Joule-Thomson Expansion | Liquefaction | Free Expansion | Adiabatic Expansion

Imagine a gas confined within an insulated container as
shown in the figure below. The gas is initially confined to a volume V_{1}
at pressure P_{1} and temperature T_{1}. The gas then is
allowed to expand into another insulated chamber with volume V_{2}
that is initially evacuated. What happens? Let’s apply the first law.

We know from the first law for a closed system that the
change in internal energy of the gas will be equal to the heat transferred
plus the amount of work the gas does, or
_{.
}
Since the gas expands freely (the volume change of the system
is zero), we know that no work will be done, so W=0. Since both chambers
are insulated, we also know that Q=0. Thus, ** the
internal energy of the gas does not change during this process**.

We would like to know what happens to the temperature of
the gas during such an expansion. To proceed, we recall that U is a *state function*.
This means that we only need to know the inital and final states of U to obtain the change in U.
The difference in U is *independent of path*. Thus we can use the powerful thermodynamic concept of
constructing an imaginary reversible path that connects the initial and final states of the gas.
Recall that Q and W by themselves are *not* state functions! Recall also that the
*actual* free expansion is *not* a reversible process, and we
can’t apply thermodynamics to the gas during the expansion. However, once
the system has settled down and reached equilibrium after the expansion,
we can apply thermodynamics to the final state.

We know that the internal energy depends upon both temperature and volume, so we write

_{
}
(1)

where we have kept the number of molecules in the gas (N) constant. The first term on the right side in equation (1) simply captures how U changes with T at constant V, and the second term relates how U changes with V and constant T. We can simplify this using Euler’s reciprocity relation, (where x,y,z are U,V,T):

_{}
(2)

to obtain an expression for the change in gas temperature

_{}
(3)

The term _{}is
a property of the gas, and is called the differential Joule coefficient.
This name is in honor of James Prescott Joule, who performed experiments
on the expansion of gases in the mid-nineteenth century. If we can either
measure or compute the differential Joule coefficient, we can then say how
temperature changes (dT) with changes in volume (dV). Let’s see how
we might compute the Joule coefficient from an equation of state. The
simplest possible equation of state is the ideal gas, where PV=nRT.
The easiest way to find the Joule coefficient is to compute _{}
and _{}.
Note that we have left off the subscript “N” for brevity, but we still
require that the number of molecules in our system is constant.

We can use the following identity

_{}
(4)

to show that _{}
so that _{ .}

If the gas is described by the van der Waals equation of state

_{}
(5)

you can show that the term in the numerator of equation (3) is given by

_{}
(6)

Think about what equation (6) is telling us. Recall
that the parameter “a” in the van der Waals equation of state accounts
for attractive interactions between molecules. Equation (6) therefore
states that the internal energy of a system expanded at constant temperature
*will* change, and this change is due to *attractive interactions
between
molecules*. Since the ideal gas equation of state neglects these
interactions, it predicts no change in the internal energy upon expansion
at constant temperature, but the van der Waals equation of state does
account for this. The term in the denominator of equation (3) is nothing
more than the constant volume heat capacity _{}.
It can be shown that C_{V} is never negative and only depends
upon temperature for the van der Waals equation of state. Since the
parameter *a* is also never negative, equations (3) and (6) tell
us that ** the temperature of a real gas
will always decrease upon undergoing a free expansion**. How much the temperature decreases depends
upon the state point and the parameter

We can compute how much the temperature is expected to decrease during a free expansion using the van der Waals equation of state. If one performs this calculation for the expansion of oxygen from 10 bar at 300 K into a vacuum, the temperature is found to be reduced by roughly 4.4 K. Therefore, free expansion is not a very efficient means of achieving low temperatures. To liquefy gases, we need to use a different process.

Joule-Thomson Expansion | Liquefaction | Free Expansion | Adiabatic Expansion