Imagine that you have a 1 cubic meter metal box with two 20 mm diameter holes. One of them is connected to an air-compressor that constantly pressurizes the inside of cube. Let's say 5 atm. The other hole is opened to air (1 atm). So my questions start at this point:

1) How can you calculate pressure inside of box at a given time? After adequate time (after there are no more changes in pressure over time) what is the last result pressure in the box ?

2) After adequate time (when is pressure stabilized), if I stop the compressor, when will the pressure inside the box be 1 atm?

Excuse me, I gave these numbers just to make it more understandable because I'm not a native English speaker. If there is lack of any data please add data by yourself.


1 Answer 1


1) Your question states that, until the compressor is switched off, a constant pressure of $5\; atm$ is maintained in the box. This answers the 1st part of your question.

(However, perhaps you mean that the compressor delivers a certain amount of gas per second while the 2nd hole is open, and you wish to know the pressure of the gas in the box when equilibrium is reached.)

2) As the Source below shows, when the compressor is switched off the rate of effusion of gas depends on : area $A$ of hole, volume $V$ of box, temperature $T$ of gas (assumed constant), the molecular mass $m$ of the gas, and the difference in pressure $P$ on either side of the hole. According to kinetic theory, the pressure inside the box will fall exponentially with time to the external pressure of $1\; atm$ :
$$P(t) = P(0)e^{-Ct}$$
where $C = \large{(\frac AV)(\frac{RT}{2πm})}$ and $R$ is the molar gas constant.

Note the restrictions for this value of $C$ to be a reasonably good estimate, as pointed out by pentane in the comments.


  • $\begingroup$ This formula is for gas effusing "through a pinhole with a diameter that is smaller than the mean free path of the gas molecules." At 1 atm that is a hole of $10^{-5}\ mm$. I guarantee you that same formula does not apply for a hole of $20\ mm$. $\endgroup$
    – pentane
    May 5, 2016 at 12:30
  • $\begingroup$ @pentane : Thank you for pointing that out. I agree, the value of C for the conditions given in the question will differ significantly from that calculated in my answer. Do you have any suggestions for a correction? Besides, the OP "gave these numbers just to make it more understandable." $\endgroup$ May 5, 2016 at 13:35

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