1. Ideal Gas in the Canonical Ensemble
Gibbs’ Paradox:
Ideal Gas in the Canonical Ensemble

2. 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
obtained from the Canonical Ensemble is:
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

3. “Sackur-Tetrode” Equation
Ideal Gas Entropy
Also, the Boltzmann Definition of Entropy is
(U,V,N)  # accessible states
Entropy in the Canonical Ensemble:
“Sackur-Tetrode” Equation
Expand the log function
(f degrees of freedom for polyatomic molecules)
Use equipartition of energy: U = (½)f kBT

4. Gas Mixing Problem For each gas:
Two cylinders (V = 1 liter each) 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 the entropy change after mixing and equilibrating.
For each gas:
The temperature
after mixing:

5. Consider two different ideal gases (N1, N2) kept in two separate volumes (V1,V2) at the same temperature. To calculate the increase of entropy in the mixing process, we can treat each gas as a separate system. In the mixing process, U/N remains the same (T will be the same after mixing). The parameter that changes is V/N:
Entropy of Mixing
if N1=N2=1/2N
V1=V2=1/2V
The total entropy of the system is greater after mixing – thus, mixing is irreversible.

6. - 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.

7. at T1=T2, S=0, as it should be (Gibbs paradox)
Problem
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)

8. An Ideal Gas: from S(N,V,U) - to U(N,V,T)
(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:

9. The “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!