Long Term Courses *    67 Selections in IIT JEE 2012 *    192 Selections in AIEEE 2012 *    19 Selections in AIPMT 2012 *    330 Students score 90% + in CBSE-XII Board 2013 *    445 Students score 10 CGPA in CBSE-X Board 2013
Attend Free Demo Class Now
Contact Number
Email ID
Contact our Toll Free Nos
  • Bahrain (National) : 973-16198627
  • Indonesia Toll Free : 1-803-015-204-5864
  • Singapore (National) : 65-31586005
  • USA/Canada Toll Free : 1-888-442-3128
  • Others : +91-9899713975
operations_kei 09718199348 Login Reg
IITians @ your home
ISO 9001:2008 Certified


The Joule–Thomson Experiment

For an ideal gas U depends only on T and n, so that H depends only on T and n:
Equation 17
Equation 18
Joule and Thomson carried out experiments in the 1850s to determine the value of (∂H/∂P)T,n for real gases. They used an apparatus equivalent to that sketched in Figure 8. This apparatus consists of two cylinders fitted with pistons and a porous
The Apparatus for the Joule–Thomson Experiment (Schematic)
Figure 8 The Apparatus for the Joule–Thomson Experiment (Schematic).

plug separating the two cylinders. Each side has a manometer to measure the pressure and a thermometer to measure the temperature of the gas. The entire apparatus is adiabatically insulated from the surroundings. During the experiment, one of the pistons is moved into its cylinder and the other piston is withdrawn in a way such that the pressure on each side remains constant as the gas flows irreversibly through the porous plug. If the process is slow enough a time-independent nonequilibrium state (a steady state) will be attained with a different constant temperature and constant pressure on each side. Several experiments are carried out on a given gas with different values of the pressure difference. The Joule–Thomson coefficient μJT is defined as the extrapolated limit
Equation 19
where the subscripts R and L indicate the right and left sides of the apparatus. We now show that the limit in this equation is a partial derivative at fixed enthalpy.
We choose as our system a sample of n moles of gas that flows through the porous plug after a steady state is established. We assume that irreversible processes take place only within the porous plug so that the initial and final states of our system can be treated as equilibrium states. The work done on the gas on the left side is given by
Equation 20
where VL1 is the initial volume of the left side of the apparatus and VL2 is the final volume of the left side of the apparatus. The work done on the gas on the right side is given by
Equation 21
Prior to the transfer, the pressure P1 of the system was equal to PL, and the initial volume of the system must have been equal to the magnitude of the change in volume of the left side:
Equation 22
The final pressure P2 must be equal to PR, and the final volume V2 must be equal to the change in volume of the right side:
Equation 23
From Eqs. (20) through (23), the total work done on the system is
Equation 24
Any change in state
Equation 25
Because the apparatus is adiabatically insulated from the laboratory, no heat is transferred to or from the laboratory. Also, no heat is transferred from the system to the apparatus after the steady state is established, because the chamber on the right is then at the same temperature as the gas that exits from the plug. Therefore,
Equations 26, 27, 28
The Joule–Thomson process therefore occurs at constant enthalpy, and the Joule–Thomson coefficient is equal to a partial derivative at constant H and n:
Equation 29
We can use the cycle rule,
Equation 30
The Joule–Thomson coefficient of an ideal gas vanishes because (∂H/∂P)T,n vanishes. Joule and Thomson found that the Joule–Thomson coefficient is measurably different from zero for ordinary gases at ordinary pressures. It depends on temperature and is positive at room temperature for most common gases except for hydrogen and helium. Even for these gases it is positive at some range of temperatures below room temperature. This means that for some range of temperature any gas cools on expansion through a porous plug. Expansion of a gas can be used to cool the gas enough to liquefy part of it, and the final step in the production of liquid nitrogen or liquid helium is ordinarily carried out in this way.

Example 23

For air at 300K and 25 atm, μJT  0.173K atm−1. If a Joule–Thomson expansion is carried out from a pressure of 50.00 atm to a pressure of 1.00 atm, estimate the final temperature if the initial temperature is equal to 300 K.



The molecular explanation for the fact that the Joule–Thomson coefficient is positive at sufficiently low temperature is that at low temperatures the attractive intermolecular forces are more important than the repulsive intermolecular forces. When the gas expands, work must be done to overcome the attractions and the potential energy increases. If no heat is added, the kinetic energy decreases and the temperature decreases.

Do you like this Topic?

Share it on

Online Classroom Program 1 on 1 Online Classes Correspondence Course Best Pool of Faculty Result Oriented Coaching Program

Welcome to Kshitij Education India

Our Guarantee:

We're so sure you'll have the time of your life with us, we back our courses with a 100% Satisfaction Guarantee.

If for any reason you aren't 100% satisfied with your classes in first 7 days, just let us know and we'll refund your fees. No questions asked.

And based on your feedback, we will take the necessary steps to ensure we never repeat any mistakes as such.

Satisfaction Guaranteed
Free Demo Live Chat