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The Ideal Gas

The basic equations:
P_{i} = (1/Z)exp(βE_{i})
Z = Σ_{i}e^{βEi}
E = ∂lnZ/∂β
K_{B}T = 1/β
S = βE + lnZ
A = K_{B}TlnZ
Consider a box with volume, V, and N identical noninteracting
particles. The states are the continous collection of the position
and momentum states for each particle. Compute the partition
function.
Z = ∫dx^{3N}dp^{3N}exp(βΣ_{n}(p_{n}^{2}/2m))
= (V^{N}/N!)[∫exp(βp^{2}/2m)dp]^{3N}
The N! is to compensate for the fact that the states in classical
mechanics are distinguishable.
Let u^{2} = βp^{2}/2m ∴ p = u√(2m/β) and dp = √(2m/β)du. The ∫
becomes:
Z = (V^{N}/N!)[√(2m/β)∫exp(u^{2})du]^{3N}
= (V^{N}/N!)(2mπ/β)^{3N/2}
Now,
lnZ = (3/2)Nlnβ + constant
E = ∂lnZ/∂β
= (3/2)NK_{B}T where β = 1/K_{B}T
This is the familiar term associated with the energy of an ideal
gas.
Pressure

Consider a piston of area, A that is moved a distance, dx, by
the gas (the gas is doing work on the piston).
dE = PAdx (work = force x distance)
= PdV
∴ P = ∂E/∂V
Consider an adiabatic process. The change in entropy associated with
an adiabatic process is 0 (ΔS = 0). Thus, we can write:
P = ∂E/∂V_{S}
Theorem:
∂E/∂V_{S} = ∂E/∂V_{T}  (∂E/∂S_{V})(∂S/∂V_{T})
Proof:
Consider following diagram:
T
 \
 \
 \
 \
 \
 \ constant S (adiabatic)

 V
How does E change along a line of constant S?
ΔE/ΔV_{T} = (∂E/∂V_{T})ΔV + (∂E/∂V_{V})ΔT
ΔE = (∂E/∂V_{T})ΔV + (∂E/∂V_{V})ΔT
ΔE/ΔV = ∂E/∂V_{T} + (∂E/∂V_{V})(ΔT/ΔV)
= ∂E/∂V_{T} + (∂S/∂T_{V})(∂E/∂S_{V})(ΔT/ΔV)
Along the S curve dS = 0. Therefore,
dS = (∂S/∂V_{T})ΔV + (∂S/∂T_{V})ΔT = 0
Therefore,
ΔT/ΔV = (∂S/∂V_{T})/(∂S/∂T_{V})
Substituting back into the ΔE/ΔV equation gives:
ΔE/ΔV ≡ ∂E/∂V_{S} = ∂E/∂V_{T}  (∂S/∂T_{V})(∂E/∂S_{V})(∂S/∂V_{T})/(∂S/∂T_{V})
_{ } = ∂E/∂V_{T}  (∂E/∂S_{V})(∂S/∂V_{T}) Q.E.D.
Now T = ∂E/∂S_{V}. Therefore,
∂E/∂V_{S} = ∂E/∂V_{T}  T∂S/∂V_{T}
_{ } = ∂(E  TS)/∂V_{T}
_{ } = ∂A/∂V_{T} where A is the Helmholtz Free energy
P = ∂E/∂V_{S}
= ∂E/∂V_{T} + (∂E/∂S_{V})(∂S/∂V_{T})
= ∂(E  TS)/∂V_{T}
= ∂A/∂V_{T}
= K_{B}T∂lnZ/∂V_{T} since A = K_{B}TlnZ
From the ideal gas we found:
Z = (V^{N}/N!)(2mπ/β)^{3N/2}
Therefore,
lnZ = NlnV + (3N/2)ln(2mπ/β)  lnN!
This leads to:
P = K_{B}T∂lnZ/∂V_{T} = K_{B}TN/V
Or,
PV = NK_{B}T