Statistical Physics Flashcards
What is a macrostate?
A thermodynamic state of the system described by macroscopic functions of state such as Temp, Pressure, Volume.
What is a microstate?
A specific quantum state of the system(not of a single particle).
Microstates specify all relevent information of all particles, spim, position, energy, velocity etc.
What does PEAP or PEEP (Peep probably better to use) mean?
PEAP- Principal of Equal A Priori Probability
PEEP- Principal of equal equilibrium probability.
What does PEAP/PEEP state?
PEEP states that when an isolated system reaches equilibrium, all microstates accessible to it are equally probable.
What does the Boltzmann equation show?
Shows the relationship between entropy and the number of ways the atoms or molecules can be arranged in a thermodynamic system.
State the Boltzmann equation and its meaning
Image shown on this side is another imortant variation of equation/shows how the W component is used rather than probability, p.
The boltzmann equation is a probability equation relating entropy S to W, the number of real microstates corresponding to the Gas’s macrostate.
IF DEALING WITH PROBABILITIES USE
S = Kb * ln (P)i
If the number of particles, N, and the number of macrostates, M, is known, how do you find the TOTAL number of microstates?
Example problem
The number of ways of placing one particular volume is V/dV (total volume/small divisions of volume finds how many small volumes can be occupied). Hence W = V/dV (for one particle)
Since Wt = W1W2W3… WN (each molecule is independent of the rest) Wt = W^N
Derivation of statistical temperature
Joule expansion entropy change calculation
Useful info on entropy and probability:
(context)
- Suppose a number of macrostates share the same energy and are therefore mutually accessible.
- Let W1, W2, … represent the number of microstates available for each of those macrostates.
- According to PEAPP, the system has a prob, Pi to be in the i-th microstate.
What is an ensemble in statistics?
An ensemble is a collection of identical systems whose statistical fluctuations are cancelled out upon averaging.
What is a micro-canonical ensemble?
What is a canonical ensemble?
- Micro-canonical: Systems with fixed total energy and number of particles N and Volume.
- Canonical ensemble: Systems with a fixed number of particles N and volume, but in contact with a thermal heat bath of temp T.
Derivation of Boltzmann distribution equation
Systems are fundamentally quantum mechanical, so if you solve the Schroedinger eq you can work out the energy levels of the system. If the system has a certain amount of energy, then the system is in a specific state (n of states = 1), hence Ws = 1 as we have specified exact information of the state.
- Ei can not change as the system is (for now) isolated, so can not lose or gain energy to change energy levels, again meaning n = 1
-Next bring into contact with heat bath with Temp, T and energy Eb, with its own number of microstates Wb. The energy of the system Ei can now change as it can exchange energy with heat bath.
- Energy conservation E = Ei + Eb
(now a microcanonical ensemble for system+heat bath)
Probability of system being in any state is (Where W it the total number of microstates, given by multiplying the W of each component of a system together)
pi ~ W = Ws*Wb = Wb
pi ~ Wb –> pi ~ Wb(E-Ei)
- pi ~ Wb and E = Ei + Eb
What is Wb?
-> Use statistical temperature definition
Rearrange, Integrate both sides of the equation with respect to dE b (T of heat bath is const). The constant of integration becomes a constant of proporionality upon exponentiating to rearange for Wb.
Since pi ~ Wb and Eb = E - Ei
pi ~ exp -Ei/KbT
(ignoring the exp constant(exp(E/HbT)–> becomes another constant of proportionality)
State the Boltzmann distribution
Differences between PEAPP and Boltzmann
PEAPP:
-Equal distribution
-Applies to Micro-canonical ensemble
-PEAPP requires everything to be considered(what a micro canonical ensenmble is), all microstates, heat bath included.
Boltzmann:
-Exponential distribution
- Applies to Canonical Ensemble- with Boltzmann we talk about the system ONLY and not the heat bath(the derivation using the heat bath does not really mean anything)
When would you use Gibbs entropy over Boltzman entropy?
Use boltzmann if Number of microstates W is known i.e. for joule expansion.
However, if we have a canonical ensemble where we can’t count W because we only have probabilities given by the Boltzmann distribution and we don’t want to worry about the heat bath we use Gibbs entropy- which is defined through porobability. Gibbs energy can ALSO be applied to micro canonical unlike Boltzmann which is just canonical.
State Gibbs Entropy equation
Don’t forget the minus!!
Derive Boltzmann entropy from the Gibbs entropy eq in a micro canonical ensemble
Hint: Remember PEAPP - All probabilities are equally likely
Gibbs entropy in Micro canonical ensemble
-In micro-canonical all W microstates are equally likely, therefore pi = 1/W (W is tot N of microstates)
-Sub into gibbs eq
- Sum of 1/W = sum of p= Ptot = 1
– > Produces Boltzmann Entropy
–> Shows Gibbs entropy is more general than Boltzmann
Shannon’s theorem
When maximised it gives he most likely distribution of pi
If there are N harmonic oscillators, each with quantum levels E = nhw, with M quanta of energy available, what is the total number of microstates possible?
What is the Partition function? What is the formula?
The partition function is the normalisation of the Boltzmann distribution.
Derive the Helholtz free energy using gibbs entropy equation.
Hint: Don’t forget (-)
Don’t forget Z
Essentially plug in boltzmann probability into gibbs and re-arrange
Derive the Helmholtz free energy and entropy of a 2 level system.
Recall eq for Z, E0 = 0, E1 = E, hence Z = e^0 + e^-E/KbT
F = -KbTln[z]
S = -dF/dT
What is the partition function for a canonical/two level system?
Hence what is p(E) and p(0)?
What is the equation for statistical temperature?
Using the equations provided for a 1D particle in a box, state the partition function.
Using the gaussian integral provided, state the Helmholtz free erergy of a particle in a 1D box.
Hint: Use partition function(found in card 26)
Given the Helholtz free energy of a particle in a 1D box, find its entropy, heat capacity and internal energy. dF = - pdV - SdT and F = E - TS will be useful.
# may not need heat capacity/can ignore.
E = F + TS
C = dQ/dT = T dS/dT #may not need C, pg.35 of STP notes.
Partition function of particle in 3D box, simply Z(1D) ^3.
What is the Helmholtz free energy for a particle in a 3D box?
F3D = -k ln (Z3D) = =3kTln(Z1D)
Given the Helmholtz free energy of a particle in a 3D box, determine the entropy of the particle, the internal energy and derive the ideal gas law for 1 particle.
Remember Nk = nR
Equipartition energy derivation:
Using the equations, find the equipartition energy for a particle in a 3D box. Hint: Z is the normalisation for the expectation energy.
Energy, is a random form of quadratic energy E = aq^2. This energy could represent KE, Spring PE etc.
What does the equipartition theorem state (card 31)?
All quadratic modes carry an energy of kT/2 at equilibrium.
Derive the ideal gas law from basic kinetic theory and equipartition energy.
Hints:
1. Find the frequency of collisions of the wall (V/2L)
- # n of collisions per second = (v/2L)
- The change in momentum upon collision = 2mv.
- The rate of change of momentum, F = dP/dT
= momentum change per collision*n of collisions = mv^2/L - Pressure = F/A = mv^2/L^3 = mv^2/V
- Using equipartition theorem, the Kinetic energy averages to kT/2, KEavg = <E> = kT/2. Hence, mv^2 = kT.</E>
- p = kT/V
When does equipartition theorem fail?
Equipartition theorem fails for large adiabatic constants, since the distribution/energy levels are more spread apart, hence more discrete rather than continuous.
What 3 energies contribute to the energy of a diatomic molecule?
What is the partition function of a diatomic molecule?
Given information on the rotational energy of a diatomic molecule, find the free energy, entropy and heat capacity due to rotational motion for a diatomic molecule.
Hint: You must do this for T«TR and T»TR individually, one requires integration, one does not.
Given information on the vibrational energy of a diatomic molecule, find the free energy, entropy and heat capacity due to vibrational motion for a diatomic molecule.
The geometric series:
1/(1-e^ax) = 1 + e^-ax + e^-2ax… may be used.
Hint: You must do this for T«TR and T»TR individually. Unlike with rotational energy neither require integration.
What is the heat capacity for a diatomic molecule for low and high temperatures. Consider contributions from translational, rotational and vibrational.
Useful info: Translational contribution is Cv = 3/2*kb for low and high temps.
