Thermal & Kinetic Lecture 13

Towards Entropy and the 2nd Law: Permutations and Combinations

Section III: Probability, Entropy

and Thermal Equilibrium

Thus far, we’ve covered the following:

Introduction to states of matter: solid, liquids, gases;Distributions, averaging;The ideal gas law;Boltzmann factors;Maxwell-Boltzmann distribution;Equipartition of energy;Specific heats, blackbody radiation;Diffusion and thermal conductivity in a gas.

…….and yet, we still don’t fully understand why heat flows from hot to cold or why molecules ‘spread out’ evenly over the available space!

Very close connection between temperature and average kinetic energy:

½ m<v2> =3kT/2

Energy transfer

Direction of heat flow, Q:

Average energy of atoms in hot block > average energy of atoms in cold block.

Therefore more likely that an atom in the hotter block will lose energy to atom in colder block.

…but how much more likely? The atoms have a distribution of energies (e.g. Maxwell-Boltzmann for molecules in a gas):

Why can’t energy flow the other way?

Why couldn’t atoms with < mean energy in hot block gain energy from atoms with > mean energy in cold block?

Speed (m/s)

Energy transfer and reversibility

The direction of heat flow shown below wouldn’t violate the conservation of energy (total energy remains the same).

Therefore 1st law of thermodynamics not violated.

Q

TT + T

The fact that heat does not flow from a cold to a hot body is equivalent to stating that it is impossible to convert heat into work at a single temperature.

If the latter were possible we could get useful work by reversing a process such as friction. That we can’t is due to a physical law of just as much importance as the conservation of energy.

Probability and statistical mechanics

Demonstration movie from “Matter & Interactions”,Ruth Chabay and Bruce Sherwood. © Ruth Chabay

We need to consider large assemblies or collections of atoms.

Statistical behaviour must be analysed in detail if we are to answer questions on previous slides.

Our model of a solid comprises a collection of simple harmonic oscillators.

“Why have we returned to a discussion of solids? Why not continue considering gases?”

We’ll return to gases soon enough! Solids are easier to consider at the moment because the atoms are fixed in place.

The Einstein model of a solid

Instead of the model of a solid shown in the movie, we’re going to consider something even simpler (first devised by Einstein in 1907).

Furthermore, replace each 3D oscillator (i.e. each atom) with three independent 1D oscillators (x, y, z).

All these assumptions mean we can’t say anything about the detailed motion of the atoms but we can consider energy states.

Consider each atom in solid as moving independently of its neighbours. BIG assumption but allowed because we’re going to determine probabilities for various distributions of energy amongst the atoms.

The Einstein model of a solid

Energies of the atomic simple harmonic oscillators comprising the solid are quantised.

U(x)

x

0)2

1( nEn

E0

E2

E1

E3

Just as we considered for the Planck model of blackbody radiation (Set 2b of the lecture notes), the energy difference between consecutive energy levels is:

In addition, we’ll again ignore zero point energy.

0

Having set up our model, we can now address the key question.

How is energy distributed amongst the available energy states?

Energy distributions

Consider bringing two identical blocks together.

What is the most probable distribution of energy amongst the two blocks?

Most probable distribution is ‘intuitively’ that where total thermal energy is shared equally between the two blocks.

However, what is the probability that the first block has more energy than the second or, indeed, ends up with all the thermal energy?

Need to consider possible arrangements of energy quanta……

Energy distributions

We have 3N independent simple harmonic oscillators (where N is the total number of atoms in the solid).

Number of ways of distributing quanta of energy amongst these oscillators?

Say we have 3 quanta of energy to distribute amongst 2 oscillators:

The arrangement above is one possible distribution of 3 quanta amongst two oscillators. Sketch the remaining possibilities. How many possibilities in total are there???

Oscillator 1 Oscillator 2

03

00

3 energy quanta distributed between 2 oscillators

Oscillator 1 Oscillator 2

03

00

Oscillator 1 Oscillator 2

02 01

Oscillator 1 Oscillator 2

01 02

ANS: Total of 4 possibilities (including arrangement shown on previous slide)

Counting arrangements

Clearly, we are not going to count by hand every arrangement of energy possible for 3N oscillators in, e.g., a mole of solid (N ~ 6 x 1023).

Need to consider permutations and combinations.

A permutation is an arrangement of a collection of objects where the ordering of the arrangement is important.

A CD reviewer is asked to choose her top 3 CDs from a list of 10 CDs and rank them in order of preference. How many different lists can be formed? (ABC ≠ BAC)

120 90 72

0

None

of the

se

7% 7%

80%

7%

a) 120

b) 90

c) 720

d) None of these

A CD reviewer is asked to choose her top 3 CDs from a list of 10 CDs and rank them in order of preference. How many different lists can be formed? (ABC ≠ BAC)??ANS: There are 10 choices for the 1st CD, 9 for the 2nd,

and 8 for the 3rd. Hence, 720 different lists.

The number of permutations of r objects selected from a set of n distinct objects is denoted by nPr where nPr = n! / (n - r)!

Counting arrangements

A CD club member is asked to pick 3 CDs from a list of 10 CDs. How many different choices are possible?

120 90 72

0

None

of the

se

78%

7%11%

4%

a) 120

b) 90

c) 720

d) None of these

A CD club member is asked to pick 3 CDs from a list of 10 CDs. How many different choices are possible???

Counting arrangements

ANS: We need to divide the previous 720 arrangements by the total number of different possible permutations of 3 choices (e.g. ABC = BAC = CAB). This is 3! permutations. Hence, 120 choices are possible.

The number of combinations of r objects selected from a set of n distinct objects is denoted by nCr where

)!(!

!

rnr

nCrn

Take a collection of 10 pool (billiard) balls, 6 of which are yellow and 4 of which are red. How many different arrangements of the coloured balls are possible (eg RRYYYYYYRR)?

210

720

120

None

of the

se

64%

30%

0%7%

a) 210b) 720c) 120d) None of these

Counting arrangements

Take a collection of 10 pool (billiard) balls, 6 of which are yellow and 4 of which are red. How many different arrangements of the coloured balls are possible (eg RRYYYYYYRR)?

??ANS: = 210 arrangements

!4!6

!10

Counting arrangements

Returning to the distribution of energy quanta amongst a collection of oscillators, we need to establish a formula for the number of possible arrangements.

Consider the case of 3 quanta of energy distributed between 2 oscillators as before. We’ll adopt the same representation as Chabay and Sherwood, p. 348…….

=1 2 1 2

Thus, we have 4 objects arranged in a certain sequence. We need N - 1 vertical bars to separate N oscillators.

=1 2 1 2

Counting arrangements

=1 2 1 2

This problem thus reduces to the pool ball problem except instead of red and yellow pool balls we have to arrange | and objects.

So, total number of arrangements of 3 quanta amongst 2 oscillators =

!1!3

!4Total number

of objects

Number of quanta

Number of boundaries between oscillators (= N-1)

Counting arrangements

Number of ways to arrange q quanta of energy amongst N 1D oscillators:

)!1(!

)!1(

Nq

Nq

How many ways can 4 quanta of energy be arranged amongst four 1D oscillators? ??

How many ways can four quanta of energy be arranged amongst four 1D oscillators?

24 48 35 121

0% 0%0%0%

1. 24

2. 48

3. 35

4. 121

ANS: = 35 ways!3!4

!7

How many ways can 4 quanta of energy be arranged amongst four 1D oscillators? ??

Counting arrangements