Roughly 20% of the molecules in the air are oxygen molecules

physicsss

Roughly 20% of the molecules in the air are oxygen molecules. The oxygen molecules are needed for life processes in a person's body. What is the probability that in one breath, one might fill his lungs with a volume of air that has no oxygen in it?

How does one go about doing this...? SpaceTiger

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physicsss said:
Roughly 20% of the molecules in the air are oxygen molecules. The oxygen molecules are needed for life processes in a person's body. What is the probability that in one breath, one might fill his lungs with a volume of air that has no oxygen in it?

How does one go about doing this...? I would first estimate the volume of someone's lungs and then find the expectation value for the number of oxygen molecules in that volume. I would then assume that the molecules follow a Poisson distribution, where the probability of having a certain number of molecules, n, is given by

$$P(n)=\frac{\nu^ne^{-\nu}}{n!}$$

where $$\nu$$ is the expectation value for the number of molecules that you calculated above. Note that the above is an approximation for large $$\nu$$, something that should be very much valid for your situation.

physicsss

Once I have the volume of a person's lung, what do I need in order to calculate the number of 02 molecules?

SpaceTiger

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The number density of all molecules in the air is given by:

$$n=\frac{P}{kT}$$

where P and T are the atmospheric pressure and temperature, respectively. Does that help?

DocToxyn

From a biological standpoint the probability is zero since one never completely empties the lungs of air, even if you exhale as hard as possible. Thus with the next influx there is already some leftover oxygen remaining in the lung and therefore no chance of "filling the lungs with a volume of air that has no oxygen in it."

But that's probably not the answer you were looking for, perhaps you can submit it for extra credit .

Gokul43201

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SpaceTiger said:
The number density of all molecules in the air is given by:

$$n=\frac{P}{kT}$$

where P and T are the atmospheric pressure and temperature, respectively. Does that help?
By number density, n, SpaceTiger is refering to the number of molecules per unit volume.

SpaceTiger, when you want all successes (n successes in n trials) in terms of the identity of the molecule that enters your lung, isn't the probability just :

$$P(n,n) = p^n$$ ?

And if it is, then wouldn't this be easier to calculate than the approximate result ?

SpaceTiger

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Gokul43201 said:
And if it is, then wouldn't this be easier to calculate than the approximate result ?
I assume you mean the binomial distribution:

$$P(0|N)=(1-p)^N$$

where p is the probability of getting an oxygen molecule and N is the total number of molecules. Yeah, that probably is easier. My astronomer brain is used to working with Poisson distributions, so that was my natural inclination. Gokul43201

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SpaceTiger said:
I assume you mean the binomial distribution:

$$P(0|N)=(1-p)^N$$

where p is the probability of getting an oxygen molecule and N is the total number of molecules. Yeah, that probably is easier. My astronomer brain is used to working with Poisson distributions, so that was my natural inclination. Yes, that's what I mean, only I was using p = 0.8 (prob of getting an O2 molecule), but it's the same thing.

In any case, I don't see the point of this problem. It hardly takes a hundred molecules or so to make the probability "tiny". This is just overkill !

Perhaps, the idea is to get you to use your head to calculate a number too small to show up on any calculator; but that would make it a math problem.

SpaceTiger

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Gokul43201 said:
Perhaps, the idea is to get you to use your head to calculate a number too small to show up on any calculator; but that would make it a math problem.
I had a few problems like this in my astro classes, the idea being to give us a feel for orders of magnitude far beyond the realm of everyday experience, as well as to drive home the validity of statistical mechanics in macroscopic systems.

In any case, I seem to remember expressing my answers in terms of exponentials, which is part of why I initially leaned towards the Poisson formalism, but the answer you get from the binomial distribution is just as simple, so it might as well be done the more precise way, as you said.

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