is the same no matter which path you take to get there. This is

true as long asx(y;z) is a smooth, single-valued function. Terminology: Ifx=x(y;z) is such a nice function then the expression dx=A(y;z)|{z} =(@x@y )zdy+B(y;z)|{z} =(@x@z )ydz is called anexact dierential. 3-15

Integrating exact dierentials.

Suppose you knowA;B(e.g.because you measured them), indx=Ady+Bdz. How can you reconstruct the state variablex(y;z)? First, in order forxto exist, it's necessary that its mixed partials be equal: @A@z y =@B@y z This is a necessary and sucient condition fordxto be exact2 If your functionsA;Bpass this check, you can proceed as follows. The overall additive constant inxisn't xed. Pick a random starting pointy0;z0and declarex(y0;z0) = 0. (Usually there will be some additional physical input that xes this ambiguity.) To getxat some other point (y;z) we integrate along a path from (y0;z0) to (y;z); since the answer is

path-independent, we can pick a convenient path. I pick this one:Taking it in two steps, we have from the rst horizontal leg:

x(y;z0) =Z y y

0dy0A(y0;z0)

Then:x(y;z) =Z

y y

0dy0A(y0;z0) +Z

z z

0dz0B(y;z0)2

Optional cultural comment for mathy people: This assumes that there is no interesting topology of the

space of values of the thermodynamic variables. We'll assume this always. 3-16 An alternative procedure, which is a little slicker (but gives the same answer, of course, up to the additive constant), is: 1. x(y;z) =Z dyA(y;z) +f(z) The last term here is like an integration constant, except it can still depend onz. We need to gure it out. To do that consider: 2.

B=@x@z

y =Z dy@ zA+dfdz 3. f(z) =Z dz BZ dy@ zA

Result:x(y;z) =Z

dyA(y;z) +Z dzB(y;z)Z Z dydz @A@z y! 3-17quotesdbs_dbs19.pdfusesText_25