Much of the early progress in thermodynamics was made in the study of the properties of gases. One of the early questions was whether or not gases cool on expansion. (Our intuition might tell us that they would, but is our intuition correct?)

Joule designed an experiment to find out whether or not gases cool on expansion and if so how much.

The Joule apparatus consisted of two glass bulbs connected by a stopcock.
One bulb was filled with gas at some *p* and *T*. The other bulb
was evacuated. The entire apparatus was insulated so that *q* = 0.
That is, the experiment would be adiabatic.

The stopcock was opened to allow the gas to expand into the adjoining
bulb. Since the gas was expanding against zero pressure no work was done,
*w* = 0. With both *q* = 0 and *w* = 0 it is clear that,

ΔThe process is at constant internal energy.U=q+w= 0.

Clearly, Δ*V* ≠
0 because the gas expanded to fill both bulbs. The question was, did *T*
change? Δ*T* was measured to be zero, no
temperature change.

(It turns out that the Joule experiment was sufficiently crude that it could not detect the difference between an ideal gas and a real gas so that the conclusions we will draw from this experiments only apply to an ideal gas.)

In effect, Joule was trying to measure the derivative,

and the result was that,(1)This particular derivative is not all that instructive, with

(2a, b, c)We know that

(3)This is an important and useful result. It says that the internal energy of an ideal gas is not a function of

for an ideal gasFor real gases, and most approximations to real gases, like the van der Waals equation of state, However, this quantity is quite small, even for real gases. We will have occasion to calculate it for the van der Waals equation of state later on.U=U(T). (4)

This result extends to the enthalpy of an ideal gas.

Thus, for an ideal gas bothH=U+pV=U(T) +nRT=H(T). (5)

Then all of the following derivatives are zero:

(6a, b, c, d)We will now use some of these results to discuss that adiabatic expansion of an ideal gas.

**Adiabatic Expansion of an Ideal Gas**

The definition of an adiabatic expansion, for now, is *dq* = 0.
That is, no heat goes in or out of the system. However, *dw* ≠
0. As the gas expands it does work on the surroundings. Since the gas is
cut off from any heat bath it can not draw heat from any source to convert
into work. The work must come from the internal energy of the gas so that
the internal energy decreases. Since the internal energy of an ideal gas
in only dependent on *T* that means that the temperature of the gas
must decrease.

From the first law with only *pV* work we have

(7a, b)because

Regarding *U* as a function of *T* and *V*. That is,
*U* = *U*(*T*,*V*), we get

(8a, b)because of the definition of

The *dU* 's in Equations (7) and (8) must be equal so that

(9a, b)Rearranging Equation (9b) we get

(10)By the same token, using enthalpy, we find

(11a, b)and

(12a, b)From which we deduce that

(13)Comparing Equations (10) and (13) we see that

(14a, b, c)Where we have written

If we regard *C _{p}* and

(15a, b, c, d, e)Equation (15e) is the equation for the adiabatic expansion of an ideal gas. You have probably seen it before.

**Adiabatic Work - Ideal Gas**

We can use Equation (15e) as the equation for an adiabatic path on a
*pV* diagram.

(16a, b, c)where

(17a, b, c, d)The constant is easily found from the knowledge of one point on the adiabatic line (path).

**The Joule-Thompson Expansion**

It soon became apparent that the result of the Joule expansion experiment was not valid for real gases. A more accurate experiment, slightly different, was carried out by Joule and J. J. Thompson to further elucidate the properties on real gases under expansion.

A sample of a gas, initially at *p*_{1}, *V*_{1},
and *T*_{1} was forced through a porous plug at constant pressure,
*p*_{1}. The gas came out of the other side of the plug at
*p*_{2}, *V*_{2}, and *T*_{2}. The
apparatus was insulated so that *q* = 0. The work has two terms, the
work done on the system to force the gas through the plug and the work
done by the system on the surroundings as it came out the other side of
the plug.

The total work is

(18a, b)Since

(19a, b, c)This process, unlike the Joule expansion, is not at constant internal energy.

The enthalpy, however, is given by,

(20a, b, c)So the Joule Thompson experiment is a process at constant enthalpy. In the experiment they could select a value for Δ

(21)

(22a, b)The numerator in Equation (22b) is zero for an ideal gas, but not necessarily zero for a real gas.

The coefficient of the Joule-Thompson effect is important in the liquefaction
of gases because it tells whether a gas cools or heats on expansion. It
turns out that this coefficient is a decreasing function of temperature
and it passes through zero at the Joule-Thompson inversion temperature,
*T*_{I}. In an expansion *dp* < 0. Whether *dT*
is positive or negative depends on the sign of *μ*_{JT}.
Looking at the definition of *μ*_{JT},

,we see that if

He 40 K

N_{2}
621 K

O_{2}
764 K

Ne 231 K

We see that N_{2} and O_{2} will cool upon expansion
at room temperature, but He and Ne will warm upon expansion at room temperature.

WRS

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