PHY 341/641 Thermodynamics and Statistical Mechanics MWF ...

PHY 341/641 Thermodynamics and Statistical Mechanics

MWF: Online at 12 PM & FTF at 2 PM

Discussion for Lecture 14:Useful thermodynamic energy functions

Reading: Chapters 5.1

1. Internal energy U

2. Enthalpy H

3. Helmholtz free energy F

4. Gibbs free energy G

Record!!!

2/26/2021 PHY 341/641 Spring 2021 -- Lecture 14 1



Reminder – lecture notes are available on the class webpagehttp://users.wfu.edu/natalie/s21phy341/lecturenote/

http://users.wfu.edu/natalie/s21phy341/lecturenote/


Your questions –

From Kristen – 1. For the heat that is free from the environment, as in Helmholtz free energy, if there is indefinite heat, will it absorb all of the energy it needs from this or will we still have to input some work? 2. Could we review a bit about the reaction for the battery as discussed on page 154, I just need a bit of clarification.

From Parker -- We will invoke the Legendre transformation mathematically for changes in V to P and TS from S to T in this class?

From Michael -- Can you further explain why fuel cell engines are not more abundant in production given their high efficiency?

From Rich – In the Gibbs Free energy equation, how can the change in entropy be negative without violating the second law of thermodynamics?

From Noah -- Can you explain more why a problem would or would not care about the +PV term? E.g. why does the example of electrolysis use Gibbs free energy instead of Helmholtz?


Internal energy U

First law of thermodynamics

dU Q WdU TdS PdV

Q TdS W PdV

= += −

⇔ ⇔ −

Note that this analysis implies that we should consider the internal energy( , )

V S

V S

U

U U S

P

V

dU

P

U UdS dV TdS dVS V

UTS V

∂ ∂+ = −

∂ ∂

∂

=

=

∂⇒ =

∂ = − ∂


Mathematical consistencySuppose ( , ) such that ( , ) ( , )

where ( , ) and ( , )

Check that second derivatives arexy

f x y df a x y dx b x y dy

a x y b x yf fx y∂ ∂∂

= +

= = ∂

2 2

consistent --

yy x x

f f f a by x y x x y y x

∂ ∂ ∂ ∂ ∂ ∂≡

∂ ∂ ∂ ∂ ∂

= ⇒ = ∂ ∂ ∂

For example: ( , )

Maxwell's relations show that

V S

V S

S V

U U SU UdS dV TdS PdVS VU

U

UTS V

TV

V

d

P

PS

=

=

=⇒ −

∂ ∂+ =

∂

= −

−∂ ∂

∂=

∂ ∂

∂ ∂

∂∂

Maxwell’s relations


Generalizing U to allow for variable N --

,

0

A

re

n

thes

e equ

Suppose and are constant; 0 a d 0

Suppose and ar

ation internally consisten

e constant; 0 and 0

?

t

0

0 0

U V

dNU V dU dV

dNS

T NS V dS dV

d

d

d

U

U

dN

T S PdV

TdS

µ

µµ

µ

µ

= =⇒ =

∂ = − ∂ = =

∂

= − +

− +

⇒ = −

=

+

,S V

UN

∂


Summary of results for internal energy and entropy --

( , , )

, 1( , )

dNPS S U V N dS dU dV dN

U

T T

U S V N dU Td

T

S PdV µµ

= = + −

= = − +

, , ,

Some first derivative relationships --

V N S N V S

PU U UTS V N

µ∂ ∂ ∂= = = − ∂ ∂ ∂

, , , , , ,

Some second derivative relationships (thanks to Maxwell) --

S N V N S V V N S V S N

T P T PV S N S N V

µ µ∂ ∂ ∂ ∂ ∂ ∂ = − = = −

∂∂

∂ ∂ ∂ ∂


What can we do to rationalize these interdependencies?

Define new function

( , )

For ,

y u

w ww u y dw du dyu y

w z ux dw dz udx xdu udx vdy udx xdudw xdu vdy

∂ ∂ ⇒ = + ∂ ∂ = − = − − = + − −

= − +

General notions of mathematical transformations for continuous functions of several variables and Legendre transforms --

and Let

),(

xy

xy

yzv

xzu

dyyzdx

xzdzyxz

∂∂

≡

∂∂

≡

∂∂

+

∂∂

=⇒

y u x

w w zx vu y y

∂ ∂ ∂ ⇒ = − = = ∂ ∂ ∂

Consider:

dz udx vdy= +


Desired thermodynamic functions --

1Entropy

G

( , , )

Internal energy ( , , )

, , )

Enthalpy Helmholtz free energy

( ( , , )

U dNPS S U V N dS dU dV dN

U S V N dU TdS PdV

S P NHF F T V

TH

N

T T

µµ

=

= = + −

=

= − +

=ibbs free energy ( , , )G G T P N=

(

Using the Legendre transformation method:( , , )

, , )

( , , )

dNF T V N U ST SdT PdV dNH S P N U PV dH TdS

NG T P N F PV dG SdT VdPd

d

VdPF

µµµ

+= − = − − += + = − +

+

+

= = +


1Entropy

n

( , , )

Internal energy ( , , )

Enthalpy

, , ) Helmholtz f

ree

(e

U dNPS S U V N dS dU dV d

d

N

P

T T T

U S V N dU TdS dV

S P N H Td PH H dNS Vd

µµ

µ=

= = −

−

+

+

= = +

= +ergy ( , , )

Gibbs free energy ( , , ) SdT PdV dN

dG SdT VdP dNF F T V N dFG G T P N

µµ

= − − += + +

== −

Summary of thermodynamic functions

Various first derivative relationships

, , , ,

, , , ,

,

Some first derivative relationships --

= =

= =

=

V N P N S N T N

V N P N S N T N

V S

U H U

V

FTS S V V

F G H GST T P P

UN

P

HN

µ

= − −

− =

∂ ∂

∂

=

∂ ∂=

∂ ∂ ∂ ∂

∂ ∂ ∂−

∂ ∂ ∂ ∂

∂ ∂∂ ∂

=, , ,

= = P S V T T P

F GN N

∂ ∂

∂ ∂


Which energy to use?Choose the one that most easily describes the circumstances of the systemAt constant T and P, the Gibbs free energy is often a good choice

Gibbs free energy ( , , )Note also that

dG SdT VdPS

G Gd

T P N dNG H ST dG H SdT Td

µ= = − + += − = − −

2 2 2 2 2 2

2 2 2

2 2 2

H H O O H O H O

H H O

2

O

2

O

H O

2

H

Example -- consider the following reaction at fixed T (298 K) and P (1 bar)1H O H O2

1In this case, 2

12

f i

f i

dN dN dN

dN dN dN

dN

G G G

dN

G G G

µ µ µ

µ µ µ

→ +

= + −

= =

= +

−

−

∆ =

≡

∆ =

−


Table of energies from your textbookfor T=298 K and

P=1 bar


2 2 2 2 2 2

2 2 2

2 2 2

H H O O H O H O

H H O

2

O

2

O

H O

2

H

Example -- consider the following reaction at fixed T (298 K) and P (1 bar)1H O H O2

1In this case, 2

12

f i

f i

dN dN dN

dN dN dN

dN

G G G

dN

G G G

µ µ µ

µ µ µ

→ +

= + −

= =

= +

−

−

∆ =

≡

∆ =

−

237.13 kJ/mole


Electrochemical reactions4

2 4

4

42

2

4 4

42

2

2 4

electrode: Pb+HSO PbSO H +2e electrode: PbO HSO 3H +2e PbSO 2H O

in solution: 2SO + 2H 2HSOnet: Pb+PbO 4H +2SO 2PbSO 2H O

−

−

− −

+ −

+

+ −

+ −

− → +

+ + + → +

→

+ → +

19 23

19 23

394 / 21.60 10 6.02 10

394000 2 Volts

For each mole

2 1.60 1

:

0 6.02 10

avo

avo

G kJ mol eNe C N−

−

∆ = − = −

= × = ×

= =× × × ×

V

V

PHY 341/641 Thermodynamics and Statistical Mechanics MWF ...

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