Fundamentals and Entropy

Question 1

Fundamentals

Classical Thermodynamics - First Law

Answer 1

dE = TdS - pdV

-this implies that any process in which energy is conserved is possible

Question 2

Fundamentals

Classical Thermodynamics - Second Law

Answer 2

dS = dQrev / T
rev = reversible
-the entropy of an isolated system can only remain constant or increase
-this implies that heat cannot flow from a cold to a hot reservoir without other changes

Question 3

Fundamentals

Direction of Time

Answer 3

• hot and cold cannot spontaneously separate

- entropy increases with time for an isolated system, or for the universe as a whole

Question 4

Fundamentals

Statistical Physics

Answer 4

• the macroscopic properties of a system are deduced from the statistical behaviour of the constituent particles
• to do this we need to consider large systems, but 1mm³ of air contains ~10^16 molecules
• all states with equal energy are equally probable
• increasing the energy of a state decreases its probability

Question 5

Microstate

Definition

Answer 5

-specified by describing the (quantum) state of each particle in the system

Question 6

Macrostate

Definition

Answer 6

• properties of a system as a whole i.e. T, p, V, E< G etc.
• no information on individual particles
• every macrostate has an enumerable number of microstates

Question 7

Postulate of equal a priori probabilities

Answer 7

-for a given macrostate, all microstates are equally probable

Question 8

The Ergodic Hypothesis

Answer 8

• the time average of a system is equal to the instantaneous ensemble average
• i.e. the average of many identical measurements made on a single system is equal to the average of a single measurement on many copies of the system

Question 9

Multiplicity

Definition

Answer 9

• the number of microstates in a macrostate

- for each macrostate there is a large number of macroscopically indistinguishable microstates

Question 10

What does the probability of a macrostate depend on?

Answer 10

• the probability Pi of a certain macrostate depends on how many microstates correspond to the macrostate
• the probability of each microstate is proportional to its multiplicity

Question 11

Probability of a Macrostate

Equation

Answer 11

Pi = ωi / Σωi

-where the sum is over all microstates in the system

Question 12

Addition Rule

Answer 12

-for mutually exclusive events

P(i or j) = P(i) + P(j)

Question 13

Multiplication Rule

Answer 13

-for independent events

P(i and j) = P(i) x P(j)

Question 14

Distinguishing Between Different Macrostates

Answer 14

-different macrostates can be distinguished by e.g. weight

Question 15

Distinguishing Between Different Microstates

Answer 15

-different microstates that correspond to the same macrostate cannot be distinguished between at the macroscopic level e.g. by weight

Question 16

Boltzmann’s Expression for Entropy

Equation

Answer 16

S = k ln Ω

S = entropy
Ω = multiplicity

Question 17

Boltzmann’s Expression for Entropy

Calculating Entropy

Answer 17

• you need to know Ω to calculate entropy
• probability is very sharply peaked around the most probable state
• the greater the number of states, N, the sharper the peak
• the distribution tends to a Gaussian as N->∞

Question 18

Boltzmann’s Expression for Entropy

Fluctuations

Answer 18

fractional magnitude of fluctuations ∝ 1/√N

• as N is increased, the fluctuations become proportionally smaller and smaller
• for macroscopic systems (in equilibrium) fluctuations are so small they are unimportant

Question 19

Stirling’s Approximation

Answer 19

-for a macroscopic system of N≈10^20, the multiplicity N! is massive
-to deal with numbers like this, take their natural logarithm:
ln(N!) = ln1 + ln2 + .. + lnN
≈ 1,N ∫ln(x) dx
= [ xlnx - x ] N,1
≈ N lnN - N

Question 20

Total Multiplicity vs Maximum Multiplicity

Answer 20

• we can replace total multiplicity Ω=2^N by maximum multiplicity Pmax = N! / (N/2)!(N/2)!
• this makes a negligible difference, especially with the natural logs

Question 21

Origin of Irreversibility

Classical Thermodynamics

Answer 21

• consider a container with a gas partitioned in one half, when the partition is removed the gas will move to fill the container
• so V1 -> V2 where V2=2*V1
• > ΔS = nR ln(V2/V1) = Nkln2

Question 22

Origin of Irreversibility

Statistically

Answer 22

-consider a container with a gas partitioned in one half, when the partition is removed the gas will move to fill the container
-there are N particles contained in half of the containers volume
-when the partition is removed, the number of states available to each particle effectively doubles
-so the multiplicity increases 2^N times
-> ΔS = Nk ln(2Ω) - Nk lnΩ
= Nkln2

Question 23

Origin of Irreversibility

Probability that all atoms end up on one side

Answer 23

P = (1/2)^N = 1/(2^N)
-let N = 3.3x10^20, the number of molecules in a cubic centimetre of air
log(2^N) = N log22 = 0.30N
P = 1/10^(0.3N) = 10^(-10^20)
-the probability that all particles are on the left side of the container is effectively zero

Question 24

Nernst’s Therorem

Answer 24

-any entropy changes in an isothermal reversible process approach zero as the temperature approaches zero
T->0 lim ΔSrev = 0

Question 25

Third Law

Answer 25

• it is impossible by any procedure to reduce any system to the zero of temperature of a finite number of steps
• it is impossible by any procedure to reduce the entropy of a system to its zero-point value in a finite number of steps

Question 26

Typical Way to Reduce Temperature

Answer 26

-to alternate between isothermal and adiabatic expansions

Question 27

Entropy at Absolute Zero

Answer 27

• a system at T=0 is in its ground state so that its entropy is determined only by the degeneracy pf the ground state
• only systems such as elementary (perfect) crystals have S(0) = 0 (Ω=0)
• S(0) is not usually zero, there is usually some residual entropy

Question 28

Residual Entropy

Answer 28

• may come from a multiplicity of possible molecular orientations in the crystal (degeneracy of the ground state) even at absolute zero
• the third law applies to equilibrium states only, glasses are not in equilibrium as the relaxation time is huge, glasses have particularly large entropy at T=0

Question 29

Clausius-Clapeyron Equation

Answer 29

∂p/∂T = ΔS/ΔV

Question 31

Consequences of the Third Law

Answer 31

-clausius-clapeyron equation: ∂p/∂T = ΔS/ΔV
-so ΔS->0 implies:
∂p/∂T->0 as T->0
-on a P-T diagram the line should asymptotically approach the x-axis at T=0

Question 32

Heat Capacity at Absolute Zero

Answer 32

0,T ∫ 𝛿Q/T’ = St - So
= 0,T ∫ Cp(T’)/T’ dT’
=> Cp(0) = 0
-to avoid divergence at T=0

Question 32

Volume Coefficient of Thermal Expansion at Absolute Zero

Answer 32

α = 1/V [∂V/∂T] |p
-using Maxwell's Relation
α = - 1/V [∂S/∂p] |p
-since [∂S/∂p] |p -> 0,
α->0 as T->0

Question 33

What is the equation that will give you the number of ways of picking n objects from N objects?

Answer 33

= N! / n!(N-n)!