Can adiabatic process be isothermal?

In summary, if we have a system which can't exchange heat with the environment, by definition, that system is an adiabatic process. The pressure may lower, the volume may increase, and the temperature may stay the same. However, if no energy enters or leaves the system, the particles must have the same energy, and the same momentum.
  • #1
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As the title says.
If we have a system which can't exchange heat with the environment that is, by definition, an adiabatic process.

Is it possible that in that situation the pressure lowers, the volume increases and the temperature stays the same?
 
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  • #2
Interesting question. I'm going to guess 'yes'. Diffusion of a drop of ink in water, for example. It's isothermal, adiabatic, and irreversible.
 
  • #3
How can you tell it's isothermal? :smile:
 
  • #4
The isenthalpic, adiabatic flow of a gas can be isothermal as long as the gas approximates an ideal gas, which all real gasses do at some point.
 
  • #5
pinsky said:
How can you tell it's isothermal? :smile:

by using a thermometer?
 
  • #6
Thank you for the answers.

Andy Resnick: I thought you made the assumption of the spreading ink as an isothermal process as something obvious by itself. Even if we were to measure it somehow , why do you think there wouldn't be a change in temperature?
 
  • #7
why should there be? I assumed the ink and water were initially at the same temperature; why would the temperature of the two combined be different?
 
  • #8
Sulfiric acid at same temperature as water when mixed willl have a higher temperature. Who is to say ink doesn't have a similar ability albeit lower magnitude?
 
  • #9
boit said:
Sulfiric acid at same temperature as water when mixed willl have a higher temperature. Who is to say ink doesn't have a similar ability albeit lower magnitude?

Sulfuric acid *reacts* chemically with water, which is the source of the heat you mention. Presumably we are talking about simple ink here .. and anyway, that is a side issue. This is a thought experiment, so we are free to stipulate that the ink does not react, and that the enthalpy of mixing for the ink-water system is zero. In that case, Andy Resnick's statement seems correct. However, it does not involve any PV work, which was part of the OP's question.

In the case of pV work being done in a closed system, I cannot see how the system can be both isothermal and adiabatic, since temperature and entropy are thermodynamic conjugates. AFAICS a hypothetical isothermal, adiabatic system from which work could be extracted would be equivalent to a perpetual motion machine, and thus violate the second law.
 
  • #10
SpectaCat is correct. If the system does thermodynamic Work and no heat flows into or out of the system (adiabatic), the temperature of the system (internal energy) cannot remain constant. Adiabatic expansion can only be isothermal if it is a free expansion (no forces opposing expansion).

AM
 
  • #11
Actually, the OP didn't mention a closed system. Gas flowing through a pipe can be adiabatic and isothermal.
 
  • #12
pinsky said:
As the title says.
If we have a system which can't exchange heat with the environment that is, by definition, an adiabatic process.

Is it possible that in that situation the pressure lowers, the volume increases and the temperature stays the same?

No.

If no energy enters or leaves the system, the particles must have the same energy, and the same momentum.

If you increase the volume, the same particles simply strike the boundary at a lower rate, and this causes the pressure to drop, but also causes the temperature measured at that same boundary to be lower too, since the energy transfer to your thermometer will be correspondingly less.

The only way to get your thermometer to register a higher reading, in the larger volume, so that you get back to the previous temperature, is to increase the energy of the particles.

But you forbade this, by saying no heat exchange with the outside world.


You could get tricky, and say, let's prevent heat exchange, but allow other types of energy exchange. Send in an electrical current, put a resistor inside the chamber, heat back up the gas etc..by converting electic energy into heat within the chamber itself. Or use a varying magnetic field penetrating the chamber to heat up a piece of iron that's inside the system, thus only sending in magnetic energy and having that turned into heat inside...and so on..but whatever you do...you still need to get "more heat" to the system somehow to get that temperature to remain the same.
 
  • #13
Q_Goest said:
Actually, the OP didn't mention a closed system. Gas flowing through a pipe can be adiabatic and isothermal.
You will have to give us an example and explain why the first law is not violated. The OP said there is an expansion. If expands against non-zero pressure it the does work. If there is no heat flow into the gas, the temperature of the gas has to decrease.

AM
 
  • #14
Andrew Mason said:
SpectaCat is correct. If the system does thermodynamic Work and no heat flows into or out of the system (adiabatic), the temperature of the system (internal energy) cannot remain constant. Adiabatic expansion can only be isothermal if it is a free expansion (no forces opposing expansion).

AM

This explained the case for me. I'we sticked to the [tex] p \cdot v = const [/tex] and couldn't see why than relation can't remain the same even in an adiabatic expansion.
I ignored that for work to be done, the pressure in the system (i visualized a piston in a closed chamber) must be higher than the pressure outside and that was the part that was blocking me.

Thank you all for your answers
 
  • #15
Hi AM,
Andrew Mason said:
You will have to give us an example and explain why the first law is not violated. The OP said there is an expansion. If expands against non-zero pressure it the does work. If there is no heat flow into the gas, the temperature of the gas has to decrease.

AM
Consider a control volume around a section of well insulated pipe through which a fluid is flowing. The first law reduces to Hin = Hout. For an ideal gas, an isenthalpic expansion is isothermal. The temperature of the gas into the section of pipe equals the temperature of the gas leaving the section of pipe, yet there is no heat transfer. Real gasses generally either warm up or cool down, but there is always a physical state that the gas goes through (I'm sure it has a name but forget now) where the gas neither warms nor cools. Granted, this isn't a closed system, though you could also consider a control mass and just examine the mass of a gas flowing through the pipe. If you rode along with this control mass, you'd find the gas expanding but remaining at the same temperature.
 
  • #16
Q_Goest said:
Hi AM,

Consider a control volume around a section of well insulated pipe through which a fluid is flowing. The first law reduces to Hin = Hout. For an ideal gas, an isenthalpic expansion is isothermal. The temperature of the gas into the section of pipe equals the temperature of the gas leaving the section of pipe, yet there is no heat transfer. Real gasses generally either warm up or cool down, but there is always a physical state that the gas goes through (I'm sure it has a name but forget now) where the gas neither warms nor cools. Granted, this isn't a closed system, though you could also consider a control mass and just examine the mass of a gas flowing through the pipe. If you rode along with this control mass, you'd find the gas expanding but remaining at the same temperature.
So where is the expansion? If the diameter of the pipe increases, the speed decreases. The density remains the same.

AM
 
  • #17
Hi AM. Not sure what you don't understand. It's an isenthalpic expansion:
- Gas flowing through a pipe drops in pressure (permanent, irreversible).
- No change in temperature.
- Density decreases.
- Enthalpy remains constant.
- Internal energy remains constant.
- PV remains constant

(for an ideal gas or any gas under the right conditions)
 
  • #18
Q_Goest said:
Hi AM. Not sure what you don't understand. It's an isenthalpic expansion:
- Gas flowing through a pipe drops in pressure (permanent, irreversible).
- No change in temperature.
- Density decreases.
- Enthalpy remains constant.
- Internal energy remains constant.
- PV remains constant

(for an ideal gas or any gas under the right conditions)
I don't understand how it expands without doing work on its surroundings. If it does work on its surroundings without heat flowing into it, the temperature has to decrease. If you disagree, perhaps you can explain how it does not violate the first law.

AM
 
  • #19
Consider the flow of an ideal gas through a horizontal pipe. We calculate the pressure drops over some length.* For the sake of argument, let's say this is perfectly insulated (doesn't really matter if there's no change in temp) and there is no work done on or by the gas on the environment.

How does the gas change state as it flows (ie: what happens to internal energy)? Apply the first law and consider what happens. One can draw a stationary control volume around the pipe, such that there is a mass flow in and a mass flow out which is the easiest way, or one can draw a control volume around a given mass that travels down the pipe (control mass).

*We can use any number of different methods to determine pressure drop through a pipe. Industry standard is to use the Darcy-Weisbach equation.
 
  • #20
Andrew Mason said:
I don't understand how it expands without doing work on its surroundings. If it does work on its surroundings without heat flowing into it, the temperature has to decrease. If you disagree, perhaps you can explain how it does not violate the first law.

AM

Note that he is talking about an ideal gas only ... the situation he is describing is just a Joule-Thompson expansion, and since an ideal gas does not exhibit a Joule-Thompson effect, he is correct ... there will be no cooling/heating in this scenario. There is no heat flow from the surroundings (insulated tube, adiabatic), and there is no work done (PV=const) in this case. The same effect can be had for a real gas if you carry out the expansion at or near the Joule-Thompson inversion temperature, which is where the J-T coefficient changes sign.
 
  • #21
SpectraCat said:
Note that he is talking about an ideal gas only ... the situation he is describing is just a Joule-Thompson expansion, and since an ideal gas does not exhibit a Joule-Thompson effect, he is correct ... there will be no cooling/heating in this scenario. There is no heat flow from the surroundings (insulated tube, adiabatic), and there is no work done (PV=const) in this case.
PV=constant does NOT mean no work is done. V = constant means no work is done. If P changes and PV does not, then work IS done (ie. V is not constant).

AM
 
  • #22
Q_Goest said:
Consider the flow of an ideal gas through a horizontal pipe. We calculate the pressure drops over some length.* For the sake of argument, let's say this is perfectly insulated (doesn't really matter if there's no change in temp) and there is no work done on or by the gas on the environment.

How does the gas change state as it flows (ie: what happens to internal energy)? Apply the first law and consider what happens. One can draw a stationary control volume around the pipe, such that there is a mass flow in and a mass flow out which is the easiest way, or one can draw a control volume around a given mass that travels down the pipe (control mass).

*We can use any number of different methods to determine pressure drop through a pipe. Industry standard is to use the Darcy-Weisbach equation.

This was good to work through. I don't exactly understand how the inversion point is calculated.

Wiki says: "The temperature of this point, the Joule–Thomson inversion temperature, depends on the pressure of the gas before expansion.", but I don't see that. Any clues?
 
  • #23
Hi Andy, good to talk to you again. There's a good discussion on the Joule-Thomson Coefficient here:
mbeychok said:
When a real gas, as differentiated from an ideal gas, expands at constant enthalpy (i.e., no heat is transferred to or from the gas, and no external work is extracted), the gas will be either cooled or heated by the expansion. That change in gas temperature with the change in pressure is called the Joule-Thomson coefficient and is denoted by µ, defined as:

µ = (dT/dP) at constant enthalpy

The value of u depends on the specific gas, as well as the temperature and pressure of the gas before expansion. For all real gases, µ will equal zero at some point called the "inversion point". If the gas temperature is below its inversion point temperature, µ is positive ... and if the gas temperature is above its inversion point temperature, µ is negative. Also, dP is always negative when a gas expands. Thus:

If the gas temperature is below its inversion temperature:
-- µ is positive and dP is always negative​
-- hence, the gas cools since dT must be negative​

If the gas temperature is above its inversion temperature:
-- µ is negative and dP is always negative​
-- hence, the gas heats since dT must be positive​

"Perry's Chemical Engineers' Handbook" provides tabulations of µ versus temperature and pressure for a number of gases, as do many other reference books. For most gases at atmospheric pressure, the inversion temperature is fairly high (above room temperature), and so most gases at those temperature and pressure conditions are cooled by isenthalpic expansion.

Helium and hydrogen are two gases whose Joule-Thomson inversion temperatures at atmospheric pressure are very low (e.g., about −222 °C for helium). Thus, helium and hydrogen will warm when expanded at constant enthalpy at atmospheric pressure and typical room temperatures.

It should be noted that µ is always equal to zero for ideal gases (i.e., they will neither heat nor cool upon being expanded at constant enthalpy).
Note that the J-T inversion temperature is not just a function of pressure, but also of temperature as mentioned above. In other words, the inversion temperature is a function of the physical state. I don't think the inversion temperature will change a whole lot though - it's relatively constant, don't know why.
 
  • #24
Thanks- I get that part, but I don't quite see how any inversion temperatures are calculated. Do they have to be measured? In other words, is an equation of state required to calculate the inversion temperature? Using the equation of state for an ideal gas I (correctly) get that the J-T coefficient is zero...
 
  • #25
Andy Resnick said:
Thanks- I get that part, but I don't quite see how any inversion temperatures are calculated. Do they have to be measured?
Sorry, but I'm probably not the best person to answer your question. As far as I know, the inversion temperature is determined emperically.
 
  • #26
I enjoyed the derivation because a measurement of [tex]\mu[/tex] was done to set the absolute temperature scale. The trick is navigating all the outdated and conflicting notation- aside from the various uses of 'd', '[tex]\delta[/tex]', '[tex]\Delta[/tex]', '[tex]\partial[/tex]', etc., there's several definitions of [tex]\mu[/tex]:

1) [tex]\mu\Lambda_{V}=\partial p/\partial \theta[/tex]
2) [tex]\mu[/tex] = J/[tex]\theta[/tex]
3) J/[tex]\mu[/tex] = [tex]\frac{JK\delta +\int pdV + p1V1-p2V2}{\int \partial p/\partial \theta dV}[/tex]

Where (not to be dull) J is a mechanical equivalent of heat, [tex]\delta[/tex] is "the observed cooling effect", etc. etc. lots of archaic terminology.

This is not entirely useless academic musings- this pertains to how a temperature scale is defined (and the range of allowable temperatures-specifically, no negative numbers), and so underlie most expressions of the first and second laws of thermo.
 
  • #27
Andrew Mason said:
PV=constant does NOT mean no work is done. V = constant means no work is done. If P changes and PV does not, then work IS done (ie. V is not constant).

AM

Hmmm ... of course you are correct in the general case. I was thinking of a closed system again, in which case

dU=dQ + dW, but dQ=0, since it is adiabatic, and dH=0, since it is isenthaplic, which led me to

dH=dU + d(pV), and since pV=const, d(pV)=0, and thus dU=0, which means that dW=0 also.

This is an open system, and so we need to add a [tex]\mu dN[/tex] term to dU, but it seems that will also be zero, if we just assume constant flow in and out of the section of pipe.

So, I definitely see what you are saying in general, but I don't see how there is any work done in this particular system, with all of the stipulations in place. Then again, my thermo is rather rusty, so it is entirely possible that I have made a silly mistake.
 
  • #28
pinsky said:
As the title says.
If we have a system which can't exchange heat with the environment that is, by definition, an adiabatic process.

Is it possible that in that situation the pressure lowers, the volume increases and the temperature stays the same?

Consider a gas in a volume in which there is a heat source. The heat cannot leak out, so whatever happens to the system, it will always be an adiabatic process. The produced heat is controlled by a thermostat which keeps the temperature the same. If the initial pressure of the system is larger than the outside pressure, the system will expand adiabatically and isothermally until the inside pressure is the same as the outside pressure.
 
  • #29
Andy Resnick said:
I enjoyed the derivation because a measurement of [tex]\mu[/tex] was done to set the absolute temperature scale. The trick is navigating all the outdated and conflicting notation- aside from the various uses of 'd', '[tex]\delta[/tex]', '[tex]\Delta[/tex]', '[tex]\partial[/tex]', etc., there's several definitions of [tex]\mu[/tex]:

1) [tex]\mu\Lambda_{V}=\partial p/\partial \theta[/tex]
2) [tex]\mu[/tex] = J/[tex]\theta[/tex]
3) J/[tex]\mu[/tex] = [tex]\frac{JK\delta +\int pdV + p1V1-p2V2}{\int \partial p/\partial \theta dV}[/tex]

Where (not to be dull) J is a mechanical equivalent of heat, [tex]\delta[/tex] is "the observed cooling effect", etc. etc. lots of archaic terminology.

This is not entirely useless academic musings- this pertains to how a temperature scale is defined (and the range of allowable temperatures-specifically, no negative numbers), and so underlie most expressions of the first and second laws of thermo.
See? I told you! I'm not the one to give you those kinds of answers... lol

I think one thing that stands out to me that seems even more interesting is that the work the gas does is not going anywhere, or more to the point, it's going back into the gas as near as I can tell.

For example; consider a well insulated cylinder fitted with a piston that's placed inside a pipe at a location where the pressure is 100 psig. Assume air is in the cylinder and also in the pipe. The cylinder will be allowed to travel along the length of the pipe to where the pipe discharges to atmosphere at 0 psig. The work done by the air in the cylinder must go into the flow stream. If the cylinder is well insulated (adiabatic) and the expansion is isentropic as we would expect it to be, the air in the cylinder cools significantly. In comparison, air in the pipe won't cool significantly at all. It's pretty close to being an ideal gas, and the change in internal energy for the air traveling down the pipe is essentially zero. There is no change unless there's heat/work added to or removed from the flowstream.

So in the first case, we have a traveling cylinder that does work PdV, and in the second case we have air that does no work on the surroundings but undergoes an identical change in pressure with no change in its internal energy.

Does the air flowing through the pipe do work on itself? If so, it does work exactly equal to the work needed to keep the internal energy change zero. At least, that's what I've always assumed is going on. It just seems strange that the work the air does exactly equals that needed to keep dU=0. The molecules always seem to 'know' how to space themselves out so that internal energy (and enthalpy) doesn't change. I guess I shouldn't be so amused by that - it's a simple conservation of energy.
 
  • #30
I do not think dropping ink (or any other substance) into water is isothermal.

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

Dropping ink into water would be adiabatic (ignoring convective losses out of the top and sides of the container) and irreversible, but not isothermal, since there will be a slight transfer of heat to the water, resulting in a temperature change.
 
  • #31
Count Iblis said:
Consider a gas in a volume in which there is a heat source. The heat cannot leak out, so whatever happens to the system, it will always be an adiabatic process. The produced heat is controlled by a thermostat which keeps the temperature the same. If the initial pressure of the system is larger than the outside pressure, the system will expand adiabatically and isothermally until the inside pressure is the same as the outside pressure.
This is not an adiabatic process. Heat is still flowing into the gas. Adiabatic means that there is no heat flow into or out of the system.

AM
 
  • #32
Andrew Mason said:
This is not an adiabatic process. Heat is still flowing into the gas. Adiabatic means that there is no heat flow into or out of the system.

AM

This depends on where you put the system boundary. I can take my system to be the gas plus the heat source. Then, even though the heat source produces heat, this does not count as heat in the thermodynamic sense.

This is analogous to free expansion process, throttling process etc. There you do not count the heat due to internal friction, because it stays in the system and does not flow across the system bondary.
 
  • #33
Count Iblis said:
This depends on where you put the system boundary. I can take my system to be the gas plus the heat source. Then, even though the heat source produces heat, this does not count as heat in the thermodynamic sense.

This is analogous to free expansion process, throttling process etc. There you do not count the heat due to internal friction, because it stays in the system and does not flow across the system bondary.
There is no such thing as internal friction in an ideal gas. All collisions at the molecular level are elastic.

If you take the system to be the gas and the heat source, there can be no creation of heat from other energy forms within the heat source (eg. chemical energy). If heat flows into the system (by converting chemical energy to heat) it is not adiabatic.

But you do point out an important issue: the conversion of heat (thermal kinetic energy) into mechanical energy (non-random kinetic energy) - in effect the gas doing work on itself. The gas that exits from a throttle in a free expansion has kinetic energy that does not follow a Maxwell-Boltzmann distribution. This results in the temperature of the gas being undefined while it is escaping. It eventually reaches a new equilbrium, however, in the expanded volume, in effect converting that non-thermal kinetic energy back into thermal energy.

AM

AM
 
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  • #34
Q_Goest said:
See? I told you! I'm not the one to give you those kinds of answers... lol

I think one thing that stands out to me that seems even more interesting is that the work the gas does is not going anywhere, or more to the point, it's going back into the gas as near as I can tell.

For example; consider a well insulated cylinder fitted with a piston that's placed inside a pipe at a location where the pressure is 100 psig. Assume air is in the cylinder and also in the pipe. The cylinder will be allowed to travel along the length of the pipe to where the pipe discharges to atmosphere at 0 psig. The work done by the air in the cylinder must go into the flow stream. If the cylinder is well insulated (adiabatic) and the expansion is isentropic as we would expect it to be, the air in the cylinder cools significantly. In comparison, air in the pipe won't cool significantly at all. It's pretty close to being an ideal gas, and the change in internal energy for the air traveling down the pipe is essentially zero. There is no change unless there's heat/work added to or removed from the flowstream.

So in the first case, we have a traveling cylinder that does work PdV, and in the second case we have air that does no work on the surroundings but undergoes an identical change in pressure with no change in its internal energy.

Does the air flowing through the pipe do work on itself? If so, it does work exactly equal to the work needed to keep the internal energy change zero. At least, that's what I've always assumed is going on. It just seems strange that the work the air does exactly equals that needed to keep dU=0. The molecules always seem to 'know' how to space themselves out so that internal energy (and enthalpy) doesn't change. I guess I shouldn't be so amused by that - it's a simple conservation of energy.

This is an excellent puzzle...
 
  • #35
If you take the system to be the gas and the heat source, there can be no creation of heat from other energy forms within the heat source (eg. chemical energy). If heat flows into the system (by converting chemical energy to heat) it is not adiabatic.

The mu dN terms count as heat?
 
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