Editing Exact differential (section)

== Some useful equations derived from exact differentials in two dimensions ==

(See also [[Bridgman's thermodynamic equations]] for the use of exact differentials in the theory of [[thermodynamic equations]])

Suppose we have five state functions <math>z,x,y,u</math>, and <math>v</math>. Suppose that the state space is two-dimensional and any of the five quantities are differentiable. Then by the [[chain rule]]

{{NumBlk|:|<math>
  dz = 
  \left(\frac{\partial z}{\partial x}\right)_y dx+
  \left(\frac{\partial z}{\partial y}\right)_x dy
  =
  \left(\frac{\partial z}{\partial u}\right)_v du
  +\left(\frac{\partial z}{\partial v}\right)_u dv
</math>|{{EquationRef|1}}}}

but also by the chain rule:

{{NumBlk|:|<math>
  dx =
  \left(\frac{\partial x}{\partial u}\right)_v du
  +\left(\frac{\partial x}{\partial v}\right)_u dv
</math>|{{EquationRef|2}}}}

and

{{NumBlk|:|<math>
  dy=
   \left(\frac{\partial y}{\partial u}\right)_v du
  +\left(\frac{\partial y}{\partial v}\right)_u dv
</math>|{{EquationRef|3}}}}

so that (by substituting (2) and (3) into (1)):

{{NumBlk|:|<math>
  \begin{align}
  dz = 
& \left[
    \left(\frac{\partial z}{\partial x}\right)_y
    \left(\frac{\partial x}{\partial u}\right)_v
    +
    \left(\frac{\partial z}{\partial y}\right)_x
    \left(\frac{\partial y}{\partial u}\right)_v
  \right] du \\
  +
& \left[
    \left(\frac{\partial z}{\partial x}\right)_y
    \left(\frac{\partial x}{\partial v}\right)_u
    +
    \left(\frac{\partial z}{\partial y}\right)_x
    \left(\frac{\partial y}{\partial v}\right)_u
  \right] dv
  \end{align}
</math>|{{EquationRef|4}}}}

which implies that (by comparing (4) with (1)):

{{NumBlk|:|<math>
  \left(\frac{\partial z}{\partial u}\right)_v
  =
  \left(\frac{\partial z}{\partial x}\right)_y
  \left(\frac{\partial x}{\partial u}\right)_v
  +
  \left(\frac{\partial z}{\partial y}\right)_x
  \left(\frac{\partial y}{\partial u}\right)_v
</math>|{{EquationRef|5}}}}

Letting <math>v=y</math> in (5) gives:

{{NumBlk|:|<math>
  \left(\frac{\partial z}{\partial u}\right)_y
  =
  \left(\frac{\partial z}{\partial x}\right)_y
  \left(\frac{\partial x}{\partial u}\right)_y
</math>|{{EquationRef|6}}}}

Letting <math>u=y</math> in (5) gives:

{{NumBlk|:|<math>
  \left(\frac{\partial z}{\partial y}\right)_v
  =
  \left(\frac{\partial z}{\partial x}\right)_y
  \left(\frac{\partial x}{\partial y}\right)_v
  +
  \left(\frac{\partial z}{\partial y}\right)_x
</math>|{{EquationRef|7}}}}

Letting <math>u=y</math> and <math>v=z</math> in (7) gives:

{{NumBlk|:|<math>
  \left(\frac{\partial z}{\partial y}\right)_x
  = -
  \left(\frac{\partial z}{\partial x}\right)_y
  \left(\frac{\partial x}{\partial y}\right)_z  
</math>|{{EquationRef|8}}}}

using (<math>\partial a/\partial b)_c = 1/(\partial b/\partial a)_c</math> gives the [[triple product rule]]:

{{NumBlk|:|<math>
  \left(\frac{\partial z}{\partial x}\right)_y
  \left(\frac{\partial x}{\partial y}\right)_z
  \left(\frac{\partial y}{\partial z}\right)_x
  =-1
</math>|{{EquationRef|9}}}}