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When an ideal gas in an adiabatic cylinder undergoes expansion, adiabatic cooling occurs and the temperature of the gas drops. This occurs because some of the gas's internal energy is used up as the system does work on the surroundings. This much I understand well. But I can't say the same for the Joule-Thomson Effect

When a real gas is passed through a porous plug for example, all the diagrams I have seen indicate that the size of the pipe before the porous plug is the same as the size of the pipe after the porous plug (https://cdn.comsol.com/cyclopedia/joule-thomson-effect/Porous-plug.ru_RU.png). So my first problem is: How does the gas's pressure decrease if it isn't even expanding after it goes through the plug? Many explanations I find say that the gas expands after it passes through a throttling valve and hence its pressure drops but clearly this can't be the case for a porous plug or any other throttling valve with constant volume on either side. So then why does the pressure drop after the gas has passed through the plug?

My second issue pertains to the fact that the Joule-Thomson effect is isenthalpic. I can't understand how this could be if the effect has anything to do with adiabatic cooling because adiabatic cooling is definitely not isenthalpic. The joule Thomson effect is more or less described by Peter Atkins as the process of allowing a real gas to "expand through a throttle (i.e porous plug) causing it to cool" without allowing heat to transfer in or out of the system (i.e. adiabatic). This description sounds almost identical to adiabatic cooling to me with the exception that the porous plug is involved. But I can't understand how a porous plug or any other throttle for that matter would lower the pressure let alone cool the gas if the pipes on either side are the same volume. The same volume on either side of the restriction indicates to me that there is no expansion occurring at all. If no expansion is occuring then the pressure shouldn't drop and neither should the temperature. Yet both these occur. So what is actually going on in the Joule-thomson effect?

Any help on this issue would be most appreciated!

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3 Answers 3

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The pressure reduces in going through a porous plug because of friction. So you need more pressure on the upstream part than on the downstream part. The work done $$ W=P_{\rm upstream} V_{\rm upstream} - P_{\rm downstream} V_{\rm downstream}$$ on forcing a gas through the plug goes into the internal energy, so, as no heat enters or leaves the system, the enthalpy $H=U+PV$ is constant. For an ideal gas $H=C_P T$, and so the temperature does not change.

The J-T effect has nothing to do with the cooling that occurs in adiabatic expansion.

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  • $\begingroup$ Thanks for the response. Is it correct of me to think as follows: Suppose we have a fluid (i.e a real gas) flowing through a pipe horizontal to the earth in the x direction. Viscous forces exert a force in the negative x direction. In order to achieve constant velocity we must have that $P_{upstream}>P_{downstream}$. Now adding the porous plug in, we still require $P_{upstream}>P_{downstream}$ but this time we need an even larger difference because now we not only need to overcome viscous drag but we also require a net force to ensure an increased velocity on the downstream side of the... $\endgroup$
    – SalahTheGoat
    Commented Feb 21, 2021 at 15:39
  • $\begingroup$ porous plug so that we can ensure a constant mass flow rate? $\endgroup$
    – SalahTheGoat
    Commented Feb 21, 2021 at 15:41
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There may be constant area before and after the plug, but, after the plug, the velocity of the gas is higher than before the plug. So the mass flow rates before and after the plug are the same.

What happens in the plug is that there is a pressure drop. This is caused by irreversible viscous shear between the plug and the flowing gas (just like the pressure drop for a fluid flowing through a capillary tube).

In the Joule-Thomson flow, the gas does not have to cool. It can also come out hotter. This depends on the equation of state of the gas. Joule-Thomson effect is strictly the result of deviation from ideal gas behavior.

Analysis of the Joule-Thomson flow using the open system (control volume) version of the first law of thermodynamics shows that, since the plug is adiabatic and no shaft work is done, the change in specific enthalpy of the gas passing through the plug is zero.

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  • $\begingroup$ Thanks for the response. So the form of the first law I think you are referring to is $q-w=h_2-h_1+\frac{V^2_2-V^2_1}{2} $ where I've left out gravity as we can assume no changes in height occur. Clearly $q=w=0$ and so we have $\Delta h =-\Delta K$. So if the velocity is greater after the porous plug as you say(which obvs must be the case else flow rates wont equal) then the process can't be truly isenthalpic. Am I correct in thinking that in practice the $\Delta K $ is so small that it is routinely neglected? That is, porous plugs aren't truly isenthalpic, only approximately isenthalpic? $\endgroup$
    – SalahTheGoat
    Commented Feb 21, 2021 at 15:20
  • $\begingroup$ Additionally, If my above comment is largely correct, why then would a temperature change occur at all? Is there a way to microscopically invision what causes the temp change? If I do the following $U_i+P_iV_i=U_f+P_fV_f \Rightarrow U_i-U_f=(P_f-P_i)V$ where I assume ($V_i=V_f$) . From this I can clearly see that if there's a pressure drop then the internal energy must increase and I can somewhat believe that is the origin of the increase in temp but I have no way of intuitively thinking about why the temperature change occurs on a molecular level? $\endgroup$
    – SalahTheGoat
    Commented Feb 21, 2021 at 15:33
  • $\begingroup$ It occurs because of change of potential energy of molecular interactions resulting from pressure decrease. Incidentally, the specific volume is not constant. Are you familiar with the equation for dH expressed in terms of dT and dP? $\endgroup$
    – Chet Miller
    Commented Feb 21, 2021 at 16:21
  • $\begingroup$ Yes $dH=-\mu C_p dp +C_p dT$ although my understanding on it is not great. $dH=-\mu C_p dp +C_p dT$. Just as a start, the reason we take the differential of H with respect to $T$ and $p$ as opposed to say V and T (which is usually the case when looking at the differential $dU$) is because we have a choice of taking the differential with respect to any two variables because 2 is enough to fix the rest? and in the case of H, taking the differential with respect to dP and dT provide more useful insights than say dV and dT would? $\endgroup$
    – SalahTheGoat
    Commented Feb 21, 2021 at 16:36
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    $\begingroup$ The correct equation is $dH=C_pdT+V(1-\alpha T)dP$, where $alpha$ is the coefficient of volumetric thermal expansion and V is the molar volume. For an ideal gas, the term in parenthesis is zero. With regard to your other question, the answer is yes, expressing dH in terms of dT and dP is often much more convenient and useful. $\endgroup$
    – Chet Miller
    Commented Feb 21, 2021 at 19:37
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So I suppose this experiment is meant to demonstrate the Joule-Thomson effect.

So in this experiment we have a big tank filled with gas, and a smaller tank with a vacuum inside. We connect the tanks with a pipe, inside of which there is the porous plug.

We ignore the slight cooling of the gas inside the big tank.

We notice that the gas in the small tank has cooled. It cooled because it was real gas that underwent free expansion.

Real gases that undergo free expansion cool because of Joule-Thomson effect. An ideal gas does not cool when undergoing free expansion.

Joule-Thomson effect has to do with attractive forces between molecules. Gas has to do work when it expands freely, if there is an attractive force between molecules.

Or, the kinetic energy of molecules in the gas do work on the potential energy of the molecules in the gas.

We must let just a small amount of gas through the pipe, otherwise the experiment does not work as intended. Or we can use a vacuum pump to keep pumping gas out of the tank with the lower pressure.

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