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Physics 2 > Thermal Physics > Flashcards

Thermal Physics Flashcards

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1
Q

system

A

a portion of the universe with certain measurable quantities such as pressure or volume which determine the equilibrium state of the system

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2
Q

surroundings

A

the rest of the universe outwith the system

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3
Q

system + surroundings =

A

universe

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4
Q

isolated system

A

a system that does not interact with its surroundings by exchanging heat energy, mechanical energy or material

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5
Q

closed system

A

energy but not material can be exchanged

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6
Q

adiabatic wall

A

system is thermally isolated and only mechanical energy (not heat) can be exchanged with the surroundings (eg vacuum flask)

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7
Q

diathermal wall

A

heat exchange is permitted, systems connected by a diathermal wall are in thermal contact (eg a metal wall)

think “dia” prefix meaning interaction, eg DIAlogue

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8
Q

equilibrium

A

all bulk physical properties are uniform throughout the system, they are time independent

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9
Q

macroscopic

A

large scale or bulk properties of a gas eg pressure, volume, temperature

one number exists

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10
Q

microscopic

A

properties on the atomic level, eg Vrms or Vmp

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11
Q

state variables

A

define the state of a system

intensive: independent of size of system (eg pressure, tension)

extensive: proportional to size of the system (eg volume or length)

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12
Q

conjugate variables

A

think of conjugate verbs as pairing up

equilibrium states of thermodynamic systems are determined by suitable pairs of conjugate variables

one variable intensive, other extensive

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13
Q

examples of conjugate variables

A

pressure and volume for a gas

tension and length for a stretched wire

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14
Q

product of all conjugate pairs…

A

has dimensions of energy

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15
Q

function of state

A

any quantity which takes unique value for each equilibrium state of a system

eg internal energy U or entropy S

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16
Q

functions of state depend only on

A

the state itself and not on how that state was produced

eg pressure, volume, temperature

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17
Q

why is heat not a function of state

A

because it is associated with a transfer of energy between states

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18
Q

if two systems are put in thermal contact…

A

generally changes occur in both

eg coffee cools down as room heats up slightly

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19
Q

two systems in thermal contact after some time

A

no further changes occur and the two systems are said to be in thermal equilibrium

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20
Q

heat

A

form of energy transferred between substances due to temperature differences between them

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21
Q

if two objects are not in thermal equilibrium, heat flows…

A

from hotter object to the colder object until thermal equilibrium is reached

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22
Q

temperature

A

property which determines whether or not a system is in thermal equilibrium with other systems

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23
Q

zeroth law of thermodynamics

A

if two systems are separately in thermal equilibrium with a third system then they must also be in thermal equilibrium with each other

three systems said to have same temperature T

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24
Q

conditions for thermal equilibrium

A
  1. thermal equilibrium - same temp
  2. mechanical equilibrium - no unbalanced forces acting
  3. chemical equilibrium - no chemical reactions occurring
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25
Q

non-uniformities in a system result in

A

a gradient and hence, momentum, heat and/or matter flows until thermodynamic equilibrium is reached

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26
Q

consequences of thermodynamic equilibrium on a macroscopic level

A

variables have constant values in time and space, ie throughout the system

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27
Q

consequences of thermodynamic equilibrium on a microscopic level

A

any process ( diffusion, collisions etc) must have an equal probability of going in the opposite direction

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28
Q

equation of state

A

f(P,V,T)=0

system described by P and V (conjugate pair). Each state has a definite temp so must be fucntion linking all 3

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29
Q

why does RHS of equation of state equal 0

A

equilibrium holds

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30
Q

how is equation of state determind?

A

determined by experimental observation or a microscopic model of the system

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31
Q

ideal gas

A

hydrostatic equilibrium holds

inward force of gravity balances outward gas pressure

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32
Q

what is assumed in ideal gas

A

atoms or molecules are non-interacting and point like

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33
Q

equation of state with universal gas constant

A

f(P,V,T)= PV-nRT=0

n is number of moles

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34
Q

equation of state with Boltzmann’s constant

A

PV-nKBT=0

n is number of molecules

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35
Q

non-ideal gas

A

has molecules with finite volume so intermolecular forces must be considered

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36
Q

equation of state for a van der Waals gas

A

adaptation of ideal gas law but includes a and b as constants specific to the gas

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37
Q

how to reduce van der waals equation back to ideal gas

A

low pressures and high volumes

n^2 a / v^2 terms approximates 0

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38
Q

indicator diagram

A

graph of an intensive variable against its conjugate extensive variable

any particular state may be represented by a single point on the diagram

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39
Q

isotherm

A

continuous line between two states on indicator diagram

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40
Q

what does isotherm being continuous allow

A

integration/ differentiation

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41
Q

isotherms for a van der waals gas

A

isotherms look same for both gas types at high temperatures

behaviour changes as temp drops

below critical temp a phase change occurs

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42
Q

when do graphs of isotherms for van der waals and ideal look approximately the same?

A

high volumes and low pressures

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43
Q

we can assign numerical values to temperatures by

A

selecting a suitable physical property of a system that varies linearly with it

X=cTx

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44
Q

explain variables in X=cTx

A

Tx is temp on X scale and c is a constant got by choosing T at a convenient fixed point

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45
Q

example of a good fixed point

A

triple point of water

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46
Q

triple point of water

A

ice, liquid water and vapour all coexist

(think going up mountain, boiling point decreases due to pressure change)

eg water boils at approx 65 celcius on everest

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47
Q

how to build thermometer

A

for small enough temperature ranges there will be some physical properties which exhibit a linear change with temp

measuring one of these allows thermometer to be built

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48
Q

gas thermometer

A

use fixed quantity of gas held within container at fixed volume

can model using IGE as long as pressure is relatively low

inconvenient and time consuming but are very accurate

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49
Q

use of gas thermometer

A

establish standard temperatures and calibrate other types of thermometers

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50
Q

liquid volume thermometer

A

liquid expands as a function of temp

small volume change leads to large height change in tube

only agree at fixed points

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51
Q

resistance thermometers

A

based on variation of electrical resistance with temp

relationship between temp and r is not straightforward and can be expensive

easy to transport

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52
Q

thermocouples

A

potential difference from a junction of two different metals

cheap to make but require careful calibration and not really linear

narrow temperature ranges

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53
Q

radiation pyrometers

A

measure black body radiation and can do at distance

H=stefan constant e AT^4

ideal for high temps

non-linear relationship with temp and require careful calibration

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54
Q

kinetic theory of gases assumptions

A

gas is ideal
collisions between molecules and walls of container are elastic
container walls are rigid and do not move when struck

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55
Q

major result from kinetic theory derivation

A

PV=1/3nN_AmVrms^2

relates bulk properties of pressure and volume to microscopic properties of gas

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56
Q

what is R/N_A

A

K_B

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57
Q

how to obtain kinetic energy equation from kinetic theory of gases derivation

A

from R/N_A=K_B

k_BT=1/3mvrms^2

times by 3 and divide by 2

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58
Q

what conclusions/results come from kinetic theory of gases derivation?

A

temperature is proportional to average translational kinetic energy

there is 1/2K_BT per degree of freedom

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59
Q

piston - work is done BY the gas if

A

it expands

i.e. energy taken out

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60
Q

piston - work done ON the gas if

A

it is compressed

i.e. energy must be added in

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61
Q

when a piston moves a small distance dx, work done is

A

dW=Fdx=PAdx=PdV

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62
Q

for finite change in x, work done is

A

integral (limits v1 and v2) of Pdv

63
Q

principle of conservation of energy

A

energy can only be changes, cannot be created/destroyed

total amount of energy in a closed system is constant with time

64
Q

conservation of energy in an ideal gas

A

no potential energy contribution, internal energy U is the sum of all energy contributions

65
Q

when a process is performed on a system…

A

there will be a change in internal energy as it moves from one equilibrium state to another

66
Q

what does thermally isolated mean

A

no heat flow (adiabatic)

67
Q

internal energy in an adiabatic process

A

deltaU=W

68
Q

when a system in not adiabatic, what must be considered

A

heat Q

think of as difference between work done in adiabatic and work done when heat allowed to flow

Q=W-Wadiabatic (can be + or -)

69
Q

piston but now walls of container are thermally conducting

A

gas expands, heat energy flows through walls into gas

work done by gas as it expands, decreases U

heat flowing dQ counteracts this change

so dQ=dU+dW

70
Q

first law of thermodynamics

A

delta U=Q-W

71
Q

sign conventions for delta U

A

positive higher energy

negative lower energy

72
Q

sign conventions for Q

A

positive heat flowing in

negative heat flowing out

73
Q

sign conventions for W

A

positive for expansion

negative for compression

74
Q

on a pv diagram, work done is

A

area under curve

because dW=PdV

75
Q

if the path changes…

A

so does area under it

deltaU must be same for each path because end points identical

deltaUA=deltaUB
QA-WA=QB-WB
QA-QB=WA-WB

75
Q

first law of thermodynamics

A

there is a finite amount of energy associated with a change of state

76
Q

monatomic

A

not bound to each other

3 dof

(translational movement)

76
Q

equipartition theorem

A

classical system in thermal equilibrium at tempT has mean energy of 1/2KBT per dof

76
Q

diatomic

A

consist of pairs of atoms

3 translational dof
2 rotational dof
2 vibrational dof (high temps only)

so depending on temp, 5 or 7 dof

**above 1000K considered high

77
Q

DOF, U per molecule and U per mole for monatomic

A

3, 3/2KBT, 3/2RT

78
Q

DOF, U per molecule and U per mole for diatomic (low T)

A

5, 5/2KBT, 5/2RT

79
Q

DOF, U per molecule and U per mole for diatomic (high T)

A

7, 7/2KBT, 7/2RT

80
Q

molar heat capacity

A

amount of heat dQ required to raise temp of one mole of gas by 1 kelvin

C=1/n dQ/dT

81
Q

two different molar heat capacities

A

CV change of temp at constant volume

CP change of temp at constant pressure

82
Q

ratio of specific heats

A

gamma = CP/CV

83
Q

isovolumetric

A

constant volume

84
Q

isothermal

A

constant temperature

(therefore can draw isotherms on PV diagram)

85
Q

isovolumetric process on pv diagram

A

jumps from one isotherm to another (vertically)

86
Q

work done in isovolumetric process

A

since work is intergral of PdV between v1 and v2 and volume is constant, no work is done

so Q=deltaU

87
Q

isobaric process

A

constant pressure

88
Q

isobaric process on pv diagram

A

jumps from one isotherm to another (horizontally)

89
Q

work done in isobaric

A

W=P deltaV (result of integration and taking P out as a constant)

90
Q

isothermal process on pv diagram

A

single isotherm

91
Q

pressure and volume in isothermal process

A

both change

92
Q

work done in isothermal

A

usual integration but sub in ideal gas law for P

gives W=nRTln(v2/v1)

93
Q

first law for isothermal

A

no change in temp so no change in internal energy

Q=W

94
Q

assumptions which allow us to say Q=0 in adiabatic process

A

system is thermally isolated from surroundings

system undergoes rapid change for which heat transfer is negligible

95
Q

consequence of PV^gamma = constant

A

on a pv diagram, adiabatic curves are steeper than isothermals

96
Q

adiabatic expansion work done

A

results in a cooling and decrease of internal energy

97
Q

adiabatic compression work done

A

work done on the gas, it heats and there is an increase in internal energy

98
Q

total pressure depends on

A

average velocity of the molecules

99
Q

total pressure exerted by a mixture of gases is

A

equal to the sum of the pressures exerted by those gases separately

each gas contributes a partial pressure

100
Q

maxwell boltzmann distribution

A

velocity distribution in a gas

plot of v against f(v)

will be a range of velocities from zero to very high

101
Q

features to note in maxwell boltzmann distribution

A

f(v) tends to zero at both velocity extremes

spread is normalised (integral over full range of v is 1)

area under curve is independent of temperature

asymmetric in shape

102
Q

most probable speed on plot of v against f(v)

A

value of v at the turning point

i.e when d/dv(f(v))=0

103
Q

peak of curve on MB distribution

A

shifts right with increasing temp

104
Q

why is shape of MB asymmetric

A

have a range of possible speeds between zero and infinity

higher velocities have more weight in the distribution

105
Q

relative location of Vrms on MB distribution

A

to the right of mean molecular speed because of higher weighting given to higher velocities

106
Q

you can change a system’s state by

A

applying heat or doing work

most bulk macroscopic processes that cause these changes are irreversible

107
Q

irreversible processes are those in which

A

energy is dissipated

108
Q

entropy S

A

macroscopic variable that quantifies the degree of disorder in a system

109
Q

entropy for processes in adiabatically isolated systems

A

can only remain constant or increase

110
Q

change in entropy =0

A

reversible process

111
Q

change in entropy >0

A

irreversible process

112
Q

units of entropy

A

joules per kelvin

113
Q

small volumetric increases will result in

A

a greater number of spatial coordinates for the gas molecules, giving rise to a greater level of disorder

114
Q

first law written for a reversible process

A

dU=Tds-Pdv

hence we only need to know initial and final states and not intermediate points

115
Q

lossy process

A

there is a loss

if you need to do more work than the bare minimum, you have wasted effort somewhere

116
Q

how to calculate change in entropy for irreversible change

A

calculate it for an equivalent reversible change as this gives minimum possible value

117
Q

change in entropy for isobaric

A

derived from integral of dQ/T and sub dQ=nCpdT

118
Q

change in entropy for isovolumetric

A

integral dQ=T

sub in dQ=nCvdT

119
Q

change in entropy for isothermal

A

again integral dQ/T

dQ=dW=Pdv

and PV=nRT

120
Q

process equivalence

A

isothermal process is equivalent to an isobaric process followed by an isovolumetric process

121
Q

pv diagrams and reversibility

A

can only plot PV diagrams for reversible processes

not possible for irreversible as do not know what is happening between states.

122
Q

free expansion of an ideal gas

A

irreversible so cannot be plotted on PV diagram

123
Q

change in entropy for for free expansion of an ideal gas

A

no work is done by the gas so no temp change should occur

124
Q

free expansion of a real gas

A

temp will decrease because internal energy depends on volume

125
Q

calculating entropy for irreversible process

A

does not depend on path between initial and final states

consider an equivalent reversible path

choose equivalent isothermal expansion so no change in temp during ideal free expansion.

126
Q

second law of thermodynamics

A

reversible: delta S=0
irreversible delta S >0
in general delta S>/=0

127
Q

natural process during a change in entropy

A
  1. transfer of heat from hotter to a cooler body
  2. overall increase in entropy
128
Q

to reach equilibrium, a system will…

A

maximise entropy and do so through irreversible microscopic processes

129
Q

two forms of the second law tell us that

A

no heat engine can be 100% efficient and that work is required in order for heat to be moved.

130
Q

pictorial representation of a heat engine

A

heat QH is extracted from a heat reservoir at temperature TH

work W is performed

waste heat QC is dumped to the environment or cold reservoir which is at temp TC

131
Q

why can’t you say TC>TH

A

heat QC will leak back from the environment into the heat resevoir

the net effect would be 100% conversion of heat QH-QC into work

cannot have a device where all heat energy is turned into work

132
Q

why you can’t take heat out of the cold resevoir and dump is elsewhere

A

this requires work to be applied to the engine which we’re also attempting to take from the heat resevoir

you cannot have a device that just moves heat from a cold body to a hot body without the need for work to be done

133
Q

plotting a heat engine’s process on a PV diagram

A

would get a loop

134
Q

efficiency of a heat engine

A

n=work done / heat input = W/QH

from conservation of energy QH=W+QC

so n=1- QC/QH

135
Q

refrigerator

A

heat engine running backwards

heat is extracted from a cold resevoir by applying work

extracted heat is then dumped into the hot resevoir

136
Q

coefficient of performance of heat pump/refrigerator

A

equivalent of a heat engine’s efficient

k=heat extracted/work done = QC/W

from conservation of energy QH=W+QC

so k=QC/ QH-QC

137
Q

efficiency of a heat pump

A

defined in terms of how much heat is produced

nHP=heat delivered/work done = QH/W

=QH/ QH-QC

138
Q

nHP-k

A

=1

139
Q

heat pumps and refrigerators work really well when

A

only a small temperature change is required

140
Q

heat pumps on their own are very effective at

A

heating buildings to a base level

eg frost protection

141
Q

why cyrogenics is very difficult

A

for refrigeration, as the ratio of hot to cooled temperature increases the amount of work required to achieve the cooling becomes very large

142
Q

carnot cycle

A

operating cycle of the most efficient heat engine possible

all transfer of heat should be between bodies of nearly equal temperature

143
Q

carnot cycle assumes

A

an engine connected by a frictionless mechanism to an external load upon which work may be done

144
Q

four processes of the carnot cycle

A

isothermal expansion
adiabatic expansion
isothermal compression
adiabatic compression

145
Q

work done in a carnot cycle

A

deltaU=Q-W=0 because reversible cycle so no change in internal energy

so W=Q

since Q=QH-QC

work done=QH-QC

146
Q

carnot’s theorem

A

no engine operating between two temperature reservoirs can be more efficient than a carnot engine operating between the same two temperatures

147
Q

proof of carnot’s theorem

A

by contradiction

assume that nengine>ncarnot and fail

use composite engine wherein engine E drives carnot refrigerator C

end up with W disappearing, cannot have this

148
Q

efficiency of a carnot engine

A

depends only on temperature

ncarnot=1-TC/TH

149
Q

kelvin defined a temperature scale based on

A

how heat flows in the ideal carnot cycle

this allowed the idea of absolute zero

150
Q
A

Physics 2 (8 decks)

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