Thermal Physics Flashcards
system
a portion of the universe with certain measurable quantities such as pressure or volume which determine the equilibrium state of the system
surroundings
the rest of the universe outwith the system
system + surroundings =
universe
isolated system
a system that does not interact with its surroundings by exchanging heat energy, mechanical energy or material
closed system
energy but not material can be exchanged
adiabatic wall
system is thermally isolated and only mechanical energy (not heat) can be exchanged with the surroundings (eg vacuum flask)
diathermal wall
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
equilibrium
all bulk physical properties are uniform throughout the system, they are time independent
macroscopic
large scale or bulk properties of a gas eg pressure, volume, temperature
one number exists
microscopic
properties on the atomic level, eg Vrms or Vmp
state variables
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)
conjugate variables
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
examples of conjugate variables
pressure and volume for a gas
tension and length for a stretched wire
product of all conjugate pairs…
has dimensions of energy
function of state
any quantity which takes unique value for each equilibrium state of a system
eg internal energy U or entropy S
functions of state depend only on
the state itself and not on how that state was produced
eg pressure, volume, temperature
why is heat not a function of state
because it is associated with a transfer of energy between states
if two systems are put in thermal contact…
generally changes occur in both
eg coffee cools down as room heats up slightly
two systems in thermal contact after some time
no further changes occur and the two systems are said to be in thermal equilibrium
heat
form of energy transferred between substances due to temperature differences between them
if two objects are not in thermal equilibrium, heat flows…
from hotter object to the colder object until thermal equilibrium is reached
temperature
property which determines whether or not a system is in thermal equilibrium with other systems
zeroth law of thermodynamics
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
conditions for thermal equilibrium
- thermal equilibrium - same temp
- mechanical equilibrium - no unbalanced forces acting
- chemical equilibrium - no chemical reactions occurring
non-uniformities in a system result in
a gradient and hence, momentum, heat and/or matter flows until thermodynamic equilibrium is reached
consequences of thermodynamic equilibrium on a macroscopic level
variables have constant values in time and space, ie throughout the system
consequences of thermodynamic equilibrium on a microscopic level
any process ( diffusion, collisions etc) must have an equal probability of going in the opposite direction
equation of state
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
why does RHS of equation of state equal 0
equilibrium holds
how is equation of state determind?
determined by experimental observation or a microscopic model of the system
ideal gas
hydrostatic equilibrium holds
inward force of gravity balances outward gas pressure
what is assumed in ideal gas
atoms or molecules are non-interacting and point like
equation of state with universal gas constant
f(P,V,T)= PV-nRT=0
n is number of moles
equation of state with Boltzmann’s constant
PV-nKBT=0
n is number of molecules
non-ideal gas
has molecules with finite volume so intermolecular forces must be considered
equation of state for a van der Waals gas
adaptation of ideal gas law but includes a and b as constants specific to the gas
how to reduce van der waals equation back to ideal gas
low pressures and high volumes
n^2 a / v^2 terms approximates 0
indicator diagram
graph of an intensive variable against its conjugate extensive variable
any particular state may be represented by a single point on the diagram
isotherm
continuous line between two states on indicator diagram
what does isotherm being continuous allow
integration/ differentiation
isotherms for a van der waals gas
isotherms look same for both gas types at high temperatures
behaviour changes as temp drops
below critical temp a phase change occurs
when do graphs of isotherms for van der waals and ideal look approximately the same?
high volumes and low pressures
we can assign numerical values to temperatures by
selecting a suitable physical property of a system that varies linearly with it
X=cTx
explain variables in X=cTx
Tx is temp on X scale and c is a constant got by choosing T at a convenient fixed point
example of a good fixed point
triple point of water
triple point of water
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 to build thermometer
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
gas thermometer
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
use of gas thermometer
establish standard temperatures and calibrate other types of thermometers
liquid volume thermometer
liquid expands as a function of temp
small volume change leads to large height change in tube
only agree at fixed points
resistance thermometers
based on variation of electrical resistance with temp
relationship between temp and r is not straightforward and can be expensive
easy to transport
thermocouples
potential difference from a junction of two different metals
cheap to make but require careful calibration and not really linear
narrow temperature ranges
radiation pyrometers
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
kinetic theory of gases assumptions
gas is ideal
collisions between molecules and walls of container are elastic
container walls are rigid and do not move when struck
major result from kinetic theory derivation
PV=1/3nN_AmVrms^2
relates bulk properties of pressure and volume to microscopic properties of gas
what is R/N_A
K_B
how to obtain kinetic energy equation from kinetic theory of gases derivation
from R/N_A=K_B
k_BT=1/3mvrms^2
times by 3 and divide by 2
what conclusions/results come from kinetic theory of gases derivation?
temperature is proportional to average translational kinetic energy
there is 1/2K_BT per degree of freedom
piston - work is done BY the gas if
it expands
i.e. energy taken out
piston - work done ON the gas if
it is compressed
i.e. energy must be added in
when a piston moves a small distance dx, work done is
dW=Fdx=PAdx=PdV
for finite change in x, work done is
integral (limits v1 and v2) of Pdv
principle of conservation of energy
energy can only be changes, cannot be created/destroyed
total amount of energy in a closed system is constant with time
conservation of energy in an ideal gas
no potential energy contribution, internal energy U is the sum of all energy contributions
when a process is performed on a system…
there will be a change in internal energy as it moves from one equilibrium state to another
what does thermally isolated mean
no heat flow (adiabatic)
internal energy in an adiabatic process
deltaU=W
when a system in not adiabatic, what must be considered
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 -)
piston but now walls of container are thermally conducting
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
first law of thermodynamics
delta U=Q-W
sign conventions for delta U
positive higher energy
negative lower energy
sign conventions for Q
positive heat flowing in
negative heat flowing out
sign conventions for W
positive for expansion
negative for compression
on a pv diagram, work done is
area under curve
because dW=PdV
if the path changes…
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
first law of thermodynamics
there is a finite amount of energy associated with a change of state
monatomic
not bound to each other
3 dof
(translational movement)
equipartition theorem
classical system in thermal equilibrium at tempT has mean energy of 1/2KBT per dof
diatomic
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
DOF, U per molecule and U per mole for monatomic
3, 3/2KBT, 3/2RT
DOF, U per molecule and U per mole for diatomic (low T)
5, 5/2KBT, 5/2RT
DOF, U per molecule and U per mole for diatomic (high T)
7, 7/2KBT, 7/2RT
molar heat capacity
amount of heat dQ required to raise temp of one mole of gas by 1 kelvin
C=1/n dQ/dT
two different molar heat capacities
CV change of temp at constant volume
CP change of temp at constant pressure
ratio of specific heats
gamma = CP/CV
isovolumetric
constant volume
isothermal
constant temperature
(therefore can draw isotherms on PV diagram)
isovolumetric process on pv diagram
jumps from one isotherm to another (vertically)
work done in isovolumetric process
since work is intergral of PdV between v1 and v2 and volume is constant, no work is done
so Q=deltaU
isobaric process
constant pressure
isobaric process on pv diagram
jumps from one isotherm to another (horizontally)
work done in isobaric
W=P deltaV (result of integration and taking P out as a constant)
isothermal process on pv diagram
single isotherm
pressure and volume in isothermal process
both change
work done in isothermal
usual integration but sub in ideal gas law for P
gives W=nRTln(v2/v1)
first law for isothermal
no change in temp so no change in internal energy
Q=W
assumptions which allow us to say Q=0 in adiabatic process
system is thermally isolated from surroundings
system undergoes rapid change for which heat transfer is negligible
consequence of PV^gamma = constant
on a pv diagram, adiabatic curves are steeper than isothermals
adiabatic expansion work done
results in a cooling and decrease of internal energy
adiabatic compression work done
work done on the gas, it heats and there is an increase in internal energy
total pressure depends on
average velocity of the molecules
total pressure exerted by a mixture of gases is
equal to the sum of the pressures exerted by those gases separately
each gas contributes a partial pressure
maxwell boltzmann distribution
velocity distribution in a gas
plot of v against f(v)
will be a range of velocities from zero to very high
features to note in maxwell boltzmann distribution
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
most probable speed on plot of v against f(v)
value of v at the turning point
i.e when d/dv(f(v))=0
peak of curve on MB distribution
shifts right with increasing temp
why is shape of MB asymmetric
have a range of possible speeds between zero and infinity
higher velocities have more weight in the distribution
relative location of Vrms on MB distribution
to the right of mean molecular speed because of higher weighting given to higher velocities
you can change a system’s state by
applying heat or doing work
most bulk macroscopic processes that cause these changes are irreversible
irreversible processes are those in which
energy is dissipated
entropy S
macroscopic variable that quantifies the degree of disorder in a system
entropy for processes in adiabatically isolated systems
can only remain constant or increase
change in entropy =0
reversible process
change in entropy >0
irreversible process
units of entropy
joules per kelvin
small volumetric increases will result in
a greater number of spatial coordinates for the gas molecules, giving rise to a greater level of disorder
first law written for a reversible process
dU=Tds-Pdv
hence we only need to know initial and final states and not intermediate points
lossy process
there is a loss
if you need to do more work than the bare minimum, you have wasted effort somewhere
how to calculate change in entropy for irreversible change
calculate it for an equivalent reversible change as this gives minimum possible value
change in entropy for isobaric
derived from integral of dQ/T and sub dQ=nCpdT
change in entropy for isovolumetric
integral dQ=T
sub in dQ=nCvdT
change in entropy for isothermal
again integral dQ/T
dQ=dW=Pdv
and PV=nRT
process equivalence
isothermal process is equivalent to an isobaric process followed by an isovolumetric process
pv diagrams and reversibility
can only plot PV diagrams for reversible processes
not possible for irreversible as do not know what is happening between states.
free expansion of an ideal gas
irreversible so cannot be plotted on PV diagram
change in entropy for for free expansion of an ideal gas
no work is done by the gas so no temp change should occur
free expansion of a real gas
temp will decrease because internal energy depends on volume
calculating entropy for irreversible process
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.
second law of thermodynamics
reversible: delta S=0
irreversible delta S >0
in general delta S>/=0
natural process during a change in entropy
- transfer of heat from hotter to a cooler body
- overall increase in entropy
to reach equilibrium, a system will…
maximise entropy and do so through irreversible microscopic processes
two forms of the second law tell us that
no heat engine can be 100% efficient and that work is required in order for heat to be moved.
pictorial representation of a heat engine
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
why can’t you say TC>TH
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
why you can’t take heat out of the cold resevoir and dump is elsewhere
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
plotting a heat engine’s process on a PV diagram
would get a loop
efficiency of a heat engine
n=work done / heat input = W/QH
from conservation of energy QH=W+QC
so n=1- QC/QH
refrigerator
heat engine running backwards
heat is extracted from a cold resevoir by applying work
extracted heat is then dumped into the hot resevoir
coefficient of performance of heat pump/refrigerator
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
efficiency of a heat pump
defined in terms of how much heat is produced
nHP=heat delivered/work done = QH/W
=QH/ QH-QC
nHP-k
=1
heat pumps and refrigerators work really well when
only a small temperature change is required
heat pumps on their own are very effective at
heating buildings to a base level
eg frost protection
why cyrogenics is very difficult
for refrigeration, as the ratio of hot to cooled temperature increases the amount of work required to achieve the cooling becomes very large
carnot cycle
operating cycle of the most efficient heat engine possible
all transfer of heat should be between bodies of nearly equal temperature
carnot cycle assumes
an engine connected by a frictionless mechanism to an external load upon which work may be done
four processes of the carnot cycle
isothermal expansion
adiabatic expansion
isothermal compression
adiabatic compression
work done in a carnot cycle
deltaU=Q-W=0 because reversible cycle so no change in internal energy
so W=Q
since Q=QH-QC
work done=QH-QC
carnot’s theorem
no engine operating between two temperature reservoirs can be more efficient than a carnot engine operating between the same two temperatures
proof of carnot’s theorem
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
efficiency of a carnot engine
depends only on temperature
ncarnot=1-TC/TH
kelvin defined a temperature scale based on
how heat flows in the ideal carnot cycle
this allowed the idea of absolute zero
