Mathematical Intermezzo 3:More Mathematical Relationships Among Partial Derivatives (PD’s)

1. Exact Differential Expression (total differential in terms of partial derivatves)

If z=z(x,y) is a state function, then when x and y change by dx and dy, z changes by:

dyy

zdx

x

zdz

xy

3. Second Partial Derivatives May Be Taken In Any Order

xy

z

x

z

yxy

2

yxy

z

xyx

z

2

=

4. Test For Exactness

2. Creating And Relating Similar Partial Derivatives

Divide by dx and set f constant for new PD’s, where f=f(x,y) also:

fxyf x

y

y

z

x

z

x

z

VP T

U

T

U

and :e.g.

xy y

zN

x

zMNdyMdxdzdz

and , : whereexact, is If

yxx

N

y

M

Then (continued)

1 © Prof. Zvi C. Koren 20.07.2010

5. Inversion Rule

z

z

x

yy

x

1

cyclic permutation:

x y z

6. Chain Rule

, ,Expression alDifferentiExact 1. From dyy

zdx

x

zdz

xy

If z=z(x,y) and x=x(y,f), then z=z(y,f).

yyyf

x

x

z

f

z

divide by df and set y constant for new PD’s

7. Euler’s Cycle Rule

1

yxzx

z

z

y

y

x

, dyy

zdx

x

zdz

xy

divide by dy and set z constant for new PD’s, and use Inversion Rule:

(Applications on next slide)

[Leonhard Euler, 1707 (Basel, Switzerland) – 1783 (St Petersburg, Russia):

http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Euler.html]

From 1. Exact Differential Equation,

Nike Problem: “Just Do It”2 © Prof. Zvi C. Koren 20.07.2010

Some Applications of the Mathematical Relationships:

Relating Similar Partial Derivatives - Examples

P

PPV

CT

H

T

H

T

H :course of ; and

2.

Strategy is to transform to known measurable coefficients:1. “ Begin with 2nd, Create the 1st ”

2. Convert P/* to 2 V/* PD’s in order to introduce coefficients and/or ,

through Cycle Rule and Inversion Rule, etc.

3. Convert H/* to CP and T/* (giving a measurable change in T property),

through Cycle Rule and Inversion Rule, etc.

Interim Results:

VπCT

U

T

U

T

UV

PVP

and

1.

P

T

μCP

H

Final Result:

κ

αμC

T

HP

V

1

Nike Problem:

“Just Do It” But we dit IT

(cont’d)

κ

α

T

P

V

Joule-Thomson

Coefficient

HP

Tμ

Nike Problem3 © Prof. Zvi C. Koren 20.07.2010

Joule-ThomsonCoefficient

HP

Tμ

Denotes the abilityof an expanding gas

to undergocooling or heating(!)

at constant H

Signs of

= (Ti,Pi)

+, cooling

–, heating

Note: All gases, except He and H2, undergo cooling from 1 atm and 25 oC

= 0 at the inversion temperature, TI

Analytical Problems: dTH = dP TH = dP

Intermolecular forces for a substance at a certain state of matter = f(T,P)

William Thomson(Lord Kelvin)1824 (Belfast, Ireland) –

1907 (Netherhall, near Largs, Ayrshire, Scotland)

Thomson designed and implemented many new devices, including the mirror-galvanometer that was used in the first successful sustained telegraph transmissions in transatlantic submarine cable between Ireland and Newfoundland. For his work on the transatlantic cable Thomson was created Baron Kelvin of Largs in 1866 by the British government. The Kelvin is the river which runs through the grounds of Glasgow University and Largs is the town on the Scottish coast where Thomson built his house.

http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Thomson.html

James Prescott Joule(1818 - 1889)

English physicist

4 © Prof. Zvi C. Koren 20.07.2010

Apparatus for Measuring the Joule-Thomson Effect:Cooling by Adiabatic Expansion

Gas expands through

the porous barrier,

which acts as a

throttle ( ֵנק ַצוָּאר, ַמשְׁ ):

Isenthalpic expansion

Whole apparatus is

thermally insulated:

Adiabatic.

This zone is maintained at a

constant high pressue.

This zone is maintained at a

constant low pressue.

(ceramic)

(steady stream)

(steady stream)

5 © Prof. Zvi C. Koren 20.07.2010

Thermodynamic Analysis of the Joule-Thomson Effect:

virtual“piston”

“ext”

these 2 parts are really on the same side; so, of course, ext piis the same as int Pi.

Consider the passage of a given amount of gas through the throttle from high Pito low Pf.

The “pistons”

represent the

upstream and

downstream

gases, which

maintain

constant

pressures

either side of

the throttle.

“ext”

these 2 parts are really on the same side

(continued)

n

n

6 © Prof. Zvi C. Koren 20.07.2010

wleft = –pexdV

= – Pi(0–Vi)

= +PiVi

wright = –pexdV

= – Pf(Vf–0)

= – PfVf

Gas is compressed

isothermally by the

upstream gas acting as a

“piston” with pressure pi:

Gas expands isothermally

against pf provided by the

downstream gas acting as

a “piston” to be driven

out:

wtotal = wleft + wright

= PiVi – PfVf

Utotal = Uf – Ui = wtotal, qadiabatic=0

= PiVi – PfVf

Uf + PfVf = Ui + PiVi

Hf = Hi

Joule-Thomson Effect is an Isenthalpic Process(continued)7 © Prof. Zvi C. Koren 20.07.2010

Measurements of

Direct (as in the previous set-up):

P

T lim

0P

HP

Tμ

Indirect (modern method):

Ti Tf

(typically: Ti >Tf)

P

T

μCP

H

Recall:

q = electrical heating required to offset the cooling

H = qP: [P=Pf is constant in right compartment]

P

H lim

0P

(continued)

knownknown

T Isothermal J-T

coefficient

(I HATE THIS NAME)

8 © Prof. Zvi C. Koren 20.07.2010

μ(T,P)P

Tμ

H

Notes:

1. is a slope

2. = 0 at TI (inversion temperature),

at an isenthalp maximum

3. There are 2 inversion temps.

at a given P, a higher TI and a

lower TI.

9 © Prof. Zvi C. Koren 20.07.2010

Selected Values at 1 atm

298K/

Katm-1

TI/K

upperTb/KGas

1.111500194.7CO2

-0.0320220.3H

-0.060404.2He

0.2562177.4N2

0.3176490.2O2

10 © Prof. Zvi C. Koren 20.07.2010

Joule-Thomson Coefficients (in deg/atm) for N2

[Table 7-1 of Maron & Lando]

P

(atm) -150°C -100°C 0°C 100°C 200°C 300°C

1 1.266 0.6490 0.2656 0.1292 0.0558 0.0140

20 1.125 0.5958 0.2494 0.1173 0.0472 0.0096

33.5 0.1704 0.5494 0.2377 0.1100 0.0430 0.0050

60 0.0601 0.4506 0.2088 0.0975 0.0372 -0.0013

100 0.0202 0.2754 0.1679 0.0768 0.0262 -0.0075

140 -0.0056 0.1373 0.1316 0.0582 0.0168 -0.0129

180 -0.0211 0.0765 0.1015 0.0462 0.0094 -0.0160

200 -0.0284 0.0087 0.0891 0.0419 0.0070 -0.0171

11 © Prof. Zvi C. Koren 20.07.2010First-Law Problems: Joule-Thomson: 32-35.

Linde Refrigerator:Liquefaction of Gases

cools

gas

low-P cold gas cools expanding high-P gas, which cools further upon expansion

Carl von Linde

(1842–1934), German

engineer, born in

Berndorf, near Baden,

Austria; devised

process of liquefying

air 1895.

http://www.chemheritage.org/EducationalServices/chemach/tpg/cvl.html

http://www.deutsches-museum.de/ausstell/dauer/physik/e_luft.htm

12 © Prof. Zvi C. Koren 20.07.2010

Final Notes Regarding J-T Coefficient

1. For an ideal gas, = 0.

So, T is unchanged for an i.g. during Joule-Thomson expansion.

However, simple adiabatic expansion does cool an ideal gas

because it does work.

2. So, why isn’t 0 as P 0 for real gases?

Because depends on derivatives

and not directly on P, V, and T themselves.

Nike Problem: “Just Do It”

13 © Prof. Zvi C. Koren 20.07.2010