1. Multiplicity, Entropy, Second & Third Law Of Thermodynamics

Question 1

Multiplicity of Einstein Solid

Answer 1

Ω(N, q) = (q + N - 1)! / q! (N- 1)!

Question 2

Multiplicity of Two-State Paramagnet System

Answer 2

Ω(N up) = N! / N up! N down!

Question 3

Fundamental Assumption of Statistical Mechanics

Answer 3

In an isolated system in thermal equilibrium, all accessible microstates are equally probable.

Question 4

Second Law of Thermodynamics

Answer 4

For spontaneous processes, the entropy of the universe increases and approaching greater disorder and randomness.
A system will naturally evolve towards the macro state with the highest multiplicity (most possible arrangements of its micro states)

Question 5

Multiplicity of Large Einstein Solid

Answer 5

lnΩ = ln((q + N)! / q!N1)

If q&raquo_space; N

Ω(N, q) = e^Nln(q/N)e^N = (eq/N)^N

If q &laquo_space;N

Ω(N, q) = (Ne/q)^q

Question 6

Multiplicity of Mono-atomic Ideal Gas

Answer 6

In 3D:

Ω = V Vp / h^3
Ω N = 1/N! (V^N/ h^3N)

1D:
L Lp/ Δx Δpx = L Lp/h

Question 7

Entropy
(Relationship between Entropy and Multiplicity)

Answer 7

S = klnΩ

Question 8

Entropy of Large Einstein Solid

Answer 8

S = kln(eq/N)^N = Nk[ln(q/N) + 1]

Question 9

Entropy of an Ideal Gas

Answer 9

S = Nk[ ln( V/N (4pimU/3Nh^2)^3/2) + 5/2] (Sackur-Tetrode Equation)

Question 10

Change in Entropy of Ideal Gas

Answer 10

Change in S = Nkln(Vf/Vi) (General)

Change in S = Q/T (any Quasistatic process)

Question 11

Entropy of Mixing

Answer 11

Change in Sa = Nkln(Vf/Vi) = Nkln2

Change in S total = Change in Sa + Change in Sb = 2Nkln2

Question 12

Relationship between Temperature, Entropy, and Potential Energy

Answer 12

1/T = (dS/dU)N, V —> T = (dU/dS) N, V

Energy will always tend to close into Theo object with stepper S vs. U graph and out of object with shallows S vs. U graph. (Steep slope is low temp, shallow is high temp)

Question 13

Thermodynamic Identity & How it’s derived

Answer 13

dU = TdS - PdV

S = f(U, V)
dS = dS/dU dU + dS/dV dV
dS = 1/T dU + P/T dV
dU = TdS - PdV

Question 14

Change in Entropy

Answer 14

dS = Q/T (constant volume, no work)

dS = CvdT/T (T is changing)
- integrate for ΔS

Question 15

Third Law of Thermodynamics

Answer 15

The entropy of a system approaches a constant value as the temperature approaches absolute zero. (System’s unique, lowest energy state)

Question 16

Relationship between Pressure, Temperature, Entropy, Volume

Answer 16

P = T (dS/dV) N, U

Question 17

Diffusive Equilibrium & Grand Thermodynamic Identity

Answer 17

dSA/ dNA = dSB/dNB

-T dSA/dNA = -TdS/dNB at equilibrium —> this is chemical potential μ = -T (dS/dN) V, U

Thus we can write GTI:
S(U, V, N)
dS = dS/dU dU + dS/dV dV + dS/dN dN
dU = TdS - PdV + μdN
dS = 1/T dU + P/T dV - μ/T dN

For Diffusive Equilibrium:

μA = μB

System that has a larger dS/dN wants more particles because it has a smaller chemical potential.
Particles tend to flow toward lower values of chemical potential.

Question 18

Total Entropy at Equilibrium

Answer 18

(dS total/dUA) NA, VA = 0 and (dS total/ dNA) UA, VA = 0

(Total entropy at maximum)

Question 19

Chemical Potential

Answer 19

U & V Fixed:
0 = TdS + μdN —> μ = -T (dS/dN) U, V

S & V fixed:
dU = μdN —> μ = (dU/dN) S, V

μ had units of energy; amount by which system’s energy changes, when you add one particle and keep the entropy and volume fixed.

Question 20

Sacker Tetrode Equation

Answer 20

Describes partition function (statistical properties) of gas with indistinguishable particles.

Partition:
Z = (V/ λt^3)^N / N!

This is proportional to multiplicity as it represents number of accessible states for system.

Ω ~ Z ~ (V/ λt^3)^N / N!

Finding entropy of this gives:

S = kBln(1/N! (V/ λt^3)^N) = e/N (V/ λt^3)^N

Stirling Approximation which is lnN! =~ NlnN - N for large can be used to simplify to:

S = Nkb(ln (V/ Nλt^3) + 5/2)

Which represents third law of thermodynamics! (If gas cooled down to 0, entropy goes to 0)