Chapter 6. The solution of thermodynamic problems:

With U, A, H and G in hand we have potentials as a functions of whichever variable pairwe want: S and V, to T and P. Additional Legendre transforms will provide us withfurther potentials in case we have other variables (such as surface area A, length L,magnetic moment M, etc.).

Thermodynamic problems always involve computing a variable of interest. It may be aderivative if it is an intensive variable, or even a second derivative. (Higher derivativesare rarely of interest.)

Example: 1st order ones like

GP

T

=V or 2nd order ones like

2GP 2

T

= VP

T

= V

The solution procedure is thus:

1. Select the derivative or variable to be computed;2. Select the potential representation that makes it easiest, or corresponds to

variables you already have in hand.3. Manipulate the thermodynamic derivative you know to get the one you want.

Easy as 1-2-3!

We now turn to two methods to manipulate the thermodynamic derivations:

Tool 1 Maxwell relations: (useful if two variables are NOT conjugate, e.g. T and V asopposed to T and S)

If z = z(x,y) dz = zx

y

dx + zy

x

dy; but 2z

xy=

2zyx

y

zx

y

= x

zy

x

Maxwells relations must hold true if z is a state function, as we saw in chapter 1. Theybasically state that the cross-second derivatives of a function must be identical.

o Example:

dG = SdT +VdP + dn SP

T ,n

=VT

P,n

= V or

Vn

P,T

=P

n,T

All four of the quantities in parenthesis are second derivatives of G. We can verify thesecond line above for an ideal gas:

P

n,T

= P

RT lnP = RTP

and Vn

P ,T

= n

nRTP

= RTP

Note: As discussed in chapter 7, only three second derivatives not involving n i areindependent; many sets can be chosen; the conventional one is , , and cP, all defined atconstant T or P (Gibbs ensemble). You can preview them at the begining of chapter 7 ifyou wish.

Tool 2 Jacobi determinants: In order to reduce thermodynamic formulas to a minimalnumber of derivatives (e.g. , , cP), the Jacobi determinants can be used in addition tothe Maxwell relations. The idea is that you can reduce ANY derivative or variable youwant to functions of T, P, n (if working in the Gibbs ensemble, other combinationsotherwise) and second derivatives , , and cP (others in different ensembles).

The Jacobi determinant is defined as

(x,y,...z)(u,v,...w)

det

xu

xv

... x

w

yu

yv

...

... zu

zv

... z

w

.

Jacobians are useful for any variables transformation (u, v, w) (x, y, z). Anythermodynamic derivative can be expressed as follows:

xu

v...w

=(x,v...w)(u,v...w)

because vu

= 0, vv

= 1, etc. in the above determinant will leave

only the desired derivative when the determinant is multiplied out.

Derivatives can then be manipulated by using the Jacobian identities:

i)

(x,y,...z)(u,v,...w)

= (y,x,...z)(u,v,...w)

= (x,y,...z)(v,u,...w)

(row or column permuation of determinant)

ii)

(x,y,...z)(u,v,...w)

= (y,x,...z)(r,s,...t)

(r,s,...t)(v,u,...w)

ex : xr

xs

yr

ys

ru

rv

su

sv

=

xr

ru

+ xs

su

, xr

rv

+ xs

sv

yr

ru

+ ys

su

, yr

rv

+ ys

sv

=xu

xr

yu

yv

iii)

(x,y,...z)(u,v,...w)

= 1(u,v,...w)(x,y,...z)

Proof : det(M1) = 1det(M)

This concludes all the formal thermodynamic tools: From S > 0 (postulate 2), to theEuler formula, Legendre transforms, and finally Maxwell relations and Jacobians, youhave all the problem-solving tools thermodynamics has to offer to tackle equilibriumproblems.

Armed with the Maxwell and Jacobian relations, we can reduce any expression to , ,and cP (plus concentration or other derivatives, depending on the extensive variables ofthe system). Let us do a few examples:

1) Heat capacity at constant volume in terms of heat capacity at constant pressure

cv dq

dT

V

= Tn

ST

V

= Tn

S,V( ) T,V( )

= Tn

S,V( ) T,P( )

T,P( ) T,V( )

= TnV

S,V( ) T,P( )

= TnV

ST

Pncp /T

VP

T

-V

SP

T

?

VT

P

V

= cp VT

2

n c

p c

v

Going through the rationale for each step:

Step 1: is is there a Maxwell relation? NO: S & T are conjugate variables, so use aJacobianStep 2: we want P held constant in the Jacobian, not V, so split the Jacobian by chain ruleStep 3: Flip columns in the second Jacobian and evaluate P/V|T, the isothermalcompressibility Step 4: Can we use Maxwell or Jacobians? NO: time to multiply out the determinantStep 5: We know three of the derivatives in terms of second derivatives of the Gibbs freeenergy. The missing one, S/P|T = ? can be obtained using Maxwell: ? = -V/T|P, justthe isobaric expansion coefficient.

Step 6: Collect everything and note that V, T, n and are all positive, as is 2, so cv mustalways be smaller than cP. This should make intuitive sense: if I keep a system atconstant pressure and put heat in it, it will also expand, losing some of the heat as work.Thus the temperature wont go up as much, so the heat capacity is bigger at constantpressure.

2) Change in enthalpy at const. P:

HT

P,n

= HS

P,n

ST

P,n

= T ST

P,n

= Tnc

p

T= nc

p

n dq =T0

T

dH =T0

T

ncp T( )d T = HT0

T

T( ) H T0( )H(0)

Thus dH = ncPdT. We derived this earlier by waving hands and using dqrev, but here it is,formally.

3) Isentropic quasistatic expansion (S = const,, so dq

rev/ T = dS = 0 ). Do the homework

problem before you check out the solution below!

This describes the pressure-temperature relationship in a molecular beam, where no heatcan flow into the expanding gas, so its temperature must drop as the pressure drops.What happens is that at the nozzle of the molecular beam, collisions preferentially ejectparticles forward, preserving their average kinetic energy, but making a stream ofparticles with a narrower relative velocity distribution. The relevant derivative is T/P|S,which looks a bit odd at first, but we are fully equipped to handle derivatives at constantentropy:

0 = dS = dq

rev/ T( )

dT = TP

S

dP T,S( ) P,S( ) =

T,S( ) P,T( )

P,T( ) P,S( ) =

S,T( ) P,T( )

1

S,P( ) T,P( )

= S

P

T

ST

P

For the numerator, S

P

T

= VT

P

= V by Maxwell. Note that just because S is

constrained to be held constant in (T/P)S does not mean that S/P = 0!

For the denominator

ST

P

=nc

p

T dT = VT

ncp

dP

For an ideal monatomic gas, = 1

T= nR

PV, and

c

p= 5

2R so we can write explicitly

dT

T= 2

5dPP

or TT

0

= P

P0

2/5

after integrating. Temperature drops in a

molecular beam, but not as dramatically as the pressure.

3) Heat capacity in terms of energy:

c

v= dq

dT

V

= Tn

ST

V

T > 0, n > 0.

ST

V

= S,V( ) T,V( ) =

S,V( ) U,V( )

U,V( ) T,V( ) =

SU

V

UT

V

= 1T

UT

V

= Tn 1

TUT

V

= 1n

UT

V

Again, we showed this before by invoking dqrev when discussing Legendre transformsfrom energy to enthalpy. 1) and 3) are just the formal way of deriving it using Maxwellsrelations and Jacobians.