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

[rogerk8]

Hi!

I wonder how a refridgerator work.

I understand that the Ideal Gas Law

pV=N_{mol}RT

is used somehow.

This is my blog description of my preliminary understanding:

"Here both N and V is constant. A compressor on the outside of the refridgerator enables high pressure of the gas there. A high pressure valve then let's some of the gas inside the refridgerator. While the pressure inside then is much lower it absorbs heat to get back to its original pressure before it is circulated back into the system(?)"

Thankful for any correction.

Roger

 

[ZapperZ]

You should look up heat pump

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/heatpump.html

Zz.

 

[russ_watters]

Vapor compression refrigeration: http://en.wikipedia.org/wiki/Vapor-compression_refrigeration

This is one of those rare cases where I prefer wikipedia to hyperphysics. Besides being thin on the details for heat pumps/air conditioners, it incorrectly defines the term "heat engine" (saying a heat pump is a heat engine, which is basically a self-contradiction).

 

[sophiecentaur]

The central part of a refrigeration cycle like this is that the expanding gas / vapour has to Do Work, which will lower its internal energy. The nozzle that the gas is allowed to pass through is the thing that does this work in the 'throttling process'. If the gas were allowed to expand without doing any work, there would be no cooling effect.

 

[Manraj singh]

I was wondering, why is it that when we open the refrigirator, we see a 'mist' kind of thing. Is it because of the moisture in the atmosphere condensing when it enters the refrigirator?

 

[dauto]

An ideal gas can be used for refrigeration in a four step process.
1st: The compressor compresses the gas adiabatically. Both pressure and temperature increase.
2nd: Heat is removed from the gas by bringing it to thermal equilibrium with the outside environment (The kitchen)
3rd: The air is allowed to expand adiabatically to a low pressure and low temperature.
4th: Heat is added to the gas by bringing it to thermal equilibrium with the inside environment (heat is removed from the inside of the refrigerator).

Two Important points.

1st: Many gases actually used in refrigeration will actually turn liquid when under high pressure, so using ideal gas equations is completely wrong. Latent heat becomes very important.

2nd: Even for gases, effects such as the Joule–Thomson_effect are very important in most cases. This effect requires treating the gas as a non-ideal gas.

 

[dauto]

The central part of a refrigeration cycle like this is that the expanding gas / vapour has to Do Work, which will lower its internal energy. The nozzle that the gas is allowed to pass through is the thing that does this work in the 'throttling process'. If the gas were allowed to expand without doing any work, there would be no cooling effect.

That's only true for ideal gases. In the real world the Joule–Thomson effect which requires no external work is the main source of cooling. In the throttling process actually used in most refrigeration systems the amount of work actually performed during the expansion is almost negligible.

 

[russ_watters]

I was wondering, why is it that when we open the refrigirator, we see a 'mist' kind of thing. Is it because of the moisture in the atmosphere condensing when it enters the refrigirator?

Yes.

 

[Manraj singh]

Thank you.

 

[jartsa]

That's only true for ideal gases. In the real world the Joule–Thomson effect which requires no external work is the main source of cooling. In the throttling process actually used in most refrigeration systems the amount of work actually performed during the expansion is almost negligible.

I still think that in common fridges evaporation is the main source of cooling. I even checked Wikipedia.

High pressure freon flows into low pressure area through a valve which throttles the flow. The freon evaporates, and the pressure of the freon gas decreases.

My intuition says that evaporation is the main cooling effect there.

 

[russ_watters]

It is.

 

[Andrew Mason]

I still think that in common fridges evaporation is the main source of cooling. I even checked Wikipedia.

High pressure freon flows into low pressure area through a valve which throttles the flow. The freon evaporates, and the pressure of the freon gas decreases.

My intuition says that evaporation is the main cooling effect there.

The working fluid does not have to change state, but you have the right idea. Due to strong inter-molecular forces an increase in the separation of molecules will require energy. Since the throttling process is adiabatic and involves no work (free expansion), the only source of this increase in potential energy has to be the internal energy. So molecular kinetic energy has to decrease, hence the temperature decreases.

AM

 

[russ_watters]

The working fluid does not have to change state...

The question wasn't whether it has to or can be done a certain way, it was whether it is done that way. In real refrigeration, the phase change absorbs/releases most of the energy.

 

[Andrew Mason]

The question wasn't whether it has to or can be done a certain way, it was whether it is done that way. In real refrigeration, the phase change absorbs/releases most of the energy.

Not quite. The phase change of the refrigerant does not absorb heat. The throttling process is adiabatic (no transfer of heat). The absorption of heat from the surroundings occurs AFTER the drop in temperature due to the JT effect as the refrigerant gas comes into thermal contact with the space being cooled.

AM

 

[russ_watters]

Not quite. The phase change of the refrigerant does not absorb heat. The throttling process is adiabatic (no transfer of heat). The absorption of heat from the surroundings occurs AFTER the drop in temperature due to the JT effect as the refrigerant gas comes into thermal contact with the space being cooled.

AM

I'm sorry, but you are really making a mess now and it seems trying to nitpick as well. Most of that is wrong:
1. The phase change in the expansion valve causes latent heat to be converted from sensible heat, dropping the temperature of the fluid as it evaporates:

The condensed liquid refrigerant, in the thermodynamic state known as a saturated liquid, is next routed through an expansion valve where it undergoes an abrupt reduction in pressure. That pressure reduction results in the adiabatic flash evaporation of a part of the liquid refrigerant. The auto-refrigeration effect of the adiabatic flash evaporation lowers the temperature of the liquid and vapor refrigerant mixture to where it is colder than the temperature of the enclosed space to be refrigerated.

2. Then in the evaporator, heat is absorbed from the surroundings and the rest of the liquid evaporates, absorbing more latent heat:

The cold mixture is then routed through the coil or tubes in the evaporator. A fan circulates the warm air in the enclosed space across the coil or tubes carrying the cold refrigerant liquid and vapor mixture. That warm air evaporates the liquid part of the cold refrigerant mixture.

And:

When the working fluid is a gas that is compressed and expanded but does not change phase, the refrigeration cycle is called a gas cycle. Air is most often this working fluid. As there is no condensation and evaporation intended in a gas cycle, components corresponding to the condenser and evaporator in a vapor compression cycle are the hot and cold gas-to-gas heat exchangers in gas cycles.

The gas cycle is less efficient than the vapor compression cycle...

Because of their lower efficiency and larger bulk, air cycle coolers are not often applied in terrestrial refrigeration.

For cooling that just uses the Joule-Thompson effect, liquifying gases for cryogenics is one of the few common applications: http://en.wikipedia.org/wiki/Joule–Thomson_effect#Applications
But notice that this is the opposite of what happens in a refrigerator: the gas expands and cools by the Joule-Thompson effect, causing it to liquify. In a refrigerator, the liquid is throttled and allowed to expand, causing it to vaporize.

 

[sophiecentaur]

For cooling that just uses the Joule-Thompson effect, liquifying gases for cryogenics is one of the few common applications: http://en.wikipedia.org/wiki/Joule–Thomson_effect#Applications
But notice that this is the opposite of what happens in a refrigerator: the gas expands and cools by the Joule-Thompson effect, causing it to liquify. In a refrigerator, the liquid is throttled and allowed to expand, causing it to vaporize.

Also, CO2 from a fire extinguisher causes a change of state as a result of the J-T cooling.

 

[Andrew Mason]

I'm sorry, but you are really making a mess now and it seems trying to nitpick as well. Most of that is wrong:
1. The phase change in the expansion valve causes latent heat to be converted from sensible heat, dropping the temperature of the fluid as it evaporates:

2. Then in the evaporator, heat is absorbed from the surroundings and the rest of the liquid evaporates, absorbing more latent heat:

The distinction between evaporative cooling and Joule-Thomson cooling is significant. In evaporative cooling the heat is absorbed from the surroundings during the evaporation process. In JT cooling the energy needed to separate the gas molecules is drawn from the internal kinetic energy of the cooling substance as it is adiabatically throttled. The heat is not absorbed during evaporation. That is the point I was making. I am sorry that you think that is nitpicking or an attempt to mess things up.

Your point that in a refrigerator there is some evaporative cooling is well taken (the liquid portion of the refrigerant that remains after throttling may evaporate during the heat exchange process).

AM

 

[rogerk8]

Hi!

I have thought some more regarding how a refrigerator might work.

I have read all your nice explanations but the most important part of understanding came from my collegue.

Now I only need to know how a compressor actually works. At least I think so.

Anyway, I am attaching a drawing of my current understanding.

Roger

 

[russ_watters]

Still a lot you aren't getting, roger:
1. A compressor increases pressure to move/circulate the fluid.
2. The valve creates a pressure drop.
3. The fluid isn't always a gas, it switches back and forth between gas and liquid.

Have you not looked at any of the links/information provided?

 

[Andrew Mason]

Hi!

I have thought some more regarding how a refrigerator might work.

I have read all your nice explanations but the most important part of understanding came from my collegue.

Now I only need to know how a compressor actually works. At least I think so.

Anyway, I am attaching a drawing of my current understanding.

You have it backwards. The compressor heats up the fluid. This is simply the first law: Q = ΔU + W where W = work done BY the gas. If work is done on the gas (W<0) and no heat is allowed in or out, then ΔU must be positive ie. T increases. The key to cooling is the expansion. PV=nRT just tells you that T is proportional to V if P is constant. When you compress the gas, the work done in compressing increases the internal energy so T increases unless heat flows out (Q<0).

If you run any thermodynamic engine cycle backward you can move heat the other way i.e from a cooler reservoir to a warmer reservoir by doing work. Work is done on the gas by compressing it, letting heat flow out of it and then letting it expand against an external pressure until it reaches a temperature just below the temperature of the cold reservoir (W > 0 means ΔU < 0 so T decreases).

But the reverse engine cycle can be slow in creating a temperature difference. Modern refrigeration requires creating that cold temperature quickly. That is where the Joule-Thomson effect comes in. A gas with a high Joule-Thomson coefficient is compressed and then allowed to expand freely and adiabatically through a throttle (narrow aperture) valve. This results a significant temperature drop but no (very little) change in internal energy. That cold refrigerant is then placed in a heat exchanger to cool the space to be refrigerated. The cold fluid absorbs heat rapidly due to the significant temperature difference. The Joule-Thomson effect is at the heart of modern refrigerators and cooling systems.

AM

 

[russ_watters]

That is where the Joule-Thomson effect comes in. A gas with a high Joule-Thomson coefficient is compressed and then allowed to expand freely and adiabatically through a throttle (narrow aperture) valve. This results a significant temperature drop but no (very little) change in internal energy. That cold refrigerant is then placed in a heat exchanger to cool the space to be refrigerated. The cold fluid absorbs heat rapidly due to the significant temperature difference. The Joule-Thomson effect is at the heart of modern refrigerators and cooling systems.

AM

You need to stop saying this because it just isn't true. The refrigerant is not a gas before going through the expansion valve, it is a liquid. The fact that it is a liquid at that point is the critical issue. And the temperature of the liquid doesn't drop due to the Joule-Thompson effect, it drops because its boiling point drops and heat is carried away by its boiling. The new temperature after the expansion valve is determined by the pressure and boiling point at that pressure, not Joule-Thompson cooling.

Think of it this way:
If you have a volume of liquid, there is no gas in it to undergo Joule-Thompson cooling. If you suddenly expand the container, you would end up with a vacuum above the liquid, but to avoid that, a certain amount of the liquid flashes to vapor, which converts sensible heat to latent heat and lowers the temperature of the entire mixture. Any joule-thompson cooling on the liquid is negligible because liquids are not very compressible.

Similarly, a can of "compressed air" is actually a liquid that is a lot like a refrigerant. There is some gas above it that blows-out when you use it. This gas cools a little, but the liquid left inside cools a lot due to the fact that it starts to boil. The liquid inside the compressed air can can't be getting colder due to the Joule-thompson effect because it starts and ends at the same pressure and never goes through the expansion valve.

If you do a sample problem, you'll see that you don't use the Joule-Thompson effect in the calculation of the new temperature after the expansion valve.

The joule-Thompson effect is relevant only at the compressor, where the refrigerant is a gas on both sides of the process.

 

[sophiecentaur]

You need to stop saying this because it just isn't true. The refrigerant is not a gas before going through the expansion valve, it is a liquid. The fact that it is a liquid at that point is the critical issue. And the temperature of the liquid doesn't drop due to the Joule-Thompson effect, it drops because its boiling point drops and heat is carried away by its boiling. The new temperature after the expansion valve is determined by the pressure and boiling point at that pressure, not Joule-Thompson cooling.

Think of it this way:
If you have a volume of liquid, there is no gas in it to undergo Joule-Thompson cooling. If you suddenly expand the container, you would end up with a vacuum above the liquid, but to avoid that, a certain amount of the liquid flashes to vapor, which converts sensible heat to latent heat and lowers the temperature of the entire mixture. Any joule-thompson cooling on the liquid is negligible because liquids are not very compressible.

Similarly, a can of "compressed air" is actually a liquid that is a lot like a refrigerant. There is some gas above it that blows-out when you use it. This gas cools a little, but the liquid left inside cools a lot due to the fact that it starts to boil. The liquid inside the compressed air can can't be getting colder due to the Joule-thompson effect because it starts and ends at the same pressure and never goes through the expansion valve.

If you do a sample problem, you'll see that you don't use the Joule-Thompson effect in the calculation of the new temperature after the expansion valve.

The joule-Thompson effect is relevant only at the compressor, where the refrigerant is a gas on both sides of the process.

Air cannot be compressed to a liquid at room temperature, afaik. Even cans of Compressed CO2 are full of gas - and the triple point of CO2 is only about -60C. Are you sure you meant that?

 

[sophiecentaur]

I think we have a dialogue of the deaf going on here. There are two distinct forms of refrigeration, it seems and the proponents of each of the two on this thread are doomed not to agree. An ordinary 'refrigerator' uses change of state in the 'evaporator' to take heat out of the fridge (or hot room). Cryogenic refrigeration doesn't start with a liquid; it can't. The liquid is only there when the JT cooling has reduced the temperature far enough to produce a liquid. The Wiki article seems to be over-emphasising JT.
The secret behind domestic refrigerators was to find a suitable fluid with an appropriate boiling point. CFCs seem to fill the bill. Unfortunately they are a bit nasty for the environment. The original Ammonia based fluid now seems to be restricted to the 'Absorption' type cycle, which is very inefficient (loose term).

 

[russ_watters]

Air cannot be compressed to a liquid at room temperature, afaik. Even cans of Compressed CO2 are full of gas - and the triple point of CO2 is only about -60C. Are you sure you meant that?

"A can of 'compressed air'" is a colloquialism people use to describe dusting products. I have a can of Endust Multi-Purpose Duster on my desk, which is compressed liquid tetrafluoroethane:

http://www.gjfood.com/pdf/msds/116_821480.pdf

There was one small error in my previous post: after you use some and it cools, it is at a lower pressure.

 

[russ_watters]

I think we have a dialogue of the deaf going on here. There are two distinct forms of refrigeration, it seems and the proponents of each of the two on this thread are doomed not to agree. An ordinary 'refrigerator' uses change of state in the 'evaporator' to take heat out of the fridge (or hot room). Cryogenic refrigeration doesn't start with a liquid; it can't. The liquid is only there when the JT cooling has reduced the temperature far enough to produce a liquid. The Wiki article seems to be over-emphasising JT.
The secret behind domestic refrigerators was to find a suitable fluid with an appropriate boiling point. CFCs seem to fill the bill. Unfortunately they are a bit nasty for the environment. The original Ammonia based fluid now seems to be restricted to the 'Absorption' type cycle, which is very inefficient (loose term).

That's all fine except that the title and OP say that the thread is about ordinary refrigerators. So it is my perception that Andrew mistakenly believes what he is saying applies here. I agree that the wiki overemphasizes it by doing the same thing. They say joule-thompson is at the heart of modern refrigeration, then go on to describe only vapor-vapor processes.

 

[russ_watters]

More...
Looking at references to thermodynamic texts, I find similar issues. Mostly it is that throttling processes are taught in one section and they are gas-gas only processes (following Joule-Thompson). Then in describing the refrigeration cycle, they label the process as a "throttling" process, which can leave people to assume it is exactly the same, when it is not. Then once you get to the cycle analysis in a thermo course, you aren't calculating the states anymore, you are just reading them from a table, so it may not occur to people that the Joule-Thompson equation no longer applies.

Wiki does also describe what is actually happening in a separate article:

Flash (or partial) evaporation is the partial vapor that occurs when a saturated liquid stream undergoes a reduction in pressure by passing through a throttling valve or other throttling device. This process is one of the simplest unit operations. If the throttling valve or device is located at the entry into a pressure vessel so that the flash evaporation occurs within the vessel, then the vessel is often referred to as a flash drum.[1][2]

If the saturated liquid is a single-component liquid (for example, liquid propane or liquid ammonia), a part of the liquid immediately "flashes" into vapor. Both the vapor and the residual liquid are cooled to the saturation temperature of the liquid at the reduced pressure. This is often referred to as "auto-refrigeration" and is the basis of most conventional vapor compression refrigeration systems.

http://en.wikipedia.org/wiki/Flash_evaporation

If I get a chance later, I'll show in an example that temperature/energy change in the liquid is approximately equal to the energy lost to evaporation (the latent heat of vaporization).

I'm not sure it is even possible to start into the math of Joule-Thompson because of the state change:
-If you apply the equation just to the liquid, you will get a ridiculously low temperature change due to the relative incompressibility of the liquid.

-Applying it to the gas is tougher because at the start of the process there is no gas. The gas that is there at the end all boiled-off at different temperatures and pressures during the process. But if you assume an arbitrary volume of gas at the starting and ending pressures, you'll find more temperature drop, but not enough even to just cool the gas to the new temperature, much less cool the liquid as well.

 

[rogerk8]

Thanks to you guys, I think I got it now!

But at first I laughed at Andrew's kind and effortful explanation because I didn't understand a single thing of what he said.

I then went back and actually read your nice link Russ, and then I read your nice link Andrew too.

I thus got to fully understand how a compressor works, then I think I got to understand part of the JT-effect.

Reading about the Joule-Thomson effect made me understand that the gas need to be at a temperature below its inversion temperature (whatever that is) for it to work according to the JT-effect which sounds simple enough in that way that throttling actually lowers the temperature of the gas.

But this topic seems to be far from simple. Your arguing is proof of that. And reading about it in Wikipedia makes me understand that there is lots to learn. One thing for instance is the definition of potential energy when it comes to gases.

The JT article f.i states (http://en.wikipedia.org/wiki/Joule–Thomson_effect#Physical_mechanism):

"expansion causes an increase in the potential energy of the gas".

To me, the further the molecules are apart would mean a lower potential energy.

But what do I know?

Roger
PS
I am attaching my current understanding of how a refrigerator may work.

 

[rogerk8]

An epiphany was that PV actually gives Joule as unit!

This along with the fact that Pressure actually is Energy Density is something totally new to me!

Amazing!

Roger

 

[SteamKing]

It's 'epiphany', not 'appifany'. The latter sounds like an app for your phone.

 

[Andrew Mason]

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

 

[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!