0
$\begingroup$

I know that $\mathrm PV=nRT$ and this equation does not take mass into consideration. However, $\mathrm P =Force/Area$. And weight is a force which is equal to $\mathrm mg$.

Will gas pressure be dependent on the mass of gas used? For eg. If you have 1 mole of Hydrogen gas (1 gm) and 1 mole of Radon gas (222 gm) then will the pressure for both be different even though the number of particles will be the same i.e. Avagadro's number.

Also is there any relation between density and pressure for gases?

Edit 1:

Can you confirm the following?

enter image description here

$\endgroup$
  • $\begingroup$ If you go through the derivation from kinetic theory en.wikipedia.org/wiki/…, it actually takes into account of mass. Temperature is the average kinetic energy, which depends on mass $\endgroup$ – K_inverse Sep 11 '18 at 10:42
  • $\begingroup$ How can temperature, being an intensive property, depend on mass? Can you elaborate on this, please? $\endgroup$ – Shah M Hasan Sep 11 '18 at 12:47
  • $\begingroup$ I can confirm unofficially confirm that your handwritten page is correct, though you must be careful when you claim proportionality, always stating what you're keeping constant – in this case temperature. I say "unofficially confirm" because checking work isn't really what we do on this site. $\endgroup$ – Philip Wood Sep 11 '18 at 17:05
  • $\begingroup$ I've added to my answer (below). $\endgroup$ – Philip Wood Sep 11 '18 at 18:07
  • $\begingroup$ BTW, hydrogen is diatomic, so 1 mole of hydrogen has a mass of (approximately) 2 grams. $\endgroup$ – PM 2Ring Mar 12 at 1:20
0
$\begingroup$

"I know that PV=nRT and this equation does not take mass into consideration. However P is also =Force/Area. And weight is a force which is equal to mg."

In a very thin horizontal layer of gas, the pull of gravity on the layer itself hardly influences its pressure. But if there is a 'column' of gas above the layer, the weight of the column will indeed increase – or even be wholly responsible for – the pressure in the layer.

This is the main reason for atmospheric pressure decreasing at increasing heights – there is less gas, experiencing the pull of gravity, above, and therefore less weight to be supported!

It would seem, though, from the rest of your question, that gravitational effects aren't your real concern, so, assuming that the gas is confined in a container that isn't excessively tall, let's turn to

"Will gas pressure be dependent on the mass of gas used?"

In general, yes it will, but not if you're comparing the pressure due to the same number of moles at the same temperature – even if the molecular masses of the gases are different. The lesson is this:

When a variable (in this case pressure) depends on more than one other variable, you must always consider exactly which variables are being kept constant.

We can put this principle into play for answering your last question: "Also is there any relation between density and pressure for gases?"

Since $n=\frac{\rho V}{M_{\text{mol}}}\ $ we have$$pV=\frac{\rho V}{M_{\text{mol}}}RT\ \ \ \ \ \ \text{that is}\ \ \ \ \ \ \ p=\frac{\rho}{M_{\text{mol}}}RT$$ So for a gas of given molecular mass at a given temperature the pressure is proportional to the density. Will that do?

$\endgroup$
  • $\begingroup$ Phillip, your derivation disguises the true relationship. $\rho = nM_{mol}/V$, and a substitution back into the $pV$ equation indicates that pressure is directly proportional to the number of moles and the temperature and inversely proportional to the volume. In other words, density is directly proportional to molar mass, so as molar mass increases, density increases by the same proportion, resulting in $\rho/M_{mol} = n/V$. $\endgroup$ – David White Sep 11 '18 at 18:55
  • $\begingroup$ @David White "your derivation disguises the true relationship. $\rho=\frac{nMmol}{V}\ $". I think not; I gave exactly this equation in the transposed form $n=\frac{ρV}{M_{mol}}\ $ ! I then substituted the right hand side for $n$ in $pV=nRT\ $, as the questioner was clearly familiar with this equation. $\endgroup$ – Philip Wood Sep 11 '18 at 21:16
  • $\begingroup$ Phillip, I can assure you from industrial experience that pressure inside a container of ideal gas is dependent ONLY on volume of the container, number of moles of gas in that container, and the temperature of the gas in that container. For a fixed volume, fixed number of moles, and fixed temperature, a different ideal gas with a different molar mass will possess a different density, but the pressure in that given container will be unchanged under those fixed conditions when you put a different ideal gas of different density in that container. $\endgroup$ – David White Sep 11 '18 at 23:37
  • $\begingroup$ Phillip, also note that the question that the OP posed indicated using the same number of moles of hydrogen vs. radon, while his calculation used 10g of each gas. The calculation used differing number of moles, which is inconsistent with the question that was posed. $\endgroup$ – David White Sep 11 '18 at 23:42
  • $\begingroup$ I have no idea exactly what you disagree with in my answer. Please state the equation(s) and/or sentence(s) that you think are wrong. As for "the question that the OP posed", you will see that there are at least two questions, and I tried to address them both (as well as what I thought might be an implied question in the OP's first paragraph) . $\endgroup$ – Philip Wood Sep 12 '18 at 8:55

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.