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If you

  1. take a container of great strength,
  2. fill it to the brim with liquid nitrogen,
  3. seal the container, and
  4. heat it to room temperature.

What will the pressure be inside?

Bonus: The same with liquid Helium.

(I posted this Reference for phase diagrams of elements , but the answer here can not be read from the phase diagram.)

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My guess is that the only way to answer this question, even remotely correct, is to do an experiment. – Pygmalion Jun 4 '12 at 21:03
@Pygmalion: A very dangerous experiment, if you do not have a rough estimate :o) – Hans-Peter E. Kristiansen Jun 4 '12 at 21:05
Hear hear!...... – Pygmalion Jun 4 '12 at 21:11

Consider that the liquid was saturated, for a density of ~808 kg/m^3. Conservation of mass requires the density of the vapor to be the same. Using NIST's property tables you can find the pressure to be about 43,000 psi or 2,900 atm. Never do this...

Both 700 atm and 82 atm are very low...

NIST Data Table:

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I think it's a good idea to make a link to NIST's property table. – Ali Jul 9 '13 at 17:03
Good idea Ali :) – ccook Jul 9 '13 at 17:52
From where did you get the liquid density? – Hans-Peter E. Kristiansen Aug 1 '13 at 13:28
I find it a little weird that they use psi on a site. – Hans-Peter E. Kristiansen Aug 1 '13 at 13:30
You can change the units beforehand, the link contains all the settings I put into generating the plot/tables, so the PSI is on me. Converting people to kPa from PSI seems to be a lot harder than F/C to K. – ccook Aug 1 '13 at 19:15

From the densities of liquid nitrogen and nitrogen gas at standard pressure the volume ratio is about 1:700.

For an ideal gas in a closed 1L container this would result in a pressure of 700 atm according to $$P V = n R T$$ From the phase diagram nitrogen is a gas at standard pressure and becomes supercritical at approximately 100 atm. The ideal gas law can therefore only be a guideline well below this pressure. To incorporate the intramolecular forces the van der Waals equation is the next best choice: $$\left(p + \frac{n^2 a}{V^2}\right)\left(V-nb\right) = nRT$$ With the coefficient $b$ for nitrogen ($3.85 \cdot 10^{-5}$ m$^3$/mole from here) the minimum volume for 1L of liquid nitrogen (31 mole) is 1.2L. So even the correction to the ideal gas law cannot capture the full range.

Only the experiment might tell us what the real pressure might be. Someone has even done it inadvertently (accident report):

enter image description here

This vessel exploded at a pressure of 1200 psi or 82 atm, so the pressure of 1L of liquid nitrogen heated up to room temperature should be between 700 and 82 atm.

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82 atm is not a very high lower limit.- but I will accept, that the answer is not simple. – Hans-Peter E. Kristiansen Jun 5 '12 at 9:56
This answer is, imho, wrong and potentially dangerous. – ccook Jul 9 '13 at 22:15
@ccook: Well, then point out what can be improved. – Alexander Jul 9 '13 at 22:20
Use the equation of state to find the pressure of the vapor at the same density of the liquid. Mass and volume are constant, fixing density, and the temperatures are the same, ambient, the state is fixed. I agree the equation of state is not likely to be accurate in this very super critical state, but the NIST data is accurate :) For that, see my short answer. – ccook Jul 9 '13 at 22:44
@ccook: the table is nice. The difference between ideal gas law and nitrogen is remarkable. I have to ask the high pressure guys if they ever tried this. – Alexander Jul 9 '13 at 23:08

As mentioned above density of liquid nitrogen at atmospheric pressure is c. 0.8g/cm3 or 800Kg/M3.

Thus 1 litre of Liquid N2 is about 800g

Warm up to room temperature and I know I should have 22.4 litres of gas at STP (Standard temperature and pressure) for each mole. Nitrgoen molecules (N2) have an atomic mass of 28, so I have 28.6 moles of gas at STP.

28.6 moles of gas at STP would like to occupy 28.6 x 22.4 litres = 640.6 litres.

I'm going to squash that in a one litre container, so the pressure comes from PV/T = constant (As long as we stay well above Tc).

T here is the same so P must be 640.6 bar-g. (about 9,300 psi)

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@Alexander summarises most of your answer, in his first sentence. For the ideal gass law to hold, we need to stay below T_c. – Hans-Peter E. Kristiansen May 30 at 14:03

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