Wednesday, November 21, 2012

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.

4 comments:

  1. Hey Daniel! Interesting post; I'd not thought to calculate the internal energy of air before. One thing I thought I'd point out: The classical ideal gas model is actually not valid for room-temperature air, as the vibrational component of the partition function is fundamentally quantum mechanical up until a temperature of a few thousand degrees Kelvin. It's actually pretty crazy to think that the air you and I breathe is behaving quantum mechanically. I might do the full calculation at some point just to see how the answer differs from the classical approximation. Although the diatomic nature of N2 and O2 limits the molecule to one vibrational mode and should make the calculation tractable, I suspect it'll take a while... Cheers!

    ReplyDelete
    Replies
    1. Huh, interesting. If you ever get around to it, let me know how it turns out!

      Delete
  2. Hey, N = 6 for nitrogen, not 5. It has six degrees of freedom (check Wiki). I'm trying a different approach.

    ReplyDelete
    Replies
    1. Indeed, diatomic gases do have 6 six degrees of freedom. In practice, at low temperatures like room temperature, the vibrational DOF can be mostly neglected. I'll make that a little clearer in the text.

      Delete

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