Page 40 - Physical Chemistry

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where the y subscripts on dz and dx indicate that these infinitesimal changes occur at Section 1.6
constant y. Division by dz gives Differential Calculus
y
0z dx y 0z 0x
1 a b a b a b
0x y dz y 0x y 0z y
since from the definition of the partial derivative, the ratio of infinitesimals dx /dz y
y
equals (
x/
z) . Therefore
y
0z 1
a b (1.32)*
0x y 10x>0z2 y
Note that the same variable, y, is being held constant in both partial derivatives in
(1.32). When y is held constant, there are only two variables, x and z, and you will
probably recall that dz/dx 1/(dx/dz).
For an infinitesimal process in which z stays constant, Eq. (1.30) becomes
0z 0z
0 a b dx a b dy z (1.33)
z
0x y 0y x
Dividing by dy and recognizing that dx /dy equals (
x/
y) , we get
z
z
z
z
0z 0x 0z 0z 0x 0z 1
0 a b a b a b and a b a b a b
0x y 0y z 0y x 0x y 0y z 0y x 10y>0z2 x
where (1.32) with x and y interchanged was used. Multiplication by (
y/
z) gives
x
0x 0y 0z
a b a b a b 1 (1.34)*
0y z 0z x 0x y
Equation (1.34) looks intimidating but is actually easy to remember because of the
simple pattern of variables:
x/
y,
y/
z,
z/
x; the variable held constant in each par-
tial derivative is the one that doesn’t appear in that derivative.
Sometimes students wonder why the
y’s,
z’s, and
x’s in (1.34) don’t cancel to
give 1 instead of 1. One can cancel
y’s etc. only when the same variable is held
constant in each partial derivative. The infinitesimal change dy in y with z held con-
z
stant while x varies is not the same as the infinitesimal change dy in y with x held
x
constant while z varies. [Note that (1.32) can be written as (
z/
x) (
x/
z) 1; here,
y
y
cancellation occurs.]
Finally, let dy in (1.30) be zero so that (1.31) holds. Let u be some other variable.
Division of (1.31) by du gives
y
dz y 0z dx y
a b
du y 0x y du y
0z 0z 0x
a b a b a b (1.35)*
0u y 0x y 0u y
The
x’s in (1.35) can be canceled because the same variable is held constant in each
partial derivative.
A function of two independent variables z(x, y) has the following four second
partial derivatives:
2
2
0 z 0 0z 0 z 0 0z
a 2 b c a b d , a 2 b c a b d
0x y 0x 0x y y 0y x 0y 0y x x
2
2
0 z 0 0z 0 z 0 0z
c a b d , c a b d
0x 0y 0x 0y x y 0y 0x 0y 0x y x