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Physics year 2 sem 1 revision > Thermal Questions > Flashcards

Thermal Questions Flashcards

1
Q

Definition: What is a thermally isolated system

A

A system which cannot exchange heat with its surroundings

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2
Q

Defintion : Firstlaw of Thermodynamics

A

Energy is conserved, neat and work are both forms of energy

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3
Q

Why are Heat and work not FoS

A

They concern the manner in which energy is delivered to ( or extracted from) the system

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4
Q

Definition: Quasistatic Process

A

A process that evolves so slowly (in infinitesimally small steps) that every point can be viewed as equilibrium during the process.

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5
Q

The relationship between a Quasistatic Process and Reversible Process

A

Reversible processes are quasistatic process in which entropy doesn’t increase

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6
Q

Why does pressure decrease in an isothermic expansion

A

Volume has increased, this causes the energy density to be reduced. Pressure and energy density are proportional hence pressure decreases (or from ideal gas eqn)

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7
Q

Clausius and Kelvins statements on the Second Law of Thermodynamics

A

Clausius: No cyclic process is possible whose sole result is the transfer of heat from a colder to a hotter body
Kelvin: No cyclic process is possible whose sole result is the complete conversion of heat into work

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8
Q

Definition: Engine

A

A system operating a cyclic process that converts heat into work. (Cyclic so that it can be continuously operated)

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9
Q

Why does heat only enter and leave only during the reversible isotherms stage of the carnot cycle

A

No heat can enter or leave during the adiabatic stage,

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10
Q

Definition: Carnots Theorem

A

Of all the heat engines working between two given temperatures, none is more efficient that a Carnot engine

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11
Q

Why is enthalpy useful

A

Represents the heat absorbed by a system for an isobaric process
Also if dH=0 then both s and p are constant

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12
Q

What is a practical drawback for both U and H

A

One of their natural variables is entropy which is not an easy parameter to vary in a lab

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13
Q

Why is Hemholtz Function useful

A

It is the maximum amount of work you can get out of a system at constant temperature since the system will do work on its surroundings intill its helmholtz function reaches a min

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14
Q

Definition: availability

A

The maximum useful work during a process which brings the system into eqm with a heat reservoir reaching maximum entropy

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15
Q

Explain why in a first order phase transition a “kink” arises in the Gibbs energy as a function of temperature

A

Transitions between states occur in the direction of lower chemical potential.

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16
Q

Difference between the terms adiathermals and adiabatic

A

Adiabatic is a special case of adiathermal processes where the process is reversible.
Adiathermal processes are processes where no heat flows.

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17
Q

A heat pump has efficiency greater than 100% does this violate any laws?

A

No, efficiency = what you want/ what you pay. Heat pump efficiency is coined as co-efficient of performance = Q pumped / W = (Q in + W)/ W= W + Qin/W

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18
Q

Prove that the change in entropy of a spontaneous process is greater than 0

A

test

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19
Q

Explain where the clausius ineq comes from

A

test

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20
Q

Summarise the meanings of thermodynamic system, thermal equilibrium, thermodynamic
equilibrium, equation of state, function of state.

A

Thermodynamic system - The part of the universe you are looking at, separated by the rest of the universe at the boundary of the system. The macroscopic properties of the system are described by state variables

Thermal equilibrium - when 2 bodies are in thermal equilibrium there is no net exchange of heat energy between them.

Thermodynamic equilibrium - When 2 bodes are in thermodynamic equilibrium there is no change in the macroscopic properties of each body , with respect to each other. I.e no change in pressure, chemical equilibrium etc

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21
Q

Define extensive, intensive and specific variables

A

Extensive property is a property which scales with the size of a system. Intensive property is a property which does not. Specific variables are intensive variables obtained by dividing an extensive property of a system by its mass/number of moles.

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22
Q

When and why do we write đ𝑄, đw

A

when the differential is path dependent

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23
Q

Define macrostate and microstate

A

Macrostate- a specification of the measurable properties of the system such as pressure volume temperature .

Microstate - A complete specification of all the variables needed to fully describe the states of all the subsystems microscopic sub systems of the system.

24
Q

Consider a system in equilibrium, why arent all microstates accessible ?

A

For a given system in equilibrium, it will have certain macroscopic properties such as internal energy, pressure, temperature etc. Not all microstates correspond with macrostates that will have the appropriate values. Some microstates will correspond to a system with a greater internal energy etc. So these microstates are forbidden and hence in accessible.

25
Q

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

A

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.

26
Q

What is a SPS

A

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

27
Q

What does distinguishability mean in thermodynamics?

What consequence does distinguishability have on the number of total microstates?

A

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.

28
Q

State the postulates of statistical mechanics

A

-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)

29
Q

What is the concept of statistcal weight? How is this related with probability?

A

Statistical weight W = Number of accessible microstates

Probability of observing a macrostate proportional to its statistical weight. / total number of microstates

30
Q

Explain the thermodynamic limit

A

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

31
Q

If 2 subsystems comprise the total system and SS1 has w1 accessible microstates and SS2 has w2 accessible microstates. What expression gives the total statistical weight of the system and why?

A

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.

32
Q

Explain why the quantity lnW behaves like entropy?

A

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

33
Q

Derive the relations between temperature, pressure and chemical potential with the statistical weight

A

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.

34
Q

Show how a closed system must evolve as it approaches :
Thermal equilibrium
Mechanical equilibrium
Diffusive equilibrium

A

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

35
Q

Why are spontaneous processes irreversible?

A

IDK yet proof

36
Q

How does the liquid-solid coexistence line differ for normal substances and water?

A

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.

37
Q

What is the difference between first-order and second-order phase transition

A

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

38
Q

What are the natural variables and equations for the thermodynamic potentials

A 
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
39
Q

What is Equipartition theory

A

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

40
Q

Types of Systems in Stat mech

A

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

41
Q

Types of Fermion Gas

A

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

42
Q

Why did we use the hemholtz free energy differential to find the expressions for p mew and S in terms of the partition function?

A

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

43
Q

What does dividing by N! to account for indistinguishabllity assume?

A

No SPS state with 2 or more particles

44
Q

What is the definition of occupancy

A

The number of particles in the specified SPS

45
Q

Two properties of ideal classical gas and what it means for the average occupancy of an SPS.

A

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

46
Q

Express U, P, S and Mew in terms of the parition function

A

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.

47
Q

In what regime is the equipartition theorem valid?

A

1/2 k T&raquo_space; splitting between quantum energy levels

48
Q

What are the properties of an ideal quantum gas

A

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

49
Q

Show that the bose einstein dis and the fermi dirac dis tend to the boltzman distribution for the hot and dilute case.

A

Recall e^E-mew/kT&raquo_space; 1

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

50
Q

What is the fermi energy and fermi temperature?

A

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

51
Q

When are negative temperatures possible and how are they physical?

A

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.

52
Q

How do you calculate fermi temperature

A

Fermi energy/ boltzman constant

53
Q

why is chemical potential zero for non-conservative particles

A

-μ/T=(∂S/∂N)=0

Equals zero as at equilibrium entropy is maximised, therefore it is equal to zero.

54
Q

Explain the heat death of the universe

A

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

55
Q

1) What is the joule free expansion coeff

2) what is the joule-kelvin coeff

A

1) ∂T/∂V

2) ∂T/∂P

56
Q

What restrictions apply to dU = TdS - pdV

A

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

Physics year 2 sem 1 revision (3 decks)

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