Page 95 - Essentials of physical chemistry

The First Law of Thermodynamics 57

We will often write differentials in terms of only two laboratory variables assuming a constant
sample size in terms of moles n and that there is some unspecified ‘‘equation of state,’’ which links
P, V, T, n. However, another very important mathematical property is obtained if we use a different
path but remember that U does not depend on the path, so that we can realize another relationship,

q qU q qU q U q U
2 2
or :
¼ ¼
qP qT qT qP qPqT qTqP
T P P T
This means that we are measuring the change in U by changing the temperature T and then changing the
pressure P, but this will be the same as changing the pressure P first and then changing the
temperature T.Check Introduction: Mathematics and Physics Review for a review of partial derivatives.
2 2
d f d 3y d f d 3y 3y
3y
Suppose f (x, y) ¼ 2x e then ¼ (6x e ) ¼ ¼ (2 e ) ¼ 6 e because this
dxdy dx dydx dy
is a continuous function of both x and y. Most beginning calculus texts teach from a subset of
continuous functions so students might expect all second derivatives to have this property but we
will find examples in later chapters on quantum mechanics where the order of differentiation does
matter in some cases. This property is shared by state variables but within the possibilities of all
functions that property is not shared by all cross derivatives. That also means, and is dependent on
the fact, that state variables are continuous functions. This relationship between the second
derivatives will be very important in later discussions of thermodynamics. We again offer the
insight in that state variables are universal variables and characteristics of the universe while q and w
are subjective variables that depend on how a process is carried out. In summary, we observe two
important characteristics of state variables, the cyclic path integrals are zero and the mixed
derivatives are equal because the state variables are continuous functions.

ISOTHERMAL PROCESSES
Example
Calculate q, w, and DU for the reversible, isothermal expansion of 10 mol of ideal gas from 1 to 0.1
atm at a constant temperature of 08C.

For the purpose of problem solving in thermodynamics the word ‘‘reversible’’ holds special
meaning. It means that PV work is carried out in a way that the internal pressure of the gas is
opposed by an external pressure that is only infinitesimally different from the internal pressure by
such a small amount that the process could easily be reversed at any time by some small fluctuation.
Later we will see that by opposing the internal pressure with the maximum (minus just a tiny
amount) external pressure the process will carry out the maximum amount of work. However, for
students facing a problem ‘‘reversible work’’ means we can equate the internal and external
pressures (P int ¼ P ext ). In addition, we encounter here the other key word ‘‘isothermal,’’ which
means DT ¼ 0. Recall previous work on gas energy where the Boltzmann average energy of a
monatomic gas is (3RT=2), which only depends on temperature. Note a monoatomic gas will have
only three degrees of freedom implying (RT=2) energy each for (v x , v y , v z ) degrees of freedom. Then
for a diatomic gas like N 2 or O 2 (air) there can also be two rotational degrees of freedom, each with
(RT=2) energy implying an energy of (5RT=2). A linear molecule can only rotate two ways but this
idea could be extended to polyatomic gases (CH 4 ), which would lead to (3RT=2) for rotational
energy and a total of (3RT )=mol. We will see later that this idea of (RT=2) energy per degree of
freedom called the law of energy equipartition is an ideal that is good for translation and rotation but
is only realized for vibration at high temperatures. Thus, if the temperature does not change, the
energy does not change either. Therefore, we have a simple realization that DT ¼ 0 and DU ¼ 0
when the temperature is constant. Thus, for this problem we see that DU ¼ 0 ¼ dq þ dw so that we