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Joule Thomson effect, Slides of Thermodynamics

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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Joule Thomson effect
Paper: Physical Chemistry
(UNIT-I Thermodynamics)
For the students of BSc II
By
Dr Lokesh Kumar Agarwal
Assistant Professor
Department of Chemistry
Mohanlal Sukhadia University, Udaipur
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Joule Thomson effect

Paper: Physical Chemistry

(UNIT-I Thermodynamics)

For the students of BSc II

By

Dr Lokesh Kumar Agarwal

Assistant Professor

Department of Chemistry

Mohanlal Sukhadia University, Udaipur

TABLE OF CONTENTS

1. Learning Outcomes

2. Introduction

3. Joule-Thomson effect

4. Joule Thomson coefficient

4.1 Joule Thomson coefficient for an ideal gas

4.2 Joule Thomson coefficient for a real gas

5. Derivation of Joule Thomson coefficient and Inversion temperature

6. Summary

B.Sc. II

(CHEMISTRY)

Paper :Physical Chemistry- III

(UNIT- Thermodynamics-I)

Topic: Joule Thomson effect

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.

4. Joule Thomson Coefficient (μJ.T.)

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.

V

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

  • P 1

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

  • P 1

V 1

= U 2

− U 1

U 2

  • P 2

V 2

= U 1

  • P 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

= − (H / P)

T

/ C

P …. (6)

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.

5. Derivation of Joule Thomson coefficient and Inversion temperature

The Joule-Thomson coefficient can be calculated with the help of the van der Waals

equation. Considering the van der Waals equation i.e.

( P + 𝑎 n

2 /V

2

( P 𝑉 +

𝑎 n

2 /V − P 𝑛𝑏 −

𝑎bn

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

R

P

V − b

T

a

RT

2

Replacing the value of R/P from Eq. 16 in Eq. 14, we have

(V / T) P =

V−b

T

a

RT

2

  • a / RT

2

6. Summary

  • The temperature below which a gas becomes cooler on expansion is known as the

inversion temperature.

  • Thus Joule-Thomson effect can be defined as the phenomenon of

temperature change produced when a gas is allowed to expand adiabatically

from a region of high pressure to a region of extremely low pressure.

  • Joule-Thomson expansion of a real gas occurs with constant enthalpy and

not with constant internal energy and thus is also known as isenthalpic

process.

  • Joule Thomson coefficient is given by (T / P) H

= − (H / P) T

/ C P

  • Joule Thomson coefficient for ideal gas is zero
  • Joule Thomson coefficient for real gas is given by

References:

1. epg Pathshala, module 5, (Classical Thermodynamics, Non-Equilibrium Thermodynamics,

Surface chemistry, Fast kinetics)

2. Atkins’ Physical Chemistry book