Brainscape's Knowledge GenomeTM

Browse over 1 million classes created by top students, professors, publishers, and experts.


See full index

Physics 2 > Thermal Physics > Flashcards

Thermal Physics Flashcards

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

surroundings

A

the rest of the universe outwith the system

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

system + surroundings =

A

universe

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

isolated system

A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

closed system

A

energy but not material can be exchanged

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

adiabatic wall

A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

equilibrium

A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

macroscopic

A

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

one number exists

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

microscopic

A

properties on the atomic level, eg Vrms or Vmp

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

examples of conjugate variables

A

pressure and volume for a gas

tension and length for a stretched wire

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

product of all conjugate pairs…

A

has dimensions of energy

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

functions of state depend only on

A

the state itself and not on how that state was produced

eg pressure, volume, temperature

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
17
Q

why is heat not a function of state

A

because it is associated with a transfer of energy between states

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
20
Q

heat

A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
22
Q

temperature

A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
26
Q

consequences of thermodynamic equilibrium on a macroscopic level

A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
29
Q

why does RHS of equation of state equal 0

A

equilibrium holds

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
30
Q

how is equation of state determind?

A

determined by experimental observation or a microscopic model of the system

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
31
Q

ideal gas

A

hydrostatic equilibrium holds

inward force of gravity balances outward gas pressure

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
32
Q

what is assumed in ideal gas

A

atoms or molecules are non-interacting and point like

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
33
Q

equation of state with universal gas constant

A

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

n is number of moles

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
34
Q

equation of state with Boltzmann’s constant

A

PV-nKBT=0

n is number of molecules

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
35
Q

non-ideal gas

A

has molecules with finite volume so intermolecular forces must be considered

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
39
Q

isotherm

A

continuous line between two states on indicator diagram

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
40
Q

what does isotherm being continuous allow

A

integration/ differentiation

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
42
Q

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

A

high volumes and low pressures

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
45
Q

example of a good fixed point

A

triple point of water

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
49
Q

use of gas thermometer

A

establish standard temperatures and calibrate other types of thermometers

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
56
Q

what is R/N_A

A

K_B

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
59
Q

piston - work is done BY the gas if

A

it expands

i.e. energy taken out

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
60
Q

piston - work done ON the gas if

A

it is compressed

i.e. energy must be added in

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
61
Q

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

A

dW=Fdx=PAdx=PdV

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
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)

Brainscape helps you reach your goals faster, through stronger study habits.
© 2024 Bold Learning Solutions. Terms and Conditions