Chapter 5Simple Applications of Macroscopic

Thermodynamics

Preliminary DiscussionClassical, Macroscopic,

Thermodynamics• Now, we drop the statistical mechanics

notation for average quantities. So that now,

All Variables are Averages Only! • We’ll discuss relationships between

macroscopic variables using

The Laws of Thermodynamics

• Some Thermodynamic Variables of Interest:Internal Energy = E, Entropy = S

Temperature = T• Mostly for Gases:

(but also true for any substance):

External Parameter = VGeneralized Force = p(V = volume, p = pressure)

• For a General System:External Parameter = xGeneralized Force = X

• Assume that the External Parameter = Volume V in order to have a specific case to discuss. For systems with another external parameter x, the infinitesimal work done đW = Xdx. In this case, in what follows, replace p by X & dV by dx.

• For infinitesimal, quasi-static processes:

1st & 2nd Laws of Thermodynamics1st Law: đQ = dE + pdV

2nd Law: đQ = TdS

Combined 1st & 2nd LawsTdS = dE + pdV

Combined 1st & 2nd LawsTdS = dE + pdV

• Note that, in this relation, there are

5 Variables: T, S, E, p, V• It can be shown that:Any 3 of these can always be expressed

as functions of any 2 others.• That is, there are always 2 independent

variables & 3 dependent variables. Which 2 are chosen as independent is arbitrary.

Brief, Pure Math Discussion• Consider 3 variables: x, y, z. Suppose we

know that x & y are Independent Variables. Then, It Must Be Possible to express z as a function of x & y. That is,

There Must be a Function z = z(x,y).• From calculus, the total differential of z(x,y)

has the form:

dz (∂z/∂x)ydx + (∂z/∂y)xdy (a)

• Suppose that, in this example of 3 variables: x, y, z, we want to take y & z as independent variables instead of x & y. Then,

There Must be a Function x = x(y,z).• From calculus, the total differential of x(y,z) is:

dx (∂x/∂y)zdy + (∂x/∂z)ydz (b)• Using (a) from the previous slide

[dz (∂z/∂x)ydx + (∂z/∂y)xdy (a)]& (b) together, the partial derivatives in (a) & those in (b) can be related to each other.

• We always assume that all functions are analytic.

So, the 2nd cross derivatives are equalSuch as: (∂2z/∂x∂y) (∂2z/∂y∂x), etc.

Mathematics Summary• Consider a function of 2 independent variables:

f = f(x1,x2).• It’s exact differential is

df y1dx1 + y2dx2 & by definition:

• Because f(x1,x2) is an analytic function, it is always true that:

2 1

2 1

1 2x x

y y

x x

• Most Ch. 5 applications use this with theCombined 1st & 2nd Laws of Thermodynamics

TdS = dE + pdV

Some MethodsMethods & Useful Math ToolsUseful Math Tools for Transforming DerivativesTransforming Derivatives

Derivative Inversion

Triple Product (xyz–1 rule)

Chain Rule Expansion to Add Another Variable

Maxwell Reciprocity Relationship

xx FyyF

1

TT SPPS

1

1

xFy Fy

yx

xF

1

THP HP

PT

TH

xxx yF

yF

TCTC

HT

TS

HS

P

P

PPP

11

y

x

x

y

x

yF

y

xF

yxxy FF

Pure Math: Jacobian Transformations•A Jacobian Transformation is often used totransform from one set of independentvariables to another.•For functions of 2 variables f(x,y) & g(x,y) it is:

yxxy

xy

xy

xg

yf

yg

xf

yg

xg

yf

xf

yxgf

,,

Determinant!

Transposition

Inversion

Chain Rule Expansion

yx

fgyxgf

,,

,,

gfyxyx

gf

,,

1,,

yx

wzwzgf

yxgf

,,

,,

,,

Jacobian TransformationsJacobian TransformationsHave Several Useful PropertiesUseful Properties

• Suppose that we are only interested in the first partial derivative of a function f(z,g) with respect to z at constant g:

gz

gfzf

g ,,

yx

gzyxgf

zf

g

,,,,

• This expression can be simplified using the chain rule expansion & the inversion property

dE = TdS – pdV (1)

First, choose S & V as independent variables:E E(S,V)

Properties of the Internal Energy E

dVV

UdS

S

UdU

SV

TS

U

V

p

V

U

S

Comparison of (1) & (2) clearly shows that

dE∂E

(2)

Applying the general result with 2nd cross derivatives gives:

VS S

p

V

T

Maxwell RelationMaxwell Relation I! I!

and∂E

∂E

∂E

If S & p are chosen as independent variables, it is convenient to define the following energy:

H H(S,p) E + pV EnthalpyEnthalpyUse the combined 1st & 2nd Laws. Rewrite them in terms of dH: dE

= TdS – pdV = TdS – [d(pV) – Vdp] ordH = TdS + Vdp

Comparison of (1) & (2) clearly shows that

(1)

(2)

Applying the general result for the 2nd cross derivatives gives:

pSS

V

p

T

But, also:

and

Maxwell RelationMaxwell Relation II! II!

If T & V are chosen as independent variables, it is convenient to define the following energy:

F F(T,V) E - TS Helmholtz Free Helmholtz Free EnergyEnergy• Use the combined 1st & 2nd Laws. Rewrite them in terms of dF:

dE = TdS – pdV = [d(TS) – SdT] – pdV or

dF = -SdT – pdV (1)

• But, also: dF ≡ (F/T)VdT + (F/V)TdV (2)

• Comparison of (1) & (2) clearly shows that (F/T)V ≡ -S and (F/V)T ≡ -p

• Applying the general result for the 2nd cross derivatives gives:

Maxwell RelationMaxwell Relation III! III!

If T & p are chosen as independent variables, it is convenient to define the following energy:

G G(T,p) E –TS + pV Gibbs Free Gibbs Free EnergyEnergy• Use the combined 1st & 2nd Laws. Rewrite them in terms of dH:

dE = TdS – pdV = d(TS) - SdT – [d(pV) – Vdp] or

dG = -SdT + Vdp (1)

• But, also: dG ≡ (G/T)pdT + (G/p)Tdp (2)

• Comparison of (1) & (2) clearly shows that

(G/T)p ≡ -S and (G/p)T ≡ V• Applying the general result for the 2nd cross derivatives gives: Maxwell RelationMaxwell Relation IV! IV!

1. Internal Energy: E E(S,V)

2. Enthalpy: H = H(S,p) E + pV

3. Helmholtz Free Energy: F = F (T,V) E – TS

4. Gibbs Free Energy: G = G(T,p) E – TS + pV

Summary: Energy FunctionsEnergy Functions

Combined 1Combined 1stst

& & 22ndnd Laws Laws

1. dE = TdS – pdV

2. dH = TdS + Vdp

3. dF = - SdT – pdV

4. dG = - SdT + Vdp

dyy

zdx

x

zdzNdyMdx

xy

yxx

N

y

M

pSS

V

p

T

VS S

p

V

T

VT T

p

V

S

pTT

V

p

S

1. 2.

3. 4.

Another Summary: Maxwell’s Relations

(a) ΔE = Q + W

(b) ΔS = (Qres/T)

(c) H = E + pV

(d) F = E – TS

(e) G = H - TS

1. dE = TdS – pdV2. dH = TdS + Vdp3. dF = -SdT - pdV4. dG = -SdT + Vdp

Maxwell Relations: “The Magic Square”?

V F T

G

PHS

E

Each side is labeled with an

Energy (E, H, F, G). The corners are labeled with

Thermodynamic Variables

(p, V, T, S). Get the

Maxwell Relations by “walking” around the square. Partial derivatives are obtained from the sides.

The Maxwell Relations are obtained from the corners.

SummaryThe 4 Most CommonMost CommonMaxwell Relations:Maxwell Relations:

PTPS

VTVS

T

V

P

S

S

V

P

T

T

P

V

S

S

P

V

T

Maxwell Relations: Table (E → U)

InternalEnergy

HelmholtzFree Energy

Enthalpy

Gibbs FreeEnergy

Maxwell RelationsMaxwell Relations from dE, dF, dH, & dG

Some Common Measureable PropertiesHeat Capacity at Constant Volume:

Heat Capacity at Constant Pressure:

∂E

More Common Measureable PropertiesVolume Expansion Coefficient:

Isothermal Compressibility:

Note!! Reif’snotation forthis is α

The Bulk Modulus is theinverse of the IsothermalCompressibility!

B (κ)-1

Some Sometimes Useful RelationshipsSummary of Results

Derivations are in the text and/or are left to the student!

Entropy:

dTRT

HdP

RT

V

RT

Gd

2

Enthalpy:

Gibbs FreeEnergy:

Typical Example• Given the entropy S as a function of temperature

T & volume V, S = S(T,V), find a convenient expression for (S/T)P, in terms of some measureable properties.

• Start with the exact differential:

• Use the triple product rule & definitions:

• Use a Maxwell Relation:

• Combining these expressions gives:

• Converting this result to a partial derivative gives:

• This can be rewritten as:

• The triple product rule is:

• Substituting gives:

Note again the definitions:• Volume Expansion Coefficient

β V-1(V/T)p

• Isothermal Compressibilityκ -V-1(V/p)T

• Note again!! Reif’s notation for theVolume Expansion Coefficient is α

• Using these in the previous expression finally gives the desired result:

• Using this result as a starting point,

A GENERAL RELATIONSHIP between the

Heat Capacity at Constant Volume CV

& the Heat Capacity at Constant Pressure Cp

can be found as follows:

• Using the definitions of the isothermal compressibility κ and the volume expansion coefficient , this becomes

General Relationship between Cv & Cp

Simplest Possible Example: The Ideal Gas

P

RTPRT

vPRT

PRT

PvPv

v

T

RTR

vPR

PRT

TvTv

v

TT

PP

1

11

1

11

2

• For an Ideal Gas, it’s easily shown (Reif) that the Equation of State (relation between pressure P, volume V,

temperature T) is (in per mole units!): Pν = RT. ν = (V/n)• With this, it is simple to show that the volume expansion

coefficient β & the isothermal compressibility κ are:

and

and

• So, for an Ideal Gas, the volume expansion coefficient & the isothermal compressibility have the simple forms:

• We just found in general that the heat capacities at constant volume & at constant pressure are related as

• So, for an Ideal Gas, the specific heats per mole have the very simple relationship:

Other, Sometimes Useful, Expressions

TCONSTANTdVV

R

T

PS

TCONSTANTdPP

R

T

VS

TCONSTANTdPT

VTVH

P

P VTV

P

P PTP

P

P PTP

0

.

0

.

0

.

More Applications: Using the Combined 1st & 2nd Laws (“The TdS Equations”)

Calorimetry Again! • Consider Two Identical Objects, each of mass m, &

specific heat per kilogram cP. See figure next page.

Object 1 is at initial temperature T1.Object 2 is at initial temperature T2.

Assume T2 > T1.• When placed in contact, by the 2nd Law, heat Q

flows from the hotter (Object 2) to the cooler (Object 1), until they come to a common temperature, Tf.

• Two Identical Objects, of mass m, & specific heat per kilogram cP. Object 1 is at initial temperature T1. Object 2 is at initial temperature T2.

• T2 > T1. When placed in contact, by the 2nd Law, heat Q flows from the hotter (Object 2) to the cooler (Object 1), until they come to a common temperature, Tf.

Object 1Initially

at T1

Object 2Initially

at T2

Q Heat Flows

221 TT

T f

• After a long enough time, the two objects are at the same temperature Tf. Since the 2 objects are identical, for this case,

For some timeafter initialcontact:

• The Entropy Change ΔS for this process can also be easily calculated:

21

21

21

2

2121

2

21

2ln2

ln2lnln

lnln1 2

TT

TTmcS

TT

Tmc

TT

Tmc

TT

Tmc

T

T

T

Tmc

T

dT

T

dTmcS

P

fP

fP

fP

ffP

T

T

T

TP

f f

• Of course, by the 2nd Law,the entropy change ΔS must be positive!! This requires that the temperatures satisfy: 0)(

02

42

2

221

212

22

1

21212

22

1

2121

TT

TTTT

TTTTTT

TTTT

Some Useful “TdS Equations”• NOTE: In the following, various quantities are

written in per mole units! Work with theCombined 1st & 2nd Laws:

Definitions:• υ Number of moles of a substance. • ν (V/υ) Volume per mole.• u (U/υ) Internal energy per mole. • h (H/υ) Enthalpy per mole. • s (S/υ) Entropy per mole. • cv (Cv/υ) const. volume specific heat per

mole. • cP (CP/υ) const. pressure specific heat per mole.

dPc

dvv

cdP

P

Tcdv

v

TcTds

dPTvdTcdPT

vTdTcTds

dvT

dTcdvT

PTdTcTds

vP

vv

PP

PP

P

vv

v

• Given these definitions, it can be shown that the Combined 1st & 2nd Laws (TdS) can be written in at least the following ways:

Internal Energy u(T,ν):

dvPvu

dTcTds

dvvu

dTTu

du

Tv

Tv

Enthalpy h(T,P):

• Student exercise to show that, starting with the previous expressions & using the definitions (per mole) of internal energy u & enthalpy h gives:

v

v

v

vvvvv

Pv

P

T

T

c

P

s

P

T

T

sT

TP

T

T

s

P

s

dvv

sdP

P

sds

vPss

1

),( Consider

P

P

P

pPPPP

Pv

v

T

T

c

v

s

v

T

T

sT

Tv

T

T

s

v

s

dvv

sdP

P

sds

1

dvv

TcdP

P

TcTds

dvv

T

T

cdP

P

T

T

cds

dvv

sdP

P

sds

PP

vv

P

P

v

v

Pv

• Student exercise also to show that similar manipulations give at least the following different expressions for the molar entropy s: Entropy s(T,ν):