1) The Joule-Thomson effect states that when a gas passes through a porous plug from a region of higher to lower pressure adiabatically, its temperature may increase or decrease depending on its initial temperature and pressure. 2) Most gases experience a decrease in temperature during Joule-Thomson expansion, but gases like hydrogen and helium may increase in temperature above their inversion points. 3) Gases are liquefied using either cascade cooling by successively lower boiling refrigerants or regenerative cooling using the Joule-Thomson effect in a closed cycle to gradually lower the temperature until liquefaction occurs.

1. 1 | U I I I , S e m e s t e r - I I ( P A N a g p u r e )
Thermodynamics - II
Joule – Thomson Effect:
It states that, “when gas under higher constant pressure is allowed to pass through a porous
plug ( or throttle ) to region of low constant pressure adiabatically, its temperature is decreased.”
The simple explanation of this effect is based on the fact that, the gas molecules are in
continuous motion and so they possess a definite amount of kinetic energy 3
2
kT
 
 
 
. When the gas
expands, it does some work in overcoming in Vander Waal’s forces of attraction exists between the
molecules. Thus part of energy of the system falls and as a result the temperature also falls.
Experimental Set Up of Joule Thomson experiment:
Experimental set up for J-T experiment is as
shown in figure. A porous plug of cotton wool or silk
fibre is kept in position between the perforated brass
plates DD. W is box wood tube surrounded by the
cotton wool contained in vessel J. A and B are copper
tubes, end A is connected to a compression pump P
through a copper spiral tube immersed in a water bath.
The initial and final temperatureare noted by means of
platinum resistance thermometers T1 and T2. The
pressure difference is measured by pressure guage.
Experimental Results:
Various gases were subjected to Joule Thomson (effect) experiment and following results
were observed,
1) All the gases show change in temperature on passing through the porous plug.
2) The change in temperature is found to be proportional to the difference in pressure on the
two sides of the porous plug.
3) At ordinary temperature, most gases suffer a fall in temperature while gases like hydrogen
and helium shows rise in temperature.
4) Every gas has inversion temperature at given initial pressure. Above this temperature, the
gas shows rise in temperature. Below this temperature, the gas shows fall in temperature.
Theory / Explanation of Joule Thomson Effect:
Consider one mole of the gas contained in the insulated cylinder divided into two parts by
means of porous plug. Before the gas passes through the porous plug, let P1 is its pressure, V1 is the
volume and U1 is the internal energy. After the gas has passed through porous plug, let P2 is its
pressure, V2 is the volume and U2 is the internal energy.
The work done on the gas at the left piston is P1V1 and work done
by the gas at the right piston is P2V2.
Hence net work done by the gas is:
W = P2V2 – P1V1

2. 2 | U I I I , S e m e s t e r - I I ( P A N a g p u r e )
Since the process was conducted adiabatically.
Δq = 0
According to first law of thermodynamics, Δq = ΔU + W
ΔU = -W
Or U2 - U1 = - (P2V2 – P1V1) = P1V1 – P2V2
 U1+ P1V1 = U2 + P2V2
H1 = H2 (Since, U + PV = H)
Or dH = 0
Thus in this process the heat content or enthalpy (U + PV = H) of the gas remains constant.
If the gas obeys Boyle’s law P1V1 = P2V2, then U2 = U1. This means that the internal energy
of the gas is the same on both sides of the porous plug. But real gas obeys the law closely only at its
Boyle temperature. Therefore for real gases there are three possible cases.
1) At the Boyle temperature, P1V1 = P2V2, then U2 = U1.
Hence in this case temperature of the gas remains same.
2) Below the Boyle temperature, P1V1 < P2V2, then U2 < U1.
Hence in this case internal energy of the gas decreases and hence there will be fall in
temperature of the gas.
3) Above the Boyle temperature, P1V1 > P2V2, then U2 > U1.
Hence in this case internal energy of the gas increases and hence there will be rise in
temperature of the gas.
Joule – Thomson coefficient:
It is defined as, the ratio of change in temperature in degrees of the gas in throttle expansion
(Joule Thomson expansion) to the pressure drop of one atmosphere under the condition of constant
enthalpy.
i,e.
H
dT
dP

 
   
For fall in temperature of the gas  is positive, for rise in temperature  is negative. The
temperature at which gas neither cools nor heats on expansion i,e. where  = 0, is called inversion
point.
Liquefaction of gases:
In general, there are two methods of liquefaction of gases.
1) Cascade cooling and 2) Regenerative (Joule-Thomson) cooling.

3. 3 | U I I I , S e m e s t e r - I I ( P A N a g p u r e )
1) Cascade cooling:
We know that, the gas cannot be liquefied until it is cooled below its critical temperature. But
once this temperature has been reached, the gas can be liquefy by applying sufficient pressure.
In this process gas is cooled below its critical temperature by passing it through the number of
stages of volatile refrigerants of successively lower boiling points. Then by applying sufficient
pressure, the gas can be liquefied.
2) Regenerative (Joule-Thomson) cooling:
According to the Joule-Thomson effect, when gas under pressure is allowed to expand
through porous plug, it suffers change in temperature. Similar effect is observed if constricted orifice
such as nozzle is used instead of porous plug.
If the initial temperature of the gas is below its inversion temperature, then on expanding
through the orifice its temperature decreases. A portion of the gas thus cooled by the Joule-Thomson
expansion is used to cool the other portion of the gas.
Thus incoming gas is cooled by the outgoing cooled gas. By continuing this process,
ultimately temperature is reached at which gas is liquefied.
Liquefaction of Hydrogen Gas:
The critical temperature of hydrogen is -2410
C
which can not be attain by any independent cooling
agent. Therefore Joule-Thomson effect is used to
liquefy the hydrogen gas. The inversion temperature
for an ordinary initial pressure is -800
C and its Boyle
temperature is -1690
C. Therefore it is necessary to
precool the gas bellow -800
C, but for appreciable
cooling by means of Joule-Thomson effect it should
be cooled to its Boyle temperature.
The apparatus used for the liquefaction of
hydrogen gas is as shown in figure.
The hydrogen gas is first carefully purified so
that it is free from O2, N2 and other impurities. It is
compressed to pressure about 200 atmospheres and
then passed through the coils as shown in figure,
before it suffers throttle expansion through the valve
N.
In first chamber it is cooled by the
surrounding solid CO2 and alcohol. In chamber A, it is further cooled by outgoing cooled hydrogen.
In chamber B, it is cooled by liquid air contained in the chamber. In chamber C, it is cooled to -2080
C
by liquid air boiled under reduced pressure (10 cm of Hg). Now cooled hydrogen at pressure 150
atmospheres, when passed through regenerative coil G and allowed to expand through nozzle N to
pressure one atmosphere. Consequently it is further cooled due to Joule-Thomson effect.

4. 4 | U I I I , S e m e s t e r - I I ( P A N a g p u r e )
The cooled hydrogen is then allowed to circulate back to the pump. This process of
regenerative cooling is continues and after few cycles the temperature of the hydrogen falls to its
boiling point -2530
C. It is then liquefied into Dewar flask.
Liquefaction of Helium Gas:
The critical temperature of helium is -2680
C and its
inversion temperature for an ordinary initial pressure is -
2430
C and its Boyle temperature is -2540
C. Therefore in
order to liquefy the helium gas it should be precooled to -
2540
C.
The apparatus used for the liquefaction of helium gas is as
shown in figure.
The Helium gas is first carefully purified so that it
is free the impurities. It is compressed to pressure about 40
atmospheres and then passed through the spiral tubes S1,S2,
S3 and S4 as shown in figure.
In the spiral tube S1 and S3 the gas cooled to -258C by
liquid hydrogen boiling under reduced pressure in chamber
E. in the spiral tubes S2 and S4, it is cooled by outgoing
cooled helium. At the nozzle N the cooled helium at -2580
C
suffers throttle expansion. Consequently it is further cooled
due to Joule-Thomson effect.
After few cycles the temperature of the helium is reached to its boiling point and it is
liquefied in chamber F.

5. 5 | U I I I , S e m e s t e r - I I ( P A N a g p u r e )
Maxwell’s Thermodynamic Relations
Thermodynamic Variables:
The thermodynamic state of a system is specified by some of its properties like pressure,
temperature, volume, internal energy, entropy. These properties undergo a change when system
passes from one state to another. These variables are known as thermodynamic variables or co-
ordinates. These are macroscopic variables.
Extensive Variable:
An extensive variable of a system is macroscopic parameter which describes system in
equilibrium and which has value equal to sum of its values in all part of the system. The extensive
variable depends on mass and size of the substance present in the system.
For example: volume, mass, internal energy, entropy, length, area, heat capacity, electric charge, etc.
Intensive Variable:
An intensive variable of a system is macroscopic parameter which describes system in
equilibrium and which has same value in all part of the system. The intensive variable is independent
on mass and size of the substance present in the system.
For example: pressure, temperature, density, surface tension, viscosity, refractive index, etc.
Consider a homogeneous system in equilibrium. Suppose system is divided into many parts.
If x is the macroscopic variable of the system and has values x1, x2, x3,…… in each part of the system,
then
x = x1 + x2 + x3 +………. if x is extensive variable;
x = x1 = x2 = x3 =………. if x is intensive variable;
Maxwell’s Thermodynamic Relations:
From the first and second law of thermodynamics, Maxwell was able to derive four
fundamental thermodynamic relations.
According to first law of thermodynamics, we have
Q dU W   as W PdV 
Q dU PdV   …… (1)
According to second law of thermodynamics, we have
Q TdS  …… (2)
From eq 1 and eq 2, we have
TdS dU PdV 
Or
dU TdS PdV  ……. (3)
Let U, S and V are the functions of two independent variables
i.e. ( , ), ( , ) & ( , )U U x y S S x y V V x y  

6. 6 | U I I I , S e m e s t e r - I I ( P A N a g p u r e )
y x
U U
dU dx dy
x y
   
         
….. (4)
y x
S S
dS dx dy
x y
   
        
….. (5)
y x
V V
dV dx dy
x y
   
        
….. (6)
Using eq 4,5 &6 in eq 3, we have
y y yx x x
U U S S V V
dx dy T dx T dy P dx P dy
x y x y x y
               
                              
Equating coefficients of dx and dy, we have
y y y
U S V
T P
x x x
       
       
       
….. (7)
x x x
U S V
T P
y y y
       
              
….. (8)
Differentiating eq 7 w.r.t. y and eq 8 w.r.t. x
U S T S V P V
T P
y x y x y x y x y x
                     
                                     
….. (9)
U S T S V P V
T P
x y x y x y x y x y
                     
                                     
….. (10)
As dU is perfect differential, therefore
U
y x
  
 
  
=
U
x y
  
   
S T S V P V S T S V P V
T P T P
y x y x y x y x x y x y x y x y
                                     
                                                                  
Also dS & dV are perfect differential, therefore
and
S S V V
y x x y y x x y
             
                     
Above eq becomes
T S P V T S P V
y x y x x y x y
                     
                                     
…… (11)
This is Maxwell’s general thermodynamic equation.
1) Let x =S and y=V, therefore from eq 11, we have
T S P V T S P V
V S V S S V S V
                  
             
                  
as 0
V S
S V
    
    
    
and 1
S V
S V
    
    
    
S V
T P
V S
    
     
    
This is known as Maxwell’s first thermodynamic relation.
2) Let x =T and y=V, therefore from eq 11, we have

7. 7 | U I I I , S e m e s t e r - I I ( P A N a g p u r e )
T S P V T S P V
V T V T T V T V
                  
             
                  
as 0
T V
V T
    
    
    
and 1
T V
T V
    
    
    
T V
S P
V T
    
    
    
This is known as Maxwell’s second thermodynamic relation.
3) Let x =S and y=P, therefore from eq 11, we have
T S P V T S P V
P S P S S P S P
                  
             
                  
as 0
P S
S P
    
    
    
and 1
S P
S P
    
    
    
S P
T V
P S
    
    
    
This is known as Maxwell’s third thermodynamic relation.
4) Let x =T and y=P, therefore from eq 11, we have
T S P V T S P V
P T P T T P T P
                  
             
                  
as 0
T P
P T
    
    
    
and 1
T P
T P
    
    
    
T P
S V
P T
    
    
    
This is known as Maxwell’s fourth thermodynamic relation.
Thermodynamic Potentials:
There are four thermodynamic potentials
i) Internal energy, U
ii) Helmholtz free energy, F=U – TS
iii) Enthalpy, H=U + PV
iv) Gibbs free energy, G = U – TS + PV
1. Internal energy, U :
The internal energy is the total en energy of a system. It is called first thermodynamic potential.
According to combined form of first and second law of thermodynamics
dU = TdS – PdV
This equation gives change in internal energy in terms of four state variables P, V, T & S.
For an isochoric-adiabatic process:
dV = 0 and dQ = 0
dU = 0 or U is constant.
2. Helmholtz free energy, F:
It is defined as F=U – TS
As U, T & S are state variables, F is also state variable. It has dimensions of energy.
( )dF dU d TS
dF dU TdS SdT
  
   

8. 8 | U I I I , S e m e s t e r - I I ( P A N a g p u r e )
As dU = TdS – PdV
dF PdV SdT   
For an isochoric-isothermal process:
dV = 0 and dT = 0
0 or is constant.dF F 
3. Enthalpy, H:
It is defined as H=U + PV
As U, P & V are state variables, H is also state variable. It has dimensions of energy.
( )dH dU d PV
dH dU PdV VdP
  
   
As dU = TdS – PdV
dH TdS VdP  
For an isobaric-adiabatic process process:
dP = 0 and dQ = TdS = 0
0dH  or H is constant
4. Gibbs free energy, G:
It is defined as G=U + PV – TS
As U, P, V, T & S are state variables, G is also state variable. It has dimensions of energy.
( ) ( )dG dU d PV d TS
dG dU PdV VdP TdS SdT
   
     
As dU = TdS – PdV
dG VdP SdT  
For an isobaric-isothermal process process:
dP = 0 and dT = 0
0dG  or G is constant
Significance of Thermodynamic Potentials:
A mechanical system is said to be in stable equilibrium when potential energy of the system is
minimum. It means that the system must proceeds in such a direction so as to acquire minimum
potential energy.
In thermodynamics, the behavior of U, F, H & G is similar to potential energy in mechanics. As the
direction of isochoric –isothermal process is to make F minimum, the direction of isobaric –adiabatic
process is to make H minimum and the direction of isobaric –isothermal process is to make G
minimum. Since four functions U, F, H & G play same role in thermodynamics as played by potential
energy in mechanics, hence they are called thermodynamic potentials.
Applications of Maxwell’s Thermodynamic Relations:
1. Joule Thomson effect:
We know that in Joule Thomson expansion enthalpy of the gas remains constant i.e.
H=U + PV = constant
( ) 0
0
dH dU d PV
dH dU PdV VdP
   
    
As dU = TdS – PdV

9. 9 | U I I I , S e m e s t e r - I I ( P A N a g p u r e )
0dH TdS VdP   
Now dS being perfect differential and S is a function of P and T i.e. S = f(P,T), we have
P T
S S
dS dT dP
T P
    
    
    
Substituting in above equation
0
P T
S S
T dT T dP VdP
T P
    
      
    
0
P T
S S
T dT T V dP
T P
     
            
0
P T
Q S
dT T V dP
T P
     
            
As P
P P
S Q
T C
T T
    
    
    
P
T
S
C dT T V dP
P
  
       
According to Maxwell’s fourth relation
T P
S V
P T
    
    
    
P
P
V
C dT T V dP
T
  
      
1
PP
dT V
T V
dP C T
  
      
As
H
dT
dP
  
  
 
is Joule Thomson coefficient
1
PP
V
T V
C T

  
      
2. Clausius - Clapeyron Latent Heat Equation:
Maxwell’s second thermodynamic relation is given by
T V
S P
V T
    
    
    
From the second law of thermodynamics
𝜕𝑆 =
𝜕𝑄
𝑇
∴
1
𝑇
(
𝜕𝑄
𝜕𝑉
)
𝑇
= (
𝜕𝑃
𝜕𝑇
)
𝑉
∴ (
𝜕𝑄
𝜕𝑉
)
𝑇
= 𝑇 (
𝜕𝑃
𝜕𝑇
)
𝑉

10. 10 | U I I I , S e m e s t e r - I I ( P A N a g p u r e )
This equation shows that the heat absorbed per unit increase in volume at constant temperature is
equal to the product of the absolute temperature and rate of increase of pressure per unit increase in
temperature at constant volume.
Let L be the latent heat of fusion (or vaporization) of the substance. Let V1 & V2 be the specific
volumes of the substance at its melting temperature (or boiling temperature). i.e.
(
𝜕𝑄
𝜕𝑉
)
𝑇
=
𝐿
𝑉2 − 𝑉1
From above equations we have
(
𝜕𝑃
𝜕𝑇
)
𝑉
=
𝐿
𝑇(𝑉2 − 𝑉1)
This is called Clausius - Clapeyron latent heat equation.