# At what temperature does air become liquid at a pressure of X bar?

... with X = 6 in my particular case.

I was trying to get a phase diagram for air, only got one for nitrogen and was slightly puzzled. Here's why:

I was wondering about a particular detail in industrial air liquification. The original Linde process used 200 bar pressure and cycled the air though several cycles to condense it, as one decompression only gives around 40 K reduction in temperature.

This process was replaced rather early by a modified process (called the Linde-Fränkle process) that compresses the air to only 6 bar and then runs it through a pre-cooled regenerator, where it liquifies in the first pass, proceeding to the distillation column immediately. That process was in use until the 1990s, so it probably had decent performance.

Now, I was wondering to what temperature the regenerator has been precooled. From the phase diagram of nitrogen, I'm estimating to somewhere not abive 150 K, most probably lower.

Since the pre cooling is done with the gaseous $N_2$ and $O_2$ boiling off the distillation column, it sounds like quite close to the liquification temperature at atmospheric pressure.

This puzzles me because it makes me wonder why such little compression is sufficient to make the whole process economic. Also, I can't quite see at what point the Joule-Thompson effect comes into play here. The air is liquefied entirely without decompression, so where dies the cooling come from?

I was also wondering if I might be confusing liquid and supercritical here, since I'm told pressurized gases in typical containers at room temperature are really supercritical already, even though laymen like me still call them liquid.

• Maybe it would suffice to point me to the right way to calculate the vapour pressure of a mixture like air? – Hanno Fietz Jul 10 '14 at 20:16