Physical Chemistry , 1st ed.

987 J(1 mol)12.47

mo

J

lK

T

T79.1 K

With an initial temperature of 273.15 K, the final temperature is about 194 K.

For an adiabatic process, the infinitesimal amount of work done can now
be determined from two expressions:

dwpextdV

dwnCVdT

Equating the two:

pextdVnCVdT

If the adiabatic process is reversible, then pextpintand we can use the ideal
gas law to substitute for pintin terms of the other state variables. We get

n

V

RTdVnC

VdT

Bringing the temperature variables to the right side, we find that



V

R

dV

C

T

VdT

The variable nhas canceled. We can integrate both sides of the equation
and, assuming that CVis constant over the change, we find (recognizing
that 1/xdxln x) that

Rln VVVifCCV ln TTTif

Using the properties of logarithms and evaluating each integral at its limits,
we get

Rln V

V

f

i

CV ln T

T

f

i

 (2.44)

for an adiabatic, reversible change in an ideal gas. Again using properties of
logarithms, we can get rid of the negative sign by taking the reciprocal of the
expression inside the logarithm:

Rln 

V

V

f

iC

V ln 

T

T

f

i



Recognizing that CpCV+ R, we rearrange it as CpCVRand substitute:

(CpCV) ln 

V

V

f

iC

V ln 

T

T

f

i



Dividing through by CV:



(Cp

C



V

CV)

ln 

V

V

f

iln T

T

f

i



The expression (CpCV)/CVis usually defined as :

(Cp

C



V

CV) (2.45)

2.8 More on Heat Capacities 49