Page 47 - Physical Chemistry

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Chapter 1 Integration with respect to x for a function of two variables is defined similarly to
Thermodynamics (1.52) and (1.59). If y(x, z) is the most general function that satisfies

0y1x, z2
c d f 1x, z2 (1.61)
0x z
then the indefinite integral of f(x, z) with respect to x is

f 1x, z2 dx y1x, z2 (1.62)

1
3
2 3
For example, if f(x, z) xz , then y(x, z) x z g(z), where g is an arbitrary func-
2
tion of z. If y satisfies (1.61), one can show [in analogy with (1.60)] that a definite in-
tegral of f(x, z) is given by
b f 1x, z2 dx y1b, z2 y1a, z2 (1.63)
a
1
1
3
2
3
6
2
3
3
For example, xz dx (6 )z g(z) (2 )z g(z) 16z .
2
2
2
The integrals (1.62) and (1.63) are similar to ordinary integrals of a function f(x)
of a single variable in that we regard the second independent variable z in these inte-
grals as constant during the integration process; z acts as a parameter rather than as a
variable. (A parameter is a quantity that is constant in a particular circumstance but
whose value can change from one circumstance to another. For example, in Newton’s
second law F ma, the mass m is a parameter. For any one particular body, m is con-
stant, but its value can vary from one body to another.) In contrast to the integrals
(1.62) and (1.63), in thermodynamics we shall often integrate a function of two or
more variables in which all the variables are changing during the integration. Such in-
tegrals are called line integrals and will be discussed in Chapter 2.
An extremely common kind of physical chemistry problem is the use of the known
derivative dz/dx to find the change z brought about by the change x. This kind of
problem is solved by integration. Typically, the property z is a function of two variables
x and y, and we want the change z due to x while property y is held constant. We
use the partial derivative 10z 0x2 , and it helps to write this partial derivative as
y
0z dz y
a b (1.64)*
0x y dx y
where dz and dx are the infinitesimal changes in z and in x, while y is held constant.
y
y
EXAMPLE 1.5 Change in volume with applied pressure
For liquid water at 25°C, isothermal-compressibility data in the pressure range 1 to
2
401 bar are well fitted by the equation k a bP cP , where a 45.259
6
1
2
8
10 bar , b 1.1706 10 bar , and c 2.3214 10 12 bar 3 . The
3
volume of one gram of water at 25°C and 1 bar is 1.002961 cm . Find the volume
of one gram of water at 25°C and 401 bar. Compare the value with the experi-
3
mental value 0.985846 cm .
We need to find a volume change V due to a change in pressure P at con-
stant T. The compressibility is related to the rate of change of V with respect to
P at constant T. The definition (1.44) of k gives
1 0V 1 dV T
k a b (1.65)
V 0P T V dP T