Page 59 - Physical chemistry eng

36 CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics

from an initial state V ,T to a final state V ,T . We first assume that U is a function of
1
2
2
1
both V and T. Is this assumption correct? Because ideal gas atoms or molecules do not
interact with one another, U will not depend on the distance between the atoms or mol-
ecules. Therefore, U is not a function of V, and we conclude that ¢U must be a function
of T only for an ideal gas, ¢U =¢U(T) .
We also know that for a temperature range over which C is constant,
V
¢U = q = C (T - T ) (2.31)
i
f
V
V
Is this equation only valid for constant V? Because U is a function of only T for an ideal
gas, Equation (2.31) is also valid for processes involving ideal gases in which V is not
constant. Therefore, if one knows C , T , and T , ¢U can be calculated, regardless of
1
V
2
the path between the initial and final states.
How many variables are required to define ¢H for an ideal gas? We write
¢H =¢U(T) +¢(PV) =¢U(T) +¢(nRT) =¢H(T) (2.32)
We see that ¢H is also a function of only T for an ideal gas. In analogy to Equation (2.31),
¢H = q = C (T - T ) (2.33)
P
f
P
i
Because ¢H is a function of only T for an ideal gas, Equation (2.33) holds for all
processes involving ideal gases, whether P is constant or not, as long as it is reasonable
to assume that C is constant. Therefore, if the initial and final temperatures are known
P
or can be calculated, and if C and C are known, ¢U and ¢H can be calculated
P
V
regardless of the path for processes involving ideal gases using Equations (2.31) and
(2.33), as long as no chemical reactions or phase changes occur. Because U and H are
state functions, the previous statement is true for both reversible and irreversible
processes. Recall that for an ideal gas C - C = nR , so that if one of C and C is
P
P
V
V
known, the other can be readily determined.
We next note that the first law links q, w, and ¢U . If any two of these quantities are
known, the first law can be used to calculate the third. In calculating work, often only
expansion work takes place. In this case one always proceeds from the equation
w =- P external dV (2.34)
L
This integral can only be evaluated if the functional relationship between P external and V
is known. A frequently encountered case is P external = constant , such that
w =-P external (V - V ) (2.35)
f
i
Z P , the work considered in Equation (2.35) is for an irreversible
Because P external
process.
A second frequently encountered case is that the system and external pressure differ
only by an infinitesimal amount. In this case, it is sufficiently accurate to write
P external = P , and the process is reversible:
nRT
w =- dV (2.36)
V
L
This integral can only be evaluated if T is known as a function of V. The most com-
monly encountered case is an isothermal process, in which T is constant. As was seen
in Section 2.2, for this case
dV V f
w =-nRT =-nRT ln (2.37)
L V V i
In solving thermodynamic problems, it is very helpful to understand the process thor-
oughly before starting the calculation, because it is often possible to obtain the value of
one or more of q, w, ¢U , and ¢H without a calculation. For example, ¢U =¢H = 0
for an isothermal process because ¢U and ¢H depend only on T. For an adiabatic
process, q = 0 by definition. If only expansion work is possible, w = 0 for a constant
volume process. These guidelines are illustrated in the following two example problems.