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I'm teaching myself some basics of the ideal gas law and working out simple density equations for atmospheric gases at various altitudes.

When I applied the gas law to Venus' atmosphere at its surface,

$$ \rho = \frac{PM_{mol}}{RT},$$

with a Venusian atmosphere of 96.5% $CO_{2}$ and 3.5% $N_{2}$, and

$$R = 8.31451 \frac{J}{mol \cdot K}$$ $$M_{mol} = 0.04344 \frac{kg}{mol}$$ $$P = 9.3 \times 10^{6} Pa$$ $$T = 735K$$

I get 66.1072 $\frac{kg}{m^{3}}$, which roughly agrees with Wikipedia's page on Venus' atmosphere (67 $\frac{kg}{m^{3}}$).

But I'm curious (and here's my question): With all the discussion on the Wikipedia pages cited above about how Venus' atmosphere is a supercritical fluid at its surface, does the ideal gas law equation still work in this exercise?

pr1268
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  • Do you know what a supercritical fluid is? – Dale Apr 24 '16 at 03:37
  • Only by its definition--i.e. what I read about in reference material, online, etc. The easiest way I can describe a supercritical fluid is that it has properties similar to both gases and liquids. Other than that, I find it difficult to conceptualize a "supercritical" fluid. – pr1268 Apr 24 '16 at 04:08
  • This is just a peeve of mine, but if pressure is given to 2 SF, you really shouldn't write the calculated density as 66.1072 - just 66 would be more correct (or perhaps 66.1(4) to indicate the approximate error in the final digit.) – Floris Apr 26 '16 at 11:56
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    this is a remarkable, antiintuitive observation! Strange, that it works so well. Keep being so curious and thorough! :) – Ilja Apr 26 '16 at 11:57
  • @Floris: Agreed, but I couldn't find any more accurate/precise figure for the surface pressure of Venus' atmosphere. :-\ – pr1268 Apr 26 '16 at 20:37
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    @pr1268 That only makes Floris's comment all the more important to pay attention to. We do not know the pressure that precisely, so you cannot calculate density that precisely either. And the standard for scientific communication is that reporting so many significant figures implies that you in fact have calculated the result so precisely. That is, reporting so many significant figures is not just cumbersome, it is misleading and almost dishonest. –  Apr 26 '16 at 22:14
  • @ChrisWhite - okay, fine, I'll concede that excessive significant digits (or, more accurately, inconsistent sig. digs.) is misleading and cumbersome, bu I respectfully disagree with the "dishonest" part. I use extra sig. digs. (often more than necessary) to demonstrate that I'm paying due diligence to detail when necessary, and also to ensure I'm doing the math correctly. "Dishonest" implies malicious intent. Certainly none of that here. – pr1268 Apr 27 '16 at 00:02
  • And also, I'm not a research scientist (at least not for pay); I'm only doing this for personal academic growth and "fun." I wasn't even a physics major in college, although I did get a math minor. Still, I do appreciate and respect the feedback. – pr1268 Apr 27 '16 at 00:04
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    I didn't mean to imply you were dishonest, certainly -- just that scientific culture could construe it that way were it to come from someone more familiar with the practices. –  Apr 27 '16 at 00:29

2 Answers2

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The ideal gas law assumes that there are no forces between molecules of a gas, and that the size of the molecules is negligible compared to the volume of the gas.

When a gas becomes a liquid, these assumptions are clearly violated; and when it becomes supercritical, the density is typically still such that the second assumption is almost certainly not valid (although the molecules will have such momentum that the intermolecular forces no longer hold them together).

Therefore I am pretty sure a supercritical fluid does not meet the definition of a (near) ideal gas. If you get the "right" answer, you were lucky. For reference, here is the phase diagram:

enter image description here

image source: wikimedia

Obviously, at the surface of Venus you are way beyond the critical point in temperature... but the density seems kind of high for an ideal gas that was liquid at a little less than half that temperature.

Afterthought: if the supercritical fluid still behaves like a nearly-ideal gas, the partial volume occupied by the molecules must be relatively small. Yet, the density of the liquid is about 1.5 g/cc; with a molar mass of 44 g, that is about $2\cdot 10^{22}$ molecules per cc. That puts the volume per molecule at $5\cdot10^{-29}~\rm{m^3}$, which would correspond to a sphere with a radius of $0.23 \rm{nm}$. According to this answer, the size of a CO2 molecule is about 232 pm (from center of O to center of O). That suggests that they are quite tightly packed at that density, and it leaves me baffled as to why it should still be so compressible. Something to ponder...

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Floris
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  • +1, but for the sake of "symmetry", rather than "The ideal gas law assumes (...) no forces (...), and that the size of the molecules is negligible (...)", IMHO you should state either "no forces, & no size" (ideal definition), or "negligible potential energy, & negligible size" (practical definition). – L. Levrel Apr 24 '16 at 09:59
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After further research, I've concluded that the ideal gas law would work for Venus' supercritical fluid atmosphere, at least reasonably well enough for my curiosity, and certainly as well as it would here on Earth.

My research took me to learning compressibility factor, equations of state, and several other real gas topics.

From what I gather, a gas behaving "ideally" will have a compressibility factor $Z = \frac{P}{nRT}$ of one. While the Wikipedia page mentions that $Z$ will deviate substantially for a gas at extreme pressures, such deviation doesn't become significant until several hundred bars of pressure (more than Venus' surface).

This web page has an online calculator of various physical properties of several gases (sorry, it's an HTTP POST so no direct URL to the calculated data), and it shows the compressibility factor is still near 1 for both gases, even at 93 bar and 735K (1.0048 for $CO_2$ and 1.0391 for $N_2$).

That page also mentions the densities of $CO_2$ and $N_2$ at Venus' surface pressure/temperature are $66.98 \frac{kg}{m^{3}}$ and $41.21 \frac{kg}{m^{3}}$, respectively. Given the fractional amount of each, this yields an overall density of $\rho = 66.08 \frac{kg}{m^{3}}$. Close enough to my original calculation!

Floris
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pr1268
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