Thermal Questions Flashcards

Question

If a system is equally likely to be in any one of its microstates, why arent all macrostates equally probable?

Answer 1

The probability of a particular macrostate being observed is proportional to the number of microstates which correspond to that macrostate. Not all macrostates have the same number of microstates which will correspond to them. Example - There are many ways to get 5 heads and 5 tails when flipping 10 coins in 1 go. There is only 1 way to get all heads or all tails. It is therefore more likely to see 5 heads and 5 tails, even though the probability of a singe flip giving heads or tails is the same.

Answer 2

A SPS is a specification of all the variables ( position and velocities), or states, of one particle, ignoring all others in the system.

Answer 3

If there is some way to tell apart the individual sub systems / particles comprising of the main system. Eg in a coin flip example, if we could label each coin with a colour. This labelling will mean we have a greater number of total microstates. Take the case of the coloured coins : R G B : coins 1 2 and 3. Now before there was only 1 way to have the macrostate HHH, now however, there are 3! ways to have HHH: ``` R G B R B G G R B G B R B G R B R G ``` So each macrostate has 3! more repeats. Or in general, N! hence we divide by this factor when trying to remove the distinguishability assumption from any derivations.

Answer 4

-Conservation of energy -Microstates exist -a closed system in eqm, in a given macrostate, is equally likely to be in any of its accessible microstates (note - for a GIVEN macrostate, so only the microstates corresponding with the observed macrostate are equally likely to be observed)

Answer 5

Statistical weight W = Number of accessible microstates | Probability of observing a macrostate proportional to its statistical weight. / total number of microstates

Answer 6

the system will appear to always choose the macrostate with the greatest statistical weight (ie, with the most accessible microstates) in the thermodynamic limit where the number of particles N --> inf

Answer 7

W1 X W2. For any given microstate of SS1 or SS2, there are all the possible microstates of the other sub system. Eg if SS1 is in a microstate, there are W2 ways the system could be in that microstate , there are W1 of these microstates so summing W2, W1 times gives W2 X W1.

Answer 8

It is additive, it is maximised in a closed system at Eqm.

Answer 9

Look at lecture notes page 11: Consider a closed system and maximise the statistical weight of the system wrt to U/V/N and match the units of the conserved quantity.

Answer 10

As a system approaches equilibrium total entropy change > 0 for spontaneous irreversible process. Express ds1 + ds2 > 0 in terms of partial derivatives and find expression in form (x_1-x_2)d(M_1)>0

Answer 11

IDK yet proof

Answer 12

Most substances expand when they melt producing a line with a positive gradient. Water shrinks when it melts causing the coexistence line to have a negative gradient.

Answer 13

First Order Involves latent heat due to the discontinuity in entropy, discontinuity also shown in volume. Both of which are first-order differentials of G. Second Order Has no latent heat as entropy doesn't show discontinuity, quantities like heat capacity and compressibility do

Answer 14

``` U = U(S,V,N) U H = H(S,P,N) H = U + pV F = F(T,V,N) F = U -TS G = G(T,P,N) G = H -TS ```

Answer 15

Energy is equally partitioned between all separate modes (degrees of freedom, n), each mode having 1/2 kT of energy, therefore mean energy of the system is n/2 kT Only true if 0.5kT is large in comparison to the splitting

Answer 16

Microconical ensemble: Constant U &N Conical ensemble: Constant T & N Grandconical ensemble: constant T and chem potential

Answer 17

Completely Degenerate Fermion Gas T--> 0 occupancy = 1 up to fermi energy level (e = mu) Degenerate Fermion Gas T<>tf All occupancies are small(less that 1/2) so mu becomes -ve. occupancy turns to m-b dist

Answer 18

Natural variables of F is N V T, these are the variables that we fixed in the canonical ensemble.

Answer 19

No SPS state with 2 or more particles

Answer 20

The number of particles in the specified SPS

Answer 21

Hot and dilute and weakly interacting = occupancy of SPS <<1

Answer 22

U = kT^2 partial ln Z / partial T, p = kt partial ln Z / partial V. mew = - KT partial ln Z / partial N, S = klnZ + U/T.

Answer 23

1/2 k T >> splitting between quantum energy levels

Answer 24

cold and dense, few SPS states relative to the total population, hence assumption of average occupancy being less than 1 not true.

Answer 25

Recall e^E-mew/kT >> 1 | E roughly equals 0 so e^-mew/KT>>1, this expression was in the notes.

Answer 26

Fermi energy is the maximum energy particles have at T=0 K or the chemical potential at T=0, beyond which n_fd ( E ) = 0. Fermi energy = k * fermi temperature

Answer 27

In a closed system with an upper bound on the possible SPS available to be occupied. Possible if you define 1/kT = dS/dU . At large internal energies, entropy actually decreases as you add more energy since the number of SPS get saturated, the system becomes more and more ordered (less and less ways to have the high energy states) and so dS/dU is negative.

Answer 28

Fermi energy/ boltzman constant

Answer 29

-μ/T=(∂S/∂N)=0 | Equals zero as at equilibrium entropy is maximised, therefore it is equal to zero.

Answer 30

For spontaneous processes dS>0 (irreversible) This results in all natural processes are trying to maximise their entropy In thermodynamic eqm entropy is maximised Once the universe reaches thermodynamic eqm, no natural processes can occur If no more natural processes occur stars etc will not burn, causing the heat death of the universe

Answer 31

1) ∂T/∂V | 2) ∂T/∂P

Answer 32

1) System is closed, there is no exchange of matter otherwise an additional term is required 2) Only considers work done by volume change