Expert: Scott Valentine - 1/30/2007

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The text above is a follow-up to ...

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Hello

How much pressure is developed in a sealed tank of liquid nitrogen at room temperature? I’m sure it is very high. Is the pressure determined by the vapor pressure of the liquid? If a quantity of dry ice is placed in a sealed container, how much pressure will be developed at room temp? How
does one calculate this?  An MSDS on liquid nitrogen gives a vapor pressure at 760 MM of mercury.  

Isn’t this atmospheric pressure? I don’t understand.

Thank you.

ruko

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In a perfectly sealed container, LN2 will approach about 1600psig. Not many containers can withstand this pressure, and most LN2 dewers are rated for about 350psig. They have special relief valves and vents to prevent overpressure.

The pressure is not determined so much from vapor pressure (which is about 100,000 pascals, or 1 atm at the boiling point), but from expansion due to boiling. The gas will take up nearly 700 times the volume of the liquid, and since LN2 boils around -320F, it is very easy to produce a gas.

Dry ice, which is liquid carbon dioxide, not liquid nitrogen, also produces a high pressure when 'boiled'. However, the process is actually called sublimation, as it goes directly from a solid to a gas. It exists as a solid at about -110F. These high pressures can create explosive situations as well.

Vapor pressure is given at equilibrium - where there is an equal exchange of molecules going to gas, and returning to the surface of the liquid. It is therfore used to describe the behaviour of both liquid and gas existing at the same time.

Thank you very much for your response.

Quote “In a perfectly sealed container, LN2 will approach about 1600psig. Not many containers can withstand this pressure”. I agree but my SCUBA tank could easily hold this pressure as it is pressurized to about 2200 psig for diving. However, more internet searching on my part has revealed that the pressure in a sealed container of LN2 can reach far higher than 1600 psig. See these web sites: http://www.powerlabs.org/ln2demo.htm  http://entropy.brneurosci.org/info/nitrogen.html  http://www.powerlabs.org/lnprop.htm

They are talking about pressure as high as 4000 psig in one case and 30,000 psig! in another. Are they that incredibly wrong?  How did you arrive at 1600 psig? I’m still very confused about this as there does not seem to be much agreement out there. Surely, there must be a way to accurately calculate the pressure based on the amount of LN2 and the volume of the container and the temerature.  

Quote: “Dry ice, which is liquid carbon dioxide”. Isn’t dry ice solid CO2?

Thanks again for your answer.

Russ


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Let's have another go...

Ideal gas law: PV = nRT
P = pressure in Pascals
V = volume in cubic meters
n = number of moles of gas
R = gas constant in J/k*mol
T = temperature in kelvin

We start by expanding one liter of LN2 into a 100-L volume. LN2 expands by 680-700 times (use 700 for simplicity) at room temperature (300 k). 700 liters of gas is 31.25 mols (22.4 liters per mol).

Our basic equation now looks like this:

[(31.25 mol)(8.314 J/mol*k)(300k)] / 0.1 m^3

= 779437.5 Pa = 780 kPa = 113 psi = 7.7 atm

Of course, this does not take into account compressibility and variations in T, nor a handful of other factors. But it should be close!

Looking at some typical real-world values, we can plug in 80 liters for the container, with 40 liters of liquid. If all of the liquid goes to gas at room temperature, the pressure in the vessel will approach 5,600 psi.

I hope that clarifies things. Again, sorry about the sloppy initial response! Looking at my notes, I was thinking about the burst pressure for some containers - usually about 4 times the rated pressure.


Scott

Thanks for all your work and the reply. That gas formula is familiar but it’s been 50 years since I used it in college. My, how useful my education was. I never could find a use for gas laws in sales. One more little question about LN2. It is something I did not make clear in my original question. My fault!

I’ll give you the scenario: Lets say I have an 80 liter container and I add to the container 75 liters of LN2 and seal it completely. With only about 5 liters of space left in the container, I don’t think all the LN2 will turn to gas. I’m quite sure the LN2, at some point, will stop boiling because of the pressure build up in the tank. Doesn’t this liquid work like water, increase the pressure on the liquid and the boiling point is raised?

So if the above is correct I will have a sealed tank with a little less than 75 liters of LN2 on the bottom and a little more than 5 liters of gas N2 on the top. Is the gas formula you supplied for calculating pressure appropriate for this scenario?    

Thanks again.

Russ


Answer
Hi Russ,

The short answer is "it depends". Without venting, there will be an equilibrium point, but the phase diagram for LN2 shows the critical point as being at 126 K (-233 degF) and 3.4 MPa, or roughly 33.5 atmospheres (493 psi). Beyond this point, the nitrogen becomes a supercritical fluid, e.g., there is no distinction between liquid and vapor, but only one phase.

This point happens so low in temperature that it is likely to affect your 80 l container, which will quickly rise above 500psi. I don't think I can give you a definitive answer without getting into compressibility and equation-of-state discussions that I'm not as well-versed in.

To be sure, the nitrogen can still exist as a liquid in a standard dewar because of insulation and venting. If the vents were to seal, pressure would build rapidly and I think you would be a in a supercritical state very quickly. This implies that the entire volume would be in this indistinct phase.

The ideal gas law may not be entirely useful in these ranges, as molecular dynamics come into play. However, some sources state the the law is valid for values up to twice the critical temperature and one-tenth below the critical pressure. For a detailed examination, you might want to check out the following title:
The Properties of Gases and Liquids (Reid, Prausnitz, Poling), published by McGraw-Hill
ISBN 0-07-051799-1

For most engineering uses, I suspect the pressure could be adequately predicted for an upper bound, given appropriate safety margins.