I have been struggling to understand how the molecules behave during the Joule–Thomson effect. I would love to get some help on this concept. Here is what I got so far.

Under adiabatic and isenthalpic conditions, when a gas expand from high pressure to low pressure, depending on the initial pressure, initial temperature, and the molecular properties, the gas can either experience cooling or heating.

Specifically, if the gas was initially in the state where attractive intermolecular forces were dominant (typically at regions of moderate temperature and low pressure), as it expand, the kinetic energy of molecules are used to overcome the attraction reach greater separation. Also, the internal energy is overall decreased as the gas need to do work to expand. Thus the temperature of the gas decreases.

If repulsive forces were dominant (at regions of very high temperature or high pressure), as the gas expand, the repulsion kind of pushes(?) the molecules to reach greater distance. In other words, the potential energy is converted to kinetic energy here. And here although internal energy may be decreased to provide work for expansion, the molecule would have gained more kinetic energy from the repulsive forces such that the temperature of the molecules increases.

My question is, how does the system being isenthalpic play a role in this effect? I get that we need the adiabatic condition so that only the internal energy is being used to provide the work required for expansion. But what about enthalpy? How is isenthalpic + adiabatic expansion different from just adiabatic expansion?

Additionally, I would appreciate if someone could also explain how $pV$ from the enthalpy formula $H=U+pV$, changes during this expansion.

  • $\begingroup$ Are you familiar with the derivation of the open system (control volume) version of the 1st law of thermodynamics? In your assessment, you are omitting a viscous heating effect due to viscous friction (in the valve or porous plug) which offsets most of the effect of expansion cooling. $\endgroup$ Apr 7, 2023 at 11:00
  • $\begingroup$ No I have never heard about that. I'm just following the textbook from Atkins which doesn't seem to explain much detail about this. $\endgroup$ Apr 7, 2023 at 11:57
  • $\begingroup$ Get yourself a decent Thermo book like Smith and Van Ness or Moran et al and read up on their sections on the open system version of the 1st law of thermodynamics. $\endgroup$ Apr 7, 2023 at 12:18

1 Answer 1


This is a throttling process due to a porous plug/constriction between the inlet and outlet. Constant flow is maintained by a pump and the pressure on either side of the plug is maintained at fixed values. The heat $q=0$ and the work done is $ W =\int_0^{V_f}pdV +\int_{V_i}^0pdV= p_fV_f-p_iV_i $ between initial and final states, therefore $0 = U_f-U_i+p_fV_f-p_iV_i$ which means that $H_i=H_f$ which does not mean that enthalpy is constant.

In response to comment I have added some details

The coefficient used to quantify cooling or heating is

$$\displaystyle \mu=\left(\frac{\partial T}{\partial P} \right)_H$$

and this is obtained from the slope between points on a pressure vs temperature plot. These points are obtained by a throttling experiment. The initial pressure and temperature is decided upon and also the final pressure, which is always less than the initial one, and the final temperature is then measured. This produces one point on a $p$ vs $T$ plot. Next only the final pressure is changed and a second temperature point obtained and so on. A set of points that approximates to a curve is produced; the isenthalp. The slope of this between two points is the JT coefficient $\mu$ and if this is positive cooling occurs and vice versa. The experiment is repeated many times with different initial $T_i$ and $p_i$ values and over the range of final pressures $p_f$. A plot is produced with many isenthalps one above the other and by jointing the points where the slope is zero $\mu=0$ a curve is produced, the inversion curve, separating heating and cooling, see the figure.

JT inversion

Joule Thompson isenthalps and the inversion curve. The cooling area is to the left of the dashed red line. The blue lines are the isenthalps.The figure is from Wikipedia commons; (https://commons.wikimedia.org/wiki/File:Isenthalpic_contours_for_temperature_versus_pressure_showing_inversion_curve.svg)

The JT coefficient can be related to heat capacity and volume for a non-ideal gas. It is always zero for an ideal gas since the the effect is dependent on intermolecular forces.

Using a Maxwell equation

$$\displaystyle dH=TdS+Vdp$$

We shall need to eliminate $TdS$ and can do this by using

$$\displaystyle TdS=C_p dT-T\left( \frac{\partial V}{\partial T}\right)_pdp$$

( this eqn. is more familiar for an ideal gas as $\displaystyle dS=\frac{C_p}{T} dT-R\frac{dp}{p}$ )

Rearranging gives

$$\displaystyle dT=\frac{1}{C_p}\left[ T\left( \frac{\partial V}{\partial T}\right)_p-V\right]dp+\frac{dH}{C_p}$$

and differentiating wrt $p$ and constant $H$ gives

$$\displaystyle \mu =\frac{1}{C_p}\left[ T\left( \frac{\partial V}{\partial T}\right)_p-V\right]$$

which for an ideal gas ($pV=RT$) is zero.

  • $\begingroup$ I'm guessing you meant that the enthalpy is constant. I did see this derivation on how enthalpy is proved to be constant throughout the throttling process. But what I'm still not getting is how constant enthalpy significantly contributes to the Joule Thomson effect. What would happen if the enthalpy isn't constant? $\endgroup$ Apr 7, 2023 at 9:50
  • $\begingroup$ @asdfasdfasdf Not restricted by dH=0, gas could do pV work against external pressure. Even H2 or He that warm up at room T because of J-T effect, would cool down because of that work. $\endgroup$
    – Poutnik
    Apr 7, 2023 at 16:13
  • $\begingroup$ No, I meant that the enthalpy initially and finally is the same not that it does not change because we can only know the initial and final equilibrium points not any intermediates ones. $\endgroup$
    – porphyrin
    Apr 7, 2023 at 19:33
  • $\begingroup$ The substitution you have used $$V = T\left( \frac{\partial V}{\partial T}\right)_p$$ implies $$\left( \frac{\partial V}{V}\right)_p =\left( \frac{\partial T}{ T}\right)_p $$ which is valid only for ideal gases, while you try to derive expressions for non-ideal gases. It should be noted and justified. $\endgroup$
    – Poutnik
    Apr 11, 2023 at 14:20
  • 1
    $\begingroup$ been similar number of years since I taught this :) $\endgroup$
    – porphyrin
    Apr 12, 2023 at 9:12

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