Lecture 8

Readings

http://umdberg.pbworks.com/w/page/68375276/Powers%20and%20exponents%20%282013%29

http://umdberg.pbworks.com/w/page/104491534/How%20energy%20is%20distributed%3A%20Fluctuations

Fluctuations in Energy Distributions

At thermal equilibrium, the average energy per degree of freedom is the same
- 2 energy packets, 10 degrees of freedom
- Average energy per degree of freedom = <E> = 0.2 energy packets
- Same for top and bottom = thermal equilibrium
The energy in a particular DoF will fluctuate
- 80% of the time, a DoF will have 0 energy, 20% of the time, a DoF will have 1 energy packet

\frac{N!}{(N - M)! * M!}

But.... you can also put 2 energy packets into a single DoF!
- It can happen randomly
- This makes the probability calculations harder
- There is no upper limit on the amount of energy in a single DoF
Actual odds are as follows

Probability	Number of Energy Packets
79%	0
19%	1
1.9%	2
0.08%	3
0.02%	4

This is the P(E) that the reading referred to, the probability of each amount of energy

Multiple Choice

Δ S = \frac{Q}{T}

| Q | t h a t l e a v e s

A is the same |Q| that enters B

Δ S_{A} = \frac{- Q}{T_{A}} = \frac{- Q}{350 K}

Δ S_{B} = \frac{+ Q}{T_{B}} = \frac{+ Q}{250 K}

Δ S_{B} = \frac{- Q}{250 K}

Δ S_{A} = \frac{+ Q}{350 K}

So B is losing more entropy than A gainsNet Entropy of AB went downTotal entropy goes upSomething else must have gained entropy more than S J/K

$|\Delta S_B| - |\Delta S_A|$

1st law = energy is conserved
entropy of AB decreases

Δ S = \frac{Q}{T}

If Q is positive, then 𝝙S is positive

Δ Q = Q_{f i n a l H_{2} O} - Q_{i n i t i a l H_{2} O}

2 molecules -> 3 moleculesEntropy of Gas > Entropy of Liquid

- 100 - (50)

Δ G = Δ H - (T * Δ S)

I f Δ G = n e g a t i v e, then reaction = spontaneous

$\Delta S$ will be positive
$\Delta G$ is negative , so spontaneous
$\Delta S$ went down , then depends on temperature