1. Classical Ideal Gas

2. In our review of undergrad Statistical & Thermal Physics, we looked briefly at the Classical Ideal Gas with N particles & volume V. We calculated the E & V dependences of the number of accessible states (N, E,V) for such a gas. We found:
(N, E,V)  VN(E)(3N/2)

3. PV = NkBT (N,E,V)  VN(E)(3N/2) p = kBT[∂ln()/∂V] So, we have simply
Using the general form for the pressure that we just talked about, the Equation of State (or P, T, V relation) can be obtained for the Classical Ideal Gas:
p = kBT[∂ln()/∂V]
So, we have simply
PV = NkBT

4. (N,E,V)  VN(E)(3N/2) The Entropy for the Ideal Gas:
S = kB ln = N ln(V) +
(3/2)N ln(E) + const
S0 = kB ln0 = N ln(V0) +
(3/2)N ln(E0) + const
S = S – S0 = N ln(V/V0) +
(3/2)N ln(E/E0)

5. (N,E,V)  VN(E)(3N/2) E = (3/2)NkBT Use along with and
Find the Energy for the Ideal Gas:
Use along with l
and
After doing this & manipulation we find:
E = (3/2)NkBT
  [1/(kBT)]

6. Gibbs’ Paradox: Entropy of Mixing

7. S = kB ln = N ln(V) + (3/2)N ln(E) + const
Entropy for the Classical Ideal Gas:
S = kB ln = N ln(V) +
(3/2)N ln(E) + const
For N non-interacting molecules the energy
E = N, where  = energy of one molecule.
In what follows, the references I used assume that each molecule is a quantum mechanical particle in a box, which introduces Planck’s Constant ħ into the otherwise classical expressions. This affects the entropy by only a constant.

8. Entropy in the Ideal Gas
Consider an Ideal, Monatomic Gas
with N molecules in volume V &
with mean energy U.
From earlier discussions, the
Entropy is (E U)

9. Entropy in the Ideal Gas its dependence on particle
As first pointed out by Gibbs,
this Entropy is WRONG!
Specifically,
its dependence on particle
number N is wrong!  Gibbs’ Paradox!
To see that this S is wrong, consider the
Mixing of 2 Ideal Gases

10. Entropy in the Ideal Gas
The Boltzmann Definition of Entropy is
Ideal Gas Entropy
in the Canonical Ensemble:
“Sackur-Tetrode” Equation

11. Entropy in the Ideal Gas
Expand the log function
(f degrees of freedom for polyatomic molecules)
Use equipartition of energy:
U = (½)f kBT

12. Gas Mixing Problem For each gas: For He: For H2:
Two cylinders (V = 1 l) are connected by a valve. In one of the cylinders – Hydrogen (H2) at P = 105 Pa, T = 200C , in another one – Helium (He) at P = 3·105 Pa, T=1000C. Find entropy change after mixing and equilibrating.
Gas Mixing Problem
For each gas:
For He:
For H2:

13. Gas Mixing Problem or: Temperature after mixing. Start with energies:
Two cylinders (V = 1 l) are connected by a valve. In one of the cylinders – Hydrogen (H2) at P = 105 Pa, T = 200C , in another one – Helium (He) at P = 3·105 Pa, T=1000C. Find entropy change after mixing and equilibrating.
Gas Mixing Problem
Temperature after mixing.
Start with energies:
So:
or:
Put this into entropies:
or:

14. Two cylinders (V = 1 l) are connected by a valve
Two cylinders (V = 1 l) are connected by a valve. In one of the cylinders – Hydrogen (H2) at P = 105 Pa, T = 200C , in another one – Helium (He) at P = 3·105 Pa, T=1000C. Find entropy change after mixing and equilibrating.
Gas Mixing Problem
if N1=N2=1/2N
V1=V2=1/2V
The total entropy of the system is greater after mixing – thus, mixing is irreversible.

15. - applies only if two gases are different !
Gibbs Paradox
- applies only if two gases are different !
If two mixing gases are of the same kind (indistinguishable molecules):
Stotal = 0 because U/N and V/N available for each molecule remain the same after mixing.
Quantum-mechanical indistinguishability is important! (even though this equation applies only in the low density limit, which is “classical” in the sense that the distinction between fermions and bosons disappear.

16. at T1=T2, S=0, as it should be (Gibbs paradox)
Two identical perfect gases with the same pressure P and the same number of particles N, but with different temperatures T1 and T2, are confined in two vessels, of volume V1 and V2 , which are then connected. find the change in entropy after the system has reached equilibrium.
- prove it!
at T1=T2, S=0, as it should be (Gibbs paradox)

17. Ideal Gas: from S(N,V,U) - to U(N,V,T)
Ideal gas: (fN degrees
of freedom)
- the “energy” equation of state 
- in agreement with the equipartition theorem, the total energy should be ½kBT times the number of degrees of freedom.
The heat capacity for
a monatomic ideal gas:

18. “Pressure” Equation of State for an Ideal Gas
The “energy” equation of state (U  T):
Ideal gas:
(fN degrees of freedom)
The “pressure” equation of state (P  T):
- we have finally derived the equation of state of an ideal gas from first principles!

19. A Perhaps Simpler Explanation of Gibbs’ Paradox?
If N molecules of an ideal monatomic gas make a free expansion from V to 2V, ΔE = 0, so that
ΔS = Sf – Si = NkB ln(2V/V)
= NkB ln 2 (1)
If the container of volume 2V is divided into equal parts, so that (N/2) molecules are in each half, then
ΔS' = Sf' – Si' = 2(N/2)kBln(V/2V)
= – NkBln (2)
This is Gibbs’ Paradox

20. NkB ln(V/N!) ≈ NkB [ln(V/N) + 1]
Gibbs’ Paradox
Gibbs’ Paradox, is removed by assuming that the molecules are indistinguishable; i.e. that the number of accessible states is
(E,V,N) = B U3N/2 VN/N!
This means that term NkBlnV in the expression for S is replaced by NkB ln(V/N!),
In the latter expression, use Stirling’s Formula:
NkB ln(V/N!) ≈ NkB [ln(V/N) + 1]