Page 98 - Essentials of physical chemistry

60 Essentials of Physical Chemistry

But according to the first law dU ¼ dq PdV allowing for the sign convention of work done by=on
gases. Thus, we see that under atmospheric conditions we have

dH ¼ dq PdV þ PdV ¼ dq,

and we can define a heat capacity under constant pressure conditions as:

qH
:
qT
C P ¼
P
In your physics text, there was probably no mention of a difference in the heat capacity whether
volume or pressure is held constant and for solids and liquids there is little difference but we will see
here that it does make a difference for gases. Students are encouraged to learn this difference and be
able to prove it on a test because it is an excellent exercise in the meaning of partial derivatives in
thermodynamics. What follows is an example of being able to perform correct algebra but not
necessarily reaching the desired simplicity. We need to pay attention to the path through the partial
derivatives and gain facility in their manipulation.

qH qU qU qV qP qU
þ P þ V :
(C P C V ) ¼ ¼
qT P qT V qT P qT P qT P qT V

qU qU
Please note that 6¼ , they are not the same. In addition, we can recognize that
qT qT
P V

qP
¼ 0by definition (there is no variation in pressure if the pressure is constant). Next we need
qT
P
to remember that P, V, T are all related by some state function so we need to write

qU qU
dT
qV qT
dU ¼ dV þ
T V
and that

qU qU qV qU qT
:
þ
qT qV qT qT qT
¼
P T P V P
With our goal of trying to make things simple we could ask why we expanded dU in terms of
T and V, leaving P to be determined by some unspecified state function. The reason is that we want

qU
to eliminate , which is the strange quantity we cannot see any easy way to measure in the
qT
P
laboratory. Thus, we expand dU in terms of the other variables and then impose constant P
conditions. While this statement seems abstract here, it is helpful when writing this derivation to
understand why we choose to expand dU in terms of T and V.

qT qU
Of course ¼ 1 and we can substitute into the (C P C V ) equation. That leads
qT P qT P
directly to the new equation as

qU qV qU qU
þ P :
(C P C V ) ¼ þ
qV qT qT qT
T P V V