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Let's say there is a pressure chamber with some sort of sample / specimen (e.g. protein crystal) in it. Now I apply a certain amount of gas pressure, e.g. 10 or 20 atm. Let's say I use xenon as a gas.

I'm wondering what will happen to the temperature inside my chamber - from the ideal gas law you get a reciprocal relationship between temperature and pressure, so given all other things stay the same (i.e. the chamber volume), what is the temperature change / final temperature as seen by my sample? I'd start at room temperature and go from there .. Obviously, the gas is compressed to start with (and has a certain temperature as well).

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3 Answers 3

We can use the Charles law to obtain your answer simply..

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Charles's law is about interdependence of volume and temperature .. i want to work at constant volume and vary the pressure. Since according to the ideal gas law varying P or V is equivalent, I could still use the same principle as in Charles's law though. However I would get that going from 1 atm at 293 K (ambient conditions) to 2 atm would mean cooling the interior of the chamber by 146.5 K in a (possibly) short amount of time, thereby shock frosting my sample .. is that correct? And does it not make a difference that the gas used for pressurization is already compressed? –  user9730 Mar 27 '13 at 17:16

Based on your responses to the previous answers, I understand that there are actually two distinct bodies of gas involved in the setup you describe:

  1. The gas which, at the start of your experiment, is already in the sample chamber (gas A)
  2. The gas which, at the start of your experiment, is stored in a separate vessel at a higher pressure (gas B)

These two bodies of gas will of course be mixed at the end of the experiment, but for the purpose of this discussion, it will be helpful to imagine these two bodies of gas separately, and then to find the final result from there.

Imagine, for example, if the sample chamber were cylindrical in shape, with a zero-mass frictionless piston dividing the two bodies of gas. The piston would then start at one end of the chamber, since the chamber's total volume is initially filled entirely with gas A:

|--------------------------------|  
| . . . . . .gas A. . . . . . ][ |  
|--------------------------------|

Then, as gas from body B is introduced on the other side of the piston, it is forced to slide towards the center of the chamber:

|--------------------------------|  
| . . gas A . . ][ . . gas B . . |  
|--------------------------------|

With this analogy, it is easy to understand that gas A will undergo adiabatic compression, heating up in the process. What happens to gas B, however, depends entirely on what's outside of the sample chamber!

If, for example, the external pressure vessel is fully sealed and rigid, then gas B will have to expend a portion of its stored heat energy in order to expand against the piston, thereby undergoing adiabatic expansion and cooling down in the process. In the case of a true ideal gas, this energy loss will exactly match the corresponding energy gain on the other side of the piston. In real life, however, we would find that the energy loss of gas B is greater than the corresponding energy gain of gas A, thanks to the Joule-Thompson Effect. As such, there would be a net temperature drop in the chamber-- although not as great a difference as you might expect.

On the other hand, if the external pressure vessel is somehow maintained at a constant pressure throughout this process, then the work necessary to expand against the piston will no longer come entirely out of the gas B! Instead, this work will be offset by whatever outside force is maintaining the pressure of the external vessel. (Examples of this kind of setup might include a balloon, a pump, a spring-piston assembly, or even just the outside atmosphere!) In that case, then the energy loss in gas B would be reduced, and we would see a net temperature rise inside the sample chamber.

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Lets look at the ideal gas law:

$$ PV=NkT $$

You seem to assume that when you increase the pressure in a pressure chamber, all other variables in this formula except the temperature stays constant. That is not how a pressure chamber typically works. Sure, you could build a pressure chamber that works like that. Simply seal the chamber and heat it. This will cause temperature and pressure to rise proportionally, while keeping the other variables constant.

In the kind of pressure chamber I think you have in mind, the pressure is typically increased by pumping more air (or other gas) into the chamber, i.e. increasing $N$. The effect on the temperature will depend on how the compressor works. Here you can read about the temperature effects of compressing gas depending on the process.

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thanks for your reply! i'm not sure we talk about the same thing - obv you influence the observable P by increasing N which is equivalent as can be seen from the ideal gas law. As for your link on gas compression, I appreciate that gas heats as you confine it - but I reckon that's different from what I want to do (sorry for not making that clearer), i.e. using already compressed gas to put it inside a pressure chamber to alter the pressure there. I think this has to happen with an associated cooling down as by increasing N (or P) and keeping V constant T has to drop. –  user9730 Mar 28 '13 at 11:28

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