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,
ΔU = q + w = 0.The process is at constant internal energy.
Clearly, ΔV ≠ 0 because the gas expanded to fill both bulbs. The question was, did T change? ΔT 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 U being held constant. We can use our version of Euler's chain relation to obtain information that is more instructive.
(2a, b, c)We know that CV for gases is neither zero nor infinity, so we must conclude that,
(3)This is an important and useful result. It says that the internal energy of an ideal gas is not a function of T and V, but of T only. That is, in equation form,
for an ideal gas U = U(T). (4)For 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.
This result extends to the enthalpy of an ideal gas.
H = U + pV = U(T) + nRT = H(T). (5)Thus, for an ideal gas both U and H are functions of T only.
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 dq = 0 for an adiabatic process.
Regarding U as a function of T and V. That is, U = U(T,V), we get
(8a, b)because of the definition of CV and because our gas is an ideal gas so that the second derivative vanishes (Equation (6a)) .
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
(12a, b)From which we deduce that
(13)Comparing Equations (10) and (13) we see that
(14a, b, c)Where we have written Cp/CV = γ .
If we regard Cp and CV as constant then Equation (14c) can be integrated to give,
(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 p1 and V1 refer to some arbitrary constant point on the path. Equation (16b) gives p as a function of V along the adiabatic line. We have added Equation (16c) just to emphasize the point that p1 and V1 refer to some fixed (constant) point on the adiabatic expansion curve. With this expression for p the work can be easily calculated,
(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 p1, V1, and T1 was forced through a porous plug at constant pressure, p1. The gas came out of the other side of the plug at p2, V2, and T2. 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 q = 0, the change in internal energy of the gas is,
(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 Δp, and then measure ΔT. The ratio of these two quantities is an approximation to a derivative,
(21)μJT is called the "coefficient of the Joule-Thompson effect." This coefficient is not zero for a real gas (or for realistic equations of state like the van der Waals equation of state), but we will now show that it is zero for an ideal gas. Applying the Euler chain rule to Equation (21) we obtain,
(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, TI. In an expansion dp < 0. Whether dT is positive or negative depends on the sign of μJT. Looking at the definition of μJT,
,we see that if μJT is positive then dT is negative upon expansion so that the gas cools. On the other hand, if μJT is negative, then dT is positive so that the gas warms upon expansion. In order to liquefy a gas by a Joule-Thompson expansion the gas must first be cooled to below the J-T inversion temperature. Some inversion temperatures are:
He 40 K
N2 621 K
O2 764 K
Ne 231 K
We see that N2 and O2 will cool upon expansion
at room temperature, but He and Ne will warm upon expansion at room temperature.
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Copyright 2004, W. R. Salzman
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