# Reason for negative Joule Thomson coefficient of Helium and Hydrogen at NTP conditions

Recently, while reading my textbook I came to know that Helium, Hydrogen and Neon are the only gases which have negative Joule Thomson coefficient at NTP conditions, i.e heating effect is observed when these gases expand from a region of high pressure to a region of low pressure. I was wandering about possible explanations for this strange behaviour (since almost all gases display cooling effect during free expansion at NTP conditions) but nothing came to my mind. Can anyone give me a reason why these gases behave differently at room temperature and 1 atm pressure?

One thing I noticed is that all the three gases in question have compressibility factor(Z) always greater than 1, i.e. negligible intermolecular forces of attractions, but how can this explain their abnormal behaviour?

• The JT coefficient changes sign at the inversion temperature, this is true for all gases. Oct 25, 2016 at 2:03
• And of course you can relate it to Z, I shall write up an answer later. Oct 25, 2016 at 2:11

Note: You can skip section I, and go straight to section II and/or the end of section III (specifically the conclusions subsection), if you are already familiar with the basic mathematical machinery/definitions

## I. Preliminaries

Feel free to skip all of this if you are familiar with it

First, we have an adiabatic expansion, i.e $\mathrm{q} = 0$.

Second, our process is carried out as follows: a gas at pressure $p_i$ flows into a region of lower pressure $p_f$ without significant change in kinetic energy. This is called the Joule–Thomson expansion.

Obviously, this process is inherently irreversible. Additionally, one can show that this expansion isenthalpic (i.e $H_i = H_f$). Unlike a free expansion, work is done, causing a change in internal energy; the exact specifics of this change will depend on the initial and final conditions and also on the nature of the working fluid.

What we measure experimentally is a change in temperature with respect to pressure at constant $H$, and we call it $\mu_{\mathrm{JT}}$ (Joule-Thomson Coefficient).

$$\mu_{\mathrm{JT}} = \left(\frac{\partial T}{\partial P} \right)_H = \frac{V(T \alpha-1)}{C_p} \tag{1}$$

where $\alpha$ is the coefficient of thermal expansion $$\alpha = \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_p$$

All real gases have an inversion point at which the value of $\mu _{\mathrm {JT}}$ changes sign. The Joule–Thomson inversion temperature, depends on the pressure of the gas before expansion.

$$\begin{array}{cc} \hline \text{Temperature} & \text{sign of}\ \partial P & \text{sign of}\ \mu _{\mathrm {JT}} & \text{thus sign of}\ \partial T \\ \hline \text{below inversion temp.} & - & + & - \\ \text{above inversion } & - & - & + \\ \hline \end{array}$$

The final pressure is always lower than the initial pressure and thus the change is always negative. Depending on whether you are above or below the inversion temperature, your change in temperature can either be positive (warming) or negative (cooling). I would like to stress that the inversion temperature depends both on the temperature and pressure.

## II. Qualitative Analysis

The arguments given below might seem a little "hand-wavy"/crude, but bear with me.

$$H = U + pV$$ $$\Delta H = \Delta U + \Delta (pV)$$

During a Joule-Thomson expansion (isenthalpic, adiabatic), the change in $pV$ represents the work done by the fluid.

If $PV$ increases,at constant enthalpy, then U must decrease as a result of the fluid doing work on its surroundings; the working material is using its own internal energy to do work. This produces a decrease in temperature and results in a positive Joule-Thomson coefficient.

Decrease in$pV$ means that work is done on the fluid and the internal energy increases. This can thus lead to an increase in temperature and the Joule-Thomson coefficient will be negative.

For an ideal gas, $pV$ does not change during a Joule-Thomson expansion.

As a result, there is no change in internal energy; also, there no change in temperature; and the joule-thomson coefficient is zero.

In real gases, $pV$ does change. The deviation from ideality (in general) is contained in $Z$, the compressibility factor. Thus, it would make sense if we could make some connection to it.

The ratio of the value of "real" $pV$ to that expected for an ideal gas at the same temperature is called the compressibility factor,

$$Z := \frac{pV}{nRT}$$

For a gas, this is typically less than unity at low temperature (attractive forces dominant) and greater than unity at high temperature.

At low pressure, the value of $Z$ always moves towards unity as a gas expands (tend towards ideality).

In general at low temperature, $Z$ and $PV$ will increase as the gas expands, resulting in a positive Joule-Thomson coefficient.

At high temperature, $Z$ and $pV$ decrease as the gas expands; if the decrease is large enough, the Joule-Thomson coefficient will be negative.

It follows, that for the $Z>1$, Joule-Thomson coefficient will be negative and the gas gets warmer. This is the case for the gases in question at the given conditions.

A graph from wikipedia showing how the Joule-Thomson Coefficient changes sign (for nitrogen):

Note: The question originally asked about gases, but I'll make a small point about condensed phases and supercritical fluids (since you can see them in the graph). At high pressures, these phases can't respond to an increase in pressure by reducing their volume, and $pV$ increases as pressure increases. The molecules are being forced together, and such the Joule-Thomson coefficient is negative.

Of course intermolecular forces, attractive and repulsive, and the limits in which they are dominant are key in explaining the changes of $Z$, and consequently of $\mu_{\mathrm{JT}}$. This point is further developed in the next section.

## III. A slightly more quantitative, but ultimately Qualitative Analysis

To see how intermolecular interactions come into play, let's look at two model systems: an ideal gas (no intermolecular interactions) and a Van der Waals gas (both attractive and repulsive interactions).

Using equation (1) and the definition of $\alpha$ we can write

$$\mu_{\mathrm{JT}} = \frac{V(T \alpha-1)}{C_p} = \frac{T \left(\left(\frac{\partial V}{\partial T}\right)_p -V\right)}{C_p} \tag{2}$$

For a perfect gast $\left(\frac{\partial V}{\partial T}\right)_p = \frac{V}{T}$ and so, $\mu_{\mathrm{JT}} = 0$

Since the Van der Waals gas equation is cubic in $V$, I will use the first few terms of the virial expansion and solve for $V$ to obtain $\left(\frac{\partial V}{\partial T}\right)_p$. I am omitting explicit computations, you may do them as an exercise.

$$Z = 1 + \frac{1}{RT} \left(b- \frac{a}{RT} \right)p + .... \tag{3}$$

Since $Z = \frac{p\tilde{V}}{RT}$ where $\tilde{V}$ is the molar volume of the gas; I am doing this for convenience's sake, I don't want to lug around a $n$ all over the place.

$$\tilde{V} = \frac{RT}{p} + \left(b- \frac{a}{RT} \right) + .... \tag{4}$$

After differentiating we obtain

$$\left(\frac{\partial V}{\partial T}\right)_p = \frac{R}{p} + \left(- \frac{a}{RT^2} \right) + .... \tag{5}$$

Substituting (4) and (5) into (2), and dropping all of the higher order terms.

$$\mu_{\mathrm{JT}} \approx \frac{\frac{2a}{RT}-b}{C_p}$$

Now, consider the low temperature limit of this expression. The $\frac{2a}{RT}$ becomes large, and $b$ is negligibly small compared to it so

$$\mu_{\mathrm{JT}} \approx \frac{2a}{RTC_p}$$

All the quantities on the right hand side of this equation are positive. so $\mu$ is positive. The attractive forces are dominant, since only the paramter $a$ appears in our expression.

Similarly, the high temperature limit is

$$\mu_{\mathrm{JT}} \approx \frac{-b}{C_p}$$

Here, $\mu$ is clearly negative. The repulsive forces are dominant, since only the parameter $b$ appears in our expression.

Conclusions

• The existence of an inversion temperature becomes obvious in a gas with both attractive and repulsive forces, since $\mu_{\mathrm{JT}}$ changes sign from positive to negative.
• The connection to attractive and repulsive forces also becomes more explicit.
• In the low temperature limit, attractive forces are dominant and the gas needs to break free from these interactions as it expands. This requires energy, which has to be supplied by $U$ the internal energy of the gas. Thus the gas cools.
• At high temperatures, the repulsions are dominant. As the gas expands, the over potential is lowered, and energy is released. Thus, the gas warms up.
• Obviously, the exact magnitude of the high and low temperature limit depends on the nature of working fluid, since the parameters $a$ and $b$ are different for each gas. The gases you are talking about, have barely an attractive interactions.

Caveats

• Our analysis based on the Van der Waals gas seems to suggest that the inversion temperature and $\mu_\mathrm{JT}$ is a function determined by temperature alone. This has experimentally proven to be false. This discrepancy arises because we dropped the higher order terms in our expansion which contained the pressure dependance.
• This model is ultimately qualitative. It will give you the wrong numbers when you try to compute something, but it explains the general trends well.
• Thanks for the detailed answer!! I think the bottom line is that.....due to decrease in interatomic/intermolecular potential energy of the gases in question.....their kinetic energy increases which results in increase in temperature.....since K+P=U and U should be constant overall (kinetic + potential). Am I correct? Oct 25, 2016 at 5:53
• @SirJMP I'm not sure I follow your argument. Oct 25, 2016 at 12:06

Since joule-thomson effect is isoenthalpic process, so everything is done at the cost of internal energy.

Taking $$a$$ as the term involving intermolecular forces of attraction and $$b$$ as the excluded volume, we get:

$$T_i = \frac{2a}{Rb}$$

As per above equation inversion temperature is the function of intermolecular forces of attraction. When the temperature is high and above $$T_i$$ intermolecular forces of attraction are weak and kinetic energy is high for particles. When in such condition gas expands the kinetic energy of particles increases rapidly and small amount of energy is spent to break forces of attraction among particles of gas and hence increase in kinetic energy is dominant over energy spent. Since temperature is the kinetic manifestation of kinetic energy($$K =\frac{1}{2}mv^2$$ and $$v =\sqrt{\frac{3}{2}RT}$$). Hence temperature of the gas increases.

However if gas is below inversion temperature the forces of attraction are huge compared to kinetic energy and expansion of such a gas involves spending much energy for breaking intermolecular forces compared to rise in kinetic energy of particles as a result of expansion. Hence use of internal energy for breaking intermolecular forces dominates over increase in kinetic energy and thus gas shows cooling.

Thus from the above arguments it can be understood that at inversion temperature both factors balance and hence gas shows neither cooling nor warming under conditions of expansion.

Taking case of $$\ce{H}$$, $$\ce{He}$$ these particles are very small and intermolecular forces of attraction are very weak and hence the inversion temperature is below room temperature (forces of attraction are stroger below room temperature). Thus when such systems are allowed to expand from region of high pressure to low pressure these show heating effect rather cooling(explained above).

• Do not add personal information at the end of your answer. this is for your own safety Apr 28, 2021 at 12:43

when the gas is expand its pressure decreases and hence volume increases so the molecules of gas are farther apart when the distance between the molecule increases their potential energy decreases so temperature(K.E) also decreases hence cooling effect produces

But in case of hydrogen gas during expansion the interaction between its molecule is more so the distance between them is less means they are most attractive as distance decreases the potential energy also decreases hence temperature(K.E) increases which produces heating effect

Another reason is that hydrogen shows cooling effect at -80 degree centigrade so at room temperature(21) it shows heating effect Hope you got the answer