1. In thermodynamics, the Joule–Thomson effect or Joule–Kelvin effect or Kelvin–Joule effect describes the
temperature change of a gas or liquid when it is forced through a valve or porous plug while kept insulated so
that no heat is exchanged with the environment. This procedure is called a throttling process or Joule–
Thomson process. At room temperature, all gases except hydrogen, helium and neon cool upon expansion by
the Joule–Thomson process.
Description
The adiabatic (no heat exchanged) expansion of a gas may be carried out in a number of ways. The change in
temperature experienced by the gas during expansion depends not only on the initial and final pressure, but
also on the manner in which the expansion is carried out.
If the expansion process is reversible, meaning that the gas is in thermodynamic equilibrium at all times, it is
called an isentropic expansion. In this scenario, the gas does positive work during the expansion, and its
temperature decreases.
In a free expansion, on the other hand, the gas does no work and absorbs no heat, so the internal energy is
conserved. Expanded in this manner, the temperature of an ideal gas would remain constant, but the
temperature of a real gas may either increase or decrease, depending on the initial temperature and pressure.
The method of expansion discussed in this article, in which a gas or liquid at pressure P1 flows into a region of
lower pressure P2 via a valve or porous plug under steady state conditions and without change in kinetic
energy, is called the Joule–Thomson process. During this process, enthalpy remains unchanged (see a proof
below).
A throttling process proceeds along a constant-enthalpy line in the direction of decreasing pressure, which
means that the process occurs from left to right on a T-P diagram. As we proceed along a constant-enthalpy
line from high enough pressures the temperature increases, until the inversion temperature. Then as the fluid
continues its expansion the temperature drops. If we do this for several constant enthalpies and join the
inversion points a line called the inversion line is obtained. this line intersects the T-axis at some temperature,
named the maximum inversion temperature. For hydrogen this temperature is -68°. In Vapour-compression
refrigeration we need to throttle the gas and cool it at the same time. This poses a problem for substances
whose maximum inversion temperature is well below room temperature. Thus hydrogen must be cooled
below its inversion temperature if any cooling is achieved by throttling.
Physical mechanism
As a gas expands, the average distance between molecules grows. Because of intermolecular attractive forces
(see Van der Waals force), expansion causes an increase in the potential energy of the gas. If no external work
is extracted in the process and no heat is transferred, the total energy of the gas remains the same because of
the conservation of energy. The increase in potential energy thus implies a decrease in kinetic energy and
therefore in temperature.
A second mechanism has the opposite effect. During gas molecule collisions, kinetic energy is temporarily
converted into potential energy. As the average intermolecular distance increases, there is a drop in the
number of collisions per time unit, which causes a decrease in average potential energy. Again, total energy is
conserved, so this leads to an increase in kinetic energy (temperature). Below the Joule–Thomson inversion
temperature, the former effect (work done internally against intermolecular attractive forces) dominates, and
free expansion causes a decrease in temperature. Above the inversion temperature, gas molecules move
faster and so collide more often, and the latter effect (reduced collisions causing a decrease in the average
potential energy) dominates: Joule–Thomson expansion causes a temperature increase.

2. The Joule–Thomson (Kelvin) coefficient
Joule-Thomson coefficients for various gases at atmospheric pressure
The rate of change of temperature T with respect to pressure P in a Joule–Thomson process (that is, at
constant enthalpy H) is the Joule–Thomson (Kelvin) coefficient μJT. This coefficient can be expressed in terms of
the gas's volume V, its heat capacity at constant pressure Cp, and its coefficient of thermal expansion α as:
See the Derivation of the Joule–Thomson (Kelvin) coefficient below for the proof of this relation. The value of
μJT is typically expressed in °C/bar (SI units: K/Pa) and depends on the type of gas and on the temperature and
pressure of the gas before expansion. Its pressure dependence is usually only a few percent for pressures up to
100 bar.
All real gases have an inversion point at which the value of μJT changes sign. The temperature of this point, the
Joule–Thomson inversion temperature, depends on the pressure of the gas before expansion.
In a gas expansion the pressure decreases, so the sign of is negative by definition.
With that in mind, the following table explains when the Joule–Thomson effect cools or warms a real gas:
If the gas temperature is then μJT is since is thus must be so the gas
below the inversion temperature positive always negative negative cools
above the inversion temperature negative always negative positive warms
Helium and hydrogen are two gases whose Joule–Thomson inversion temperatures at a pressure of one
atmosphere are very low (e.g., about 51 K (−222 °C) for helium). Thus, helium and hydrogen warm up when
expanded at constant enthalpy at typical room temperatures. On the other hand nitrogen and oxygen, the two
most abundant gases in air, have inversion temperatures of 621 K (348 °C) and 764 K (491 °C) respectively:
these gases can be cooled from room temperature by the Joule–Thomson effect.
For an ideal gas, μJT is always equal to zero: ideal gases neither warm nor cool upon being expanded at
constant enthalpy.

3. Applications
In practice, the Joule–Thomson effect is achieved by allowing the gas to expand through a throttling device
(usually a valve) which must be very well insulated to prevent any heat transfer to or from the gas. No external
work is extracted from the gas during the expansion (the gas must not be expanded through a turbine, for
example).
The effect is applied in the Linde technique as a standard process in the petrochemical industry, where the
cooling effect is used to liquefy gases, and also in many cryogenic applications (e.g. for the production of liquid
oxygen, nitrogen, and argon). Only when the Joule–Thomson coefficient for the given gas at the given
temperature is greater than zero can the gas be liquefied at that temperature by the Linde cycle. In other
words, a gas must be below its inversion temperature to be liquefied by the Linde cycle. For this reason, simple
Linde cycle liquefiers cannot normally be used to liquefy helium, hydrogen, or neon.
[edit] Proof that enthalpy remains constant in a Joule–Thomson process
In a Joule–Thomson process the enthalpy remains constant. To prove this, the first step is to compute the net
work done by the gas that moves through the plug. Suppose that the gas has a volume of V1 in the region at
pressure P1 (region 1) and a volume of V2 when it appears in the region at pressure P2 (region 2). Then the
work done on the gas by the rest of the gas in region 1 is P1V1. In region 2 the amount of work done by the gas
is P2 V2. So, the total work done by the gas is
The change in internal energy plus the work done by the gas is, by the first law of thermodynamics, the total
amount of heat absorbed by the gas (here it is assumed that there is no change in kinetic energy). In the Joule–
Thomson process the gas is kept insulated, so no heat is absorbed. This means that
where E1 and E2 denote the internal energy of the gas in regions 1 and 2, respectively.
Using the definition of enthalpy H = E + PV, the above equation then implies that:
where H1 and H2 denote the enthalpy of the gas in regions 1 and 2, respectively.
Derivation of the Joule–Thomson (Kelvin) coefficient
A derivation of the formula
for the Joule–Thomson (Kelvin) coefficient.
The partial derivative of T with respect to P at constant H can be computed by expressing the differential of
the enthalpy dH in terms of dT and dP, and equating the resulting expression to zero and solving for the ratio
of dT and dP.
It follows from the fundamental thermodynamic relation that the differential of the enthalpy is given by:
(here, S is the entropy of the gas).

4. Expressing dS in terms of dT and dP gives:
Using
(see Specific heat capacity),
we can write:
The remaining partial derivative of S can be expressed in terms of the coefficient of thermal expansion via a
Maxwell relation as follows. From the fundamental thermodynamic relation, it follows that the differential of
the Gibbs energy is given by:
The symmetry of partial derivatives of G with respect to T and P implies that:
where α is the coefficient of thermal expansion. Using this relation, the differential of H can be expressed as
Equating dH to zero and solving for dT/dP then gives:
It is easy to verify that for an ideal gas the thermal expansion coefficient α is 1 / T, and so an ideal gas does not
experience a Joule–Thomson effect. The cooling of a gas by pure isentropic expansion is not Joule-Thomson
cooling, although it is sometimes erroneously called J-T cooling by some laboratory practitioners.