Thermodynamics: Chapter 03

Interacting systems: The connection between

the microscopic and the macroscopic

September 18, 2014

Sept. 11

*******

How entropy governs the direction of the change in interacting

systems;

How entropy is related to other bulk properties, such as temper-

ature, heat capacity, pressure;

How to use entropy to predict the thermal behavior of thermal

systems.

To answer those questions, we need to look at systems that are

not only in equilibrium state but also interacting with each other

to evolve toward equilibrium state. We will systematically look at

1

different equilibriums in this table which I showed you in Lecture01:

Equilibrium ExchangesThermal Thermal energy

Mechanical VolumeDiffusive ParticlesChemical Chemical reactions

H2O ↔ H+ +OH−

H2SO4 ↔ 2H+ + SO2−4

§1. Entropy and temperature

let’s start our discussion from something we are fa-

miliar: the Carnot cycle we talked about at the end

of Chapter 1. The four steps:

Step 1: isothermal expansion, (V1, P1, Th)→ (V2, P2, Th)

∆Q1 = NkThlnV2

V1Step 2: adiabatic expansion, (V2, P2, Th)→ (V3, P3, Tc)

∆Q2 = 0,∆W2 = ∆U2 = CV (Tc − Th)

Step 3: isothermal compression, (V3, P3, Tc)→ (V4, P4, Tc)

∆Q3 = −∆W3 = NkTclnV4

V3Step 4: adiabatic compression, (V4, P4, Tc)→ (V1, P1, Th)

∆Q4 = 0,∆W4 = ∆U4 = CV (Th − Tc)

From this cycle, we arrived at

∆Q1

Th+

∆Q3

Tc≡ 0 (1)

We also talked that, if we divide one big cycle into

many micro-Carnot cycles, we should have∮closed cycle

δQT = 0 which is true for all reversible cycles.

Obviously, since ∆Q1Th

+ ∆Q3Tc≡ 0, for the reversible

cycles, we can define a variable dS = dQT , the inte-

gration of which is constant for all reversible cycles -

constant with respect to different cycles. In physics,

we are very much interested in such variables: energy,

momentum, etc.

In a Ideal Gas system, we have proved ∆S = δQT .

(One of the Homework problems, 2.34) where S is

the ”entropy”: the logarithm of multiplicity, simply

speaking.

So, you can imagine the entropy (multiplicity of mi-

crostates) is closely related with temperature (macro-

property).

In the meantime, we also know temperature is closely

related to the zeroth law of thermodynamics: sys-

tems in thermal equilibrium with one system are also

in thermal equilibrium with each other - there should

exist one parameter (temperature) that characterizes

the ”thermal equilibrium”. So, you can imagine, en-

tropy should also closely related to the zeroth law of

thermodynamics.

Now, let’s look into the details with a quantitative

example - a lit bit different from what you see in the

textbook.

Einstein Solid: at q N approximation (the high

temperature situation, Chapter 2), Ω =(eqN

)N. As-

sume two Einstein Solids: NA = 300, NB = 200, q =

qA + qB = 5100 N . (Equation 2.21 on p.64 in the

textbook)

SA = 300.0ln(2.718q/300.0)

SB = 200.0ln(2.718(5100.0− q)/200.0)

Stotal = 300.0ln(2.718q/300.0)

+ 200.0ln(2.718(5100.0− q)/200.0)

Let’s take a look at these three functions on a S − q

diagram:

Aq

500 1000 1500 2000 2500 3000 3500 4000 4500 5000

tota

lS

0

200

400

600

800

1000

1200

1400

1600

1800

2000

Let’s look at this plot closely. We have learned in

Chapter 2 the second law of thermodynamics, in

terms of entropy:

”Any large system in equilibrium will be found in the

macrostate with the greatest entropy (aside from fluc-

tuations that are normally too small to measure).”

Or, if we start from a non-equilibrium system, this

can be said as:

”Entropy tends to increase as time goes.”

Let’s look at several points on this entropy plot,

At equilibrium Pa+b Pa Pb

Away from equilibrium P’a+b P’a P’b

(1) The equilibrium state: ∂SA∂qA

= −∂SB∂qA= ∂SB

∂qB. So ∂SA

∂qA

is related to temperature TA.

(2) The non-equilibrium state: ∂SA∂qA

> 0 means qA

(the energy!) will keep increasing according to the

2nd law. So, system-A absorbs energy ⇒ TA < TB.

In the meantime, ∂SA∂qA

> ∂SB∂qB

. So, it should be TA ∝(∂SA∂qA

)−1. Indeed, it is simply

TA ≡∂SA∂qA

−1

(2)

We do not need another α or β to be the proportional

factor. The units also fit on both side. We will see

this in the example below.

To be precise, the relation is

T ≡(∂U

∂S

)N,V

(3)

1

T≡(∂S

∂U

)N,V

(4)

This is very useful. We can calculate Temperature if

we know the entropy as a function of energy. Let’s

take a look at an example: the Einstein Solid (with

N oscillators and total q energy units).

S = klnΩ = kln

(eq

N

)N(5)

S = kN

[lnq

N+ 1

]= kN

[lnU

εN+ 1

]= kNlnU − kNln(εN) + kN (6)

What is the T?

T =(∂S

∂U

)−1

=(kN

U

)−1=

U

kN(7)

Or,

U = kNT . Or, U =1

2kT · (2N) (8)

This is the equipartition theorem!

Note-1: Hharmonic oscillator = p2

2m + 12Kq

2, the phase

space is 2-d, ie. the oscillator in Einstein solid has

2-d of freedom.

Note-2: In a 3-d Einstein solid, if we have N oscilla-

tors (particles in the formular, each with one or more

energy unit ~ω), we will have N/3 atoms.

We also know the entropy of ideal gas is

S = kN

lnVN

(4πmU

3Nh2

)3/2 +

5

2

S = kNln

VN

(4πm

3Nh2

)3/2· U3/2

+5

2kN

S = kNlnU3/2 + kNln

VN

(4πm

3Nh2

)3/2 +

5

2kN

T =(∂S

∂U

)−1=

(kN

1

U3/2

3

2U1/2

)−1

T =(

3

2kN

1

U

)−1or U =

3

2kNT (9)

Again, this agrees with the equipartition theorem for

ideal gas.

Eq. (8) and (9) also tell us that the definition of tem-

perature Eq. (4) does NOT need other coefficients.

We can connect the temperature (T), thermal en-

ergy (U) with entropy or multiplicity - the microscopic

structure, in such an elegant (precise as well) way.

Very impressive!

§2. Entropy and heat

Heat capacity is defined as CV ≡(∂U∂T

)N,V

. We have

learned how U and T are connected with microscopic

properties. How to calculate CV ? It looks quite

straight forward, in principle one can just follow the

following 5 steps:

(1) Quantum mechanics (microstate) and combina-

torics to calculate multiplicity Ω as function as U, V, T,N .

(2) S = klnΩ.

2

(3) T =(∂S∂U

)−1. To get T(U, N, V)

(4) Solve for U as a function of T

(5) Calculate CV using the definition CV ≡(∂U∂T

)N,V

.

But this procedure can be quite intricate for many sys-

tems. Some simplest examples are, say the Einstein

solid (either at high temperature (q N) or low tem-

perature (q N) ) or ideal gas, the thermal energy U

were given by UE.s. = kNT and Ui.g. = 32kNT . The re-

sult of step (4). So, the heat capacity CV ≡(∂U∂T

)N,V

of them will be,

CV,E.S. = kN (10)

CV,I.G. =3

2kN (11)

where k is Boltzmann’s constant, 1.381×10−23 J/K.

Too simple to learn anything new! Let’s have some

fun with these 5-step exercise by doing a system that

is a lit bit more complicated: Paramagnetism.

B

Energy

State Up Down

+μB

0

-­‐μB

Two-­‐state paramagnet: Magne:c dipoles in an external magne:c field (leB) and energy levels of a single dipole (right)

Step-(1):

The total energy:

U = µB(N↓ −N↑) = µB(N − 2N↑) (12)

The magnetization:

M = µ(N↑ −N↓) = −U

B(13)

The multiplicity:

Ω = CN↑N =

N !

N↑!N↓!(14)

where µ is the magnetic momentum. When N is not

so big, we can calculate Ω and carry out numerically

calculations for the 5-steps - Please read the textbook.

Here, I will show you the analytic solution when N is

very large, when we can use the Stirling’s approxima-

tion lnN ! ≈ NlnN −N

Step-(2):

S = klnΩ = k[lnN !− lnN↑!− ln(N −N↑)!]

≈ kNlnN − kN↑lnN↑ − k(N −N↑)ln(N −N↑) (15)

Now, we are at

Step-(3):

1

T=

(∂S

∂U

)N,B

=∂S

∂N↑

∂N↑∂U

= −1

2µB

∂S

∂N↑(16)

1

T= −

k

2µB

−lnN↑ − 1 + ln(N −N↑) + 1

1

T=

k

2µBln

N↑N −N↑

∵N↑ =N − U/µB

2, from (12)

1

T=

k

2µBln

N − U/µBN + U/µB

(17)

Step-(4): Solving for U as following,2µBkT = ln

(N−U/µBN+U/µB

)e

2µBkT = N−U/µB

N+U/µB

Ne2µBkT + (U/µB)e

2µBkT = N − U/µB

N

(1− e

2µBkT

)= (U/µB)

(1 + e

2µBkT

)

So,

U = NµB

1− e2µB/kT

1 + e2µB/kT

= NµB

e−µB/kT − eµB/kTe−µB/kT + eµB/kT

= −NµBtanh

(µB

kT

)(18)

using Eq. (13):M = −U

B

M = Nµtanh

(µB

kT

)(19)

Note-1: Basic hyperbolic functions:

tanhx = (sinhx)/(coshx),

sinhx = (ex − e−x)/2, coshx = (ex + e−x)/2.

Note-2: tanhx =? when x is small? Several useful

expressions for approximations:

sinhx = x+ x3

3! + x5

5! + ...

coshx = x+ x2

2! + x4

4! + ...

tanhx = x− x3

3 + 2x5

15 + ...

Step-(5): Solve for CB We have to calculate ∂∂T tanh

µBkT .

Let’s first see what ∂tanhX∂X is.

∂sinhX∂X = coshX

∂coshX∂X = sinhX

∂tanhX∂X = ∂

∂XsinhXcoshX = cosh2X−sinh2X

cosh2X= 1

cosh2X

So, ∂∂T tanh

µBkT = − 1

cosh2µBkT

µBkT2

∂∂TU = NµB 1

cosh2µBkT

µBkT2

We now have, solving for CB:

CB =(∂U

∂T

)N,B

= Nk(µB/kT )2

cosh2(µB/kT )(20)

For electronic two-state paramagnet the value of the

constant µ is the Bohr magneton:

µB =eh

4πme= 9.274× 10−24 J/Tessla

= 5.788× 10−5 eV/Tessla (21)

where e and me are electron charge and mass.

Note: At a few Kelvin and above, µB kT . At this

limit, tanhx ≈ x. Therefore, by Eq. (18) we have

M ≈Nµ2B

kTwhen µB kT. (22)

The relation M ∝ 1T was discovered by P. Curie and it

is called Curie’s Law. It is valid at high temperatures

for all paramagnets. In this limit, the heat capacity

is, from Eq. (20), ∝ 1T2

Before we move on, let’s say a few more words about

entropy.

(1) How to measure entropy?

1T ≡

(∂S∂U

)N,V

Keep V and N in the experiment, add a small amount

of heart Q without causing large temperature change

(in phase transition, for example), this can be written

as:

dS =Q

T(23)

In cases where temperature does change, we can use

the heat capacity to write down:

dS =CV dT

T(24)

which can be used to calculate the total entropy change

after the system changes its temperature from Ti to

Tf

∆S =∫ TfTi

CVTdT (25)

When CV can be treated as constant as it is in most

cases, the integration becomes ∆S = CV lnTfTi

.

In-Class Quiz/Exercise 3-01:

(1) What is the heat capacity of 200 gram of water?

(2) By how much the entropy will change if we heat

200 grams of water from 10o C to 60o C

(2) Macroscopic view of entropy.

Originally the entropy was introduced by Rudolf Clau-

sius (1865) to be the thing that increases by Q/T

when heat Q enters a system at temperature T.

This also leads to the fact that when interacting sys-

tems evolve toward equilibrium, the entropy increases.

∆Scold = QTc

∆Swarm = QTw

∆Scold + ∆Swarm = QcTc

+ QwTw

∆Stotal ≥ 0 → QcTc

+ QwTw≥ 0, or QcTw +QwTc ≥ 0

Since |Qc| ≡ |Qw| (energy conservation). So, there are

two options:

+Tw − Tc ≥ 0 assuming Qc > 0.

or

−Tw + Tc ≥ 0 assuming Qw > 0.

Since Tw > Tc, therefore,

Qc ≥ 0: Cold object absorbs heat,

and

Qw ≤ 0: Warm object releases heat.

And, at equilibrium where ∆S = 0, we have Tc = Tw.

Sept. 16

*******

§3. Mechanical Equilibrium and Pressure

energy exchange governed by temperature

volume exchange governed by pressure

particles exchange governed by chemical potential

Let’s see how ”entropy” may help describe the pro-

cess.

3

3.4 Mechanical Equilibrium and Pressure 109

UA, VA, SA UB , VB , SB

Figure 3.13. Two systems that can exchange both energy and volume with eachother. The total energy and total volume are fixed.

UAVA

Stotal

Figure 3.14. A graph of entropy vs. UA and VA for the system shown in Fig-ure 3.13. The equilibrium values of UA and VA are where the graph reaches itshighest point.

UA and VA, as shown in Figure 3.14. The equilibrium point is where Stotal attainsits maximum value. At this point, its partial derivatives in both directions vanish:

∂Stotal

∂UA= 0,

∂Stotal

∂VA= 0. (3.36)

We studied the first condition already in Section 3.1, where we concluded that thiscondition is equivalent to saying that the two systems are at the same temperature.Now let us study the second condition in the same way.

The manipulations are exactly analogous to those in Section 3.1:

0 =∂Stotal

∂VA=

∂SA∂VA

+∂SB∂VA

=∂SA∂VA

− ∂SB∂VB

. (3.37)

The last step uses the fact that the total volume is fixed, so dVA = −dVB (anyvolume added to A must be subtracted from B). Therefore we can conclude

∂SA∂VA

=∂SB∂VB

at equilibrium. (3.38)

3.4 Mechanical Equilibrium and Pressure 109

UA, VA, SA UB , VB , SB

Figure 3.13. Two systems that can exchange both energy and volume with eachother. The total energy and total volume are fixed.

UAVA

Stotal

Figure 3.14. A graph of entropy vs. UA and VA for the system shown in Fig-ure 3.13. The equilibrium values of UA and VA are where the graph reaches itshighest point.

UA and VA, as shown in Figure 3.14. The equilibrium point is where Stotal attainsits maximum value. At this point, its partial derivatives in both directions vanish:

∂Stotal

∂UA= 0,

∂Stotal

∂VA= 0. (3.36)

We studied the first condition already in Section 3.1, where we concluded that thiscondition is equivalent to saying that the two systems are at the same temperature.Now let us study the second condition in the same way.

The manipulations are exactly analogous to those in Section 3.1:

0 =∂Stotal

∂VA=

∂SA∂VA

+∂SB∂VA

=∂SA∂VA

− ∂SB∂VB

. (3.37)

The last step uses the fact that the total volume is fixed, so dVA = −dVB (anyvolume added to A must be subtracted from B). Therefore we can conclude

∂SA∂VA

=∂SB∂VB

at equilibrium. (3.38)

Let system A and B exchange their energy and vol-

ume, with the total volume and energy fixed. Then

the entropy will be

Stotal = S(VA, UA). The equilibrium point is where

Stotal reaches the maximum:

∂Stotal∂UA

= 0 (26)

∂Stotal∂VA

= 0 (27)

Eq. (26) leads to the thermal equilibrium when TA =

TB: (T ≡(∂U∂S

)N,V

1T ≡

(∂S∂U

)N,V

)

You can imagine what Eq. (27) tells us:

∂Stotal∂VA

=∂SA∂VA

+∂SB∂VA

(28)

Because V = VA + VB, V = constant.∂Stotal∂VA

=∂SA∂VA

−∂SB∂VB

(29)

So, at mechanical equilibrium:∂SA∂VA

=∂SB∂VB

(30)

The actual pressure is P = T

(∂S

∂V

)U,N

.

Note:

(1) PA = PB at mechanical equilibrium ;

(2) [T(∂SB∂VB

)U,N

]units = K(J/K)/m3 = N ∗ m/m3 =

N/m2 = [P ]units

For Ideal Gas, let’s see if we can get the expression

for pressure P .

Using Eq. (2.41) Ω(U, V,N) = f(N)V NU3N/2,

S = NklnV +3

2kNlnU + klnf(N) (31)

P = T

(∂S

∂V

)U,N

(32)

= T∂S

∂V(kNlnV ) =

kNT

V(33)

That is PV = kNT !! (34)

This is exactly the Ideal Gas law!

How does the total entropy S = S(U, V,N, T ) change

when the system has a tiny amount of change in vol-

ume and energy?

dS =(∂S

∂V

)U,N

dV +(∂S

∂U

)V,N

dU (35)

dS =P

TdV +

1

TdU (36)

which can be written as:

dU = TdS − PdV (37)

The thermodynamic identity.

True for any infinitesimal change in any system as

long as T and P are well defined and no other relevant

variables are changing.

The first law of thermodynamics: dU = Q + W (en-

ergy conservation). Compare with dU = TdS − PdV

(with changes of Q and V). So, Q = TdS.

Three conditions for this equation are:

(1) Change in volume is quasi-static: pressure is uni-

form (slow change allowed!) throughout the system;

Quasistatic process system goes through a sequence

of states that are infinitesimally close to equilibrium

the process is typically reversible.

A slow process that is NOT

quasistatic: Gas particles ex-

pands into vacummn.

Another example: fast com-

pression or expansion.

(2) No other forms of work but −PdV ;

(3) No N or other relevant variables are changing.

When Q = 0, dS = 0 - No entropy increase! - This

is called ”isentropic” process. So, ”isentropic = adi-

abatic + quasistatic” ! The volume is NOT required

to be fix!

Two interesting examples:

1: Entropy change when P is fixed: heat exchange Q

↔ T changes (V may also change). ∆S =∫TfTi

CPT dT .

2: Entropy change during phase transition: T is fixed,

heat exchange Q ↔ V changes. ∆S = Q/T .

An example for case-1In-Class Quize/Exercise 3-02:

(1) How much heat is needed to have 1 liter water

boiled at 100oC and 1 atmosphere pressure ?

Hint: From Table on page 167.

(2) What is the change in entropy of 1 liter after it is

boiled?

In-Class Quize/Exercise 3-03: (Problem 3.33)

(1) Use the thermodynamics identity to derive the

heat capacity formula CV = T(∂S∂T

)V

.

(2) The heat capacity at constant pressure CP =(∂S∂T

)P

+ P(∂V∂T

)P

. The 2nd term comes from work

done at constant P. If we define a new quantity

H = U + PV , called ”enthalpy”, the heat capacity

at constant pressure can be written as CP =(∂H∂T

)P

because P is fixed. Derive a similar formula for CP by

first write dH in terms of dS and dP .

§4. Diffusive Equilibrium and Chemical Potencial

Now let energy and particle exchange between the two

systems:

System-A: (UA, PA, VA, SA, NA, TA)

System-B: (UB, PB, VB, SB, NB, TB).∂Stotal∂UA

NA,VA

= 0 and

∂Stotal∂NA

UA,VA

= 0 (38)

∵NA +NB = N fixed∂SA∂NA

UA,VA

=

∂SB∂NB

UB,VB

at equilibrium (39)

To see what change in energy caused by particle ex-

change only, we assume the two system are at the

same temperature and have no work done to or be-

tween these two systems. So, we also have

T(∂SA∂NA

)UA,VA

= T(∂SB∂NB

)UB,VB

at equilibrium.

The quantity µ = −T(∂S∂N

)U,V

defines another quan-

tity: Chemical Potential.

Note-1: The ”-” sign: Particles tend to flow from

the system with higher chemical potential to the sys-

tem with lower chemical potential.

Note-2: µ has units of energy.

Let’s look at an example, the Ideal Gas: We have obtained the

famous Sackur-Tetrode Equation:

S ≈ kN[ln

(V

N

(4πmU

3Nh2

)3/2)

+5

2

](40)

S ≈ kN[ln

(V

(4πmU

3h2

)3/2)− lnN5/2 +

5

2

](41)

Take the derivative with respect to N, we get:

4

∂S

∂N=k

[ln

(V

(4πmU

3h2

)3/2)− lnN5/2 +

5

2

]+ kN

(−

5

2N

)∂S

∂N=k

[ln

(V

(4πmU

3h2

)3/2)− lnN5/2

]

∂S

∂N=kln

VN

4πm32NkT

3Nh2

3/2

∂S

∂N=kln

VN

(2πmkT

h2

)3/2

So, we have

µ = −T(∂S

∂N

)U,V

µ = −kT[ln

(V

(4πmU

3h2

)3/2)− lnN5/2

](42)

µ = −kT

lnV

4πm32NkT

3h2

3/2− lnN5/2

µ = −kT

lnV N3/2

(2πmkT

h2

)3/2− lnN5/2

µ = −kT

lnVN

(2πmkT

h2

)3/2 (43)

Summarize this chapter by this table:

What is the thermodynamic identity to include processes

in which N also changes?

For a system, if we allow the exchange of energy, volume, and

particles at the same time, we can describe the change by entropy

as

dS =(∂S

∂U

)N,V

dU +(∂S

∂V

)N,U

dV +(∂S

∂N

)U,V

dN (44)

dS =1

TdU +

P

TdV −

µ

TdN (45)

dU = TdS − PdV + µdN (46)

Since different particles have different chemical potential, if a

system consists of different types of particles, the generalized

5

thermodynamic identity is

dU = TdS − PdV +∑i

µidNi (47)

Note-1: In a process with fixed U and V, one has 0 = TdS+µdN ,

µ = −T(∂S∂N

)U,V

. (3.59) in the textbook.)

Note-1: In a process with fixed S and V, one has dU = µdN ,

µ =(∂U∂N

)S,V

. (3.60) in the textbook.)

In-Class Quize/Exercise 3-04: (Problem 3.37)

Consider a monatomic ideal gas that lives at height z above the

sea level, so each molecule has potential energy Ep = mgz in

addition to its kinetic energy. Show that the chemical potential

is the same as if the gas were at sea level, plus an additional

term mgz:

µ(z) = −kT ln[VN

(2πmkTh2

)3/2]

+mgz.

About mid-term exam.

• Sept 23 (exam, to be posted at class-time, 4 problems/2

hours.)

6