Physical Chemistry
Lecture 12
Adiabatic Processes, Joule
Expansion, Joule-Thomson
Expansion
Isothermal, reversible
expansion of an ideal gas
Work on the system
w = −
∫ PdV
m
path
Vm 2
1
= − RT ∫
dVm
V
Vm1 m
Internal energy change
∆U m
 Vm 2 

= − RT ln
 Vm1 
= 0
Heat transferred found by difference
q = ∆U m
 Vm 2 

− w = RT ln
 Vm1 
Adiabatic process
A process with no heat transfer between
the system and the surroundings
q=0
Differential changes
dU m
= dw = − Pext dVm
Adiabatic, reversible process
for an ideal gas
q=0
Reversibility requires Pext = P
Differential change
dU m
= − PdVm
For an ideal gas, the energy does not
depend on Vm
RT
CVm dT = −
dVm
Vm
Adiabatic, reversible process
for an ideal gas
Rearrange differential and integrate to
give T
 Vm 2 
CVm
∫T T dT = − R ln Vm1 
For a monatomic gas, CVm is usually
independent of temperature (or nearly
so)
2
1
 Vm 2 
 T2 

CVm ln  = − R ln
 T1 
 Vm1 
Joule expansion
Expansion at constant internal energy
To determine the temperature change, one must
know
 ∂T 


 ∂Vm U m
 1  ∂U m 


= − 
 CVm  ∂Vm T
For an ideal gas, this is zero and the temperature
does not change
For a real gas, this is nonzero and the
temperature does change during the expansion
Joule expansion of a van der
Waals gas
Determine the appropriate derivative
 ∂U m 


 ∂Vm T
a
 ∂P 
= T
 −P =
Vm2
 ∂T Vm
Integrate over volume, assuming a
volume-independent heat capacity
∆T
 ∂T 
 dVm
= ∫ 
∂Vm U
Vm1 
m
Vm 2
a
= −
CVm
Vm 2
1
∫V Vm2 dVm
m1
=
1 
a  1


−
CVm  Vm 2 Vm1 
Joule expansion of CO2
Treat as a van der Waals gas
Information


a = 0.3649 Pa-m6/mol
CVm = 28.09 joule/K-mol
Consider a doubling of the volume from 22.4 L/mol to
44.8 L/mol
∆T
=

0.3684 Pa − m 6 
1
1

 = − 0.29 K
−
3
3 
28.09 joule / K  0.0448 m 0.0224 m 
A tiny, but measurable increase
Joule-Thomson experiment
Adiabatic expansion of a
gas through an orifice
Treat as a sequence of two
steps


Compress gas at constant pressure,
P1, to V=0
Expand gas at constant P2 to V2
Because of adiabaticity of
the process
w1 = P1V1
w2 = − P2V2
∆U = U 2 − U1 = w1 + w2
= P1V1 − P2V2
U 2 + P2V2 = U1 + P1V1
An isenthalpic process
H 2 = H1
Joule-Thomson experiment
Adiabatic expansion of a
gas through an orifice
Isenthalpic process

H1 = H2
Temperature change
given by
µ JT
 ∂T 
= 

 ∂P  H
Ideal gas: ∆T = 0
Real gas: ∆T ≠ 0
∆T
=
P2
∫
P1
 ∂T 
  dP
 ∂P H
Joule-Thomson coefficient
Joule-Thompson
coefficient
µ JT
 ∂T 
= 

 ∂P  H
The size of the
coefficient determines
temperature change for
a pressure drop
One gets µJT from
measurements in JouleThompson experiments
∆T
=
=
 ∂T 
∫P  ∂P  H dP
1
P2
P2
∫µ
P1
JT
dP
Joule-Thomson effect
Used for air
conditioning, cooling
Joule-Thomson
coefficient, µJT,
determines size of
effect
 Usually positive –
cooling upon
expansion
 Can be negative
 Typical size – 1
K/bar
Material
µJT (K bar-1)
Helium
-0.0623
Argon
0.366
Xenon
1.823
Nitrogen
0.215
Hydrogen
-0.0299
Carbon dioxide
1.093
Methane
0.438
Ethane
1.100
Propane
1.666
Butane
2.533
Ammonia
2.811
1,2-dichloro-1,1,2,2-tetrafluoroethane
1,1,1-trifluoroethane
1.973
1.8815
Joule-Thomson effect
Ranque-Hilsh vortex tube
Mechanical device that separates a
compressed gas into hot and cold streams
Two explanations


Outer air is under higher pressure than the inner air (because of
centrifugal force); the temperature of the outer air is higher than of
the inner air
Both vortices rotate at the same angular velocity and direction, the
inner vortex has lost angular momentum, the decrease of which is
transferred as kinetic energy to the outer vortex
Standard states
Can calculate changes in variables with
thermodynamic mathematics
To determine an energy scale, one must
define the value at a particular set of
conditions
Standard state – a state where the
properties are defined
Standard states
Solids and liquids

The material at the temperature under a pressure
of 1 bar
Gases

The equivalent ideal gas at the temperature and
at a pressure of 1 bar
May see other standard states defined in
some circumstances
Must know the standard state to make use of
absolute values of state variables
Summary
Can calculate changes in state variables
under various conditions




Isothermal
Adiabatic
Joule expansion
Joule-Thomson experiment
To define scales for thermodynamic
parameters, one must specify standard
state