THE JOULE THOMSON EXPERIMENT - cryoelectric THE JOULE THOMSON EXPE… · THE JOULE – THOMSON EXPERIMENT ... the valve V was opened allowing the gas in A to expand in ... the Joule-Thomson

Roberto Giardelli . THE JOULE – THOMSON EXPERIMENT - pag 1

THE JOULE – THOMSON EXPERIMENT

During the first half of the nineteenth century, Joule attempted to measure the temperature

change that occurs when a gas expands in a vacuum. Joule designed an experiment to find

out if the gas cools in expansion and if so how much.

Figure 1. Joule Apparatus 1

The equipment Joule consisted of two glass bulbs connected by a tap.

A bulb of copper, the A to the left, was filled with gas at a certain pressure P and

temperature T and isolated from another bulb B contiguous, previously emptied, via an

adjustment valve V. Both bulbs were immersed in a water bath equipped with a

thermometer sensitive. After thermal equilibrium had been reached, after appropriate

mixing (see stirrer on the right), the valve V was opened allowing the gas in A to expand in

the bulb B adjacent.

Joule did not notice any change in temperature and therefore q = 0 where the experiment

took place under adiabatic conditions. Because the gas is expanded against zero pressure

was not performed no work, that is w = 0. Since, therefore, q = 0 and w = 0 is clear that,

then, ΔU = q + w = 0. The process takes place at constant internal energy U. Clearly,

ΔV ≠ 0 because the expanding gas has filled both bulbs. The question was:


the temperature T changes? The change ΔT was measured to be zero, that is no heat

exchange. (In fact, the Joule experiment was sufficiently rough so that it was not able to

detect the difference between an ideal gas and a real one so that the conclusions that can be

drawn from this experiment apply only to an ideal gas.)

In fact Joule tried to measure the variation of T with respect to V while maintaining the

internal energy U constant namely he set out to know experimentally the partial derivative

( ) and the result observed was ( ) = 0 implying, given the

relationship ( ) = ( ) ( ) , that ( ) = 0 i.e. that

the internal energy U is independent of the volume (and therefore of the pressure) at

constant temperature. Its measurement apparatus was not very sensitive to having a large

heat capacity than air, and was not in condition to detect, as has been said, no change of

temperature T within the range of error of the measurements. In reality ( ) 0

since the gas in the bulb A warmed slightly and that expanded in B was somewhat colder

and when thermal equilibrium was finally reached the gas was at a temperature slightly

different from that earlier expansion. Resulting, therefore, ( ) we get,

consequently, that ( ) 0. It is only when the gas pressure, before expansion, is progressively reduced that the change

of temperature gradually decreases until the effect would be zero as soon as the pressure, at

the limit of its reduction phase, is annulled. In that case U would be independent of the

volume i.e. U = U(T) but this is the case of the ideal gas and not the real gas with U =

U(T,V) for which the result of the Joule experiment is invalid.

The study of the dependence of the energy and enthalpy of real gas by volume

(pressure) was made by Joule in collaboration with Thomson following a different

procedure. They allowed the gas to expand freely through a porous baffle. This is an

irreversible process, but the thermodynamic considerations apply to this system by simply

considering the initial and final state of equilibrium before and after the execution of the

process. To study the expansion gas through the filter constituted by the porous baffle, we

will focus on a given mass of gas. You can consider this as a mass of gas enclosed between

two movable pistons imaginary (that are outlined in the figure) so as to maintain constant

pressures P1 and P2 .

Figure 2. Principle of the Joule-Thomso 1


As shown in Fig. 2, the gas expands from the pressure P1 to pressure P2 by the action of

throttling due to the porous baffle. The entire system is thermally isolated i.e. cylinder and

pistons are supposed impermeable to heat; thus it does not allow the system to receive or

lose heat so that the expansion occurs by performing a adiabatic transformation, which

corresponds to

q=0

The gas is allowed to flow continuously through the porous baffle, and when the stable

conditions are reached the temperatures gaseous, before and after expansion, T1 and T2, are

measured by thermocouples sensitive. It will verify that the expansion of the gas takes

place at constant enthalpy. Consider the expansion of a given mass of gas through the

porous baffle. In the experiment made by Thompson and Joule (1852) the porosities were

quite large compared to the molecular mean free path, such as to enable the gas passing

through the plug like a normal flow, rather than like a diffusion process; at the same time,

however, they were small enough to ensure that the high viscous resistance encountered by

the gas would reduce its kinetic energy to negligible values.

The gas occupies a volume V1 a at pressure P1 and temperature T1 and before the

expansion to a volume V2 at P2,T2 after the expansion. What is the work done during this

process? The compression of the piston to the left side provides a work (executed from the

surrounding area of the system) equivalent to

─ P1 where

= - V1 or wSX = + P2 · V2

Similarly, the expansion of the imaginary piston to the right side provides a work on the

surrounding environment by the system equal to

─ P2 where = V2 ─ 0 = V2 or w DX = ─ P2 · V2

As the gas is separated from any heat source it cannot absorb any amount of heat to convert

it to work. This must come from the energy of the internal gas U so that this decreases and

since U is a function of T this means that the gas temperature must decrease.

The total change in internal energy of the gas during the adiabatic expansion is then

ΔU = q + w = 0 + w = + w

or

ΔU = P1·V1 ─ P2V2 = U2 ─ U1 0


The process does not occur at constant internal energy U. By rearranging we get

U2 + P2V2 = U1 + P1V1

but recalling the definition of enthalpy

H = U + PV we get

H2 = H1

This is therefore an expansion isenthalpic and the experiment directly measures the change

in temperature of a gas with the pressure at constant enthalpy which is called the Joule-

Thomson coefficient µJT

µJT = (

H =

For an ideal gas since the process is isenthalpic we can write

( ) = ( ) ( ) = ─ Cp µJT

but ( ) = + P·V))/ = ( ) + /

then ( ) = ( ) = 0 since for an ideal gas, as mentioned above,

H and U are functions of T only. Since the heat capacity at constant pressure Cp is other

than zero, the Joule-Thomson coefficient should be zero for an ideal gas.

Being µJT = 0 for an ideal gas where there are no molecular interactions, unlike that in the

real gas, there is no Joule-Thomson effect. It is concluded, therefore, that the one described

above depends on the interaction molecules.

The coefficient µJT for a real gas is different from zero. The expansion of the gas implies a

change of temperature ΔT dependent on changes in kinetic energy directly influenced by

the conditions of throttling i.e. by the porous structure of the septum but also by the nature

of the gas that is not ideal, that is owing of molecular interactions as already highlighted. A

real gas can heat up or cool down. Referring to Fig.2 T2 T1 and, in certain conditions,

is higher and in others is lower. Recalling the relationship µJT =

( in the


expansion phase ΔP is negative and therefore a positive value for µJT corresponds, in

that phase, to cooling and vice versa.

It is noted, feature not always stressed in the literature, that while the enthalpy H

remains unchanged in the expansion of the Joule-Thomson effect, by its nature

isenthalpic, one cannot say the same for the entropy S: the entropy change Δ S

associated to the expansion of the gas must be greater than that associated with the

reduction of its temperature, otherwise the gas not would flow through the porous

baffle.

Let us examine, now, the above physical behavior for real gases.

If it's performed a Joule-Thompson experiment , the corresponding pairs of values of

pressure and temperature, that is, P1 and T1, P2 and T2, P3and T3, etc. determine a

number of points on the pressure-temperature diagram as in Figure 3a and given that

H1 = H2 = H3 etc.., the enthalpy is the same in all these points i.e. combining all these

points we shall be able to obtain a constant enthalpy line (Fig. 3a).

Such a plot of T versus P is called isenthalpic curve and its slope is given by the Joule-

Thomson coefficient

µJT = (

H


Note that this curve is not the process executed by the gas in the passage through the porous

baffle, since the process is irreversible and the gas does not pass through a series of

equilibrium states. The final temperature and pressure should be measured at a certain

distance from the porous baffle so as to allow cancellation of lacks in homogeneity present

in the flow, and the gas passes from a point on the curve to another by an irreversible

process.

The coefficient µJT 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 (TINV). In order to liquefy a gas by a Joule-

Thompson expansion the gas must first be cooled to below the Inversion Temperature.

Performing another series of experiments, keeping unchanged the initial pressure and

temperature in each series, but then varying both from a series to another, you can

obtain a family of curves corresponding to different values of H. One such family is

shown in Fig. 3b that is typical of all real gases.

If the temperature is not too high the curves pass through a maximum named reversal point.

The slope of a curve at any point is isenthalpic (

H and the maximum of the curve, or

turning point, corresponds to µJT = 0.

The inversion curve represents the points of the P-T diagram for which the Joule-Thomson

coefficient µJT is null and separates the two zones in which the temperature rises (negative

coefficient) or decreases (positive coefficient) with decreasing pressure.

Below this curve the fluid cooling takes place by expansion.

Note that, below a certain temperature (maximum inversion Temperature (TINVMAX)), the curves have a maximum isenthalpic: this means that if you put initially the

pressure and temperature corresponding to the maximum of one of the isenthalpic

curves and the gas undergoes a Joule-Thomson expansion, the temperature decreases.

To this we refer when we say that normally a gas cools when expanding.

When using the Joule-Thompson effect in the liquefaction of gas for expansion, it is

obvious that conditions must be chosen such as to lead to the decrease of temperature. This

is possible only if the initial temperature and pressure are within the curve inversion. Then

a temperature drop would be produced in the expansion from point a to b, and then c to that

c, but a rise in temperature would be expanding from d to e, points outside the inversion

curve. It is not obvious that any gas with µJT = 0 is ideal; from the above it is obvious that

real gases have many temperatures at which µJT = 0.


Below it is presented a graph of the Joule-Thomson coefficient for various gases at

atmospheric pressure.

Many gases at room temperature and moderate pressures are within the "cooling" area Fig.

3b; it should be noted that hydrogen and helium exhibit abnormal behavior having a

temperature inversion well below the room temperature, and at the room one they behave as

in the transformation from d to e, that is, expanding they warm up.

Let us try to understand the physical mechanisms involved. When a gas expands, the

average distance between its molecules increases. Given the presence of

intermolecular attractive forces, the expansion causes an increase in potential energy

of the gas. If you do not extract work from the system during the expansion process

("free expansion") and heat is not transferred, the total energy of the gas remains the

same for the conservation of energy. The increase in potential energy then produces a

decrease of the kinetic energy and thus a decrease of the gas temperature. Another

mechanism has rather the opposite effect: during collisions between the molecules of

the gas, the kinetic energy is temporarily converted into potential energy. While the

average intermolecular distance increases, there is a decrease in the number of

collisions per unit time, which in turn causes a decrease of the potential energy

average. Given that the total energy is conserved it involves an increase of the kinetic

energy (and therefore of the temperature). Inside the Joule-Thomson inversion curve,

the first effect (internal work done against the intermolecular attractive forces)

dominates, and free expansion causes a decrease in temperature. Outside the

inversion curve dominates the second effect (decrease of the potential energy

associated with collisions) and the free expansion causes an increase in temperature.


CALCULATION OF THE JOULE-THOMSON COEFFICIENT

It's often necessary to express the Joule – Thomson coefficient in terms of other partial

derivatives. Being the enthalpy function can depending on the state of the system that is the

temperature and pressure i.e. H = H(T,P), we can write the total differential as

In the experiment of the Joule-Thompson H = Cost i.e. dH = 0 and, therefore,

rearranging the previous expression equated to zero, is obtained

but (∂H/∂T)P , partial derivative of enthalpy, is Cp, heat capacity at constant pressure.

Bearing in mind that in a closed system dU = dQ – dw where dw is the work done by the

system and also by differentiating the expression of the enthalpy H = U + PV, we obtain

dH = dU + d(PV) but dU = dQ – dw = Tds –PdV being a reversible process and

therefore dH = dU + d(PV) = Tds –PdV + PdV +VdP = TdS +VdP from which

(∂H/∂P)T = T (∂S/∂P)P + V ; at last, by the relation of Maxwell applied to the

differential of Gibbs free energy dG = d(U-TS +PV) = VdP – SdT, it is obtained

(∂S/∂P)T = (∂V/∂T) P .

Substituting this result in the expression of the equivalent differential enthalpy H with

respect to T with constant pressure is obtained

( = V ─

Therefore


µJT =

For a real gas can be obtained from any of the state equation as shown

below with the equation of Van der Waals forces.

VAN DER WAALS EQUATION OF STATE

The equation of state concerning Van der Waals forces for one mole of real gas is the

following

(P +

) (V – b) = RT

where the constants a and b, always positive, depend on the nature of the gas. The

corrective term

takes account of the forces of attraction between the molecules which

tends to keep them as far away from the walls by decreasing the pressure on the same,

decrease proportional to the number of pairs of particles present i.e. ∝ n2 while b refers to

the volume physically occupied by the molecules of a mole of gas (intrinsic molar volume),

that is (V-b) is the volume "free" and this parameter increases the pressure.

At low temperatures the interactions between molecules may be significant compared to

the thermal energy and therefore the term a becomes important while at high temperatures,

however, the term b becomes significant as the thermal energy becomes greater than any

interaction.

The above equation for real gases in standard form can be rewritten, after multiplying the

terms in brackets and rearranging, in order to obtain PV in function of other quantities

PV = RT

bP

The whole expression can be rewritten by neglecting the term

given that a and b are

small and by replacing the term

with

in the form specified under


V =

–

b

Differentiating this expression with respect to the temperature T with P, pressure,

maintained constant is obtained

− V =

– b

By replacing µJT =

is obtained

µJT =

This equation does not appear so simple but is treated by theorists looking at extreme

conditions. At temperatures T tending to zero the term

becomes much larger than b,

which can, therefore, be overlooked. It has, therefore, as a result the Joule-Thomson

coefficient

, positive expression since each term contained therein is

positive. This accords with the experiment.

A positive Joule-Thomson coefficient means cooling at very low temperatures.

At high temperatures, the term containing T in the denominator goes to 0, and therefore

µJT =

This result agrees with the experiment since it is negative, which means warming. Note that

the term with the coefficient a has disappeared, remaining only b. The measurement of the


Joule-Thompson coefficient allows the reconstruction of the state function of the gas and is

essential in the design of heat pumps.

EXPERIMENTAL PROCEDURE

The Joule-Thomson apparatus is shown in Fig. 3. Since the porous baffle needs a rather

long time to reach a steady thermal state, the gas will be on about two hours before the

beginning of the laboratory to ensure that the temperature difference across the porous

baffle has attained a constant value. This is indicated by the constancy of f.e.m. between the

thermocouple leads.

Figure 3. Joule-Thomson apparatus 1 1) To use and operate the digital pressure gauge, we need about 5 minutes. First, we have

to wait about 90 seconds to go to 780 Tor, then reset by pressing and holding down the

zero button on the gauge for two seconds.

2) The values will change for a few seconds, but, in this case, it's not a big problem.

Secondly, after the reset to zero, we will have to adjust the desired pressure value

VERY SLOWLY opening the needle valve on the gas cylinder and by operating a

control pressure to a value of about 250 Tor and taking the reading off the voltmeter.


3) We take the readings as described above at intervals of 5 minutes until four readings

(f.e.m., ∆P) don't show significant differences (i.e. no systematic drifts)

4) We take arithmetic mean of the four readings and assign to it the limit of confidence

interval

5) VERY SLOWLY (about 90 seconds) we increase the pressure difference, ∆P, of

about 100 Tor across the porous baffle very slowly opening the needle valve on the

gas cylinder. We have to take readings 5 minutes after it was made the change in

pressure and then gradually at intervals of five minutes until, as before, four readings

do not show a significant difference. In this way we get 8 experimental points. Using

the calibration graph is calculated ∆T, the change of temperature across the porous

baffle.

CALCULATIONS

For each point we determine the average values of ∆P and ∆T. We determine the

uncertainties in the values of T and P and draw a graph of T versus to P enclosing each

point in a box of indeterminacy. We draw the best line that fits through these experimental

points and determine its slope. We draw, too, the lines of maximum and minimum slopes.

Then we calculate the slope m and the intercept b for the line applying the method of least

squares to the spreadsheet data. Finally we calculate m and b and compare all with our

analysis made graphically.

It is determined by the slope Joule-Thompson coefficient, µJT , in °C/atm. and the

uncertainty±µJT. We calculate the Joule-Thompson coefficient µJT from the gas equation

of state Van der Waals using the equation

µJT =


Below shows the values of the constants a and b relating to carbon dioxide, helium and

nitrogen from the equation of van der Waals

Values of constants (in MKS units)

CO2 He N2

Van derWaals

a(j m3mole

-2) 0.364 3.457x 10

-3 0.141

b(m3 mole

-1) 4.267 x 10

-5 2.370 x 10

-5 3.913 x 10

-5

CP (joule mole-1

deg-1

) 37.085 20.670 26.952

N.B. 1 atm / 760 mm Hg /760 Torr/ 101.32 kPa.

ROBERTO GIARDELLI

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THE JOULE THOMSON EXPERIMENT - cryoelectric THE JOULE THOMSON EXPE… · THE JOULE – THOMSON EXPERIMENT ... the valve V was opened allowing the gas in A to expand in ... the Joule-Thomson