Gibbs Free Energy

Suppose, finally, that $T$ and $p$ are the are the two independent parameters that specify the system. Because

$$T dS = d(T S)-S dT,$$ (6.110)

we can rewrite Equation (6.91) in the form

$$dG = -S dT + V dp,$$ (6.111)

where

$$G = H - T S = E-T S+p V$$ (6.112)

is termed the Gibbs free energy.

Proceeding as before, we write

$$G= G(T,p),$$ (6.113)

which implies that

$$dG= \left(\frac{\partial G}{\partial T}\right)_p dT + \left(\frac{\partial G}{\partial p}\right)_T dp.$$ (6.114)

Comparison of Equations (6.111) and (6.114) yields

$$\left(\frac{\partial G}{\partial T}\right)_p = -S,$$ (6.115)

$$\left(\frac{\partial G}{\partial p}\right)_T = V.$$ (6.116)

We also know that

$$\frac{\partial^{ 2} G}{\partial p \partial T} = \frac{\partial^{ 2}G}{\partial T \partial p},$$ (6.117)

or

$$\left(\frac{\partial}{\partial p}\right)_T\left(\frac{\partial G}... ...rac{\partial}{\partial T}\right)_p\left(\frac{\partial G}{\partial p}\right)_T.$$ (6.118)

Thus, it follows from Equations (6.115) and (6.116) that

$$-\left(\frac{\partial S}{\partial p}\right)_T = \left(\frac{\partial V}{\partial T}\right)_p.$$ (6.119)

Equations (6.89), (6.99), (6.109), and (6.119) are known collectively as Maxwell relations.