PROBLEMS
2.7 THE JOULE AND JOULE–THOMSON EXPERIMENTS
In 1843 Joule tried to determine (U/V)T for a gas by measuring the temperature change after free expansion of the gas into a vacuum. This experiment was repeated by Keyes and Sears in 1924 with an improved setup (Fig. 2.6).
Initially, chamber A is filled with a gas, and chamber B is evacuated. The valve between the chambers is then opened. After equilibrium is reached, the temperature change in the system is measured by the thermometer. Because the system is sur-rounded by adiabatic walls, q is 0; no heat flows into or out of the system. The expansion into a vacuum is highly irreversible. Finite unbalanced forces act within the system, and as the gas rushes into B, there is turbulence and lack of pressure equilibrium. Therefore dw P dV does not apply. However, we can readily cal-culate the work w done by the system. The only motion that occurs is within the system itself. Therefore the gas does no work on its surroundings, and vice versa.
Hence w 0 for expansion into a vacuum. Since U q w for a closed system, we have U 0 0 0. This is a constant-energy process. The experiment mea-sures the temperature change with change in volume at constant internal energy, (T/V)U. More precisely, the experiment measures T/V at constant U. The method used to get (T/V)Ufrom T/V measurements is similar to that described later in this section for (T/P)H.
We define the Joule coefficient mJ(mu jay) as
(2.62) How is the measured quantity (T/V)U mJrelated to (U/V)T? The variables in these two partial derivatives are the same (namely, T, U, and V ). Hence we can use
mJ⬅ 10T>0V2U
CP CV c a0U 0Vb
T
Pd a0V 0Tb
P
a0U 0Tb
P
a0U 0Tb
V
a0U 0Vb
T
a0V 0Tb
P
dUP
dTP a0U 0Tb
V
a0U 0Vb
T
dVP dTP
Section 2.7 The Joule and Joule–Thomson Experiments
55
Figure 2.5
Specific heat of H2O(g) plotted versus T and versus P.
Figure 2.6
The Keyes–Sears modification of the Joule experiment.
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The First Law of Thermodynamics 56
(x/y)z(y/z)x(z/x)y 1 [Eq. (1.34)] to relate these partial derivatives.
Replacement of x, y, and z with T, U, and V gives
(2.63) where (z/x)y 1/(x/z)y, (U/T)V CV, and mJ (T/V)U[Eqs. (1.32), (2.53), and (2.62)] were used.
Joule’s 1843 experiment gave zero for mJand hence zero for (U/V)T. However, his setup was so poor that his result was meaningless. The 1924 Keyes–Sears experi-ment showed that (U/V)T is small but definitely nonzero for gases. Because of experimental difficulties, only a few rough measurements were made.
In 1853 Joule and William Thomson (in later life Lord Kelvin) did an experiment similar to the Joule experiment but allowing far more accurate results to be obtained.
The Joule–Thomson experiment involves the slow throttling of a gas through a rigid, porous plug. An idealized sketch of the experiment is shown in Fig. 2.7. The system is enclosed in adiabatic walls. The left piston is held at a fixed pressure P1. The right piston is held at a fixed pressure P2 P1. The partition B is porous but not greatly so.
This allows the gas to be slowly forced from one chamber to the other. Because the throttling process is slow, pressure equilibrium is maintained in each chamber.
Essentially all the pressure drop from P1to P2occurs in the porous plug.
We want to calculate w, the work done on the gas in throttling it through the plug.
The overall process is irreversible since P1 exceeds P2 by a finite amount, and an infinitesimal change in pressures cannot reverse the process. However, the pressure drop occurs almost completely in the plug. The plug is rigid, and the gas does no work on the plug, or vice versa. The exchange of work between system and surroundings occurs solely at the two pistons. Since pressure equilibrium is maintained at each pis-ton, we can use dwrev P dV to calculate the work at each piston. The left piston does work wLon the gas. We have dwL PLdV P1dV, where we use subscripts L and R for left and right. Let all the gas be throttled through. The initial and final vol-umes of the left chamber are V1and 0, so
The right piston does work dwR on the gas. (wR is negative, since the gas in the right
chamber does positive work on the piston.) We have The
work done on the gas is w wL wR P1V1 P2V2.
wR
兰
0V2 P2dV P2V2. wL冮
V01
P1 dV P1
冮
V01
dV P110 V12 P1V1 a0U
0Vb
T
CVmJ a0U
0Vb
T
c a0T 0Ub
Vd1c a0V 0Tb
Ud1 a0U 0Tb
Va0T 0Vb
U
a0T 0Ub
V
a0U 0Vb
T
a0V 0Tb
U
1
Porous Plug
(a)
P1 P2
P1, V1, T1
(b) Adiabatic Wall
B
P1 P2
P2 P1
(c)
P1 P2
P2, V2, T2
Figure 2.7
The Joule–Thomson experiment.
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The first law for this adiabatic process (q 0) gives U2 U1 q w w, so U2 U1 P1V1 P2V2or U2 P2V2 U1 P1V1. Since H⬅ U PV, we have
The initial and final enthalpies are equal in a Joule–Thomson expansion.
Measurement of the temperature change T T2 T1 in the Joule–Thomson experiment gives T/P at constant H. This may be compared with the Joule experi-ment, which measures T/V at constant U.
We define the Joule–Thomson coefficient mJTby
(2.64)*
mJTis the ratio of infinitesimal changes in two intensive properties and therefore is an intensive property. Like any intensive property, it is a function of T and P (and the na-ture of the gas).
A single Joule–Thomson experiment yields only (T/P)H. To find (T/P)H val-ues, we proceed as follows. Starting with some initial P1and T1, we pick a value of P2less than P1and do the throttling experiment, measuring T2. We then plot the two points (T1, P1) and (T2, P2) on a T-P diagram; these are points 1 and 2 in Fig. 2.8.
SinceH 0 for a Joule–Thomson expansion, states 1 and 2 have equal enthalpies.
A repetition of the experiment with the same initial P1and T1but with the pressure on the right piston set at a new value P3 gives point 3 on the diagram. Several repe-titions, each with a different final pressure, yield several points that correspond to states of equal enthalpy. We join these points with a smooth curve (called an isen-thalpic curve). The slope of this curve at any point gives (T/P)Hfor the tempera-ture and pressure at that point. Values of T and P for which mJTis negative (points to the right of point 4) correspond to warming on Joule–Thomson expansion. At point 4, mJTis zero. To the left of point 4, mJTis positive, and the gas is cooled by throt-tling. To generate further isenthalpic curves and get more values of mJT(T, P), we use different initial temperatures T1.
Values of mJTfor gases range from 3 to 0.1°C/atm, depending on the gas and on its temperature and pressure. Figure 2.9 plots some mJTdata for N2gas.
Joule–Thomson throttling is used to liquefy gases. For a gas to be cooled by a Joule–Thomson expansion (P 0), its mJTmust be positive over the range of T and P involved. In Joule–Thomson liquefaction of gases, the porous plug is replaced by a narrow opening (a needle valve). Another method of gas liquefaction is an approxi-mately reversible adiabatic expansion against a piston.
A procedure similar to that used to derive (2.63) yields (Prob. 2.35a)
(2.65) We can use thermodynamic identities to relate the Joule and Joule–Thomson coeffi-cients; see Prob. 2.35b.
a0H 0Pb
T
CPmJT mJT⬅ a0T
0Pb
H
H2 H1 or ¢H 0
Section 2.7 The Joule and Joule–Thomson Experiments
57
Figure 2.8
An isenthalpic curve obtained from a series of Joule–Thomson experiments.
Figure 2.9
The Joule–Thomson coefficient of N2(g) plotted versus P and versus T.
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The First Law of Thermodynamics 58