Say you have liquid nitrous oxide in a standard gas cylinder (about 55 bar) at room temperature. Normally the cylinder sits upright and gas is released via valve.

Now let's assume you released liquid instead of gas from the cylinder (either via built-in siphon tube or turning cylinder upside down). What would happen to the released liquid N2O at atmospheric pressure?

Looks like it should start boiling? But boiling can occur just at -88 °C. So will it cool itself down to this temperature on its release? Or will the liquid immediately vaporize upon release?


1 Answer 1


Since the ambient temperature is above the boiling point, the vapour pressure is above ambient pressure and thus, vapour bubbles will start appearing everywhere (say "explosively"). As soon as this happens, the vapour production draws internal energy from the surrounding liquid, cooling it locally. This happens until the remaining liquid has cooled sufficiently that it has just reached the boiling point, and the produced vapour has excaped the bulk liquid. So, just as you have assumed.

  • $\begingroup$ Thanks for answer. So if I drain 1L of fluid to a glass, it will have 20C and start boiling, however cooling itself from 20C to -88C. How long could this take for that 1L? How much about will be left from that 1L then? $\endgroup$
    – Kozuch
    Mar 27, 2021 at 19:46
  • 1
    $\begingroup$ This will happen almost immediately, because if the temperature remained somewhere above the boiling point in the liquid, new bubbles would appear. The left amount of liquid will depend on the state equations of nitrous oxide, which I don't know. The heat capacity of nitrous vapour will depend on temperature because it is not an ideal gas. This complicates the calculations because the temperature difference is so large. $\endgroup$
    – oliver
    Mar 27, 2021 at 20:28

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