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Gas molecules go at an insane velocity, and though they are miniscule, yet there is a LOT of them. Of course, because of all these molecules hurtling around, there is air pressure; yet if you envision a lot of bullets flying around, they don't really "apply pressure": they smash stuff. So why aren't things being destroyed by these mini-torpedoes?

I sense the reason they don't wreak havoc is because they are not coordinated, i.e. they are random. Also things may not work out microscopically as they do macroscopically.

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    $\begingroup$ Think about the momenta of the gas particles. $\endgroup$ – Chris2807 Apr 22 '15 at 13:21
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    $\begingroup$ that isn't really relevant. Air pressure is reasonably uniform so there are no nett forces on bodies in general. If you really want to understand your question look at the average momentum for a Boltzmann distributed gas and momentum of a bullet. $\endgroup$ – Chris2807 Apr 22 '15 at 13:34
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    $\begingroup$ Bullets would not hurt as much if you were made of the same stuff as the bullets. You are made of the same stuff as the air. $\endgroup$ – Solomon Slow Apr 22 '15 at 18:23
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    $\begingroup$ We all know that air pressure (from all those tiny collisions) can destroy stuff, but to make it concrete here's a picture of a train car crushed when the air one the inside was taken out: i.ytimg.com/vi/Zz95_VvTxZM/hqdefault.jpg $\endgroup$ – QuadmasterXLII Apr 22 '15 at 19:26
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    $\begingroup$ @jameslarge: Uh... no? Both of those sentences are incorrect, or at the very least misleading. A bullet made of flesh would do quite a bit of damage to flesh, bullets made of metal do quite a bit of damage to targets made of metal, and the composition of a human is quite different from the composition of air. $\endgroup$ – user2357112 Apr 23 '15 at 0:14
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When you say "why aren't things being destroyed", you presumably mean "why aren't the chemical bonds that hold objects together being broken". Now, we can determine the energy it takes to break a bond - that's called the "bond energy". Let's take, for example, a carbon-carbon bond, since it's a common one in our bodies.

The bond energy of a carbon-carbon bond is $348\,\rm kJ/mol$, which works out to $5.8 \cdot 10^{-19}\,\rm J$ per bond. If an impacting gas molecule is to break this bond, it must (in a simplified collision scenario) have at least that much energy to break the bond. If the average molecule has that much energy, we can calculate what the temperature of the gas must be:

$$E_\text{average} = k T$$ $$T = \frac{5.8 \cdot 10^{-19}\,\rm J}{1.38 \cdot 10^{-23}\,\rm m^2 kg\, s^{-2} K^{-1}}$$ $$T = 41,580\rm °C$$

That's pretty hot!

Now, even if the average molecule doesn't have that energy, some of the faster-moving ones might. Let's calculate the percentage that have that energy at room temperature using the Boltzmann distribution for particle energy:

$$f_E(E) = \sqrt{\frac{4 E}{\pi (kT)^3}} \exp\left(\frac{-E}{kT} \right)$$

The fraction of particles with energy greater than or equal to that amount should be given by this integral:

$$p(E \ge E_0) = \int_{E_0}^{\infty} f_E(E) dE$$

In our situation, $E_0 = 5.8 \cdot 10^{-19}\,\rm J$, and this expression yields $p(E \ge E_0) = 1.9 \cdot 10^{-61}$.

So, the fraction of molecules at room temperature with sufficient kinetic energy to break a carbon-carbon bond is $1.9 \cdot 10^{-61}$, an astoundingly small number. To put that in perspective, if you filled a sphere the size of Earth's orbit around the sun with gas at STP, you would need around 16 of those spheres to expect to have even one gas particle with that amount of energy.

So that's why these "torpedoes" don't destroy things generally - they aren't moving fast enough at room temperature to break chemical bonds!

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    $\begingroup$ But isn't this bonding force much stronger than the weak inter-molecular force which holds many substances together? If I cut something soft, I'm not usually cutting strong bonds, but rather weaker inter-molecular-bonds? So this would still count as destroying something, if an air molecule could break off a whole molecule off some bigger structure? $\endgroup$ – Falco Apr 23 '15 at 9:46
  • $\begingroup$ Yes, but even intermolecular forces are strong enough to resist most collisions with air molecules. I mean, I don't know the numbers offhand, but it must work out that way, otherwise objects as we know them wouldn't exist ;-) So perhaps you only need a sphere of air the size of, I don't know, the moon to get one air molecule moving fast enough to break an intermolecular bond, instead of 16 Earth-orbit-sized spheres. And besides, breaking one molecule off something hardly counts as destroying it. $\endgroup$ – David Z Apr 23 '15 at 11:11
  • $\begingroup$ @DavidZ Not to mention that many such cases would probably simply cause the separated molecule to attach nearby - it would still tend to minimize the total energy of the system. $\endgroup$ – Luaan Apr 23 '15 at 11:19
  • $\begingroup$ @Brionius Amazing answer. Now I have some follow up questions: Why do they apply so much pressure (14.7 pascals at sea-level) then? I suppose they don't break stuff, but this is a lot of pressure. Does it come from gravity pulling the air down, or is it from its speed? And could you explain more in-depth about that equation of how much power you need to break a carbon-carbon bond? $\endgroup$ – HyperLuminal Apr 23 '15 at 12:10
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    $\begingroup$ @HyperLuminal even a cupful of air at room temperature contains somewhere around a billion trillion gas molecules; the collisions that make up the pressure are tiny but incredibly frequent. As to your question about whether the pressure originates from gravity or speed, the answer is "both and neither". It's best treated as another question entirely. $\endgroup$ – hobbs Apr 24 '15 at 0:24
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Things actually do get destroyed by what those air molecules pick up and throw around.

Take look at this example rock

[image from here: http://en.wikipedia.org/wiki/File:Arbol_de_Piedra.jpg ]

Just like their bigger sized brothers, it's the load of those mini-torpedos that brings the destruction.

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    $\begingroup$ These are destroyed by microscopic rock flying around, not by the gas molecules themselves. $\endgroup$ – Mooing Duck Apr 22 '15 at 17:56
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    $\begingroup$ @MooingDuck He did say exactly that $\endgroup$ – MauganRa Apr 22 '15 at 21:38
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    $\begingroup$ @MauganRa: But the OP did not ask that. $\endgroup$ – Mooing Duck Apr 22 '15 at 21:38
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    $\begingroup$ @Schilcote I'm not sure it's about being in a closed space per se but the question is certainly asking about effects caused by the random motion of individual molecules, not effects caused by the bulk motion of the gas picking up bits of rock. $\endgroup$ – David Richerby Apr 23 '15 at 8:25
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    $\begingroup$ @Paddling Ghost I will point out one mistake you made. Though it does seem logical that it is the air providing the power to blow the rocks,in reality it is NOT. Even in STILL air gas molecules travels at insane rates. Their collisions is what effectively negates them and evens out their force to provide a constant pressure. The force picking up the rocks on the contrary is WIND. My question was that why are we not torn to shreds by the force of the molecules in STILL AIR. Did I make myself clear. $\endgroup$ – HyperLuminal Apr 23 '15 at 21:03
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Another way of looking at this is that things that would be destroyed by the environment (be it heat, light, etc) have already been destroyed (like ice on a hot summer day). The things that you see around you are the ones where the bond energy was high enough that they survived.

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  • $\begingroup$ Does this mean that everything contacting the air has a high bond energy? $\endgroup$ – HyperLuminal Apr 23 '15 at 20:52
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In fact, they do!!

Watch what happens to an ice cube that is left in the air... trillions of particles of its exterior are torn out of their stable arrangement, and soon they cascade down the sides—a microscopic waterfall!

So in this case you are right, but it is just the very exterior surface of an object that is exposed to the air and thus affected by it.

Remember that substances already at room temperature are composed of tiny particles which move at very high velocities. If this isn't enough to tear apart the substance, the air won't do much.

That said, I suspect that when one cuts through an object, the air molecules do in fact tear holes in the tiny peaks and crags on the newly exposed surface, until they are torn down and smoothed—but this would probably happen within milliseconds of being exposed to the air. I wonder whether doing this in a vacuum or in a more viscous medium like oil would change the effects.

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  • $\begingroup$ Incorrect. Ice cubes melt because of the heat. If you place an ice cube in below freezing temperatures, it will be intact even though gas molecules still go at extremely high speeds. Plus if only the exterior is exposed, we should expect everything to start being worn out and destroyed around us according to your logic. $\endgroup$ – HyperLuminal Apr 24 '15 at 12:14
  • $\begingroup$ "Ice cubes melt because of the heat." Well, they melt because of heat transfer. One mechanism of heat transfer is conduction. And in the case of an ice cube (which is quite dense compared to air), conduction is quite nicely visualised as a cloud of bullets whizzing around and crashing into a wall of vibrating magnetic bricks, knocking them out of place and making them vibrate more violently until they start being able to wander well away from their original position. The birth of a liquid! $\endgroup$ – Artelius Apr 26 '15 at 8:12
  • $\begingroup$ "If you place an ice cube in below-freezing temperatures", then the gas molecules are still travelling very fast, but not as fast as at room temperature. At a given velocity, bullets do less damage the lighter they are, and gas molecules are VERY light. Below freezing point, the damage done is not enough to cause melting. Also note that above freezing, the temperature of the gas affects the rate of melting, so as expected the faster the "bullets" travel, the more "damage" occurs. $\endgroup$ – Artelius Apr 26 '15 at 8:13
  • $\begingroup$ "if only the exterior is exposed, we should expect everything to start being worn out and destroyed around us"--I never suggested that! In fact the point of my post is that the "smashing bullets" analogy works well in certain circumstances, like when the object in question is colder than the air. I intended this to be a stepping-stone to understanding why air doesn't just turn all objects to dust. I address that in my 2nd-last paragraph. $\endgroup$ – Artelius Apr 26 '15 at 8:13
  • $\begingroup$ So, why does water, with such a high attraction to each other, melt, when other cold things do not? $\endgroup$ – HyperLuminal Apr 26 '15 at 23:37
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Brionius has the right answer, but there is more to be said. Water at room temperature in air will slowly evaporate. Water at room temperature in a vacuum will boil, as is shown here. So these mini torpedos can prevent damage to chemical bonds.

Water molecules are polar. The O's are a little negatively charged. The H's are a little positive. The H's and O's are attracted to each other. Water molecules are sticky.

This is how ice forms. The molecules arrange themselves so that H's and O's are near each other and form relatively weak bonds. The molecules are vibrating at insane speeds. But at low temperatures, not enough to break the bonds.

At higher temperatures, the speed of the faster molecules is enough to disrupt bonds. The ice melts. In liquid water, nearby molecules still tend to arrange themselves so that H's and O's are near each other. This holds the water together as a liquid.

Air around the water also helps. Some of the faster molecules have enough energy to fly completely apart. They would, except that they promptly bump into air molecules. This helps hold the liquid together.

Exactly how well water molecules stick together is determined by temperature and pressure. In some cases, water does go directly from solid to gas. If the pressure is high, water remains liquid even at temperatures of hundreds of degrees. This happens on the ocean floor at volcanic hydrothermal vents.

This phase diagram shows behavior at different regions.

phase diagram

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  • $\begingroup$ So basically just like I said earlier right? I said instead of a concentrated impact it is a spread out pressure. $\endgroup$ – HyperLuminal Apr 24 '15 at 18:09
  • $\begingroup$ Yes and no. If you are talking about a spread out area, yes. If you are talking about keeping individual molecules from boiling away, they are kept back by individual impacts. But you are right that bullets is too forceful an analogy for that. $\endgroup$ – mmesser314 Apr 24 '15 at 21:32

protected by Qmechanic Apr 24 '15 at 10:02

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