Statistical Mechanics - Physics at Oregon State University
Statistical Mechanics - Physics at Oregon State University
Statistical Mechanics - Physics at Oregon State University
You also want an ePaper? Increase the reach of your titles
YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.
4.2. BOLTZMANN GAS AGAIN. 75<br />
Although the distribution functions depend on µ, it is often possible to use<br />
this approach<br />
to obtain functions of N,T,V. Also note th<strong>at</strong> for the ideal gas<br />
∂U<br />
∂V = 0 and hence th<strong>at</strong> the pressure in an ideal gas is completely due to<br />
T,N<br />
the deriv<strong>at</strong>ive of the entropy, as we have st<strong>at</strong>ed before. This is due to the fact<br />
th<strong>at</strong><br />
<br />
∂F<br />
∂U<br />
∂S<br />
p = −<br />
= −<br />
+ T<br />
(4.30)<br />
∂V T,N ∂V T,N ∂V T,N<br />
In a real gas system, of course, a change in volume will change the average<br />
distance between particles and hence the inter-particle interactions. This will<br />
give a small correction to the formula for U, and hence a small contribution to<br />
the pressure. In solids, on the other hand, the pressure is mainly due to the<br />
change in interaction energy.<br />
Summ<strong>at</strong>ion of the chemical potential.<br />
We have transformed our independent variables to the set T, V, N and therefore<br />
need to find the corresponding free energy. The Helmholtz free energy is<br />
the important quantity and can be calcul<strong>at</strong>ed as soon as µ(T, N, V ) is found, because<br />
µ = <br />
∂F<br />
∂N . The Helmholtz free energy for a system with zero particles<br />
T,V<br />
is zero, and hence we have<br />
N<br />
0<br />
F (N, T, V ) =<br />
N<br />
0<br />
µdN ′ =<br />
′ N<br />
kBT log<br />
dN<br />
V nQ(T )<br />
′ <br />
= NkBT<br />
<br />
n<br />
log( ) − 1<br />
nQ(T )<br />
(4.31)<br />
where we used N<br />
0 log(cN ′ )dN ′ = N log(c) + N log(N) − N.<br />
Formally, there is an objection against using an integral over the number of<br />
particles when this number is small. Remember th<strong>at</strong> the chemical potential is<br />
the free energy needed to add one particle to the system, and hence a correct<br />
expression for F is<br />
F (N, T, V ) =<br />
N<br />
µ( ˆ N, T, V ) =<br />
ˆN=1<br />
N<br />
<br />
ˆN<br />
kBT log<br />
= kBT log(N!) − N log(nQ(T )V ) (4.32)<br />
V nQ(T )<br />
ˆN=1<br />
Here we used the fact th<strong>at</strong> N ˆN=1 log(c ˆ N) = N log(C) + log(N!). If N is very<br />
large we can again use Stirling’s formula to show th<strong>at</strong> this form really reduces