Statistical Physics, Second Revised and Enlarged Edition

70 Maxwell–Boltzmanngases

6 .3.3 Free energy and pressure

Now let us turn to the free energyF,and Method 3 of section 2.^5 , the ‘royal route’.
Without reference as yet to any specific gas, thehardworkwasdoneinthe previous
section in the derivation of ( 6 .14). One can see that a verysimple expression forF
emerges, namely

F≡U−TS=−NkkkBT(lnZ−lnN+ 1 ) (6.1 6 a)

UsingStirling’s approximationin reverse, thismaybe neatlywritten

F=−NkkkBTlnZ+kkkBTlnN! (6.16b)

The answeristhe same as (2.28)forlocalizedparticlesbut withtheaddition ofthe
N! term. Our arguments from ( 6 .1 6 ) branch alongthree rather distinct lines.

1 .Pressure.First we can readilyuseF(T,V,N)to calculate the pressure asfollows:

P=−(∂F/∂V)T,N

since

dF=−SdT−PdV+μdN

Hence

P=NkkkBT(∂lnZ/∂V)T,N

using( 6 .1 6 ).

For the monatomic MB idealgas, sinceZ∝V(equation ( 6. 6 )), this becomes

P=NkkkBT/V (6.17)

The result ( 6 .17) remains true for polyatomic MBgases, since we shall see in

Chapter 7 thatZ ∝V for these also; the box volume onlyentersZvia the

translational motion of the molecules. Since this is identical to the ideal gas law,

PV=RT,thisjustifies completelyour statisticaldefinitions oftemperature and

entropy. By calculatingPand comparing with the ideal gas law we have verified

β=− 1 /kkkBT (2.20)

and

S=kkkBln (1.5)

withthe constantkkkB=R/N,the gas constant per molecule,i.e. Boltzmann’s

constant.