# 10.1: Thermodynamic Relationships from dE, dH, dA and dG

In Chapter 9, we substitute $$dq^{rev} = TdS$$, from the second law, into

$dE = dq + PdV$

from the first law, to obtain, for any closed system undergoing a reversible change in which the only work is pressure–volume work, the fundamental equation, $$dE = TdS + PdV$$. In view of the mathematical properties of state functions that we develop in Chapter 7, this result means that we can express the energy of the system as a function of entropy and volume, $$E = E\left(S,V\right)$$. With this choice of independent variables, the total differential of $$E$$ is

$(dE = {\left({\partial E}/{\partial S}\right)}_VdS + {\left({\partial E}/{\partial V}\right)}_SdV.$

Equating these expressions for $$dE$$, we find

$\left[{\left(\frac{\partial E}{\partial S}\right)}_V + T\right]dS + \left[{\left(\frac{\partial E}{\partial V}\right)}_S + P\right]dV\mathrm{=0}$

for any such system. Since $$S$$ and $$V$$ are independent variables, this equation can be true for any arbitrary state of the system only if the coefficients of$$\mathrm{\ }dS$$ and $$dV$$ are each identically equal to zero. It follows that

${\left(\frac{\partial E}{\partial S}\right)}_V = T$

and

${\left(\frac{\partial E}{\partial V}\right)}_S\mathrm{=-}P$

Moreover, because dE is an exact differential, we have

$\frac{\partial }{\partial V}{\left(\frac{\partial E}{\partial S}\right)}_V = \frac{\partial }{\partial S}{\left(\frac{\partial E}{\partial V}\right)}_S$

so that

${\left(\frac{\partial T}{\partial V}\right)}_S\mathrm{=-}{\left(\frac{\partial P}{\partial S}\right)}_V$

Using the result $$dH = TdS + VdP$$, parallel arguments show that enthalpy can be expressed as a function of entropy and pressure, $$H = H\left(S,P\right)$$, so that

${\left(\frac{\partial H}{\partial S}\right)}_P = T$

and

${\left(\frac{\partial H}{\partial P}\right)}_S = V$ and ${\left(\frac{\partial T}{\partial P}\right)}_S = {\left(\frac{\partial V}{\partial S}\right)}_P$

Since $$dA\mathrm{=-}SdT + PdV$$, the Helmholtz free energy must be a function of temperature and volume, $$A = A\left(T,V\right)$$, and we have

${\left(\frac{\partial A}{\partial T}\right)}_V\mathrm{=-}S$ and ${\left(\frac{\partial A}{\partial V}\right)}_T\mathrm{=-}P$ and ${\left(\frac{\partial S}{\partial V}\right)}_T = {\left(\frac{\partial P}{\partial T}\right)}_V$

Likewise, $$dG\mathrm{=-}SdT + VdP$$ implies that the Gibbs free energy is a function of temperature and pressure, $$G = G\left(P,T\right)$$, so that

${\left(\frac{\partial G}{\partial T}\right)}_P\mathrm{=-}S$

and

${\left(\frac{\partial G}{\partial P}\right)}_T = V$ and ${\left(\frac{\partial V}{\partial T}\right)}_P\mathrm{=-}{\left(\frac{\partial S}{\partial P}\right)}_T$