Internal Energy

Equation (6.5) can be rearranged to give

$$dE=T dS - p dV.$$ (6.82)

This equation shows how the internal energy, $E$ , depends on independent variations of the entropy, $S$ , and the volume, $V$ . If $S$ and $V$ are considered to be the two independent parameters that specify the system then

$$E = E(S,V).$$ (6.83)

It immediately follows that

$$dE = \left(\frac{\partial E}{\partial S}\right)_V dS + \left(\frac{\partial E}{\partial V}\right)_S dV.$$ (6.84)

Now, Equations (6.82) and (6.84) must be equivalent for all possible values of $dS$ and $dV$ . Hence, we deduce that

$$\left(\frac{\partial E}{\partial S}\right)_V = T,$$ (6.85)

$$\left(\frac{\partial E}{\partial V}\right)_S = -p.$$ (6.86)

We also know that

$$\frac{\partial^{ 2} E}{\partial V \partial S} = \frac{\partial^{ 2}E}{\partial S \partial V},$$ (6.87)

or

$$\left(\frac{\partial}{\partial V}\right)_S\left(\frac{\partial E}... ...rac{\partial}{\partial S}\right)_V\left(\frac{\partial E}{\partial V}\right)_S.$$ (6.88)

In fact, this is a necessary condition for $dE$ in Equation (6.84) to be an exact differential. (See Section 4.5.) It follows from Equations (6.85) and (6.86) that

$$\left(\frac{\partial T}{\partial V}\right)_S = -\left(\frac{\partial p}{\partial S}\right)_V.$$ (6.89)