14.5: Mechanics of Manipulating a Function of State

Given that \(f(x, y, z)\) is any state function that characterizes the system and \((x, y, z)\) is a set of independent variable properties of that system, we know that any change \(\Delta \mathrm{f}\) will be only a function of the value of " \(f\) " at the final and initial states,

\[\Delta f=f_2-f_1=f\left(x_2, y_2, z_2\right)-f\left(x_1, y_1, z_1\right) \nonumber \]

Since \(f=f(x, y, z)\), we can mathematically relate the total differential change (df) to the partial derivatives \(\dfrac{\partial f}{\partial x}\), \(\dfrac{\partial f}{\partial y}\), and \(\dfrac{\partial f}{\partial z}\) of the function, as follows:

\[d f=\left(\dfrac{\partial f}{\partial x}\right)_{y, z} d x+\left(\dfrac{\partial f}{\partial y}\right)_{x, z} d y+\left(\dfrac{\partial f}{\partial z}\right)_{x, y} d z \nonumber \]

where, in general: \(\left(\dfrac{\partial f}{\partial x}\right)_{y, z}=\) the change of f with respect to x , while y and z are unchanged.

If we want to come up with the total change, \(\Delta f\), of a property (we want to go from \(\PageIndex{2}\) to \(\PageIndex{1}\), we integrate the expression in \(\PageIndex{2}\) to get:

\[\Delta f=f_2-f_1=\int_{x_1}^{x_2}\left(\dfrac{\partial f}{\partial x}\right)_{y, z} d x+\int_{y_1}^{y_2}\left(\dfrac{\partial f}{\partial y}\right)_{x, z} d y+\int_{z_1}^{z_2}\left(\dfrac{\partial f}{\partial x}\right)_{x, y} d z \nonumber \]

Let us visualize this with an example. For a system of constant composition, its thermodynamic state is completely defined when two properties of the system are fixed. Let us say we have a pure component at a fixed pressure ( P ) and temperature ( T ). Hence, all other thermodynamic properties, for example, enthalpy \((H)\), are fixed as well. Since \(H\) is only a function of \(P\) and \(T\), we write:

\[H=H(P, T) \nonumber \]

and hence, applying 6.2, any differential change in enthalpy can be computed as:

\[d H=\left(\dfrac{\partial H}{\partial P}\right)_T d P+\left(\dfrac{\partial H}{\partial T}\right)_P d T \nonumber \]

The total change in enthalpy of the pure-component system becomes:

\[\Delta H=H_2-H_1=\int_{P_1}^{P_2}\left(\dfrac{\partial H}{\partial P}\right)_T d P+\int_{T_1}^{T_2}\left(\dfrac{\partial H}{\partial T}\right)_P d T \nonumber \]

Now we are ready to spell out the exactness condition, which is the mathematical condition for a function to be a state function. The fact of the matter is, that for a function to be a state function — i.e., its integrated path shown in \(\PageIndex{3}\) is only a function of the end states, as shown in \(\PageIndex{1}\) — its total differential must be exact. In other words, if the total differential shown in \(\PageIndex{2}\) is exact, then f(x,y,z) is a state function. How do we know if a total differential is exact or not?

Given a function Ψ(x,y,z),

\[d \Psi=M(x, y, z) d x+N(x, y, z) d y+Q(x, y, z) d z \nonumber \]

where:

\[M(x, y, z)=\left(\dfrac{\partial \Psi}{\partial x}\right)_{y, z} \nonumber \]

\[N(x, y, z)=\left(\dfrac{\partial \Psi}{\partial y}\right)_{x, z} \nonumber \]

\[Q(x, y, z)=\left(\dfrac{\partial \Psi}{\partial z}\right)_{x, y} \nonumber \]

we say that \(d \Psi\) is an exact differential and consequently \(\psi(x, y, z)\) a state function if all the following conditions are satisfied:

\[\left(\dfrac{\partial M}{\partial y}\right)_{x, z}=\left(d\frac{\partial N}{\partial x}\right)_{y, z} \nonumber \]

\[\left(\dfrac{\partial N}{\partial z}\right)_{x, y}=\left(\dfrac{\partial Q}{\partial y}\right)_{x, z} \nonumber \]

\[\left(\dfrac{\partial M}{\partial z}\right)_{x, y}=\left(\dfrac{\partial Q}{\partial x}\right)_{y, z} \nonumber \]

The above three Equations are called the exactness condition.