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Hi to all of you enthusiasts,

I have a 1 liter compressed air tank with around 207 bar of air pressure inside. It has been sitting out for a while and I am assuming its temperature has almost reached the surrounding temperature which is 40 degrees Celsius. I am also assuming that the surrounding is at around 1 bar (atmospheric pressure). I am not sure how to go about calculating the air temperature as it leaks out of the tank so I can predict any potential frosting.

I did use the ideal gas law of $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$ where $P_1$, $V_1$ and $T_1$ belong to the tank and $P_2$, $V_2$ and $T_2$ belong to the surrounding. I thought if any volume of air leaks to the surrounding its volume would increase by 207 times (i.e. $207\times V_1 = V_2$), because the tank can contain 207 times more air than the atmosphere (which has roughly 1 bar of pressure). However, the resulting temperature ends up being 40 degrees Celsius again. Am I doing this right? Any help or hint is appreciated.

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Your mistake: the volume is not 207 times greater.

This would be only true for constant temperature; so it's no wonder if you get constant temperature :)

The volume increases more than this. It is governed by the laws for adiabatic expansion. See wikipedia. The relevant equation is $$T^\kappa p^{1-\kappa} = const.$$ Here $\kappa$ is a parameter of the gas. It can be measured, and, in some approximations, computed. It changes with temperature, but it's probably sufficient for you to know that it is near 1.4 for air.

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  • $\begingroup$ I got something around -206 degrees Celsius which is really low, which makes me suspicious. All that aside, is my approach correct you think? I mean the stuff I explained before (except the 207 times part of course). $\endgroup$
    – Jamy codes
    Apr 22, 2016 at 12:04
  • $\begingroup$ yes, if you use the appropriate formula (which would be $pV^\kappa = const$), you can also compute it with your approach with the volume. $\endgroup$
    – Ilja
    Apr 22, 2016 at 12:10
  • $\begingroup$ the result should be okay, too. 207 bar is really a lot :) I get a factor of 6 in the temperature. $\endgroup$
    – Ilja
    Apr 22, 2016 at 12:12
  • $\begingroup$ I see. May I ask how you decided to use adiabatic expansion? Is it you assumption or inherent in the problem? $\endgroup$
    – Jamy codes
    Apr 22, 2016 at 13:19
  • $\begingroup$ well, adiabatic means, that there is no heat exchange. This is fulfilled automatically, if your process is very fast. So it's a good guess in practice. If it's not completely protected from heat exchange (e.g if you consider the air mixing with room air on its way out) then you get something in between adiabatic and isothermic (which is valid for really slow processes). Also, if you want the temperature to drop, you can't take the isothermic case :) $\endgroup$
    – Ilja
    Apr 22, 2016 at 14:41

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