Joule-Thomson effect

Illustration of the Joule-Thomson effect: a gas is expanded through a throttle.

Close-up view of a double chamber separated by a porous membrane. The gas is expanded via this. In the chambers you can see the thermal sensors with which the temperature difference is measured.

Experimental setup for measuring pressure and temperature in the Joule-Thomson effect.

The Joule-Thomson effect (not to be confused with the Thomson effect ) describes the temperature change of a gas with an isenthalpic pressure reduction. The direction and strength of the effect is determined by the strength of the attractive and repulsive forces between the gas molecules. Under normal conditions , most gases and gas mixtures , e.g. B. for air, that the temperature drops during relaxation. In contrast, it increases z. B. with hydrogen , helium , neon . In an ideal gas there are no molecular forces and consequently it does not show a Joule-Thomson effect. The Joule-Thomson effect plays an important role in the thermodynamics of gases.

Examples of occurrence and applications:

• Cooling of soda water , whipped cream and soft ice cream while foaming from a pressure bottle
• Freezing water in a snow cannon
• Production of dry ice at the dentist's or with dry ice blasting (see sandblasting )
• Gas liquefaction using the Linde process

history

In connection with the discovery of the first and second law of thermodynamics around the middle of the 19th century, the balance of work and heat in the compression and expansion of air was investigated with increasing measurement accuracy. Previously, in the Gay-Lussac experiment, no temperature change was found when air expands into a larger volume with falling pressure without heat flow or work. Accordingly, their internal energy does not depend on the volume. This was verified with better accuracy in 1852 by James Prescott Joule and Sir William Thomson (later Lord Kelvin ) by subjecting flowing air to a controlled pressure reduction. To do this, the air was continuously pumped through a thick, long, thermally insulated pipe with a small screen halfway, and showed a slight cooling at the other end. According to their publication, this temperature change is referred to as the Joule-Thomson effect. Since the equation of state of the ideal gas would result in a constant temperature, it was also established that air is not exactly an ideal gas. Van der Waals' equation of state presented in 1873 fits better and results in a weak volume dependence of the internal energy. The Joule-Thomson effect was made by Carl von Linde in 1895 as the basis for technical air liquefaction and the production of pure oxygen and nitrogen.

Isenthalpic relaxation

In the ideal model of isenthalpic relaxation, a thermally insulated gas flows due to a pressure difference through an obstacle on which it does no work. It is a continuous, irreversible process that is exchanged in which no heat and pressures and should remain constant before and after the obstacle. If a quantity of gas has the volume in front of the obstacle, the volume work must be supplied to it in order to push it through the obstacle, and it releases the work behind the obstacle, where it occupies the greater volume . According to the 1st law, your internal energy changes from to ${\displaystyle p_{1}}$ ${\displaystyle p_{2}}$ ${\displaystyle V_{1}}$ ${\displaystyle p_{1}V_{1}}$ ${\displaystyle V_{2}}$ ${\displaystyle p_{2}V_{2}}$ ${\displaystyle U_{1}}$

${\displaystyle U_{2}=U_{1}+p_{1}V_{1}-p_{2}V_{2}}$

Hence the enthalpy

${\displaystyle H=U_{1}+p_{1}V_{1}=U_{2}+p_{2}V_{2}}$

constant.

If, as with the ideal gas, the enthalpy only depends on the temperature and not on the volume (for 1 mol:) , the temperature remains the same with isenthalpic relaxation. For real gases, however, neither the internal energy nor the product pV is independent of the volume. This can be precisely observed with the continuously running experiment by Joule and Thomson, because all possible disruptive effects such as heat transfers or various kinetic energies in the gas flow can be easily controlled. This is an advantage over the Gay-Lussac experiment , in which a compressed gas expands once into a larger volume. Initially, only the gas near the opened passage makes a free expansion into the vacuum, while the gas in the middle areas of the container first has to be accelerated towards the passage, which is caused by the gas in the rear area through adiabatic expansion and cooling. The total internal energy remains constant, but is now distributed unevenly in the amount of gas. Only after all differences have been compensated is there a uniform temperature again, from which any deviations from the ideal gas can be read. ${\displaystyle H=C_{V}T+RT}$

thermodynamics

Joule-Thomson coefficient

The strength and direction of the temperature change is described by the Joule-Thomson coefficient μ :

${\displaystyle \mu _{\mathrm {JT} }=\left({\frac {\partial T}{\partial p}}\right)_{H}}$

It represents the partial derivative of the temperature T with respect to the pressure p with constant enthalpy H. The index H means that the enthalpy must be kept constant when the pressure changes. A positive value shows that the temperature falls when the pressure is reduced, a negative value that it rises. ${\displaystyle \mu _{\mathrm {JT} }{\mathord {>}}0}$

Joule-Thomson inversion curve for a van der Waals gas in a pT diagram

The cause of the Joule-Thomson effect lies in the interaction of the gas particles. Since the particles attract each other at a greater distance (see Van der Waals forces ), work has to be done to increase the distance between the particles. The particles slow down, the gas cools down. However, if the distance between them is small, the particles repel each other, accelerating them if they can move away from each other, and the gas heats up. The relationship between the two effects depends on the temperature and the pressure. The predominantly effective effect determines the sign of the Joule-Thomson coefficient. For example, the cooling with nitrogen from normal temperature and not too high a pressure is 0.14 K per 1 bar pressure reduction.

In order to be able to use the Joule-Thomson effect for cooling the gas, it must be in a state with . These states are all below a certain temperature, referred to as the inversion temperature T _inv , and below a certain maximum pressure (see figure). For nitrogen, for example, the corresponding values ​​are T _inv = 607 K (334 ° C) and p = 40 MPa (400 bar). If you want to cool gases such as hydrogen, helium or neon in the Linde process , you have to pre-cool the gases because their inversion temperatures are 202 K, 40 K and 228 K. ${\displaystyle \mu _{\mathrm {JT} }{\mathord {>}}0}$

Apart from hard, elastic collisions, no interactions between the particles are taken into account in the model of the ideal gas. Ideal gases therefore have no Joule-Thomson effect.

Calculation of the Joule-Thomson coefficient

If the enthalpy defined by is expressed as a function of its natural variables entropy and pressure , then its total differential is given by ${\displaystyle H=U+pV}$ ${\displaystyle S}$ ${\displaystyle p}$

${\displaystyle \mathrm {d} H=T\mathrm {d} S+V\mathrm {d} p}$

In order to be able to calculate the Joule-Thomson coefficient, the entropy must be expressed by the variables and . The differential is then: ${\displaystyle T}$ ${\displaystyle p}$ ${\displaystyle \mathrm {d} S}$

${\displaystyle \mathrm {d} S=\left({\frac {\partial S}{\partial T}}\right)_{p}\mathrm {d} T+\left({\frac {\partial S}{\partial p}}\right)_{T}\mathrm {d} p}$

This means that the change in enthalpy can be expressed using the variables pressure and temperature:

${\displaystyle \mathrm {d} H=T\left({\frac {\partial S}{\partial T}}\right)_{p}\mathrm {d} T+T\left({\frac {\partial S}{\partial p}}\right)_{T}\mathrm {d} p+V\mathrm {d} p}$

Now is in the first summand

${\displaystyle T\left({\frac {\partial S}{\partial T}}\right)_{p}=C_{p}}$

the heat capacity at constant pressure (because with and there ). ${\displaystyle T\Delta S=Q=\Delta U=C_{p}\Delta T}$ ${\displaystyle \Delta U=Q+W}$ ${\displaystyle W=0}$ ${\displaystyle dV=0}$

In the second summand the following applies (according to the Maxwell relation for the free enthalpy with the total differential ): ${\displaystyle \mathrm {d} G=-S\mathrm {d} T+V\mathrm {d} p}$

${\displaystyle \left({\frac {\partial S}{\partial p}}\right)_{T}=-\left({\frac {\partial V}{\partial T}}\right)_{p}}$

This is given by the thermal expansion coefficient (at constant pressure) : ${\displaystyle \alpha }$

${\displaystyle \alpha V=\left({\frac {\partial V}{\partial T}}\right)_{p}}$

Insertion into the equation for the differential of enthalpy yields:

${\displaystyle \mathrm {d} H=C_{p}\mathrm {d} T+V\left(-T\alpha +1\right)\mathrm {d} p}$

By setting the left-hand side to zero and solving for , the Joule-Thomson coefficient is obtained: ${\displaystyle \mathrm {d} T/\mathrm {d} p}$

${\displaystyle \mu _{\mathrm {JT} }=\left({\frac {\partial T}{\partial p}}\right)_{H}={\frac {V}{C_{p}}}(T\alpha -1)}$

For an ideal gas , and thus : the Joule-Thomson effect does not exist, in accordance with the fact that the internal energy of the ideal gas does not depend on the volume. In the case of real gases , on the other hand, the effect is also present in the areas of pressures and temperatures that are not too high, where they otherwise behave in a very good approximation (such as nitrogen or air under normal conditions). ${\displaystyle \alpha =1/T}$ ${\displaystyle \mu _{\mathrm {JT} }=0}$

Joule-Thomson coefficient for a van der Waals gas

In an approximate consideration, the Van der Waals equation (for 1 mole)

${\displaystyle p={\frac {RT}{V-b}}-{\frac {a}{{V}^{2}}}}$

for to ${\displaystyle V\gg b}$

${\displaystyle p={\frac {RT}{V}}\left(1+{\frac {b}{V}}-{\frac {a}{VRT}}\right)}$

simplified. Solving for V gives

${\displaystyle V={\frac {RT+{\sqrt {-4ap+RT(4bp+RT)}}}{2p}}}$

From a development in as well as consistent in follows ${\displaystyle a/(RTV)\ll 1}$ ${\displaystyle b/V\ll 1}$

${\displaystyle V={\frac {RT}{p}}+b-{\frac {a}{RT}}}$

This gives the coefficient of thermal expansion ( and assumed to be constant) ${\displaystyle a}$ ${\displaystyle b}$

${\displaystyle \alpha ={\frac {1}{V}}\left({\frac {\partial V}{\partial T}}\right)_{p}={\frac {1}{V}}\left({\frac {R}{p}}+{\frac {a}{RT^{2}}}\right)}$

For the Joule-Thomson coefficient it follows:

${\displaystyle \mu _{\mathrm {JT} }={\frac {1}{C_{p}}}\cdot \left({\frac {2a}{RT}}-b\right)}$

It can be seen that the effect of the attractive forces (van der Waals parameters ) and the repulsive (van der Waals parameters ) are influenced in opposite ways. The coefficient is positive when the temperature is below a critical value ${\displaystyle a}$ ${\displaystyle b}$

${\displaystyle T_{\mathrm {inv} }={\frac {2a}{Rb}}}$

Inversion and critical temperatures

gas	T crit  [K]	T inv  [K]	T crit / T inv
air	132.6	≈760	≈5.7
hydrogen	33.18	≈200	≈6
helium	5.19	≈40	≈7.5

and varies with the temperature dependence according to . Both reflect the observations well and support the given physical interpretation in a qualitative manner. Compared with the critical temperature T _crit = 8a / (27Rb) , the inversion temperature is 6.75 times higher, namely ${\displaystyle 1/T}$

${\displaystyle T_{\mathrm {inv} }={\frac {27}{4}}T_{\mathrm {crit} }}$.

However, the agreement of this simple formula with the inversion temperatures observed for various gases is not good, and the observed dependence of the inversion temperature on the pressure is also missing. For more precise values ​​and a reproduction of the entire inversion curve, the approximation of the Van der Waals equation must be supplemented by a term: ${\displaystyle p(T_{\mathrm {inv} })}$

${\displaystyle V={\frac {RT}{p}}+b-{\frac {a}{RT}}+{\frac {abp}{R^{2}T^{2}}}}$

Then the coefficient of thermal expansion

${\displaystyle \alpha ={\frac {R}{p}}+{\frac {a}{RT^{2}}}-2{\frac {abp}{R^{2}T^{3}}}}$

and the Joule-Thomson coefficient

${\displaystyle \mu _{\mathrm {JT} }={\frac {1}{C_{p}}}\cdot \left({\frac {2a}{RT}}-b-3{\frac {abp}{R^{2}T^{2}}}\right)}$

This equation gives a value for nitrogen at room temperature and Pa (100 bar) of , close to the measured value. ${\displaystyle p=10^{7}}$ ${\displaystyle \mu _{\mathrm {JT} }=0{,}188\ \mathrm {K/bar} }$

The inversion curve in the pT diagram is obtained by setting in the last equation : ${\displaystyle \mu _{\mathrm {JT} }=0}$

${\displaystyle T_{\mathrm {inv} }^{2}-{\frac {2a}{Rb}}T_{\mathrm {inv} }+{\frac {3ap}{R^{2}}}=0}$

Rearranged after , results for a parabola open at the bottom. Physically meaningful values belong to the temperature range . This limit is the value of the inversion temperature found above, which therefore only applies in the range of low initial pressures. If you want to work at higher initial pressures for the purpose of stronger cooling, the temperature must be lower, whereby the maximum possible pressure requires the temperature . ${\displaystyle p}$ ${\displaystyle p\left(T_{\mathrm {inv} }\right)}$ ${\displaystyle p>0}$ ${\displaystyle T_{\mathrm {inv} }<2a/Rb}$ ${\displaystyle p=a/(3b^{2})}$ ${\displaystyle T_{\mathrm {inv} }=a/Rb}$

Technical aspects

The Linde-Fränkl process (low pressure process) see: Linde process

The Linde process for gas liquefaction requires a positive Joule-Thomson coefficient. This is the only way to dissipate the energy of the compressed gas even though the ambient temperature is higher than that of the gas. In the Linde machine, air is expanded from around 200  bar to around 20 bar through a throttle valve . It cools down by about 45 Kelvin. The cooled air is now used to cool down further compressed air before expansion ( countercurrent heat exchanger ). Over several compression and expansion stages, the gas can be cooled down to the point where it condenses and thus becomes liquid.

A gas that has a negative Joule-Thomson coefficient at room temperature must, in order for the Linde process to work, to be pre-cooled using other processes until it falls below its inversion temperature. Only then does it cool down further with isenthalpic throttling because of the now positive Joule-Thomson coefficient. For example, helium has to be cooled to about −243 ° C (30 K).

literature

• Peter W. Atkins: Physical Chemistry . Wiley-VCH, Weinheim 2001, ISBN 3-527-30236-0 .
• Refah Ayber: Thomson-Joule effect of methane-hydrogen and ethylene-hydrogen mixtures (VDI research booklet ; Vol. 511). VDI-Verlag, Düsseldorf 1965.
• Lew Dawidowitsch Landau and Evgeni Michailowitsch Lifschitz : Textbook of Theoretical Physics . Akademie-Verlag, Berlin
  • 5. Statistical Physics . 1987, ISBN 3-05-500069-2 .

Web links

• Video: JOULE-THOMSON effect and LINDE method - How do you create liquid air? . Jakob Günter Lauth (SciFox) 2013, made available by the Technical Information Library (TIB), doi : 10.5446 / 15652 .

Individual evidence

1. ↑ JP Joule, W. Thomson: On the thermal effects experienced by air in rushing through small apertures , Philosophical Magazine, Series 4, Volume 4, Issue 284, pages 481-492 (1852) [1]
2. ^ ^{A b} Walter Greiner, Neise, L., Stöcker, H .: Thermodynamics and statistical mechanics . Verlag Harri Deutsch, 1993, p. 154 ff .
3. ^ Klaus Stierstadt: Thermodynamics . Springer Verlag, 2010, ISBN 978-3-642-05097-8 , pp. 466 .
4. ^ Fran Bošnjaković, Karl-Friedrich Knoche: Technical Thermodynamics Part 1 . 8th edition. Steinkopff Verlag, Darmstadt 1998, ISBN 3-642-63818-X , 15.2.1. Inversion of a throttle effect.
5. ↑ Hans-Christoph-Mertins, Markus Gilbert: Examination trainer experimental physics . Elsevier, Munich 2006, ISBN 978-3-8274-1733-6 .