Joule-Thomson effect

Notation

We differentiate between state and process variables. State variables can be determined from a state function and do not depend on the path that was used to realize the state. Process variables, in contrast, depend on the path that was used to realized the state and, thus, do cannot be derived from a state function.

Infinitesimal changes of state variables can be determined from the total differential of a state function. Since this is not possible for process variables, we use a different symbol to improve clarity. We keep this notational difference also for macroscopic changes:

Change	State variable $Z$	Process variable $P$
inifinitesimal	$\mathrm d Z$	$\delta P$
makroskopic	$\Delta Z$	$P$

Note that these symbols should not be confused with the symbol for partial derivatives $\partial$ .

Theory

Ideal Gas

An ideal gas is a model that assumes the gas particles do not have a volume and there are no interactions besides elastic collisions between the gas particles. It is described by the ideal gas equation:

pV_\text m = RT

The product of pressure $p$ and molar volume $V_\text m$ is proportional to the temperature $T$ . The proportionality constant $R$ is called universal gas constant, with $R=8{.}314\,\text J/(\text{mol}\,\text K)$ . Like every molar quantity the molar volume $V_\text m$ is defined as ratio between volume $V$ and amount of substance $n$ .

Real gases — the Van der Waals equation

Next, we consider both the individual volume of the gas particles (repulsion) as well as attractive interactions between the gas particles. A possible model of such gases is given by the Van der Waals equation.

\left(p + \frac{a}{V_\text m^2}\right)\left(V_\text m - b\right) = RT

Compared to the ideal gas equation, both the pressure and volume term have been modified:

The Van der Waals coefficient $a$ describes attractive interactions between the gas particles, which reduce the momentum transfer on the containing walls. This reduces the pressure of the gas.
$\begin{aligned} p &= p_\text{ideal} - \frac{a}{V_\text m^2}\\ \quad\Leftrightarrow\quad p_\text{ideal} &= p + \frac{a}{V_\text m^2} \end{aligned}$
The term scales with a factor of $V_\text m^{-2}$ since the interactions are proportional to the particle density within the gas and proportional to the number of particles which collide with the containing walls . This results in a quadratic dependency on the inverse molar volume.
The Van der Waals coefficient $b$ describes the volume of the particles and is thus subtracted from the volume (as the measured volume is increased due to the particle volume).
$\begin{aligned} V_\text{m} &= V_\text{m,ideal} + b\\ \quad\Leftrightarrow\quad V_\text{m,ideal} &= V_\text{m} - b \end{aligned}$

At small pressure and large temperature gases can be described by the ideal gas equation rather well, since at small pressure the particle density is small enough to justify neglecting the particle volume. Similarly, at large temperature the kinetic energy of the gas particles is significantly larger than their attractive interaction, which justifies neglecting the attractive contribution.

The critical point

Figure shows isotherms of an ideal and a real (Van der Waals) gas at various temperatures in a $p$ - $V$ plot. The isotherms of the ideal gas exhibit a hyperbolic shape. The temperature $T_{\mathrm{ideal},1}$ is larger than the temperature $T_{\mathrm{ideal},2}$ . The pressure of the gases at large volume is small. In the limit of a volume approaching zero, the pressure rises approaches infinity. For the ideal gas, each pressure corresponds to a single volume value.

Interactive visualization (1) of this figure Interactive visualization (2) of this figure

Isotherms of a real and and an ideal gas at different temperatures. It is assumed that $T_{\text{ideal},1}>T_{\text{ideal},2}$ and $T_{\text{real},1} > T_{\text{k}} > T_{\text{real},2}$ .

As expected, at large temperatures $T_{\mathrm{real},1}$ a real gas shows a similar trend as in the $p$ - $V$ plot as the ideal gas. However, when considering lower temperatures the real gas equation shows a significant deviation from the ideal gas equation. At a temperature $T_{\mathrm{k}}$ there is an inflection point. This point is also referred to as critical point and is given by a corresponding critical pressure $p_\mathrm k$ , critical volume $V_\mathrm k$ and critical temperature $T_\mathrm k$ . The critical quantities can be determined by using that the critical point is an inflection point:

\begin{aligned} p_\mathrm k &= \frac{RT}{V_\mathrm{m,k} - b} - \frac{a}{V_\mathrm{m,k}^2}\\ \frac{\partial p_\mathrm k}{\partial V_\mathrm{m,k}} &= 0\\ \frac{\partial^2 p_\mathrm k}{\partial V_\mathrm{m,k}^2} &= 0 \end{aligned}

From this we can compute the critical quantities as functions of the Van der Waals parameter $a$ and $b$ :

\begin{aligned} p_\mathrm k &= \frac{a}{27 b^2}\\ V_\mathrm{m,k} &= 3 b\\ T_\mathrm k &= \frac{8 a}{27 b R} \end{aligned}

Maxwell correction

At temperatures less than the critical temperature $T_\mathrm k$ the isotherms in the $p$ - $V$ plot in figure take on a shape shown for temperature $T_{\mathrm{real},2}$ . In certain volume and pressure intervals the pressure seems to increase when increasing the volume. Additionally, a fixed pressure can yield up to three different volumes. Thus, in these intervals the Van der Waals equation does not produce physically relevant results.

A correction to this problem was proposed by Maxwell. A straight line is drawn such that the areas (integrals) with the isotherm below and above the line are equal. Here, the gas condensates and is in thermodynamic equilibrium with its liquid phase. The equal area shown in figure ensures that the chemical potential of liquid and gas phase are equal.

Interactive visualization of this figure

Van der Waals gas with Maxwell correction.

The simplified Van der Waals equation

The Van der Waals equation can be simplified when assuming that the pressure is small (up unto approximately 10 hPa) . This is useful, as the mathematical effort required to deal with the Van der Waals equation can be quite high. In this script we will state explicitly if the simplified form is used.

The expanded form of equation is

pV_\text m + \frac{a}{V_\text m} - pb - \frac{ab}{V^2_\text m} = RT.

It is obvious that the last term on the left-hand side becomes negligible for large molar volume $V_\text m$ — i.e. at small pressure. Furthermore, the molar volume in the second term of the left-hand side can be approximated by the ideal gas equation (equation ). We thus obtain the simplified Van der Waals equation

V_{\text m} = \frac{RT}{p} +b - \frac{a}{RT}

The same result can be obtained if the Van der Waals equation is expanded using a Virial expansion on $V_{\text m}^{-1}$ up to the linear term and with some further approximation.

For ideal gases, their inner energy $U$ depends only on temperature, but not on volume and pressure. If there are interactions between the gas molecules, their behavior is not described by the above ideal gas equations. The interactions cause a change in temperature if they are expanded or compressed. This change in temperature is subject of the Joule-Thomson experiment.

The Joule-Thomson coefficient $\mu$ describes the change in temperature of a gas as a reaction of a change in pressure.

\mu = \left(\frac{\partial T}{\partial p}\right)_H

If attraction dominates the interaction between the gas particles the gas cools down upon expansion, as energy is required to distribute the particles within the larger volume. The Joule-Thomson coefficient is positive in this case. If repulsion dominates the interaction between the gas particles, the gas heats up upon expansion and the Joule-Thomson coefficient is negative. For the ideal gas $\mu = 0$ , as the inner energy of the ideal gas only depends on temperature and not on pressure or volume.

Experimental setup

The Joule-Thomson effect can be measured using the following apparatus.

Using two pistons a gas is expanded over a porous wall from a constant pressure $p_1$ to a different constant pressure $p_2$ . Since the two chambers are thermally isolated, the process is adiabatic, meaning that $\delta Q = 0$ . Naturally, the pressure $p_1$ is larger than the pressure $p_2$ as otherwise the gas would not expand.

Energy balance

We consider the sum of the volume work $W$ on both pistons

W = W_1 + W_2 = \int_{V_1}^0 -p_1 \mathrm d V + \int^{V_2}_0 -p_2 \mathrm d V

Since the pressures are constant they can be moved in front of the integral. Flipping the bounds of the integration changes the sign of the integral. Integration yields

W = p_1 \int^{V_1}_0 \mathrm d V - p_2 \int^{V_2}_0 \mathrm d V = p_1 V_1 - p_2 V_2

The process is adiabatic since no heat is exchanged with the surroundings. This means that the change in inner energy $\Delta U$ equals the volume work $W$ .

\begin{aligned} \mathrm d U &= \delta W\\ \quad\Rightarrow\quad \Delta U &= W \end{aligned}

Replacing this expression for the volume work $W$ (equation ) in equation yields the expression

\Delta U = U_2 - U_1 = p_1 V_1 - p_2 V_2

Rearranging yields

U_1 + p_1 V_1 = U_2 + p_2 V_2

The expression on the right and the left-hand side in this equation corresponds to the definition of the enthalpy $H = U + pV$ , so that we can write

H_1 = H_2 \quad\Leftrightarrow\quad \Delta H = 0 \quad\Leftrightarrow\quad H = \text{konst.}

This shows that the Joule-Thomson effect is isenthalpic, meaning that the enthalpy does not change. This relationship is used in the next section to determine the Joule-Thomson coefficient of a Van der Waals gas.

Determination of the Joule-Thomson coefficient

The definition of the Joule-Thomson coefficient in equation can be straightforwardly determined experimentally. However, it is difficult to determine the Joule-Thomson coefficient analytically for a Van der Waals gas. For this reason, we will use fundamental thermodynamic relation by Gibbs in the following.

For $\mathrm d H~=~0$ the total differential of the enthalpy is

\mathrm d H = \left(\frac{\partial H}{\partial p}\right)_T \mathrm d p + \left(\frac{\partial H}{\partial T}\right)_p \mathrm d T = 0

This expression can be rearranged to yield $\frac{\mathrm d T}{\mathrm d p}$ and can directly be placed into the definition of the Joule-Thomson coefficient.

\mu = \left(\frac{\partial T}{\partial p}\right)_H = -\frac{\left(\frac{\partial H}{\partial p}\right)_T}{\left(\frac{\partial H}{\partial T}\right)_p}

The numerator is the heat capacity at constant pressure $C_p$ .

The denominator can be rewritten as well. From the fundamental thermodynamic relation by Gibbs follows for the enthalpy

\mathrm d H = T \mathrm d S + V \mathrm d p

follows for the partial derivative of the enthalpy with respect to pressure at constant temperature

\left(\frac{\partial H}{\partial p}\right)_T = T \left(\frac{\partial S}{\partial p}\right)_T + V

It follows for the Joule-Thomson coefficient $\mu$ from equations and that

\mu = \frac{-T \left(\frac{\partial S}{\partial p}\right)_T - V}{C_p}

The partial derivative of the entropy $S$ with respect to pressure $p$ at constant temperature can be expressed using the fundamental thermodynamic relation for the free enthalpy

\mathrm d G = V\mathrm d p - S\mathrm d T.

For this we compute the partial derivatives of the free enthalpy with respect to pressure and pressure.

\quad \left(\frac{\partial G}{\partial T}\right)_p = -S,\quad \left(\frac{\partial G}{\partial p}\right)_T = V

These partial derivatives are again derived, so that the mixed derivatives of the free enthalpy with respect to pressure and temperature are obtained.

\left(\frac{\partial \left(\frac{\partial G}{\partial T}\right)_p}{\partial p}\right)_T = -\left(\frac{\partial S}{\partial p}\right)_T\quad \text{.}\left(\frac{\partial \left(\frac{\partial G}{\partial p}\right)_T}{\partial T}\right)_p = \left(\frac{\partial V}{\partial T}\right)_p

Since the partial derivatives can be exchanged (using Schwarz' theorem) it follows that

-\left(\frac{\partial S}{\partial p}\right)_T = \left(\frac{\partial V}{\partial T}\right)_p

It follows for the Joule-Thomson coefficient $\mu$ from equations and that

\mu = \frac{T \left(\frac{\partial V}{\partial T}\right)_p - V}{C_p} = \frac{T \left(\frac{\partial V_\text m}{\partial T}\right)_p - V_\text m}{C_{\text{m},p}}

In the absence of a model for the heat capacity~ $C_p$ it is at this point impossible to determine the Joule-Thomson coefficient for arbitrary temperatures and pressures. Table provides some experimentally obtained Joule-Thomson coefficients for various gases.

Experimentally determined Joule-Thomson coefficients of various gases at normal conditions ( $T = 298{.}15\,\text K$ , $p = 101325\,\text{Pa}$ ).
Gas	$\mu / (\text K\,\text{Pa}^{-1})$
H₂	-0.03 · 10^-5
He	-0.06 · 10^-5
N₂	0.25 · 10^-5
CO₂	1.10 · 10^-5

Inversion curve of the Van der Waals gas

It is possible to determine the conditions at which the Joule-Thomson coefficient changes it sign. For this it is useful to consider the pressure and temperature at which the Joule-Thomson coefficient is zero. This relationship is called inversion curve, since here the Joule-Thomson coefficient changes it sign. Equivalently, the numerator in equation must be zero.

\mu = 0 \quad\Rightarrow\quad T \left(\frac{\partial V}{\partial T}\right)_p - V = 0

Further rearrangement yields the condition for the inversion curve

\left(\frac{\partial V_\text m}{\partial T}\right)_p = \frac{V_\text m}{T}

The partial derivative of the volume with respect to temperature at equal pressure can be determined from the Van der Waals equation by considering that the molar volume ~ $V_\text m$ is as well a function of temperature.

p = p(T, V_\text m(T)) = \frac{RT}{V_\text m-b}-\frac{a}{V_\text m^2}

Computing the total derivative of the pressure with respect to temperature at constant pressure yields according to the chain rule

\left(\frac{\mathrm d p}{\mathrm d T}\right)_p \stackrel{!}{=} 0 = \underbrace{\frac{R}{V_\text m-b}}_{\left(\frac{\partial p}{\partial T}\right)_p} + \underbrace{\left(-\frac{RT}{(V_\text m-b)^2}+\frac{2a}{V_\text m^3}\right)}_{\left(\frac{\partial p}{\partial V_\text m}\right)_p} \left(\frac{\partial V_\text m}{\partial T}\right)_p

Further rearrangment yields the quantity

\left(\frac{\partial V_\text m}{\partial T}\right)_p = \frac{\frac{R}{V_\text m-b}}{\frac{RT}{(V_\text m-b)^2}-\frac{2a}{V_\text m^3}} = \frac{V_\text m-b}{T-\frac{2a(V_\text m-b)^2}{RV_\text m^3}}

This expression can be inserted into the condition for the inversion curve, equation . Further rearrangement yields

\begin{aligned}\frac{V_\text m-b}{T-\frac{2a(V_\text m-b)^2}{RV_\text m^3}} &= \frac{V_\text m}{T} \\ \quad\Leftrightarrow \quad \frac{V_\text m-b}{V_\text m} &= \sqrt{\frac{bRT}{2a}} \end{aligned}

The rearrangement of equation ,

V_\text m-b=\sqrt{\frac{bRT}{2a}}V_\text m

\frac{1}{V_\text m} = \frac{1}{b} - \frac{\sqrt{\frac{bRT}{2a}}}{b}

can be replaced in the Van der Waals equation so that with equation we obtain

p_\text{inv} = \frac{RT}{\sqrt{\frac{bRT}{2a}}V_\text m}-\frac{a}{V_\text m^2}

and with equation we finally obtain

p_\text{inv} = p_\text{inv}(T) = -\frac{a}{b^2}+ \sqrt{\frac{8aRT}{b^3}}-\frac{3RT}{2b}

The pressure is termed inversion pressure $p_\text{inv}$ and is a function of temperature.

Setting equation to zero and resolving for the temperature yields the upper and lower inversion temperature $T_\text{inv,u}$ and $T_\text{inv,o}$ .

T_\text{inv,u} = \frac{2a}{9Rb}\qquad T_\text{inv,o} = \frac{2a}{Rb}

The pressure $p_\text{max}$ is the maximum pressure at which a gas can exhibit a Joule-Thomson coefficient of zero. For any larger pressure than $p_\text{max}$ the gas heats upon expansion. The temperature corresponding to $p_\text{max}$ is called $T_{p_\text{max}}$ . Both quantities can be obtained by setting the first derivatives with respect to pressure of equation to zero.

p_\text{max} =\frac{a}{3b^2}\qquad T_{p_\text{max}} = \frac{8a}{9Rb}

Figure shows the inversion curve of a Van der Waals gas. Below the inversion curve the Joule-Thomson coefficient is positive and the gas cools upon expansion. On the curve itself the Joule-Thomson coefficient is zero; above the curve negative. Additionally, the graph shows the upper and lower inversion temperature as well as the maximum pressure and corresponding temperature.

Go to interactive visualization of the inversion curve

Inversion curve of the Van der Waals gases — Inversion curve of a Van der Waals gas with upper and lower inversion temperature. The maximum pressure $p_\text{max}$ and its corresponding temperature $T_{p_\text{max}}$ , at which the Joule-Thomson coefficient changes it sign are shown as well.

The upper inversion temperature is the limit above which the attractive interactions of a real gas are too weak in relation to its thermal energy to cool it down by isenthalpic expansion. Below the lower inversion temperature, however, the system is in the liquid state. (Please note that, strictly speaking, the Van der Waals equation is no longer valid from liquefaction onwards (see also Maxwell's construction and Van der Waals loop).)

Table shows some experimentally determined Van der Waals coefficients and heat capacities of various gases.

Van der Waals coefficients and molar heat capacities of various gases at normal conditions ( $T = 298{.}15\,\text K$ , $p = 101325\,\text{Pa}$ ).
Gas	$a / (\text{Pa}\,\text m^6\,\text{mol}^{-2})$	$b / (\text m^3\,\text{mol}^{-1})$	$C_{\text{m},p} / (\text J\, \text{mol}^{-1}\, K^{-1})$
H₂	196.7 · 10^-3	21.9 · 10⁻⁶	28.83
He	34.6 · 10^-3	23.7 · 10⁻⁶	20.79
N₂	136.5 · 10^-3	38.5 · 10⁻⁶	29.12
CO₂	364.9 · 10^-3	42.7 · 10⁻⁶	37.13

Go to interactive visualization of this figure

Inversion curve of Nitrogen, Hydrogen and Helium gas — Using equations and the parameters from table the calculated inversion curve of nitrogen, hydrogen and helium gas is shown. In this view, the temperature axes can not be differentiated from a line at normal pressure.

Joule-Thomson coefficient of a Van der Waals gas

The Joule-Thomson coefficient can be computed using equation and the Van der Waals equation (equation ). However, since the computation is rather elaborate, we consider in the following the simplified Van der Waals equation (equation ). The partial derivative of the volume with respect to temperature is then given by

\left(\frac{\partial V_{\text m}}{\partial T}\right)_p = \frac{R}{p}+\frac{a}{RT^2}

Replacing this derivative and the simplified Van der Waals equation in equation yields the Joule-Thomson coefficient

\mu = \frac{\frac{2a}{RT}-b}{C_{\text{m},p}}

Application of the Joule-Thomson effect

The Joule-Thomson effect is often encountered in refrigeration techniques. By compressing a gas the gas can be liquefied. Any created heat is led away from the system. Consequently, the gas is expanded again. If the expansion is performed within a pressure and temperature range at which the gas exhibits a positive Joule-Thomson coefficient, the gas cools down and liquefies again. The cold, liquefied gas can be used as a countercurrent to cool the incoming gas.

Figure shows the inversion curve of various gases. By expanding nitrogen gas it is possible to cool hydrogen gas to a temperature of less than approximately 210 K. The expansion of hydrogen gas can then be used at this temperature to cool helium to a temperature of 30 K. The consequent expansion of helium can then be used to reach very little temperatures.

Experimental setup

The experimental setup used in the lab course is different from the one shown in figure . Instead of pistons we use a pressure reducer to expand gas from a gas cylinder at a constant pressure against the (constant) environmental pressure. A porous frit ensures a slow expansion. Using two thermometers the temperature in front and directly after the frit is determined. Since the gas is taken from a gas cylinder at large pressure, the pressure reduction across the supply line causes a temperature change due to the Joule-Thomson effect as well. In order to allow for all gases to be expanded at room temperature, the gases pass a heat exchanger (made up of a thin cylindrically metal tube).

Instructions [Laboratory experiment]

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Lab course instructions

The aim is to determine the temperature difference (using two thermometers) as function of the adjustable pressure difference of the gases carbon dioxide and nitrogen.

Measurements

Start with carbon dioxide first and repeat the same measurement steps listed below with nitrogen gas.

Adjust the three-way cock for the corresponding gas.
Open the (upper) main valve by turning counter-clockwise until the red mark is replaced by the green mark.
Now slowly open the (lower) fine valve by turning it clockwise until a pressure difference of approximately 1.00 bar is reached. Make sure the pressure difference does not change over time.
Wait for approximately four minutes until the pressure difference is constant. Note down the pressure difference and the Celsius temperature at both thermometers.
Reduce the pressure at the (lower) fine valve slowly in by approximately 0.15 bar. Wait again approximately four minutes and note down the pressure and the Celsius temperature at both thermometers.
Repeat the last step until the pressure difference has reached approximately 0.25 bar. The measurement series for the gas is hereby ended.
Close first the (upper) main valve until the red mark appears. Wait until the pressure built up in the system has been reduced and then close the (lower) fine valve. Now the three-way cock can be switched to the next gas.

End of the measurement

Make sure that all valves are closed (the red mark must be visible).

Instructions [Digital experiment]

Set the internal pressure from ca. 125 kPa to ca. 250 kPa in steps of about 25 kPa. Wait about 10 seconds for the temperatures to stabilise and record the readings. Carry out the measurement for carbon dioxide, nitrogen and helium.

Analysis