Exact Differential - Some Useful Equations Derived From Exact Differentials in Two Dimensions

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, and . Suppose that the state space is two dimensional and any of the five quantities are exact differentials. Then by the chain rule

(1)~~~~~ 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

but also by the chain rule:

(2)~~~~~ dx = \left(\frac{\partial x}{\partial u}\right)_v du +\left(\frac{\partial x}{\partial v}\right)_u dv

and

(3)~~~~~ dy= \left(\frac{\partial y}{\partial u}\right)_v du +\left(\frac{\partial y}{\partial v}\right)_u dv

so that:

(4)~~~~~ 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

which implies that:

(5)~~~~~ \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

Letting gives:

(6)~~~~~ \left(\frac{\partial z}{\partial u}\right)_y = \left(\frac{\partial z}{\partial x}\right)_y \left(\frac{\partial x}{\partial u}\right)_y

Letting gives:

(7)~~~~~ \left(\frac{\partial z}{\partial y}\right)_v = \left(\frac{\partial z}{\partial y}\right)_x + \left(\frac{\partial z}{\partial x}\right)_y \left(\frac{\partial x}{\partial y}\right)_v

Letting, gives:

(8)~~~~~ \left(\frac{\partial z}{\partial y}\right)_x = - \left(\frac{\partial z}{\partial x}\right)_y \left(\frac{\partial x}{\partial y}\right)_z

using (\partial a/\partial b)_c = 1/(\partial
b/\partial a)_c gives the triple product rule:

(9)~~~~~ \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

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