Untitled - Aerobib - Universidad Politécnica de Madrid

18 CHAPTER 1. THERMOCHEMISTRY
function is determined. Then, the problem lies in expressing the thermodynamic functions
of the gas by means of its partition function. According to Statistical Mechanics
the solution is as follows.
a) Internal energy.
b) Enthalpy.
c) Entropy.
h = R g T
( ) ∂ ln Q
u = R g T 2 . (1.65)
∂T
V
[( ) ∂ ln Q
+
∂ ln T
V
s = R g
[ln Q + T
( ) ] ∂ ln Q
. (1.66)
∂ ln V
T
( ) ] ∂ ln Q
. (1.67)
∂ ln T
V
d) Free energy.
d.1) Gibbs:
g = R g T
[
ln Q −
( ) ] ∂ ln Q
. (1.68)
∂ ln V
T
d.2) Helmholtz: f = −R g T ln Q. (1.69)
e) Chemical Potential.
G = M g g = RT
[
ln Q −
( ) ] ∂ ln Q
. (1.70)
∂ ln V
T
Heat capacities at constant volume, c v , and at constant pressure, c p , are expressed
by u and h respectively in the form
c v = ∂u
∂T ,
c p = ∂h
∂T . (1.71)
This enables the calculation of their values as a function of Q by means of (1.65) and
(1.66).
Thus, the problem reduces to the determination of the energy levels for the formation
of Q. For the purpose, one must analyze the ways in which a gas molecule can
store energy. Except when working at very high temperatures, where it is necessary to
consider the states of electronic excitation in the molecule, this can be considered as
a system of material points which are its atoms. Then, the energy of each molecule is
the summation of the kinetic energies of its atoms plus the potential energy of the field
of forces that hold them together. Be n the number of atoms in each molecule. Then,
its number of degrees of freedom is 3n. Of which three are external degree of freedom
corresponding to the translational motion of the center of gravity of the molecule. The