Page 43 - Physical Chemistry

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Chapter 1 The relation (
z/
x) 1/(
x/
z) [Eq. (1.32)] shows that three of these six are the re-
y
y
Thermodynamics
ciprocals of the other three:
0T 1 0T 1 0P 1
a b , a b , a b
0P 10P>0T2 0V m P 10V >0T2 0V m T 10V >0P2
V m V m m P m T
(1.41)
Furthermore, the relation (
x/
y) (
y/
z) (
z/
x) 1 [Eq. (1.34)] with x, y, and z
z x y
replaced by P, V , and T, respectively, gives
m
0P 0V m 0T
a b a b a b 1
0V m T 0T P 0P V m
0P 0P 0V m 10V >0T2 P
m
a b a b a b (1.42)
0T 0V m T 0T P 10V >0P2
V m m T
where (
z/
x) 1/(
x/
z) was used twice.
y
y
Hence there are only two independent partial derivatives: (
V /
T) P and
m
(
V /
P) . The other four can be calculated from these two and need not be measured.
T
m
We define the thermal expansivity (or cubic expansion coefficient) a (alpha) and the
isothermal compressibility k (kappa) of a substance by
1 0V 1 0V m
a1T, P2 a b a b (1.43)*
V 0T P,n V m 0T P
1 0V 1 0V m
k1T, P2 a b a b (1.44)*
V 0P T,n V m 0P T
a and k tell how fast the volume of a substance increases with temperature and de-
creases with pressure. The purpose of the 1/V factor in their definitions is to make
them intensive properties. Usually, a is positive; however, liquid water decreases in
volume with increasing temperature between 0°C and 4°C at 1 atm. One can prove
from the laws of thermodynamics that k must always be positive (see Zemansky and
Dittman, sec. 14-9, for the proof). Equation (1.42) can be written as
0P a
a b (1.45)
0T k
V m

EXAMPLE 1.3 A and k of an ideal gas

For an ideal gas, find expressions for a and k and verify that Eq. (1.45) holds.
To find a and k from the definitions (1.43) and (1.44), we need the partial
derivatives of V . We therefore solve the ideal-gas equation of state PV RT
m
m
for V and then differentiate the result. We have V RT/P. Differentiation with
m
m
respect to T gives (
V /
T) R/P. Thus
m
P
1 0V m 1 R P R 1
a a b a b (1.46)
V m 0T P V m P RT P T
1 0V m 1 0 RT 1 RT 1
k a b c a bd a b (1.47)
V m 0P T V m 0P P T V m P 2 P
0P 0 RT R
a b c a bd (1.48)
0T 0T V m V m
V m V m
1
1
But from (1.45), we have 10P>0T2 a/k T /P 1 P/T nRTV /T R/V ,
m
V m
which agrees with (1.48).