I have been doing some pneumatics recently, and it came to my mind that volume of some tank could be calculated by measuring airflow ($\mathrm m^3/\mathrm s$) and the speed of pressure raise ($\mathrm dp/\mathrm dt$). When volume of tank is big enough, inside geometry of tank is hard to measure and tank is not bending much in such a small pressures, I think this should be somehow reasonable way to make rough estimations of volume. Let's think that this is rather slow process and temperature is always same than environment temperature.

The challenge is in equations. I think there's something wrong with my thinking. I apply ideal gas law $pV=nRT$ and take derivatives on both sides $p\frac{dV}{dt} = RT\frac{dn}{dt}$. Then I use relation between mass, mole mass and amount of substance $n=m/M$ and the relation between volume, density and mass $m=\rho V$. Then I get $V=\frac{dV}{dt}/\frac{dp}{dt}\rho RT$.

Is this equation correct to use, or did I make a mistake in my thinking?

I tested this and got quite weird results..

Best regards,



Well given your conditions, it is possible to do, at least theoretically. Since the volume does not change, and we assume a process slow enough so that temperature is constant:

$$\frac{dp}{dt} V = \frac{dn}{dt}RT$$

which leads to what you want, if you can measure with enough precision the rate of air flow $\frac{dn}{dt}$ and track the rate of pressure drop:

$$V = RT \frac{dn}{dt} / \frac{dp}{dt}$$

This means of course you need to know the temperature of the gas, which I assume can be taken as ambient temperature if the tank is not particularly prepared for insulation.

Imp: This relation makes sense for a container from which air is escaping, but the amount of escaping air is small enough compared to the total volume. I am not sure what you mean with "pressure raise", but to be clear, this does not apply to a a container which had no air and is being filled, since here the change in pressure and air contained are not small compared to $\frac{dp}{dt}$ and $\frac{dn}{dt}$.

  • $\begingroup$ By this you propapbly mean that if there is vacuum, ideal gas law does not apply well? This is not the case. Container is in 1 bar pressure. In experiment, I would raise pressure up until 1,5 bars. $\endgroup$ – dr_mushroom Sep 14 '15 at 6:11

When you take derivatives on both sides, isn't it
$V\frac{dp}{dt} + p\frac{dV}{dt} = \frac{dn}{dt}RT$

Setting $\frac{dV}{dt}=0$ and proceeding with the calculation, you get

$V = \frac{\frac{\rho RT}{M}} {\frac{dp}{dt}}$

  • $\begingroup$ Thanks for your aswer Sarat.Kant. Hmm.. I think in this situation dV/dt should be zero. At least volume change of the system is not very big even though chamber walls are not exactly rigid $\endgroup$ – dr_mushroom Sep 14 '15 at 6:15

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