2
$\begingroup$

Lets say I have an aluminum canister with an interior volume of $V$ cubic feet (under normal circumstances and pressure, that's how much air it can hold). If I pressurize the canister to $x$ psi, the air in the canister would weigh $w$ pounds.

How can I calculate $w$ by plugging in different values to $v$ and $x$?

(i.e. $v=15 ft^3$, $x=7000 psi$, find $w$)

Is there a formula?

$\endgroup$
8
  • $\begingroup$ Here is an ideal gas formula calculator. $\endgroup$
    – peterh
    Jul 16 '16 at 22:08
  • $\begingroup$ Have you ever heard from that density * volume = mass, and that mass * g = weight? $\endgroup$
    – peterh
    Jul 16 '16 at 22:19
  • $\begingroup$ Anyways, a quick tutorial on the SI system would be also useful for you. feet -> meter, psi -> Pa, pound -> kg, so do it. $\endgroup$
    – peterh
    Jul 16 '16 at 22:21
  • $\begingroup$ @peterh Oh great thank you! I have been using the imperial system all of my life (because I live in the US), so I'm new to the SI system. I had a quick look at the calculator and thought that the units were wrong... I see my mistake now. If you could answer this in the "answer" section, I can mark it as the best answer and the question will be answered. Thank you again and sorry about my confusion. $\endgroup$
    – SuperMaku
    Jul 16 '16 at 22:27
  • $\begingroup$ I feel I can't survive if I don't give you a beautiful answer. See soon. $\endgroup$
    – peterh
    Jul 16 '16 at 23:02
1
$\begingroup$

(NASA lost a Mars Probe)

What you can do:

There is a well-known $pV=\frac{m}{M}RT$ universal gas formula. Here:

  • $p$ is the pressure
  • $V$ is the volume
  • $m$ is the mass
  • $M$ is the molar mass of the gas
  • $R$ is the universal gas constant ($8.314$)
  • $T$ is the absolute temperature (in Kelvin)

Although your question literally interpreted asks for the weight of the gas, from the context it is clearly visible that you are asking for its mass. These are different things:

  • weight means the force, with what a body would pull its suspension, or pushes the things below it. It is measured in Newtons ($N$).
  • mass is... well, the quantity of the matter. It is in $kg$s or in pounds. 1 pound is around 0.45 $kg$.

Now we know the $V$, it is clear. Convert the cubic feets to cubic meters. We also know the pressure, but don't forget to recalculate the psis to $Pa$s (Pascals).

The molar mass is a little bit more problematic. Every gas has a molar mass ($M$), it essentially means the mass of its $6*10^{23}$ molecules in grams. ($1kg = 1000 g$). It depends only on the gas, and it is rougly the same if you calculate all of the protons and the neutrons in its atoms.

For example, in methane, it is 16 (1 carbon atom, with atomic mass 12, and 4 hydrogene atoms, with atomic mass 1). I have a strong impression that you are experimenting with methane :-). It means, that the mass of $1 mol$ of methane is $16g$.

Now you want to know $m$. So, a little bit of reordering happens on our equation:

$m=\frac{pVM}{RT}$

You know $p$, you know $V$, I now said $R$. I suspect $T$ is room temperature, which is rougly $\mathrm{300K}$.

The only what we don't know is $M$, but if you say which type of gas is it, I will say it to you on the spot.

From that point, you need only unit conversions and elemental unit multiplications-divisions.


Extension: (it is not needed for a rough calculation) This formula is not always enough accurate, particularly on low temperature or on big pressure. There is a more accurate Van der Waals gas formula:

$(p+\frac{n^2a}{V})(V-bn)=\frac{m}{M}RT$

which is essentially an extension of the universal gas formula. Here there are 2 new constants, they belong to the actual gas type (just as the molar mass), and you can find it by googling for the Van der Waals constants.

Calculating with these would be much harder, you need to solve a 3rd degree equation numerically for that. You probably can't do that. But it works.

On my experience, until around $\mathrm{200 bar}$ pressure and on room temperature, the difference from the universal gas formula is around 10-15%.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.