Chapter 4 Joule Thomson Effect Chemical Potential Fugacitychemistry.tcd.ie/staff/people/duesberg/ASIN/lecture not… · · 2016-08-24Joule Thomson Effect Chemical Potential Fugacity

Chemical Thermodynamics : Georg Duesberg Chemical Thermodynamics : Georg Duesberg

Chapter 1 : Slide 1

Chapter 4

Joule Thomson Effect

Chemical Potential

Fugacity

Chapter11111 1 : Slide

1

Chemical Thermodynamics : Georg Duesberg 2

Joule – Thomson Effect

Refrigeration developed by Carl von Linde in 19th Century, in response to a request from Guinness in Dublin for a new cooling technique.

Based upon the fact that gases cool as they expand: Joule-Thomson effect (William Thomson, later Lord Kelvin, born in Belfast),

The Linde refrigerator combines the JT process

with a counter-flow heat exchanger.

The gas is re-circulated and it cools on expansion

through the throttle. The cooled gas cools the high-

pressure gas, which cools still further as it

expands. Eventually liquefied gas drips from the

throttle.

Heat capacities

V

VT

UC

P

PT

HC

PdVdWwith

From first law of thermodynamics:

dVV

UdT

T

UdU

TV

pVUH

dpp

HdT

T

HdH

Tp

dVV

UdTCdU

T

V

dp

p

HdTCdH

T

p

Heat capacity at

constant volume

(isochore)

Heat capacity at

constant pressure

(isobar)

dWdQdU

pVdQdU

Chemical Thermodynamics : Georg Duesberg

Change of Internal Energy at constant temperature

for ideal gases No interactions between molecules.

the internal energy of gas depends on temperature only.

• Z = 1; PV = RT

• U = U (T)

V

VT

UC

dVV

UdT

T

UdU

TV

T

TV

U

dVdTCdU TV

Internal Pressure

of a gas

(0 for ideal gas)

Heat capacity at

constant volume

[Note from VdW: a(n2/V2)= πT ]


Internal Pressure, πT

For ideal gas, πT = 0 because U

independent of molecular separation

for volume changes at constant T

• Implies ideal gas law

For real gas,

If dU >0 as dV increases with T

constant, attractions between

molecules dominate and πT > 0

If dU <0 as dV increases with T

constant, repulsions between

molecules dominate and πT <0

[Note from VdW: a(n2/V2)= πT ]


Joule Experiment: Measure πT

pex = 0, therefore w = 0

dT=0, therefore q = 0

DU = q +w = πT dV= 0

πT must =0 since dV>0

w pexdV

V1

V2

“No change in temperature occurs when

air is allowed to expand in such a manner

as not to develop mechanical power” J. P.

Joule, Phil. Mag., 26, 369 (1845)

Joule experiment incapable of detecting

small effects since the water calorimeter he

used had a large heat capacity pex = 0


THE JOULE-THOMPSON EXPERIMENT A further test of intermolecular forces in real gases.

Imagine a sample of gas pushed through a porous

plug, in an isolated tube (adiabatic system). The

temperature is measured on each side of the plug. Analysis

w = piVi - pfVf Since DU = Uf - Ui = w (because q = 0),

Uf + pfVf = Ui + piVi

Hf = Hi i.e. DH = 0

This is a constant enthalpy (isenthalpic) process.

T

P

P P

HC

T

H

???

What is

0D

D

D P

P

HT

T

HH

TP


Thermodynamic Consequences of State Functions

Enthalpy, H(p,T)

To evaluate (dH/dp)T we apply Euler chain rule:

Cp= (dH/dT)p and one defines (dT/dP)H as µ as Joule Thomson Coefficient

then (dH/dp)T =- Cpµ, is the Isothermal Joule Thomson Coefficient

JT

HP

T

T

PJTP

HC

0D

D

D P

P

HT

T

HH

TP

Hp

p

H

TP

T

T

H

H

T

P

T

p

H

TP

H

???

P

P

CT

H


Chapter99999 1 : Slide

9

Joule Thompson Experiment

Modern methods measure µCP (Isothermal)

not µ

Isothermal J-T coefficient, µT = (dH/dp)T

= -Cpµ

Practical applications: Gas liquification,

Isotope separation, artificial snow


µ - Physical consensus

µ is a function of p and T (see Fig. for CO2)

µ can be either (+) or (-)

• Positive µ means dT is negative when dp is

negative

Gas cools on expansion

• Negative µ means that means dT is

positive when dp is negative

Gas warms on expansion

• Transition between positive and negative

µ is called the Joule-Thompson inversion

temperature (Ti =2TB )

Gases have both high and low inversion

temperatures

Joule-Thomson Process for the vdW Gas

0D

D

D P

P

HT

T

HH

TP

The JT process corresponds to an isenthalpic expansion:

PP

HTCC

T

H

T

PP

P

D

D

VP

ST

P

H

PVSTH

TT

DDD

relation) (MaxwellPT T

V

P

S

This is a pretty general (model-independent) result. By applying this result to the vdW equation,

one can qualitatively describe the shape of the inversion curve (requires solving cubic

equations...).

p

P

p

T

p

T

H C

VT

VT

C

P

H

T

H

P

H

P

T

PT

V

May be obtained form the equation of state


We’ll consider the vdW gas at low densities: NbVV

aNP

2

2

( ) ( )PBBT

TNkPNbV

aNPVTNkNbV

V

aNP ...

2

2

2

2

22

2

V

aNP

Nk

T

VNk

T

V

V

aN

T

VP B

P

B

PP

PP

B

C

bRT

a

C

NbTk

Na

22

p

P

H C

VT

VT

P

T



The upper inversion temperature:

(at low densities)

BC

B

INV TTbk

aT 2

4

272

Cooling: 02

bTk

a

B

JT

If b = 0, T always decreases in the JT process: an increase of Upot at the expense of K.

If a = 0, T always increases in the JT process (despite the work of molecular forces is 0):

Heating: 02

bTk

a

B

JT

Thus, the vdW gas can be liquefied by

compression only if its T < 27/4TC.

•Tc =8a/27Rb •the critical temperature of the vdW gas •TB = a/Rb

The JT Process in Real Gases

PVUUH potkin

In real gases, molecules interact with each

other.

At low densities, the intermolecular forces

are attractive. When the gas expands

adiabatically, the average potential energy

increases, at the expense of the kinetic energy.

Thus, the temperature decreases because of

the internal work done by the molecules

during expansion.

Upot x

expansion

vdW gas

0JT


The JT Process in Real Gases

For T < TINV, the

drop in pressure

(expansion) results

in a temperature

drop.

Gas boiling T

(P=1 bar)

inversion T

@ P=1 bar

CO2 195 (2050)

CH4 112 (1290)

O2 90.2 893

N2 77.4 621

H2 20.3 205

4He 4.21 51

3He 3.19 (23)

isenthalpic curves

(H =const) for

ideal and

real gases

cooling

heating

All gases have two inversion temperatures: in the

range between the upper and lower inversion

temperatures, the JT process cools the gas, outside

this range it heats the gas.

Upot x

expansion

At high densities, the effect is reversed: the free

expansion results in heating, not cooling. The overall

situation is complicated: the sign of DT depends on initial T

and P.

0JT


Liquification of Gases

For air, the inversion T is above

RT. In 1885, Carl von Linde

liquefied air in a liquefier based

solely on the JT process: the gas

is recirculated and, since T is

below its inversion T, it cools on

expansion through the throttle.

The cooled gas cools the high-

pressure gas, which cools still

further as it expands. Eventually

liquefied gas drips from the

throttle.

Most gas liquifiers combine the expansion engines with JT process: the expansion

engine helps to pre-cool the gas below the inversion T. The expansion engines are a

must for He and H2 liquefiers (the inversion T is well below RT).

Linde

refrigerator


Virial Equation of State and Joule Thompson

For Low pressures (0-15 bar) B is

sufficient

for moderate pressures (<100 bar) B

and C describe Z well

High Pressure better use other EOS

Also allow derivation of exact correspondence

between virial coefficients and intermolecular

interactions.

Virial coefficients:

Z = p Vm / RT = (1 + B’p + C’p2 + ...)

Z = p Vm / RT = (1 + B/Vm + C/Vm2 + ...)

dT

TdBTTB

pJT

)()(

0


The Perfect gas and T, a, kT,

Property Acr Value for Perfect Gas Info:

T

TV

U

Internal

Pressure T=0 Strength/nature of

interactions between

molecules

pT

V

V

1a

Expansion

Coefficient a =1 / T

The higher T, the less

responsive is its volume

to a change in

temperature

T

Tp

V

V

1k

Isothermal

Compressibility kT=1 / p The higher the p, the

lower its compressibility

Hp

T

Joule-Thomson

Coefficient =0 Another indication of

molecular interactions.


Chapter 1 : Slide 19

Chemical Potential

Chapter1919191919 1 :

Slide 19


Chemical potential (m)

• Energy status of molecules in a system (e.g. Benzene in

water)

– Internal energies

• Chemical Bonds, vibrations, flexations, rotations.

– External Energies

• Whole molecule transitions, orientations

• Interactions of molecule with surroundings

• Energy status is a function of:

– Temperature

– Pressure

– Chemical composition

• “Average energy per molecule”


Chemical Potential – definition

For a pure substance


mG

• No way to directly measure chemical potential.

• Can only determine differences in , based on the tendencies of a chemical to move from one situation to another.

• Need a reference point, like sea level or absolute zero.

• often: select pure liquid chem. as reference

Chemical Potential dG = (μi

β – μiα) dni

In a spontaneous processes, dG < 0 at constant T and P. dni moves

from phase α to phase β to have negative change in free energy.

The spontaneous transfer of a substance takes place from a region with

a higher μi to a lower μi. The process continues to equilibrium where

dG = 0, and μi and μi become equal.

Phase β Phase α


If two populations of a chemicals (for example, the chemical

coexists in two separate phases): each will have its own value of

1 and 2

1: liquid 2: gas

1 2

Start: liquid benzene (1) and very little vapor benzene (2)

Initial disequilibrium: 1 2 (1 > 2)

Open stopcock. Benzene volatilizes. Net movement of benzene to the right.

1 decreases, 2 increases until they are equal – till there is no net movement

of benzene



Chemical Potential

1

,,1

jnpTn

G

Partial molar Gibbs energy Chemical potential for component 1

...2

,,2

1

,,1

dn

n

Gdn

n

GdG

jj npTnpT

If T and p are kept constant (i.e. dT and dp = 0) then

Or:

...2211 nnG

...2211 dndndG

iii nnTPG ),,(

The total free energy is the sum of the contributions from all the

different components present:


Chemical potential = free energy added to system with each added

increment of i

i

i

i nPTn

G,,

Where = chemical potential (kJ/mol)

DG = free energy (kJ)

ni = moles of component (i)

iiii STHG

For a single component system containing 1 mole of substance:

mG


26

Properties of the Gibbs energy

G = H - TS

dG = dH –TdS - SdT

dG = dU + pdV + Vdp –TdS - SdT

dU = TdS –pdV dG = TdS – pdV + pdV + Vdp –TdS - SdT

dG = Vdp - SdT G = f ( p, T )

dH = dU +pdV + Vdp

H = U + pV

dqrev =TdS and dwrev = - pdV

Both are not state functions but the sum

dq + dw can be treated to be reversible

because they only depend on the state

functions


27

Properties of the Gibbs energy

dG = Vdp - SdT

ST

G

p

V

p

G

T

V is positive so G is

increasing with

increasing p

G

T (constant p)

Slope = -S G

P (constant T)

Slope = V

S is positive (-S is negative)

so G is decreasing with

increasing T


28

Dependence of G on p

It would be useful to determine the Gibbs energy at one

pressure knowing its value at a different pressure.

dG = Vdp - SdT

We set dT = 0 (we make sure that the temperature is steady

and integrate:

f

i

)()( if

p

pVdppGpG

f

i

f

i

p

p

G

GVdpGd

Df

i

p

pVdpG



29


f

i

)()( if

p

pVdppGpG

Liquids and Solids.

Only slight changes of volume with pressure mean that we can effectively

treat V as a constant.

)()()( ifif ppVpGpG

pVpGpG D )()( if

Often V Dp is very small and may be neglected i.e. G for solids and liquids

under normal conditions is independent of p. Exception: Geography


30


f

i

)()( if

p

pVdppGpG

Ideal Gases.

For gases V cannot be considered a

constant with respect to pressure.

For a perfect gas we may use:

i

fi

if

ln)(

)()(f

i

p

pnRTpG

p

dpnRTpGpG

p

p



Ideal Gases.

i

fif ln)()(

p

pnRTpGpG

We can set pi to equal the standard pressure, p ( = 1 bar).

Then the Gibbs energy at a pressure p is related to its

standard Gibbs energy, G, by:

p

pnRTGpG f

f ln)(


Fugacity (lat.) = “urge to flee” – same

units as pressure

The fugacity is a parameter which enables us to apply the perfect gas

expression to real gases….

In the thermo dynamical treatment of the Chemical Potential


33


Real Gases.

p

fRTGpG ln)( mfm

For real gases we modify the

expression for a perfect gas and

replace the true pressure by a new

parameter, f, which we call the

fugacity.

The fugacity is a parameter we

have simply invented to enable us

to apply the perfect gas expression

to real gases.



Real Gases.

We may show that the ratio of fugacity to pressure is called the fugacity

coefficient:

p

f

Where is the fugacity coefficient

is related to the compression factor Z and the Viral coefficients:

dpp

Zp

0

1ln

1

2lnf

fRTG DWe may then write

..)'' 2 PCpBepfViral coefficients


A real gas is in its standard state when its fugacity is equal to 1 bar and it is

behaving as if it were an ideal gas at some specified temperature (the

diagram shown below is exaggerated to make this point):

In the limit of zero pressure real

gases behave more and more like

ideal gases and the fugacities of

real gases approach their partial

pressures: lim fi --> Pi

Pi --> 0

ideal gas fi

= Pi

real gas

fugacity

pressure

f = 1 bar

P = 1 bar

hypothetical

standard state

not the standard state

The standard state of a pure liquid or solid at a fugacity of 1 bar and some

specified temperature. Since the molar volumes of solids and liquids are

generally small and relatively insensitive to pressure, the activities of solids and

liquids at pressures that are not too far removed from 1 bar remain close to

unity.


Pressure region Z f a

I (very Low) 1 P 0 0

II (moderate) <1 <P >0 >0

III (high) >1 >P <0 <0


37

Dependence of G on T

ST

G

p

Using the same procedure as for the

dependence of G on p we get:

TSdGd

To go any further we

need S as a function of T

?

Instead we start with: G = H - TS

-S = (G – H)/T


38


T

HGS

ST

H

T

G

Let G/T = x ST

Hx

2T

H

T

x

p

2

)/(

T

H

T

TG

p

This is the Gibbs-Helmholtz

Equation

2

)/(

T

H

T

TG

p

D

D


39


2

)/(

T

H

T

TG

p

D

DS

T

G

p

D

DTwo expressions:

Gibbs-Helmholtz Equation

Changes in entropy or, more commonly, changes in enthalpy can be used

to show how changes in the Gibbs energy vary with temperature.

For a spontaneous (DG < 0) exothermic reaction (DH < 0) the change in

Gibbs energy increases with increasing temperature.

DG/T

T (constant p)

Slope = -DH/T2 = positive for exothermic reaction

Very negative

Less negative



• Comparing the chemical potential of the real gas to the chemical

potential of an ideal gas at the same pressure

( )( )( )

PPlnRT

PflnRTidreal

• The chemical potential of a real gas is

written in terms of its fugacity

• The activity coefficient (J) relates the

activity to the concentration terms of interest.

• In gaseous systems, we relate the fugacity (or activity) to the ideal

pressure of the gas via

Jo

J fRT ln

JJJ fP

Change in Internal Energy @ Constant P • Suppose we want to know how the internal energy, U, changes with temperature at

constant pressure

– But the change in volume with temperature

at constant pressure is the related to the

thermal compressibility of the gas, a

• Large a means sample responds strongly to changes in T

• For an ideal gas, T = 0 so

dU TdV C

VdT

U

T

p

TdV C

VdT( )

T

p

U

T

p

T

V

T

p

CV

a 1

V

V

T

p

orV

T

p

aV

soU

T

p

TaV C

V

U

T

p

CV U

T

p

definition


Enthalpy, H(p,T)

• Like U, H is a state function

• For closed system with constant composition

• At constant volume

• By the chain rule, we can find (dp/dT)V

dH H

p

T

dp H

T

p

dT

butH

T

p

C p

so dH H

p

T

dp CpdT

H

T

V

H

p

T

p

T

V

Cp

Chain Rule : if z z(x,y) theny

x

z

x

z

y

z

y

x

1

Re call p(V,T)

p

T

V

1

T

V

p

V

p

T

By analogy to a earlier, the isothermal

compressibility, kT,

k T 1

V

V

p

T

orV

p

T

kTV

so

p

T

V

1

T

V

p

kTV

by reciprocal identity ,y

x

z

1x

y

z

,

T

V

p

1V

T

p

and we knowV

T

p

aV

soT

V

p

1

aV

This makesp

T

V

a

kT

andH

T

V

H

p

T

a

k TC p


Enthalpy, H(p,T)

• To evaluate (dH/dp)T apply chain rule and reciprocal identity again

• But, Cp= (dH/dT)p and if one defines (dT/dP)H as µ, then (dH/dp)T =- Cpµ

• Thus,

H

p

T

1

T

H

p

p

T

H

Euler chain rule

H

p

T

H

T

p

T

p

H

reciprocal relation 2x

H

T

V

H

p

T

a

kT Cp

H

T

V

Cpa

kT Cp

or

H

T

V

C p 1 a

k T

45

pT

v

p T

U

V

UC

T

U

dVV

UdT

T

UdU

TV

PT

V

V

1

T

v

p V

UVC

T

U

pJT

V

CT

H)/1( k

TP

V

V

1k

A Word about kT • We know that if we increase the pressure the volume decreases so if

dp is positive, dV is negative

– Since kT(1/V)(dV/dp)T, kT is always positive

• For and ideal gas V=nRT/p so

(dV/dp)T= -nRT/p2 and kT(1/V)(-nRT/p2 )

Simplifying kT,

kT (nRT /V)(1/p2 )

kT (p)(1/p2 )

kT (1/p )

• This means that as the pressure increases, the compressibility

of a gas decreases

Documents

Chapter 4 Joule Thomson Effect Chemical Potential Fugacitychemistry.tcd.ie/staff/people/duesberg/ASIN/lecture not… · · 2016-08-24Joule Thomson Effect Chemical Potential Fugacity