Why can’t you feel the collision of each air molecule?

Question

The molecules in a gas move very fast because of its thermal motion. But why can’t we feel the hit of each of the fast gas molecules?

Answer 1

A single atomic impact you can't feel. The energy is too small for that. But the combined impact of all air particles is felt as air pressure (like what you feel in an airplane when the pressure drops). Air pressure has not enough force to compress your body significantly. Even deep in the sea, your body resists.
Because the energy of one air molecule is too small, you cannot feel many of them impacting either. It's like a zillion too small marbles for causing damage are thrown at you. If one doesn't damage they all don't. This would only be the case if the marbles were thrown in a very specific way and if there is one thing that's not specific it's the motion of air particles. One example of a very specific motion is the wind though. Even for small velocities, you can feel the wind. Or can a house be blown down (contrary to air pressure which can't blow a house down).

Answer 2

Because air molecules are extremely small, such that their kinetic energy even at very high velocities is really, really tiny.

How tiny? Let's do some math and find out.

According to the kinetic theory of gases, the average kinetic energy of a gas molecule is $\frac12 f k_B$ times the temperature of the gas in Kelvins, where $k_B ≈ 1.380649 × 10^{-23}\,{\rm \frac JK}$ is the Boltzmann constant and $f$ is the effective number of thermal degrees of freedom the molecule has.

Most air molecules are diatomic, and thus have $5$ effective degrees of freedom (three translational and two rotational; the vibrational modes that would add two more degrees of freedom being negligible at room temperature). At $300\,{\rm K} = 26.85\,{\rm °C} = 80.33\,{\rm °F}$, i.e. approximately room temperature, a typical air molecule thus has a kinetic energy of $$\frac 52 k_B × 300\,{\rm K} ≈ 1.035 × 10^{-20}\,{\rm J}.$$

---

OK, that $-20$ in the exponent sure seems small, but how small? Let's compare it to something that you might be just barely able to feel, like a feather falling onto your skin.

According to the Internet, the mass of a typical chicken feather is about $m = 8.2\,{\rm mg}$. (Honestly that seems like a very small feather to me, but then, most of the feathers on a bird presumably are quite small. The big feathers in the wings and the tail are the exception, not the rule.)

I'm going to assume that a small feather floating very slowly down in still air might be moving at about $v = 1\,{\rm \frac{cm}s}$ (that about 0.4 inches per second for Muricans). Thus, its kinetic energy equals $$E = \frac12 mv^2 = \frac12 × 8.2\,{\rm mg} × \left( 1\,{\rm \frac{cm}s} \right)^2 = 4.1 × 10^{-10}\,{\rm J}.$$

Thus, a feather floating down at $1\,{\rm \frac{cm}s}$ has about $4.1 × 10^{10} = 41{,}000{,}000{,}000$ (that's $41$ billion) times as much kinetic energy as a typical air molecule at room temperature.

Answer 3

Because an individual gas molecule doesn't carry much energy/momentum, even if it could be moving fast when hitting us.

Answer 4

The force of a single molecule is very weak because, while such a molecule is typically moving quite fast, is has very little mass.

The most probable velocity of a gas molecule is $$v = \sqrt{\frac{2k_\mathrm{B}T}{m}},$$ where $k_\mathrm{B}$ is the Boltzmann constant (which just relates the energy units we use to the temperature units we use), $T$ is the temperature, and $m$ is the mass of the molecule. Air is mostly nitrogen (N₂) gas, so each molecule has a mass of 28.01 Da = 4.652 × 10⁻²⁶ kg. At room temperature (298 K), the most probable speed is 420.6 m/s (940.9 miles per hour for the Americans).

When such a molecule collides with the lipids on the surface of your skin, it will impart a momentum of about $2mv$ (assuming an elastic collision, with the molecule rebounding in the opposite direction with the same speed). Looking at a molecular dynamics simulation I ran, it seems like the collision takes place over about 500 femtoseconds. So the force should be something like $$F \approx \frac{2mv}{\tau},$$ where $\tau = 500~\mathrm{fs}$.

So, I get: $$F_\mathrm{gas~molecule}\approx 88~\mathrm{pN} = 88\times10^{-12}~\mathrm{N}.$$

This is on the order of the weight of an average human cell. I have a hard time feeling the weight of a mosquito on my skin ($5\times10^{-5}~\mathrm{N}$), so I don't think I can feel something a million times smaller. Can you feel it when a single cell detaches from your skin?

To feel something, the membrane of a cell has to be distorted enough to open a mechanosensitive channel that allows the flow of ions which eventually makes a nerve signal to that goes to your brain (very roughly). All these things are much too slow to pick out a 500 fs event (the time resolution of your sensory nervous system is too slow) and the sensors are likely not sensitive enough to pick out this small force anyway.

A better question is, why would you want to?

It would be difficult for your body to have a sensor that could sense the collision of a single N₂ molecule since all the water and other molecules in your body are jiggling around with a similar velocity. However, you could imagine a sensor cooled to cryogenic temperatures that could detect the collision of a single molecule. Although it might be possible for your body to sense the force of a single molecule, there would be no reason for your body to evolve this capability because the information is useless. If your cheek has an area of $A = $ 4 cm by 4 cm, then you'd constantly experience $6\times10^{24}$ collisions per second ($\frac{N_\mathrm{collisions}}{\Delta t} = \frac{P A}{2 m v}$). What would you do with all that information? On the other hand, our senses do sometimes give us useless information. For example, the end of my nose is always in my field of vision.

Answer 5

Individual molecules have very small mass, and they are not moving relativistically also, to give them any significant kinetic energy.(lorentz factor is almost '1')

Answer 6

If you were a smoke particle, and smoke particles were allowed to post on StackExchange, you might reasonably have asked the question "When I'm floating in the air minding my own business, what are those invisible particles that keep crashing into me and sending me all over the place?"

This is the basis behind "Brownian Motion". If you look at smoke under a microscope you will see that the individual particles move in a haphazard manner, due their collisions with molecules in air. We don't feel the collisions because the effects are so small.

There's a video showing Brownian Motion at https://www.youtube.com/watch?v=gPMVaAnij88