The Internal Energy of Air

You know how sometimes, as you're going about your daily life, a completely random thought leads to you suddenly being intensely curious about something and unable to rest until your curiosity has been sated? This weekend I was thinking about nothing in particular while setting up for the morning at work, when I got the burning desire to know how much internal energy a cubic meter of air contained.

The nice part of being a physicist is that I can satisfy these urges, and the nice part of having a blog is that I can share it with other people! So without further ado, let's attempt to calculate the internal energy of 1 cubic meter of air at standard atmospheric pressure and room temperature (around 80 °F, or specifically for ease of calcuation, 300 kelvin).

This is actually fairly simple in theory. There exists a simple equation in thermodynamics for the internal energy of an ideal gas: \begin{equation}U=\frac{N}{2}nRT\tag{1}\end{equation} In this equation, U is the internal energy locked up in each of the N degrees of freedom of the gas, n is the number of moles of gas, the constant R is the ideal gas constant with value \(8.3144621\ \text{J}/(\text{mol}\cdot\text{K})\), and T is the absolute temperature in kelvins.

Now, at this point I should probably elaborate on what internal energy is and what it has to do with degrees of freedom. As I'm sure you know, the temperature of a system is merely a measure of the average energy of its constituent particles. This energy is called the internal energy of the system and can be stored in several different ways, each of which is known as a degree of freedom: in the motions of particles (atoms or molecules), in the rotation or vibration of molecules, or in the excitation and relaxation of electrons in the atoms (not all of these actually apply to all systems, as we shall see).

In thermodynamics there is a theorem known as the equipartition theorem that states that the available internal energy of a system is equally divided among all of its degrees of freedom. If we look at an ideal monatomic gas (such as any of the noble gases), it has only three degrees of freedom, corresponding to the three ways the particles making up the gas can move in three dimensions. Technically, the energy stored by electrons being excited in the atoms could count as another degree of freedom, but at the relatively low temperatures we're considering for this problem there is very little excitation going on and we are free to ignore this effect.

This would be fine if we were considering a monatomic gas, but we are interested in air, which is primarily composed of two diatomic gases: nitrogen (78%) and oxygen (21%). (The monatomic gas argon makes up about 90% of the remaining ~1% of the atmosphere, so we will simply assume that it is all argon for simplicity.) Diatomic molecules bring a new factor into the equation, as they can rotate in two dimensions around their long axis, and since energy can be stored in their rotational motion, this gives them another two degrees of freedom. Diatomic molecules can also store energy in the bond between them, and while this could count as another degree of freedom, in practice it takes temperatures much higher than we are considering here for this to be a significant effect.

So, in summary, monatomic gases have 3 degrees of freedom, representing the kinetic energy associated with their motion through three-dimensional space; diatomic gases – at the temperatures we are interested in – have 5 degrees of freedom since they have the ability to rotate as well. (If I was looking at much higher temperatures I'd have to take into account that vibrational mode I neglect here.)

The upshot of that lengthy diversion is that equation (\(1\)) above looks like \(\frac{3}{2}nRT\) for monatomic gases and \(\frac{5}{2}nRT\) for diatomic ones. Since R is a constant and we have a temperature in mind already, all that remains is to find n, the amount of each type of gas.

The n in that equation refers to moles of gas. The mole (abbreviated mol) is a unit used in chemistry and physics to represent a quantity of substance in terms of the number of particles (atoms or molecules) that make it up. A closely related concept is that of Avogadro's Number, \(6.022\times10^{23}\) (named after the Italian scientist Amadeo Avogadro). One mole of a substance is simply the amount of that substance that contains Avogadro's number of particles in it (it's slightly more complicated than that, but this will suffice for our purposes). Avogadro's number may seem arbitrary, but it is actually measured and defined such that if you have an amount of a substance in grams equal to its mean atomic mass in daltons, then you have one mole of that substance. (The name dalton is given to the unit of mass formerly known as the atomic mass unit, a handy measure for measuring the weight of atoms. It is roughly equivalent to the mass of a nucleon.)

For example: a hydrogen atom has a mean atomic mass of \(1.01\) daltons. Hydrogen typically combines with itself to form dihydrogen gas, H\(_2\). Thus dihydrogen gas has a mean molecular mass of \(2.02\) daltons. If you have \(2.02\) grams of dihydrogen gas, you then have one mole (\(6.022\times10^{23}\)) of dihydrogen gas molecules. Oxygen (mean atomic mass \(16.00\) daltons) likewise combines to form dioxygen (O\(_2\)) with a mean atomic mass of \(32.00\) daltons. If you have \(32.00\) grams of dioxygen gas, you then have one mole of dioxygen molecules. Combining the two to make water, H\(_2\)O, gives water a mean atomic mass of \(18.02\) (\(2\times1.01+16.00\)), so if you have \(18.02\) grams of water, you have one mole of water molecules.

Anyway, this lengthy preface should hopefully enable you to follow what should be a fairly straight-forward calculation, which we are finally ready to begin.

First off, we need to find the number of moles of oxygen, nitrogen, and argon in one cubic meter of our theoretical approximation of air. We can do that by first finding the density of air at our specified conditions (300 K, ~80 °F and 1 standard atmosphere of pressure, 101.325 kPa), then multiplying by the fractions established before to find out how much mass of each gas exists, before converting that mass into moles of each gas to fit the equation.

There is a equation for the density of dry air (which we are assuming it is) given by \[\rho=\frac{P}{R_{\text{specific}}T}\] In this equation, \(\rho\) (the Greek letter rho) stands for density (in kg/m\(^3\)), P stands for pressure, R\(_{\text{specific}}\) is a version of the ideal gas constant specifically for dry air equal to 287.058 J/(kg\(\cdot\)K), and T is again the temperature in kelvins.

Putting in the numbers and doing the math, we get:
\begin{align}\rho&=\frac{101,325 \frac{\text{N}}{\text{m}^2} }{287.058 \frac{\text{N}\cdot\text{m}}{\text{kg}\cdot\text{K}} \cdot300.00\ \text{K}}\\
&=1.1766\frac{\text{kg}}{\text{m}^3}
\end{align}
Since we are assuming a single cubic meter of air, our mass of air consists of 1.1766 kg (about 2.6 pounds of air). Multiplying by the fractions we assumed for each of the ingredients, we get:
\begin{align}
m_{\text{N}_2}&=0.78\cdot1.1766\ \text{kg}=0.9177\ \text{kg}=917.7\ \text{g}\\
m_{\text{O}_2}&=0.21\cdot1.1766\ \text{kg}=0.2471\ \text{kg}=247.1\ \text{g}\\
m_{\text{Ar}}&=0.01\cdot1.1766\ \text{kg}= 0.0118\ \text{kg}=11.8\ \text{g}
\end{align}
Now that we have the masses involved, we can convert to moles using their mean atomic masses:
\begin{align}
n_{\text{N}_2}&=917.7\ \text{g}/28.013\frac{\text{g}}{\text{mol}}=32.762\ \text{moles}\\
n_{\text{O}_2}&=247.1\ \text{g}/31.9988\frac{\text{g}}{\text{mol}}=7.7222\ \text{moles}\\
n_{\text{Ar}}&= 11.8 \ \text{g}/39.948\frac{\text{g}}{\text{mol}}=0.29538 \ \text{moles}
\end{align}
Having now obtained the number of moles of each gas in our hypothetical approximation to air, we can now use equation (\(1\)) to calculate the amount of internal energy each gas contributes to the whole.
\begin{align}
U _{\text{N}_2}&=\frac{5}{2}\cdot32.762\ \text{mol}\cdot8.314\frac{\text{J}}{\text{K}\cdot\text{mol}}\cdot300.00\ \text{K}=204.3\ \text{kJ}\\
U_{\text{O}_2}&= \frac{5}{2}\cdot7.7222\ \text{mol}\cdot8.314\frac{\text{J}}{\text{K}\cdot\text{mol}}\cdot300.00\ \text{K}=48.12\ \text{kJ} \\
U_{\text{Ar}}&=\frac{3}{2}\cdot0.29538\ \text{mol}\cdot8.314\frac{\text{J}}{\text{K}\cdot\text{mol}}\cdot300.00\ \text{K}=1.105\ \text{kJ}
\end{align}
This gives us a total of
\[ U _{\text{N}_2}+ U_{\text{O}_2}+ U_{\text{Ar}}=253.5\ \text{kJ}\]
That...actually turns out to be a bit more than I was expecting. That's a quarter of a million joules of energy contained in the motion and rotation of the air molecule in a single cubic meter of air.

To put this number in perspective, let's do some conversions to units you may be more familiar with. That many kilojoules is almost exactly 60 kilocalories (or Calories), the unit the energy in food is measured in. Put another way, the normal energy needs of an adult human are typically pegged at around 2,000 Calories per day. If you could somehow extract the energy from air, you'd need only about 33 cubic meters of air per day to survive, a volume smaller than the amount of air in most average-sized homes. Alternatively, the average amount of solar power over a 1 square meter area at the Earth's surface is about 1 kilojoule per second (1 kilowatt), so the amount of energy we calculated is equivalent to the amount hitting an area of 253,500 square meters (a quarter of a square kilometer) every second during full daylight. There's a lot of energy locked up in the air around you.

In a sense, though, I suppose I really shouldn't be too surprised. Gas molecules in the air whiz about at great speed, and this speed comes from the kinetic energy they have. In fact, we can estimate the root-mean-square speed of a typical nitrogen molecule fairly easily (the M is the molar-mass of the gas, in kg/mol):
\begin{align}v_{\text{rms}}&=\sqrt{\frac{3RT}{M}}\\
&=\sqrt{\frac{3\cdot 8.314\frac{\text{J}}{\text{K}\cdot\text{mol}}\cdot300.00\ \text{K}}{0.028013\frac{\text{kg}}{\text{mol}}}}\\
&=516.8\frac{\text{m}}{\text{s}}
\end{align}In case you're wondering, that a whopping 1,156 miles per hour. Those nitrogen molecules are, on average, moving about that fast (oxygen and argon move a bit slower, since they're more massive). So I guess when you consider billions upon billions of tiny atoms all zooming around at speeds comparable to this, it makes sense that there's a lot of energy tied up in their motion as kinetic energy. Wow. Amazing stuff.

Experiments in sleep-deprived post-writing.

As I write this, it is 4 o`clock Saturday morning (Hawaiian Standard Time), and I have been observing for 10 and a half hours, and awake for a little over 19. Unlike Wednesday night (well, Thursday morning), when wind speeds in excess of 50 mph at the summit forced us to close up early, the weather tonight has been more amenable, with moderate seeing (hovering around 1 arcsecond, which is quite good for most observatories on Earth, but poor for Mauna Kea with its often 0.5 arcsecond seeing).

(Seeing is simply a measure of how well you can resolve two objects at the telescope. If you have seeing of one arcsecond, you can resolve two objects one arcsecond apart, assuming an optimally built and focused telescope [an arcsecond is 1/3600 of a degree of arc, a very small angular measurement].)

Observing can be pretty boring work. I expect to be here for nearly two more hours, which makes almost 11 and a half hours of observing in a row. [Edit: it turned out to be a little less than an hour and a half, I forgot twilight came so early in the summer] We're using the SNIFS camera on the UH 2.2-meter telescope, which has a lot of automation built-in that results in large portions of down time for us. The good part of this is that I will soon have more data to work with, because the data I've been analyzing has all come from SNIFS. With the scripts I've written, it is a simple matter to run the data through and catch it up to the rest of the data.

Writing posts while sleep-deprived to this extant is turning out to be harder than I anticipated, so I'm going to stop now and focus on staying awake. We have one more night scheduled for the current run, on Sunday, which I plan to be at. Due to a misunderstanding, the WFGS2 camera that we might have used did not have its dewar of liquid nitrogen refilled, so we'll be using SNIFS again (being a spectrometer, it does not have such a critical need to be kept cold). The upside, once again, is more data for me to work with!

Burning questions of a Saturday morning.

So there I was lying in bed this morning, still half-asleep, when I got to pondering the fact that I, like the atoms that compose me, am made up of 99.99% empty space. I then idly contemplated the fact that most of this mass was in the nuclei of my atoms, at which point I was seized by a burning desire to find out how much all the electrons in all the atoms of my body put together weighed.

I know that last I checked, I had a mass of approximately 60 kg (my weight is left as an exercise for the reader), and a little searching on the 'net produced a table with the percent-by-mass composition of an average person. Starting with the first entry in the list, I have around \(61.35\text{%}\cdot 60\ \text{kg} = 36.81\ \text{kg}\) of oxygen in me. That's 36,810 grams of oxygen. A mole of oxygen being 16.00 grams, that's almost exactly 2,300 moles of oxygen in me.

(A mole, in chemistry, is simply a very large number \(-\ 6.022\times10^{23}\), to be exact. The conversion between grams and moles is pretty easy; if you have an amount of a substance equal in grams to the atomic weight of one atom (or molecule) of that substance, you have one mole of that substance, i.e., \(6.022\times10^{23}\) atoms (or molecules) of the substance. Since the atomic weight of oxygen is 16.00 atomic mass units, one mole of oxygen is 16.00 grams of oxygen. One more of oxygen at \(0^\circ\)C and normal atmospheric pressure would occupy about 22.4 liters, or just under 6 gallons.)

Now, each oxygen atom comes with 8 electrons in tow, \(8\) electrons \(\times\) \(2,300\) moles \(\times\) \(6.022\times10^{23}=1.11\times10^{28}\) electrons.

That...is a very large number. (11.1 billion billion billion).

And that's just the first element on the list! (Although, by virtue of its nature, it is likely to provide the largest number of electrons.) Next up is carbon, which, at 22.83%, comprises 13.70 kg of me. Dividing by 12.00 grams per mole, and multiplying by 6 electrons to each carbon atom, we arrive at \(3.09\times10^{27}\) electrons from carbon, a factor of 10 less than from oxygen (not surprisingly).

Since the procedure is relatively boring, and quite simple, I will spare you further talk and simply list the amounts in the table below:

---

oxygen        36.81 kg  \(1.11\times10^{28}\) electrons

carbon        13.70 kg  \(3.09\times10^{27}\) electrons

hydrogen      6.00 kg   \(3.57\times10^{27}\) electrons

nitrogen      1.54 kg   \(4.64\times10^{26}\) electrons

calcium       0.86 kg   \(2.58\times10^{26}\) electrons

phosphorus    670 g     \(1.94\times10^{26}\) electrons

potassium     120 g     \(3.51\times10^{25}\) electrons

sulfur        120 g     \(3.61\times10^{25}\) electrons

sodium        84 g      \(2.42\times10^{25}\) electrons

chlorine      84 g      \(2.43\times10^{25}\) electrons

magnesium     18 g      \(5.35\times10^{24}\) electrons

iron          6.0 g     \(1.68\times10^{24}\) electrons

fluorine      2.4 g     \(6.85\times10^{23}\) electrons

zinc          1.8 g     \(5.00\times10^{23}\) electrons

silicon       0.60 g    \(1.80\times10^{23}\) electrons

rubidium      0.60 g    \(1.56\times10^{23}\) electrons

strontium     0.30 g    \(7.84\times10^{22}\) electrons

bromine       0.24 g    \(6.33\times10^{22}\) electrons

lead          0.12 g    \(2.86\times10^{22}\) electrons

copper        0.06 g    \(1.65\times10^{22}\) electrons

aluminum      60 mg     \(1.74\times10^{22}\) electrons

cadmium       60 mg     \(1.54\times10^{22}\) electrons

cerium        60 mg     \(1.50\times10^{22}\) electrons

barium        18 mg     \(4.42\times10^{21}\) electrons

iodine        18 mg     \(4.53\times10^{21}\) electrons

tin           18 mg     \(4.57\times10^{21}\) electrons

titanium      18 mg     \(4.98\times10^{21}\) electrons

boron         18 mg     \(5.01\times10^{21}\) electrons

nickel        12 mg     \(3.45\times10^{21}\) electrons

selenium      12 mg     \(3.11\times10^{21}\) electrons

chromium      12 mg     \(3.34\times10^{21}\) electrons

manganese     12 mg     \(3.29\times10^{21}\) electrons

arsenic       6.0 mg    \(1.59\times10^{21}\) electrons

lithium       6.0 mg    \(1.56\times10^{21}\) electrons

cesium        5.4 mg    \(1.35\times10^{21}\) electrons

mercury       5.4 mg    \(1.30\times10^{21}\) electrons

germanium     4.2 mg    \(1.11\times10^{21}\) electrons

molybdenum    4.2 mg    \(1.10\times10^{21}\) electrons

cobalt        2.4 mg    \(6.62\times10^{20}\) electrons

antimony      1.8 mg    \(4.54\times10^{20}\) electrons

silver        1.8 mg    \(4.72\times10^{20}\) electrons

niobium       1.2 mg    \(3.19\times10^{20}\) electrons

zirconium     0.60 mg   \(1.58\times10^{20}\) electrons

lanthanum     0.60 mg   \(1.48\times10^{20}\) electrons

gallium       0.60 mg   \(1.61\times10^{20}\) electrons

tellurium     0.60 mg   \(1.47\times10^{20}\) electrons

yttrium       0.54 mg   \(1.42\times10^{20}\) electrons

bismuth       0.42 mg   \(1.00\times10^{20}\) electrons

thallium      0.42 mg   \(1.00\times10^{20}\) electrons

indium        0.36 mg   \(9.25\times10^{19}\) electrons

gold          0.18 mg   \(4.35\times10^{19}\) electrons

scandium      0.18 mg   \(5.06\times10^{19}\) electrons

tantalum      0.18 mg   \(4.37\times10^{19}\) electrons

vanadium      0.12 mg   \(3.26\times10^{19}\) electrons

thorium       0.060 mg  \(1.40\times10^{19}\) electrons

uranium       60 μg     \(1.40\times10^{19}\) electrons

samarium      42 μg     \(1.04\times10^{19}\) electrons

beryllium     30 μg     \(8.02\times10^{18}\) electrons

tungsten      18 μg     \(4.36\times10^{18}\) electrons

---

(if you're wondering about the units, conveniently, 0.06 g = 60 mg and 0.06 mg = 60 μg)

If you add all those electrons up, you come up with a total of \(1.88\times10^{28}\) electrons. That's a lot of electrons -- 18.8 billion billion billion, to be exact. But how much mass do they have? And how much do they weigh?

Finding the mass is easy. The mass of an electron is \(9.11\times10^{-31}\) kg so \(1.88\times10^{28}\) of them together have a mass of 17.1 grams, which has a weight of approximately 1/25 of a pound.

Wow. All the electrons in my body make up a measly 0.0000029% of my mass. Everything else is concentrated in the nuclei of my atoms. Wild, huh?