The Maxwell Relations | Chemistry Optional Notes for UPSC PDF Download

Introduction

• Modeling the dependence of the Gibbs and Helmholtz functions behave with varying temperature, pressure, and volume is fundamentally useful. But in order to do that, a little bit more development is necessary. To see the power and utility of these functions, it is useful to combine the First and Second Laws into a single mathematical statement. In order to do that, one notes that since
  dS = dq/T
  for a reversible change, it follows that
  dq = TdS
  And since
  dw = TdS − pdV
  for a reversible expansion in which only p-V works is done, it also follows that (since  dU = dq + dw
  dU = TdS − pdV
• This is an extraordinarily powerful result. This differential for  dU can be used to simplify the differentials for H, A, and G. But even more useful are the constraints it places on the variables T, S, p, and V due to the mathematics of exact differentials!

Maxwell Relations

The above result suggests that the natural variables of internal energy are  S and  V (or the function can be considered as  U(S,V). So the total differential (dU) can be expressed:

Also, by inspection (comparing the two expressions for  dU) it is apparent that:
(22.3.1)
and

But the value doesn’t stop there! Since  dU  is an exact differential, the Euler relation must hold that

By substituting Equations 22.3.1 and 22.3.2, we see that

or

This is an example of a Maxwell Relation. These are very powerful relationship that allows one to substitute partial derivatives when one is more convenient (perhaps it can be expressed entirely in terms of  α and/or  κT for example.)

A similar result can be derived based on the definition of  H.
H ≡ U + pV
Differentiating (and using the chain rule on d(pV)) yields
dH = dU + pdV + Vdp
Making the substitution using the combined first and second laws (dU = TdS – pdV) for a reversible change involving on expansion (p-V) work

This expression can be simplified by canceling the  pdV terms.
dH = TdS + Vdp (22.3.3)
And much as in the case of internal energy, this suggests that the natural variables of  H are  S and  p. Or
(22.3.4)

Comparing Equations 22.3.3 and 22.3.4 show that
 (22.3.5)
and
(22.3.6)
It is worth noting at this point that both (Equation 22.3.1)

and (Equation 22.3.5)

are equation to  T. So they are equation to each other

Morevoer, the Euler Relation must also hold

so

This is the Maxwell relation on H. Maxwell relations can also be developed based on A and G. The results of those derivations are summarized in Table 6.2.1..
Table 6.2.1: Maxwell Relations

The Maxwell relations are extraordinarily useful in deriving the dependence of thermodynamic variables on the state variables of p, T, and V.

Solved Example

Example: Show that

Ans: Start with the combined first and second laws:
dU = TdS − pdV
Divide both sides by  dV and constraint to constant  T:

Noting that

The result is

Now, employ the Maxwell relation on  A (Table 6.2.1)

to get

It is apparent that

Note: How cool is that? This result was given without proof in Chapter 4, but can now be proven analytically using the Maxwell Relations!