1.2 Systems and States in Physical Chemistry 15
df(
∂f
∂x)
y,zdx+(
∂f
∂y)
x,zdy+(
∂f
∂z)
x,ydz (1.2-5)where (∂f /∂x)y,z,(∂f /∂y)x,z, and (∂f /∂z)x,yarepartial derivatives. If the function is
represented by a formula, a partial derivative with respect to one independent variable
is obtained by the ordinary procedures of differentiation, treating the other indepen-
dent variables as though they were constants. The independent variables that are held
constant are usually specified by subscripts.
We assume that our macroscopic equilibrium state functions are differentiable except
possibly at isolated points. For an equilibrium gas or liquid system of one phase and
one substance, we can writedV(
∂V
∂T
)
P,ndT+(
∂V
∂P
)
T,ndP+(
∂V
∂n)
T,Pdn (1.2-6)This equation represents the value of an infinitesimal change in volume that is produced
when we impose arbitrary infinitesimal changesdT,dP, anddnon the system, making
sure that the system is at equilibrium after we make the changes. For an ideal gasdVnR
PdT−nRT
P^2dP+RT
P
dn (ideal gas) (1.2-7)Amathematical identityis an equation that is valid for all values of the variables
contained in the equation. There are several useful identities involving partial deriva-
tives. Some of these are stated in Appendix B. An important identity is thecycle rule,
which involves three variables such that each can be expressed as a differentiable
function of the other two:
(
∂z
∂x)
y(
∂x
∂y)
z(
∂y
∂z)
x−1 (cycle rule) (1.2-8)If there is a fourth variable, it is held fixed in all three of the partial derivatives.
Some people are surprised by the occurrence of the negative sign in this equation. See
Appendix B for further discussion.EXAMPLE 1.4
Takezxyand show that the three partial derivatives conform to the cycle rule.
Solution
(
∂z
∂x)yy
(
∂x
∂y)z−
z
y^2
(
∂y
∂z)x
1
x
(
∂z
∂x)y(
∂x
∂y)z(
∂y
∂z)x−y
z
y^21
x
−
z
xy
− 1