Simple Nature - Light and Matter

it make sense that the whole thing is proportional ton? .Answer, p.

1057

To show consistency with the macroscopic approach to thermo-

dynamics, we need to show that these results are consistent with

the behavior of an ideal-gas thermometer. Using the new defini-

tion 1/T= dS/dQ, we have 1/T = dS/dE, since transferring an

amount of heat dQinto the gas increases its energy by a correspond-

ing amount. Evaluating the derivative, we find 1/T = (3/2)nk/E,

orE= (3/2)nkT, which is the correct relation for a monoatomic

ideal gas.

A mixture of molecules example 20

.Suppose we have a mixture of two different monoatomic gases,

say helium and argon. How would we find the entropy of such a

mixture (say, in terms ofV andE)? How would the energy be

shared between the two types of molecules, i.e., would a more

massive argon atom have more energy on the average than a

less massive helium atom, the same, or less?

.Since entropy is additive, we simply need to add the entropies of

the two types of atom. However, the expression derived above for

the entropy omitted the dependence on the massmof the atom,

which is different for the two constituents of the gas, so we need

to go back and figure out how to put thatm-dependence back in.

The only place where we threw awaym’s was when we identified

the radius of the sphere in momentum space with

√

2 mE, but

then threw away the constant factor ofm. In other words, the final

result can be generalized merely by replacingEeverywhere with

the productmE. Since the log of a product is the sum of the logs,

the dependence of the final result onmandE can be broken

apart into two different terms, and we find

S=nklnV+

3

2

nklnm+

3

2

nklnE+...

The total entropy of the mixture can then be written as

S=n 1 klnV+n 2 klnV+

3

2

n 1 klnm 1 +

3

2

n 2 klnm 2

+

3

2

n 1 klnE 1 +

3

2

n 2 klnE 2 +...

Now what about the energy sharing? If the total energy isE =

E 1 +E 2 , then the most ovewhelmingly probable sharing of energy

will the the one that maximizes the entropy. Notice that the depen-

dence of the entropy on the massesm 1 andm 2 occurs in terms

that are entirely separate from the energy terms. If we want to

maximizeSwith respect toE 1 (withE 2 =E−E 1 by conservation

of energy), then we differentiateSwith respect toE 1 and set it

equal to zero. The terms that contain the masses don’t have any

332 Chapter 5 Thermodynamics