# Can adiabatic process be isothermal?

1. Mar 11, 2010

### pinsky

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?

2. Mar 11, 2010

### Andy Resnick

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. Mar 11, 2010

### pinsky

How can you tell it's isothermal?

4. Mar 11, 2010

### Q_Goest

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. Mar 11, 2010

### Andy Resnick

by using a thermometer?

6. Mar 12, 2010

### pinsky

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. Mar 12, 2010

### Andy Resnick

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. Mar 12, 2010

### boit

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. Mar 12, 2010

### SpectraCat

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. Mar 12, 2010

### Andrew Mason

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. Mar 12, 2010

### Q_Goest

Actually, the OP didn't mention a closed system. Gas flowing through a pipe can be adiabatic and isothermal.

12. Mar 12, 2010

### matwiz

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, lets 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. Mar 12, 2010

### Andrew Mason

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. Mar 13, 2010

### pinsky

This explained the case for me. I'we sticked to the $$p \cdot v = const$$ and couldn't see why than relation cant 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.

15. Mar 13, 2010

### Q_Goest

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.

16. Mar 13, 2010

### Andrew Mason

So where is the expansion? If the diameter of the pipe increases, the speed decreases. The density remains the same.

AM

17. Mar 13, 2010

### Q_Goest

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. Mar 14, 2010

### Andrew Mason

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. Mar 14, 2010

### Q_Goest

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. Mar 14, 2010

### SpectraCat

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. Mar 14, 2010

### Andrew Mason

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. Mar 14, 2010

### Andy Resnick

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. Mar 14, 2010

### Q_Goest

Hi Andy, good to talk to you again. There's a good discussion on the Joule-Thomson Coefficient here:
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. Mar 15, 2010

### Andy Resnick

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. Mar 15, 2010

### Q_Goest

Sorry, but I'm probably not the best person to answer your question. As far as I know, the inversion temperature is determined emperically.