Thermodynamics and Chemistry

CHAPTER 6 THE THIRD LAW AND CRYOGENICS

6.3 CRYOGENICS 157

side of a porous plug and at a constant lower pressurep^00 at the right side. Because of the
pressure difference, the gas flows continuously from left to right through the plug. The flow
is slow, and the pressure is essentially uniform throughout the portion of the tube at each
side of the plug, but has a large gradient within the pores of the plug.
After the gas has been allowed to flow for a period of time, a steady state develops in
the tube. In this steady state, the gas is assumed to have a uniform temperatureT^0 at the left
side of the plug and a uniform temperatureT^00 (not necessarily equal toT^0 ) at the right side
of the plug.
Consider the segment of gas whose position at timest 1 andt 2 is indicated by shading
in Fig.6.2. This segment contains a fixed amount of gas and expands as it moves through
the porous plug from higher to lower pressure. We can treat this gas segment as aclosed
system. During the interval between timest 1 andt 2 , the system passes through a sequence
of different states, none of which is an equilibrium state since the process is irreversible. The
energy transferred across the boundary by heat iszero, because the tube wall is insulated
and there is no temperature gradient at either end of the gas segment. We calculate the
energy transferred by work at each end of the gas segment from∂wD pbAsdx, where
pbis the pressure (eitherp^0 orp^00 ) at the moving boundary,Asis the cross-section area of
the tube, andxis the distance along the tube. The result is

wDp^0 .V 20 V 10 /p^00 .V 200 V 100 / (6.3.1)

where the meaning of the volumesV 10 ,V 20 , and so on is indicated in the figure.
The internal energy changeÅUof the gas segment must be equal tow, sinceqis zero.
Now let us find the enthalpy changeÅH. At each instant, a portion of the gas segment is in
the pores of the plug, but this portion contributes an unchanging contribution to bothUand
Hbecause of the steady state. The rest of the gas segment is in the portions on either side
of the plug, with enthalpiesU^0 Cp^0 V^0 at the left andU^00 Cp^00 V^00 at the right. The overall
enthalpy change of the gas segment must be

ÅHDÅUC.p^0 V 20 Cp^00 V 200 /.p^0 V 10 Cp^00 V 100 / (6.3.2)

which, when combined with the expression of Eq.6.3.1forwD ÅU, shows thatÅH
iszero. In other words, the gas segment has the same enthalpy before and after it passes
through the plug: the throttling process isisenthalpic.
The temperaturesT^0 andT^00 can be measured directly. When values ofT^00 versusp^00
are plotted for a series of Joule–Thomson experiments having the same values ofT^0 and
p^0 and different values ofp^00 , the curve drawn through the points is a curve of constant
enthalpy. The slope at any point on this curve is equal to theJoule–Thomson coefficient
(or Joule–Kelvin coefficient) defined by

JTdefD



@T

@p



H

(6.3.3)

For an ideal gas,JTis zero because the enthalpy of an ideal gas depends only onT(Prob.

5. 1 );T cannot change ifH is constant. For anonidealgas,JTis a function ofT and
   pand the kind of gas.^6 For most gases, at low to moderate pressures and at temperatures

(^6) See Sec.7.5.2for the relation of the Joule–Thomson coefficient to other properties of a gas.