Physics Revision Notes – Thermodynamics Page 1 of 34
Classical Thermodynamics
Ideal gases
Assumptions
o Identical particles in random motion.
o Small solid spheres – occupy negligible volume.
o Elastic collisions.
o No long range forces – only forces during collisions only
energy is KE, ½mv2 per particle.
Results
o Pressure is given by p = 13 nm v 2 .
o Flux is given by J = 14 n v .
The Maxwell-Boltzmann Distribution
4pv 2 exp (-mv 2 /2kT )
3/2
o P (v ) = ( 2pmkT )
o This gives
8kT
v =
pm
3kT
v2 =
m
Energy…
o U = 23 RT = 23 N AkT
Definitions
A system is in Thermal Equilibrium when all its macroscopic
observables have ceased to change with time.
A function of state is any physics quantity that has a well-defined
value for each equilibrium state of the system. They are represented by
exact differentials.
Functions of state can be either
o Extensive (proportional to system size) – eg: volume.
o Intensive (independent of system size) – eg: pressure.
o Intensive and extensive variables form conjugate pairs, the
product of which is energy.
© Daniel Guetta, 2008
, Physics Revision Notes – Thermodynamics Page 2 of 34
The heat capacity is the amount of heat we need to supply to raise the
temperature by dT. This can be measured at constant volume or at
constant pressure, so
æ dQ ÷ö æ dQ ö÷
CV = çç C p = çç
è dT ÷÷øV è dT ÷÷øp
For an ideal gas
pV = nRT
Stirling’s Approximation
ln N ! » N ln N - N
The First Law
The First Law of Thermodynamics states that
Energy is conserved, and heat and work
are both forms of energy
DU = DQ + DW
Convention:
DQ and DW are POSITIVE when
energy is given TO the system
For a differential change
d U = d Q + dW
The work done compressing a gas is given by
dW = -pdV
And so for a gas, the first law can be written
dU = dQ - p dV
Heat Capacities
In general, the internal energy will be a function of temperature and
volume, so
æ ¶U ö÷ æ ¶U ö÷
dU = ççç ÷÷ dT + ççç ÷ dV
è ¶T øV è ¶V ÷øT
Using the form of the first law for a gas, above, we can write
d Q = d U + p dV
æ ¶U ö÷ éæ ¶U ö ù
dQ = ççç ÷÷ dT + êêççç ÷÷÷ + p úú dV
è ¶T øV ëè ¶V øT û
© Daniel Guetta, 2008
, Physics Revision Notes – Thermodynamics Page 3 of 34
And dividing by dT:
dQ æç ¶U ö÷ éæ ¶U ö ù
= çç ÷÷ + êêççç ÷÷ + p ú dV
dT è ¶T øV ëè ¶V ø÷T ûú dT
By taking the equation above at constant volume and constant
pressure (only the dV/dT term will matter), we obtain
æ ¶U ö÷
CV = ççç ÷
è ¶T ø÷V
éæ ¶U ö ù
C p = CV + êççç ÷ + p ú ççæ ¶V ÷÷ö
÷
êëè ¶V ÷øT ç ÷
ûú è ¶T øp
For an ideal gas:
C p = CV + R
We define
Cp
g=
CV
Reversibility
A change is reversible if an infinitesimal change in external conditions
would reverse the direction of the change.
Reversible changes are typically very slow and quasi-static.
They are also frictionless – no viscosity, turbulence, etc…
Isothermal expansions
When an expansion is isothermal, the temperature of the system does
not change. Therefore, the internal energy of the system does not
change, and
d W = - dQ
Therefore, when a gas is expanded isothermally from V1 to V2 at a
temperature T, the heat absorbed by the gas is given by
DQ = ò dQ
= -ò dW
V2
= ò
V1
p dV
=RT /V
V2 1
= RT ò dV
V1 V
© Daniel Guetta, 2008
, Physics Revision Notes – Thermodynamics Page 4 of 34
V2
DQ = RT ln
V1
An adiabatic process is both adiathermal (no flow of heat) and
reversible, so
dQ = 0
And
dU = dW
However, for an ideal gas
dU = CV dT
Therefore
CV dT = dW
CV dT = -p dV
RT
CV dT = - dV
V
dT R dV
=-
T CV V
Obvious from
def n of g
dT dV
= (1 - g )
T V
TV g -1 = constant
Other versions can easily be generated using pV µ T .
The Second Law
The Second Law of Thermodynamics can be stated in two different
ways
The Clausius Formulation
No process is possible whose sole result is
the transfer of heat from a hotter to a cooler
body.
The Kelvin Formulation
No process is possible whose sole result is
the complete conversion of heat into work.
The equivalence of these two formulations can be shown in two steps
o Violating Kelvin Violating Clausius
© Daniel Guetta, 2008
Classical Thermodynamics
Ideal gases
Assumptions
o Identical particles in random motion.
o Small solid spheres – occupy negligible volume.
o Elastic collisions.
o No long range forces – only forces during collisions only
energy is KE, ½mv2 per particle.
Results
o Pressure is given by p = 13 nm v 2 .
o Flux is given by J = 14 n v .
The Maxwell-Boltzmann Distribution
4pv 2 exp (-mv 2 /2kT )
3/2
o P (v ) = ( 2pmkT )
o This gives
8kT
v =
pm
3kT
v2 =
m
Energy…
o U = 23 RT = 23 N AkT
Definitions
A system is in Thermal Equilibrium when all its macroscopic
observables have ceased to change with time.
A function of state is any physics quantity that has a well-defined
value for each equilibrium state of the system. They are represented by
exact differentials.
Functions of state can be either
o Extensive (proportional to system size) – eg: volume.
o Intensive (independent of system size) – eg: pressure.
o Intensive and extensive variables form conjugate pairs, the
product of which is energy.
© Daniel Guetta, 2008
, Physics Revision Notes – Thermodynamics Page 2 of 34
The heat capacity is the amount of heat we need to supply to raise the
temperature by dT. This can be measured at constant volume or at
constant pressure, so
æ dQ ÷ö æ dQ ö÷
CV = çç C p = çç
è dT ÷÷øV è dT ÷÷øp
For an ideal gas
pV = nRT
Stirling’s Approximation
ln N ! » N ln N - N
The First Law
The First Law of Thermodynamics states that
Energy is conserved, and heat and work
are both forms of energy
DU = DQ + DW
Convention:
DQ and DW are POSITIVE when
energy is given TO the system
For a differential change
d U = d Q + dW
The work done compressing a gas is given by
dW = -pdV
And so for a gas, the first law can be written
dU = dQ - p dV
Heat Capacities
In general, the internal energy will be a function of temperature and
volume, so
æ ¶U ö÷ æ ¶U ö÷
dU = ççç ÷÷ dT + ççç ÷ dV
è ¶T øV è ¶V ÷øT
Using the form of the first law for a gas, above, we can write
d Q = d U + p dV
æ ¶U ö÷ éæ ¶U ö ù
dQ = ççç ÷÷ dT + êêççç ÷÷÷ + p úú dV
è ¶T øV ëè ¶V øT û
© Daniel Guetta, 2008
, Physics Revision Notes – Thermodynamics Page 3 of 34
And dividing by dT:
dQ æç ¶U ö÷ éæ ¶U ö ù
= çç ÷÷ + êêççç ÷÷ + p ú dV
dT è ¶T øV ëè ¶V ø÷T ûú dT
By taking the equation above at constant volume and constant
pressure (only the dV/dT term will matter), we obtain
æ ¶U ö÷
CV = ççç ÷
è ¶T ø÷V
éæ ¶U ö ù
C p = CV + êççç ÷ + p ú ççæ ¶V ÷÷ö
÷
êëè ¶V ÷øT ç ÷
ûú è ¶T øp
For an ideal gas:
C p = CV + R
We define
Cp
g=
CV
Reversibility
A change is reversible if an infinitesimal change in external conditions
would reverse the direction of the change.
Reversible changes are typically very slow and quasi-static.
They are also frictionless – no viscosity, turbulence, etc…
Isothermal expansions
When an expansion is isothermal, the temperature of the system does
not change. Therefore, the internal energy of the system does not
change, and
d W = - dQ
Therefore, when a gas is expanded isothermally from V1 to V2 at a
temperature T, the heat absorbed by the gas is given by
DQ = ò dQ
= -ò dW
V2
= ò
V1
p dV
=RT /V
V2 1
= RT ò dV
V1 V
© Daniel Guetta, 2008
, Physics Revision Notes – Thermodynamics Page 4 of 34
V2
DQ = RT ln
V1
An adiabatic process is both adiathermal (no flow of heat) and
reversible, so
dQ = 0
And
dU = dW
However, for an ideal gas
dU = CV dT
Therefore
CV dT = dW
CV dT = -p dV
RT
CV dT = - dV
V
dT R dV
=-
T CV V
Obvious from
def n of g
dT dV
= (1 - g )
T V
TV g -1 = constant
Other versions can easily be generated using pV µ T .
The Second Law
The Second Law of Thermodynamics can be stated in two different
ways
The Clausius Formulation
No process is possible whose sole result is
the transfer of heat from a hotter to a cooler
body.
The Kelvin Formulation
No process is possible whose sole result is
the complete conversion of heat into work.
The equivalence of these two formulations can be shown in two steps
o Violating Kelvin Violating Clausius
© Daniel Guetta, 2008
