How is pressure related to the rate of momentum change in a gas?

[rogerk8]

It appears that Russ and I have slightly different interpretations of what is and what is not a Joule-Kelvin or Joule-Thomson expansion.

Russ takes the position that an isenthalpic rapid flash evaporation/expansion or auto-refrigeration of a compressed fluid in passing through a throttle valve is not an example of Joule-Thomson expansion. I would say that it is. There appears to be support for both points of view.

I think we both agree that the temperature reduction occurs because the internal kinetic energy of the fluid itself is converted to potential energy (the farther apart the molecules are, on average, the more potential energy they have because of the attractive forces between molecules). The process does not involve heat flow to or from the fluid during expansion/evaporation so the increase in potential energy is at the expense of internal kinetic energy, hence temperature decreases. The heat exchange occurs later in the evaporator once the fluid has been cooled by its passage through the throttle valve. That subsequent heat exchange does not involve the Joule-Thomson effect (but keep in mind that the subsequent heat exchange could not occur without the prior cooling).

Here is a passage from my 1968 5th Ed. of Zemansky Heat and Thermodynamics at p. 181-2:

"In order to gain a little more insight into the working of a refrigerator, let us consider some of the details of a modern refrigeration plant that are reflected in most home refrigerators. The schematic diagram in Fig. 7-7a shows the path of a constant mass of refrigerant as it is conveyed from the liquid storage, where it is at the temperature and pressure of the condenser, through the throttling valve, through the evaporator, into the compressor, and finally back to the condenser.

In the condenser the refrigerant is at a high pressure and at as low a temperature as can be obtained with air or water cooling. The refrigerant is always of such a nature that, at this pressure and temperature, it is a saturated liquid. When a fluid passes through a narrow opening (a needle valve) from a region of constant high pressure to a region of constant lower pressure adiabatically, it is said to under go a throttling process (see Prob. 4-8), or a Joule-Thomson or Joule-Kelvin expansion. This process will be considered in some detail in Chapter 11. It is a property of saturated liquids (not of gases) that a throttling process always produces cooling and partial vaporization. In the evaporator the fluid is completely vaporized, with the heat of vaporization being supplied by the materials to be cooled. The vapor is then compressed adiabatically, thereby increasing in temperature. In the condenser, the vapor is cooled until it condenses and becomes completely liquefied."

There appears to be a lot of confusion on this point and maybe Zemansky is confused. But, as I say, whatever one wants to call it, this auto-refrigeration effect due to the pressurized refrigerant suddenly expanding and cooling without exchanging heat with its surroundings is at the heart of modern refrigerators.

AM

I now totally and finally understand the fact marked with bold above (along with energy conservation, E=KE+PE=constant).

While kinetic energy (KE) strongly relates to temperature, potential energy (PE) may be viewed as the work done (dW=Fdx) moving a particle gravitationally or even against cloumbous force.

This means that the further things are apart, the higher the potential energy is.

Sounds almost too basic

Roger

 

[russ_watters]

OK, now I understand at least that this is partly an issue of terminology: you DO apparently understand that almost all of the temperature drop is due to the latent heat of the phase change, not gas-gas expansion - you just consider that part of the Joule-Thompson effect. It doesn't change much of the problem though since you hadn't emphasized the contribution of the phase change before.

More to the point, while terminology is arbitrary/by convention, to me a process that is labeled as a Joule-Thompson effect should be describable via the Joule-Thompson coefficient and related equations. Otherwise, students could be led into a problem they can't solve because of the insufficient definition.

 

[rogerk8]

OK, now I understand at least that this is partly an issue of terminology: you DO apparently understand that almost all of the temperature drop is due to the latent heat of the phase change, not gas-gas expansion - you just consider that part of the Joule-Thompson effect. It doesn't change much of the problem though since you hadn't emphasized the contribution of the phase change before.

More to the point, while terminology is arbitrary/by convention, to me a process that is labeled as a Joule-Thompson effect should be describable via the Joule-Thompson coefficient and related equations. Otherwise, students could be led into a problem they can't solve because of the insufficient definition.

I think I showed too much self confidence above.

I only think I understand the gas-gas expansion part of the Joule-Thompson effect. It sounds reasonable to me. But the magic part is that temperatures has to be lower than the inversion temperature for it to happen.

I do however know the basic fact that a gas under high enough pressure will be transformed into liquid.

Anyway, why is the phase change so important? To me it seems like it suffices that the refrigerant always is at it's gas state (but at two very different pressures). I do not see the need for it to become liquified to create a cooling effect. Maybe as a matter of efficiency, though.

I have read some hyperphysics about thermodynamics (obviously not so well though). I thought I learned that the internal energy (U) of a gas is composed of KE+PE. Two different gases of the same temperature has the same KE but may have different U (even if pressure is the same?). Now, the first law of thermodynamics is from above Q=\Delta U+W where Q is total energy(?) and W is work done by the gas (W>0). So maybe you may view this as if total energy has to remain constant, and work is done by the gas, U will have to decrease. But according to my above "understanding" either KE or PE may decrease. But if you at the same time can say that pressure is lower, PE is by definition higher and conservation of energy thus gives lower KE and hence lower T.

How far from the truth am I now?

Roger

 

[voko]

I do however know the basic fact that a gas under high enough pressure will be transformed into liquid.

No, that's not true. There is a so-called critical temperature, above which no pressure can convert gas into liquid (but can convert it into solid, if really big enough). In helium-4, for example, the critical temperature is 5.2 K.

Anyway, why is the phase change so important? To me it seems like it suffices that the refrigerant always is at it's gas state (but at two very different pressures). I do not see the need for it to become liquified to create a cooling effect. Maybe as a matter of efficiency, though.

It is. The latent heat of vaporisation is very large.

 

[rogerk8]

No, that's not true. There is a so-called critical temperature, above which no pressure can convert gas into liquid (but can convert it into solid, if really big enough). In helium-4, for example, the critical temperature is 5.2 K.

Is it correct to say that below this critical temperature gases may be converted into liquid?

Yet I find it fascinating that above this critical temperature gases may be converted into solid instead (if pressure is high enough). Skipping the liquid phase so to speak(?)

It is. The latent heat of vaporisation is very large.

This I obviously need to study some more.

Roger

 

[voko]

Is it correct to say that below this critical temperature gases may be converted into liquid?

It is only below the critical temperature that they can. Check out "supercritical fluid".

 

[rogerk8]

What is the exact relationship between pressure and potential energy of a gas?

 

[dauto]

What potential energy? do you mean the amount of work that can be extracted from the gas if allowed to expand? that depends on the exact expansion process. For instance, an adiabatic expansion will give a different result than an isothermic expansion. The work is not a state function.

 

[rogerk8]

Jesus, I know nothing but please consider this:

U=KE+PE

and

KE=kT≈\frac{mv^2}{2}

and

\frac{PV}{N}=kT

Then

U=kT+kT

?

 

[voko]

You seem to be looking for the internal energy of gas. For an ideal gas, whose molecules do not interact, it is just the kinetic energy. For non-ideal gas, there are many models of varying complexity. Perhaps the simplest one is the van der Waals model.

 

[Andrew Mason]

Jesus, I know nothing but please consider this:

U=KE+PE

and

KE=kT≈\frac{mv^2}{2}

and

\frac{PV}{N}=kT

Then

U=kT+kT

?

The average translational kinetic energy is NkT/2 for each translational degree of freedom, so <KE> = 3NkT/2

But that is not necessarily the entire internal kinetic energy. For non-monatomic gases, there may be additional degrees of freedom of movement such as rotational or vibrational modes. These contribute to internal energy BUT NOT to temperature. There is an average KE = NkT/2 associated with each additional degree of freedom.

So, letting f = no. of degrees of freedom, the internal average kinetic energy is <KE> =fNkT/2

But internal energy can include potential energy if there are forces between molecules. It takes energy to increase the average separation of the molecules. So when potential energy increases due to increased average separation of molecules, internal energy increases.

According to the equipartition theorem, which is an important part of the kinetic theory, there is an energy NkT/2 associated with each degree of freedom for potential energy as well.

So U = KE + PE = fNkT/2 where f is the total no. of degrees of freedom for kinetic and potential energy.

AM

 

[rogerk8]

Thank you once again Andrew Mason!

I am thankful for all your help!

Now I know that there is a kinetic energy of NkT/2 per degree of freedom and that this is associated with potential energy as well.

My brand new understanding of internal energy is thus

U=KE+PE=fNkT/2
But what about PE and pressure?

Roger

 

[Andrew Mason]

Thank you once again Andrew Mason!

I am thankful for all your help!

Now I know that there is a kinetic energy of NkT/2 per degree of freedom and that this is associated with potential energy as well.

My brand new understanding of internal energy is thus

U=KE+PE=fNkT/2
But what about PE and pressure?

Roger

Do you have a specific question about PE and pressure?

First of all you have to have a non-ideal gas with attractive forces between molecules. For such a gas PE increases with volume (as the space between molecules grows, on average, so does PE). If T remains constant (i.e. KE is constant) what happens to the pressure? (hint: how is pressure related to the rate at which molecules change momentum in striking the container walls? If KE is constant, but the distance between walls increases, what happens to the frequency of molecules colliding with the wall?).

AM

 

[rogerk8]

Do you have a specific question about PE and pressure?

First of all you have to have a non-ideal gas with attractive forces between molecules. For such a gas PE increases with volume (as the space between molecules grows, on average, so does PE). If T remains constant (i.e. KE is constant) what happens to the pressure? (hint: how is pressure related to the rate at which molecules change momentum in striking the container walls? If KE is constant, but the distance between walls increases, what happens to the frequency of molecules colliding with the wall?).

AM

I tried understanding this reading sentence by sentence but I felt stupid until I read your hint.

High pressure has to do with a high rate of momentum change, right?

So if the distance to the walls increases, the frequency must lessen (at the same KE), right?

The answer to your riddle must then be that pressure lessens

Roger
PS
While I obviously do not understand physics in spite of beautiful and simple equations here is a couple of naive thoughts:

p=U/V=(KE+PE)/V=fnkT/2...[1]
and we are almost back at the Ideal Gas Law. The difference being f=2 (instead of 3 for mono-atomic gases, right?)

But I should probably not stare so much at equations and try to understand instead. Here is by the way another thought

KE_{av}=fkT/2...[2]
which comes from my Plasma Physics book where f is said to be 3 but there's no N here!