Very simple concept for some, but quite honestly I’m striving to grasp:


  • You have a $20 \text{ m}^3$ pressure vessel.
  • You start filling it with gas.
  • You see a rise in the gauge pressure of $1 \text{ bar}$.

I’m told that this gauge pressure rise means you have injected a total of precisely $20 \text{ m}^3$ of gas when the reader shows exactly $1 \text{ bar}$. And that, therefore, knowing the gas density (lets say for example $2 \text{ kg/m}^3$ at normal conditions, this meaning $0^\circ\text{C}$ and absolute pressure of $1,01325 \text{ bar}$) you deduce the total amount of mass injected in the pressure vessel:

$20 \text{ m}^3 \cdot 2 \text{ kg/m}^3 =40 \text{ kg}$

Doubt 1: Where may I read anything proving that direct relationship between pressure rise and tidal volume?

Doubt 2: how may the stated density apply if this happens inside the vessel, where you have relative pressure and the value is given for absolute pressure?

Many thanks in advance!


1 Answer 1


Using Ideal Gas Law:

$\frac{p_1V_1}{n_1RT_1} = \frac{p_2V_2}{n_2RT_2}$

The volume stays the same, so we cancel those out. R is obviously the same, so that goes too. If we assume the temperature stays the same throughout the process,

$\frac{p_1}{n_1} = \frac{p_2}{n_2}$ , or

$p_2 = p_1\frac{n_2}{n_1}$

Since at a constant temperature, all (ideal) gases have the same molar volume, if you add a volume of gas equal to the initial volume (add a number of moles equal to the original number), the pressure will double.

  • $\begingroup$ what to do if the temperature is not constant? $\endgroup$
    – Innovine
    Aug 2, 2019 at 12:00

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