have simple forms:

dU = TdS − PdV (1)

dH = TdS + V dP (2)

dA = −SdT − PdV (3)

dG = −SdT + V dP. (4)

0.2 Maxwell Relations

The first equation implies that(

∂U

∂S

)

V

= T (5)

and(

∂U

∂V

)

S

= −P. (6)

The second derivatives are equal, however:

∂2U

∂S∂V=

∂2U

∂V ∂S, (7)

so(

∂T

∂V

)

S

= −

(

∂P

∂S

)

V

. (8)

In a similar way, we find(

∂T

∂P

)

S

=

(

∂V

∂S

)

P

, (9)

(

∂S

∂V

)

T

=

(

∂P

∂T

)

V

. (10)

and(

∂S

∂P

)

T

= −

(

∂V

∂T

)

P

. (11)

from the differential equations for the other three potentials. These four equations arecalled the Maxwell relations. From them we can deduce immediately some interestingfacts. For example, for most (but not all) systems V increases on heating at constantpressure: (∂V/∂T )P ≥ 0. This means that (∂S/∂P )T ≤ 0. So the entropy normallydecreases on compression at constant temperature, not nearly so obvious as the firststatement. A second example relates to the specific heat at constant volume, Cv.

Cv =

(

dQ

dT

)

V

=

(

∂U

∂T

)

V

= T

(

∂S

∂T

)

V

. (12)

Hence∂Cv

∂V|T = T

(

∂

∂T

)

V

(

∂S

∂V

)

T

= −T

(

∂2P

∂T 2

)

V

. (13)

2

The right-hand-side depends only on the equation of state of the system. Oncethis is known, then the volume dependence of the specific heat is known.More complicated examples of uses of the Maxwell relations abound. One of the

most often encountered is the equation relating the specific heat at constant volumeCv = T (∂S/∂T )v (which is easily calculated) to the specific heat at constant pressureCp = T (∂S/∂T )P (which is easily measured). This requires a short digression intothe subject of Jacobians. Consider two functions w(y, z) and x(y, z). The area dw dxin the w − x plane is related to the area dy dz in the y − z plane by

dw dx =∂(w, x)

∂(y, z)dy dz, (14)

where

∂(w, x)

∂(y, z)=

∣

∣

∣

∣

∣

∣

∣

(

∂w∂y

)

z

(

∂w∂z

)

y(

∂x∂y

)

z

(

∂x∂z

)

y

∣

∣

∣

∣

∣

∣

∣

= J. (15)

J is the Jacobian of the transformation from w, x to y, z. The straight bracketsindicate the determinant. Note that J = J(y, z). A special case is

∂(w, x)

∂(y, x)=

(

∂w

∂y

)

x

. (16)

In case there is yet a third set of variables s, t, the Jacobian satisfies a product rule

∂(w, x)

∂(y, z)=

∂(w, x)

∂(s, t)

∂(s, t)

∂(y, z). (17)

We wish to make a change of variables. Cv = T(

∂S∂T

)

Vis expressed in terms of

temperature and volume. We wish to transform to temperature and pressure. Write

Cv = T∂(S, V )

∂(T, V )= T

∂(S, V )

∂(T, P )

∂(T, P )

∂(T, V )(18)

= T

[(

∂S

∂T

)

P

(

∂V

∂P

)

T

−

(

∂S

∂P

)

T

(

∂V

∂T

)

P

](

∂P

∂V

)

T

(19)

= T

(

∂S

∂T

)

P

− T

(

∂S

∂P

)

T

(

∂V

∂T

)

P

(

∂P

∂V

)

T

(20)

Now for any substance, we define the thermal expansion coefficient α ≡ 1

V

(

∂V∂T

)

P=

fractional change in volume per degree and κT ≡ −1

V

(

∂V∂P

)

T= isothermal compress-

ibility. Both depend only on the equation of state. Putting these results togetherwith the Maxwell relation

(

∂S

∂P

)

T

= −

(

∂V

∂T

)

P

, (21)

3

we find

Cv = T

(

∂S

∂T

)

P

+ T

(

∂V

∂T

)2

P

(

∂P

∂V

)

T

. (22)

Substituting the definitions above gives

Cv = Cp − TVα2

κT

, (23)

or, as is more often seen,

Cp − Cv = TVα2

κT

, (24)

Since κT > 0,(as we shall see below), Cp > Cv. For an ideal gas

=1

V

(

∂V

∂T

)

P

=1

V

(

∂

∂T

)

P

NkT

P=

Nk

PV=1

T, (25)

and

κT = −1

V

(

∂V

∂P

)

T

= −1

V

∂

∂P

NkT

P=

NkT

P 2V= 1/P. (26)

Hence Cp − Cv = TV T−2P = PV/T = Nk. In the dimensionless heat capacities perparticle cp = Cp/Nk and cv = Cv/Nk, we have cp − cv = 1. For solids, it is almostalways the case that the volume changes little in, say, doubling the temperature atconstant pressure. (Think of the metal grill in an oven). Hence (T/V )(∂V/∂T )P << 1and cp ≈ cv.

0.3 The Third Law

This law of nature is not as firmly based or as universal as the first two laws ofthermodynamics. It states that the entropy of any system at zero temperature iszero: S(T = 0) = 0. We will discuss the conditions of validity of this law later on.For now, let us just mention some of the consequences of the law.Consider heating a substance at constant volume starting at T = 0 and ending

up in some final state at temperature Tf . The final entropy is

S(T ) =∫ Tf

0

CV

TdT.

Now if CV = a + bT + cT 2 + ..., then the third law implies that a = 0. The sameholds true for a process at constant pressure. All heat capacities must vanish atT = 0. This is not consistent with the relation we just deduced for ideal gases thatcp − cv = 1. At very low temperatures, ideal gases cannot exist. Indeed, gases do notexist at very low temperatures, as experiment has shown. The lowest boiling point forany substance at atmospheric pressure is 4.2K for helium. Near absolute zero, onlyhelium even remains liquid; all other substances solidify. For liquids and perfectly

4