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Transcript
MAXWELL’S THERMODYNAMIC
RELATIONSHIPS AND THEIR APPLICATIONS
Submitted By
Sarvpreet Kaur
Associate Professor
Department of Physics
GCG-11, Chandigarh
James Clerk Maxwell (1831-1879)
http://en.wikipedia.org/wiki/James_Clerk
_Maxwell
• Born in Edinburgh,
Scotland
• Physicist well-known
for his work in
electromagnetism and
field theory
• Also known for his
work in
thermodynamics and
kinetic theory of gases
Why Use Maxwell Relations?
• Certain variables in
thermodynamics are hard to
measure experimentally
such as entropy
• Some variables like
Pressure, Temperature are
easily measureable
• Maxwell relations provide a
way to exchange variables
Maxwell relations derived by the method
based on Thermodynamic Potentials
Why are thermodynamic potentials useful
Thermodynamic potentials give the complete knowledge about any
thermodynamic system at equilibrium
e.g. U=U(T,V) does not give complete knowledge of the system
and requires in addition
P=P(T,V)
equation of state
U=U(T,V) and P=P(T,V)
complete knowledge of equilibrium properties
U(T,V) is not a thermodynamic potential
However
We are going to show: U=U(S,V)
complete knowledge of equilibrium
properties
U(S,V): thermodynamic potential
The thermodynamic potential U=U(S,V)
Now Consider first law in differential notation
dU  dQ  dW
dQ  TdS
dQ
inexact differentials
expressed by
dW
dU  TdS  PdV
So dU is an exact potental.
2nd law
exact differentials
dW  PdV
Note: exact refers here to the
coordinate differentials dS and dV.
TdS and PdV are inexact
By Legendre transformation
(T,V):
dU  TdS  PdV  d(TS)  SdT  PdV
d(U  TS)  SdT  PdV
from (S,V)
: F
to
(T,P):
Helmholtz free energy
dF  SdT  PdV  SdT  d(PV )  VdP
d(F  PV )  SdT  VdP
: G
Gibbs free energy
G  F  PV  U  TS  PV  H  TS
dU  TdS  dPV   VdP
easy check:
Product
rule
 dPV   VdP  VdP  PdV  VdP  PdV
dU  dPV   TdS  VdP
dU  PV   TdS  VdP
=:H (enthalpy)
Enthalpy
H=H(S,P) is a thermodynamic potential
dH  TdS  VdP
Now dU, dF, dG and dH are exact differentials e.g
Internal Energy dU  TdS  PdV
Helmholtz Free Energy dF  d (U  TS )   SdT  PdV
Gibb' s Function dG  d ( F  PV )   SdT  VdP
Enthalpy dH  TdS  VdP
Using these exact differentials we derive maxwell’s relations .
Maxwell’s Thermodynamic Relations
 T 
 P 

   
 V  S
 S V
 S   P 

  
 V T  T V
 T   V 
  

 P  S  S  P
 S 
 V 


 


 P T
 T  P
Deriving Maxwell Relations Using
thermodynamic Potentials
First, start with a known equation of
state such as that of internal energy
Next, take the total derivative of
with respect to the natural
variables. For example, the
natural of internal energy are
entropy and volume.
 U 
 U 
dU  
 dS  
 dV
 S V
 V  S
Deriving Maxwell Relations Continued
Now that we have the total derivative with
respect to its natural variables, we can refer
back to the original equation of state and
define, in this example, T and P.
 U 
 U 
dU  
 dS  
 dV
 S V
 V  S
 U 

 T
 S V
 U 

  P
 V  S
Deriving Maxwell Relations Continued
We must now take into account a
rule in partial derivatives
When taking the partial derivative
again, we can set both sides equal
and thus, we have derived a
Maxwell Relation
Similarily using dF,dG and dH other Maxwell
Relations are
Mnemonic Device for Obtaining Maxwell
Relations
T
P
V
S
Write T,V,S,P in a clockwise manner by Remembering the line
TV Special Programme..
•Four relations are obtained by starting either from T or S Clockwise or anticlockwise
direction. A negative Sign must appear in the resulting equation
Using Maxwell Relations
Maxwell Relations can be derived from basic equations
of state, and by using Maxwell Relations, working
equations can be derived and used when dealing with
experimental data.
Application of Maxwell’s Relations
The Four Maxwell relations have a very wide range
of applications . They apply to all kind of substances
(solids,liquids,gases)under all type of conditions of
Pressure, volume and temperature. Before Discussing applications
We define some thermodynamic terms e.g
i)
Specific Heat at Constant Volume
ii)
Specific Heat at Constant Pressure
 U 
Cv  

 T  v
 U 
CP  

 T  p
iii) Pressure and Volume Coffecient of Expansion
1. Cooling Produced By Adiabatic Expansion of Any Substance
1. Cooling Produced By Adiabatic Expansion of Any Substance
In adiabatic process entropy S remains constant. Therefore by considering
the Thermodynamic relation
 T

 V
We can prove

 p 
   
S
 S  V
T 
Tp 
mCv
Most of the substances expand on heating , they have +ve beta value. T Will be –ve
i.e all the substances will cool down. A few substances like rubber have –ve beta value.
They will get heat up..
2. Adiabatic Compression of A Substance
By considering the Thermodynamic relation
 T 
 V

  
 S
 p  S
We can prove


p
TV
T  C
p

Above result shows that if
is +ve, then adiabatic increase in pressure
causes the temprature to rise.
** Similarily using other maxwell’s equations we can explain the stretching
of wires and thin films
3. Change of internal energy with Volume,
Using the third Maxwell’s relation
 s   P 
  

 v T  T v
 u 
 P 

T
 

 P
 v T
 T v
Since
19
For Ideal Gas
This result helps to show that the internal energy of an ideal gas does not depend
upon specific volume. This is known as Joule’s Law.
For Vander Waal’s / real gases
( p
a
)(V b )  RT
2
V
 
U
V
T

a
V
2
V
Thus Vander Waal’s gas expands isothermally as its
internal energy increases.
20
4. Cp – Cv = R for ideal gases.
Other relations for the specific heats are given below.
where  is the volume expansivity and  is the isothermal compressibility, defined as
21
The difference Cp – Cv is equal to R for ideal gases and to zero for incompressible
substances (v = constant).
5. Variation of Cv with specific volume.
5. Variation of Cv of an ideal gas does not depend upon specific volume.
For an ideal gas
22
Therefore, the specific heat at constant volume of an ideal gas is independent of
specific volume.
For Vander Waal’s gas also it is independent of volume.
23
6. Change of state and clapeyron’s equation
In ordinary phase transition of matter(solid phase to liquid phase,
liquid to vapour, and solid to vapour) take place under constant
Temperature and pressure. During the transition a certain amount of
heat, known as latent heat must be supplied to the substance for a change
Of phase. During this change temperature remains constant. Therefore
using maxwell relation
   
S
p

V T
T
V
using ΔU  L, the latent heat, Wehave
 
p
T
V
L

T (V f Vi )
This equation is known as Clausius-Clapeyron’s latent heat
Thank You