Physical Chemistry Third Edition

68 2 Work, Heat, and Energy: The First Law of Thermodynamics

If the initial valuesV 1 andT 1 are specified, this equation givesT 2 as a function ofV 2.

If we drop the subscripts onV 2 andT 2 , we can write Eq. (2.4-21a) in the form:

TT(V)T 1

(

V 1

V

)R/CV, m

(reversible adiabatic process,

ideal gas,CVconstant) (2.4-21b)

Equation (2.4-21) can be used for a reversible adiabatic compression as well as for an

expansion. It is an example of an important fact that holds for any system, not just an

ideal gas: For a reversible adiabatic process in a simple system the final temperature

is a function of the final volume for a given initial state.All of the possible final state

points for reversible adiabatic processes starting at a given initial state lie on a single

curve in the state space, called areversible adiabat. This fact will be important in our

discussion of the second law of thermodynamics in Chapter 3.

EXAMPLE2.17

A system consisting of 2.000 mol of argon expands adiabatically and reversibly from a volume

of 5.000 L and a temperature of 373.15 K to a volume of 20.00 L. Find the final temperature.

Assume argon to be ideal withCVequal to 3nR/2.

Solution

TT 1

(

V 1

V

)nR/CV

(373.15 K)

(

5 .000 L

20 .00 L

) 2 / 3

 148 .1K

Figure 2.7 shows the reversible adiabat that represents the final temperature as a func-

tion of the final volume for this example. For each initial state, there is only one such

curve.

200

300

400

100

0

T/K

V/L

(V 1 ,T 1 )

(V 2 ,T 2 )

01015205

For given (V 1 ,T 1 ), the

final state point must

lie on this curve.

Figure 2.7 Final Temperature as a
Function of Final Volume for the
Adiabatic Expansion of an Ideal Gas.

Exercise 2.13

Find the volume to which the system of the previous example must be adiabatically and reversibly

expanded in order to reach a final temperature of 273.15 K.

Exercise 2.14

Find the final temperature,∆U, andwfor a reversible adiabatic compression of 1.000 mol of

helium gas (assumed ideal withCV 3 nR/2) from a volume of 20.00 L and a temperature of

298.15 K to a volume of 10.00 L.

An equation analogous to Eq. (2.4-19) can be written for a real gas if CV

is independent ofVand if an adequate equation of state is available. For a gas obeying

the van der Waals equation of state it can be shown (see Problem 4.10) that

(

∂U

∂V

)

T,n



(

∂Um

∂Vm

)

T,n



a

Vm^2

(van der Waals gas) (2.4-22)

This can be used to derive an equation analogous to Eq. (2.4-21). For each such

equation, there is a unique curve in theV–Tplane containing all of the points that can

be reached by adiabatic reversible processes from a given initial state.