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This effect describes the change in temperature of a real gas or liquid when it is forced to pass through porous plug or valve while it is kept insulated so ...
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TABLE OF CONTENTS
which it expands is zero. From the first law equation we can say that U is also equal to
zero. Thus, when an ideal gas undergoes expansion under adiabatic conditions in vacuum,
no change takes place in its internal energy. In other words, the internal energy of a given
quantity of an ideal gas at a constant temperature is independent of its volume, i.e., (U / V) T
An ideal gas may, therefore, be defined thermodynamically by the following two
equations:
(i) PV = constant, at constant temperature
(ii) (U / V) T
The quantity (U / V) T
is called the internal pressure and it is zero for ideal gas.
Joule and Thomson derived the relationship between fall of pressure of gas on expansion
and resulting lowering of temperature by performing the following technique:
Fig. 1 The porous experiment
A tube made of a non-conducting material is fitted with a porous plug G in the middle
and two pistons A and B on the sides, as shown. The tube is thoroughly insulated to ensure
adiabatic conditions. Let the volume of gas enclosed between the piston A and the porous
plug G at pressure P 1 is V 1
. This volume is forced to pass through porous plug by moving
the piston A inwards. At the same time the volume of gas enclosed between porous plug G
and piston B i.e.
2 is allowed to expand at a lower pressure P 2
by moving the piston B
outward, as shown.
Therefore, work done on the system at the piston A = +P 1
V 1
and work done by the
system at the piston B = – P 2
V 2
Then, net work done by the system = – P 2
V 2
V 1
Since the expansion of the gas is done adiabatically i.e. no exchange of heat takes place
between the system and surroundings. Thus the work is done by the system at the expense of
internal energy only. Let the internal energy of the system changes from U 1
to U 2
−P 2
V 2
V 1
= U 2
− U 1
U 2
V 2
= U 1
V 1
H 2
= H 1
H 2
− H 1
= 0
i.e. H = 0 …(2)
This states that the Joule-Thomson expansion of a real gas occurs with constant enthalpy
and not with constant internal energy. According to this the process is known as isoenthalpic
process.
Taking enthalpy as the function of temperature and pressure, then the total differential of
the enthalpy H can be written as
dH = (H / P) T
dP + (H / T) P
dT ...(3)
But (H / T) P
= C P
Substituting equation (4) in (3) we get
dH = (H / P) T
dP + C P
dT
Since for adiabatic expansion, dH = 0, hence
(H / P) T
dP + C P
dT = 0
Rearranging the above equation dT/dP = (H / P) T
/ C P…..(5)
i.e., (T / P) H
The quantity (T / P) H
is called Joule-Thomson coefficient and it is denoted as J.T.
4.2 Joule-Thomson Coefficient for a Real Gas
There is no Joule Thomson effect for ideal gas but real gases do have Joule Thomson effect. So
we can conclude that the Joule Thomson effect depends internal energy is dependent on change
in volume. (U / V) T
is positive for real gas. This can be explain as follows:
Consider the expansion of real gas in vacuum. In such expansion no external work is done since
the external pressure is zero. But some work will definitely be done in separating the gas
molecules against the intermolecular forces of attraction (the van der Waals forces). This work
is stored in the gas in the form of potential energy, In other words, the potential energy of the
gas increases and hence kinetic energy decreases by an equivalent amount if no heat is
exchanged between the system and the surroundings. Due to this temperature of the gas falls.
The Joule-Thomson coefficient can be calculated with the help of the van der Waals
equation. Considering the van der Waals equation i.e.
2 /V
2
2 /V − P 𝑛𝑏 −
3 /V
2
In this equation, both terms a and b are small, hence the term ab/V
2
can be neglected when
the pressure of the system is low. By rearranging this equation for 1 mole and neglecting the
term ab/V
2
we get,
PV =RT – a/V + bP …(11)
Substituting V by RT/P (approximation) we have
PV = RT – aP/RT + bP …(12)
Dividing the whole equation by P, we get
V = RT/P – a/RT + b ...(13)
Differentiating with respect to temperature at constant pressure, we get
(V / T) P
Rearranging Eq. 12, we have
= R / P + a / RT
2
RT = P(V – b) + aP/RT …(15)
Dividing both sides of the above equation by PT, we get
2
Replacing the value of R/P from Eq. 16 in Eq. 14, we have
(V / T) P =
V−b
T
a
RT
2
2
inversion temperature.
temperature change produced when a gas is allowed to expand adiabatically
from a region of high pressure to a region of extremely low pressure.
not with constant internal energy and thus is also known as isenthalpic
process.
= − (H / P) T
/ C P
References:
1. epg Pathshala, module 5, (Classical Thermodynamics, Non-Equilibrium Thermodynamics,
Surface chemistry, Fast kinetics)
2. Atkins’ Physical Chemistry book