Maxwell Relations

 

Starting with the first law

 

dU = TdS - PdV,

 

we can express the enthalpy as

 

dH = dU + PdV + VdP = TdS + VdP,

 

the Helmholtz free energy as

 

dA = dU - TdS - SdT = -SdT - PdV,

 

and the Gibb's free energy as

 

dG = dH - TdS - SdT = -SdT + VdP.

 

These four relations are exact differentials.

 

dU = TdS - PdV

dH = TdS + VdP

dA = -SdT - PdV

dG = -SdT + VdP

 

For each of these we can derive an expression based on the equality of the second cross derivative.  These relations, also known as Maxwell relations are fundamentally useful for thermodynamics.

 


 


Let us begin the with the internal energy.  We can write the internal energy as a total derivative

Note that this implies that

 


 


Now take the second derivative (i.e. the cross derivative)

 


 


And since the second cross derivatives are equal (for an exact differential)

 


 


This symmetry leads us to state that the natural variables of the internal energy are S and V.  You might ask why I do not consider V and T.  After all we can write

 


 


And we can see immediately that

 


 


So


 


But then we are left with the second term.  It is important to note that

 


 


But rather

 


 


The difference between these two partial derivatives is in the variable that is held fixed.  How do we find the value of (U/V)T?

 

We will show a Maxwell relation can be used to find an expression for (U/V)T.  However, it is clear that the variables V and T are going to lead to a more complicated expression for U than the variables V and S did.

Let us consider the Maxwell relations for A.

 


 


This implies that

 


 


Now take the second cross derivative of each of these.

 


 


Because of the equality of the second cross derivatives we have

 


 


This is perhaps the most useful of the Maxwell relations.  Let us use it first to find (U/V)T.

Start with A = U - TS.  Take the derivative with respect to V at constant temperature.

 


 


Use the Maxwell relation for A above to substitute for (S/V)T.

Note that (A/V)T = -P so we have

 


 


Which rearranges to

 


 


While this would be inconvenient to use in the first law, it is a useful equation and is sometimes known as a thermodynamic equation of state.  For example, for an ideal gas P = nRT/V and therefore (P/T)V = nR/V.  If we substitute into the thermodynamic equation of state we find

 


 


This is not a surprise.  We have shown that for an ideal gas U = 3/2nRT and therefore the internal energy depends only on the temperature and not on the volume.

 

          McQuarrie and Simon show that the same is true for a hard sphere gas P(V-nb) = nRT (page 891), but not for van der Waal's or other equations of state that include "a" parmeters that depend on the intermolecular potential energy.

          We now use the Maxwell relation for A to derive a general relationship between Cp and Cv.  Recall that we have stated that for an ideal gas Cp - Cv = nR.  Let us further examine the relationship.

Even though V and T are not the natural variables for U we can start with

 


 


 


Likewise, we can express U in terms of the variables P and T


We can find a natural relationship between these two using the differential expression for the volume.

 


 


Rearranging we have

 


 


We see that

 


 


Using the fact that U = H - PV we have

 


 


 And noting that

 


 


 we obtain

 


 


The Maxwell relation allows us to write this as

 


 


Noting that

 


 


We can substitute in for (P/T)V to obtain

 


 


We define the isothermal compressibility

 


 


And the coefficient of thermal expansion

 


 


Using these definition the difference in the heat capacities becomes

 


 


Although we have focused on the Maxwell relations for U and A, the same procedure leads to Maxwell relations for H and G.  These are described in Chapter 22.

 

    The Maxwell relations and associated machinery are NOT something you memorize.  Everything can be derived from dU = TdS – PdV which is a combined statement of the First and Second Laws.

 

Since you can get dH, dG, and dA from this expression, you can obtain the Maxwell relations from the expression of each of these as a total derivative.

 

State function

Total derivative

Maxwell relation

dU

 TdS – PdV

(T/V)S = -(P/S)V

dH

TdS + VdP

(T/P)S = (V/S)P

dA

–SdT – PdV

(S/V)T = (P/T)V

dG

–SdT + VdP

-(S/P)T = (V/T)P

 

 

The Standard State for a Gas

 


 


The standard state for a gas at any temperature is the hypothetical ideal gas at one bar.  To calculate corrections to the entropy due to deviations from ideality we begin with the Maxwell relation for dG.

 

This expression allows us to calculate the pressure dependence of the entropy.

 


 


Suppose we wish to determine the deviation from ideality at 1 bar of pressure (i.e. at the standard state).  At some very low pressure all gases behave ideally so we consider an ideal gas at pressure Pid. 

 


 


The quantity (V/T)P can be determined from the equation of state for the gas.  For a hypothetical ideal gas this same expression is

 


 


Adding these last two equations together gives the difference from ideality at 1 bar of pressure

 


 


Now we need only find an equation of state that we can use to determine (V/T)P.  We can use the virial equation of state for example

 


 


From the virial expansion we have

 


 


Which gives a particularly simple expression for the deviation from the ideal entropy

 


 


The temperature and pressure dependencies of the Gibb's free energy

 


 


We shown that dG = -SdT + VdP.  This differential can be used to determine both the pressure and temperature dependence of the free energy.  At constant temperature SdT = 0 and dG = VdP.  The integrated form of this equation is

For one mole of an ideal gas we have

 


 


The same formula can be obtained from

 


 


We can use the above expression to indicate the free energy at some pressure P relative to the pressure of the standard state P = 1 bar.

 


 


G0(T) is the standard molar Gibb’s free energy for a gas.  As discussed above the standard molar Gibb’s free energy is the free energy of one mole of the gas at 1 bar of pressure.  The Gibb’s free energy increases logarithmically with pressure.  This is entirely an entropic effect.

 

If we are dealing with a liquid or a solid the molar volume is more or less a constant as a function of pressure.  Actually, it depends on the isothermal compressibility, k = -1/V(V/dP)T, but since k is typically a number of the order 10-4 atm-1 for liquids and 10-6 atm-1  for solids.  For our purposes we can treat the volume as a constant and we obtain

 


 


The Gibb’s free energy also depends on temperature.  The easiest way to determine the temperature dependence is to begin with G = H – TS and divide through by T.

 


 


We differentiate this expression with respect to T keeping P fixed.

 


 


The last two terms cancel since (S/T)P = CP/T.

 


 


Which can be applied to any process in which case it is written as

 


 


You might think that we could calculate the temperature dependence of the free energy directly from DG = DH – TDS and, in fact, we can.  We can calculate the enthalpy referenced to the enthalpy at absolute zero and the absolute entropy.  However, as discussed on page 903 and 904 of McQuarrie and Simon, this is a cumbersome procedure and involves some duplication of effort since DfusS = DfusH/Tfus and DvapS = DvapH/Tvap so the calculations done for entropies and enthalpies of phase transitions will cancel.  As we have already seen this is consistent with the fact DfusG = 0 and DvapG = 0.   A plot of G(T) vs. T has a constant negative slope with discontinuities at the phase transitions.  The reason for the negative slope (see Figure 22.7) is that

 


 


Of course, the entropies of solid, liquid, and vapor are different giving rise to the changes in slope at the phase transitions.