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In Chemistry class yesterday, I learned that real gases with low atomic masses behave like an ideal gas at high temperatures and low pressure. Since on earth at sea level the pressure is close to $1\ \mathrm{atm}$ (which is a low pressure) and temperature is close to $288.15\ \mathrm K$ (high temperature), I can use the equation $PV=nRT$ to approximate properties of the gas. (I got these values from Wikipedia, under Standard Sea Level).

$$\left(\text{pressure}\right)\cdot \left(\text{volume}\right)=\left(\text{number of moles}\right)\left(R\right)\left(\text{temperature}\right)$$

In physics we learned that pressure is the force per unit area. On earth the atmospheric pressure is earths gravitational pull on the air. The gravitational pull of any object on earth is $G_\text{pull}=\left(\text{mass}\right)\cdot \left(9.8\ \frac{\mathrm m}{\mathrm s^2}\right)$. Because the gravitational pull is $G_\text {pull}=\left(\text{mass}\right)\left(\text{acceleration due to gravity}\right)$.

Since different planets have different masses the acceleration due to gravity on these planets are different. Below I listed the values:

$G_\text{Mercury}=3.59\:\frac{\mathrm m}{\mathrm s^2}$

$G_\text{Venus}=8.87\:\frac{\mathrm m}{\mathrm s^2}$

$G_\text{Mars}=3.37\:\frac{\mathrm m}{\mathrm s^2}$

For planets like Mercury, Venus, Mars the approximation of $PV=nRT$ still can be used since they would have even lower pressures than Earth. But what about planets like Jupiter and Neptune

$G_\text{Jupiter}=25.95\:\frac{\mathrm m}{\mathrm s^2}$

$G_\text{Neptune}=14.07\:\frac{\mathrm m}{\mathrm s^2}$

Jupiter and Neptune have more pressure at ground level than that of Earth. So can $PV=nRT$ still be used? If yes, how well? If not, is there a different formula that can be used to find the volume and number of moles on these other planets?

Any input is appreciated. I am just being curious and didn't find anything on Google.

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    $\begingroup$ You can use it, it just becomes less accurate. There are many other equations of state that try to account for deviation from ideality. en.wikipedia.org/wiki/Equation_of_state By the way, there is no difference whether you are on Mars or Jupiter or Venus. A gas at high pressure on Earth is the same as a gas at high pressure in Jupiter. $\endgroup$
    – orthocresol
    Oct 17, 2015 at 20:17
  • $\begingroup$ @orthocresol Thanks, I didn't know about the other equations of state. Is their a correction factor to make it more accurate that can be added to the equation? Or is $PV=nRT$ still the best approximation formula I can use to calculate properties of gasses on other planets? $\endgroup$
    – James Smith
    Oct 17, 2015 at 20:26
  • $\begingroup$ @orthocresol Oh and Boyles and Charles Law seem to be a simpler version of $PV=nRT$ If $nRT=k$, you get $PV=k$ which is Boyles Law. And if $nR/P$ is a constant you get $nR/P=k$, $V/T=k$. Which is Charles Law $\endgroup$
    – James Smith
    Oct 17, 2015 at 20:29
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    $\begingroup$ If you're interested about the parameterisation (i.e. correction factors), you might want to read up particularly about the van der Waals equation of state hyperphysics.phy-astr.gsu.edu/hbase/kinetic/waal.html And you are right about Boyle's and Charles's laws. They are both encompassed by the ideal gas law. $\endgroup$
    – orthocresol
    Oct 17, 2015 at 20:45
  • $\begingroup$ Venus might actually have a little higher surface pressure than Earth. $\endgroup$
    – Oscar Lanzi
    May 29 at 14:31

3 Answers 3

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The differences in acceleration due to gravity is not the main factor in comparing how accurate the approximation is for each planet. The main factor is the mass of gas each planet's atmosphere contains.

Mercury has almost no atmosphere. According to the NSSDCA — Mercury Fact Sheet (accessed 2023-05-22) the total mass of all gas in Mercury’s atmosphere is less than $\pu{E4 kg}$ and the pressure is less than $\pu{E-14 bar}.$ The approximation will be very accurate.

According to the NSSDCA — Venus Fact Sheet (accessed 2023-05-22) the mass of Venus’s atmosphere is $\pu{4.8E20 kg}$ and the pressure is $\pu{92 bar},$ so $pV = nRT$ will be much worse of an approximation than on Earth.

Mars has a small atmosphere, less than Earth and more than Mercury. Jupiter, Saturn, Uranus and Neptune are gas giants with no distinct surface.

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    $\begingroup$ This only addresses the question for "open air" scenarios, right? If we are looking at closed systems then the answer is that the planet's atmosphere has no relevance, and gravity is only relevant to the degree that it can create a measurable pressure gradient within the system in question. $\endgroup$
    – feetwet
    Oct 18, 2015 at 15:38
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    $\begingroup$ @feetwet Right, it only applies to the conditions at the surfaces of the planets, "outdoors". $\endgroup$
    – DavePhD
    Oct 18, 2015 at 16:00
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    $\begingroup$ This answer is so wrong. PV=nRT on earth, but gas density at high altitudes isn't the same as low altitudes because of gravity. The simple equation ignores differences in altitude. If you are using a sphere with a volume of one cubic meter sphere that is fine. But if you container is one $1 mm^2$ then it would be a million meters high and gravity would have an effect if the tube was vertical. If the tube was horizontal then gravity could be ignored. $\endgroup$
    – MaxW
    Nov 3, 2015 at 5:58
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    $\begingroup$ Another interesting factoid along this line. Helium boils off the earth because at high altitude collision frequency is reduced. So a helium atom gets knocked with enough energy to reach escape velocity and if it doesn't hit a nitrogen or oxygen molecule which are trapped by the earth's gravity, then it escapes into outer space. $\endgroup$
    – MaxW
    Nov 3, 2015 at 6:00
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    $\begingroup$ Obviously when there is a great range of altitudes there is also a change in temperature due to height. Imagine the atmosphere with zillions of layers. How much sunlight reaches each layer varies. Since the density of the gas varies due to gravity. Thus temperature of layer depends on how much sunlight gets to that layer and how much gas is in the layer to absorb that sunlight. $\endgroup$
    – MaxW
    Nov 3, 2015 at 6:48
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Just adding some reasoning:

The equation is better applied when the intermolecular forces can be neglected, this is when (i) the concentration of gas is small, (ii) pressure low and (iii) chemical potential who considers gas proclivity to react with same/other species (inert $\ce{N2}$ vs. oxidant $\ce{O2}$). For intermolecular interactions there are virial corrections and so on (Compressibility Factor, Van der Waals forces).

Planets with no atmosphere and relatively small in size (i.e. rocky planets like Mercury, Mars) don't have high concentration of gases over their surface, besides pressure and attraction exerted by gravity are low, which make the gases behave as ideal. Bigger gas planets (Jupiter, Neptune) are often more turbulent and the gas behavior depends more on the distance of these to center of the planet, and corrections should be talking into account.

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Apropos:

Comparing this equation with the similar equation of state for the Earth's atmosphere, you will notice that the gas constants are different, 1149 for Mars and 1718 for Earth. These numbers are different because the Martian atmosphere is almost entirely composed of carbon dioxide, while the Earth's atmosphere is a mixture of 78% nitrogen and 21% oxygen.

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