Well, here I am, about to head off to my last day of work. Of course the only reason I'm working all day today instead of just tonight as I was originally scheduled is that my co-worker who was supposed to work today got stranded in the Mainland because of a canceled flight and couldn't make it home in time. Thus I'm filling in.
So, yeah. At the end of this week I'll be leaving my current job for my new one at the James Clerk Maxwell Telescope, which starts in January. Big changes are afoot.
Thursday, December 20, 2012
Tuesday, December 18, 2012
One Ring Nebula to Rule Them All
Today I have something besides another globular cluster picture for your perusal. It's a picture of a nebula fairly famous in astronomical circles that I've seen prob-ably hundreds of times in the telescope (it's a popular target during the summer) but have never actually imaged before.
Perhaps it's appropriate that I have this picture less than a week after The Hobbit came out, as this object, Messier 57, is popularly known as the Ring Nebula. It's a small planetary nebula (small on the sky, not physically) found in the constellation Lyra, the Lyre, best seen during the summer and autumn. When I say small, it's only about 1.5 by 1 arc-minutes in diameter; compare that with Messier 55 from my last post, at 19 arc-minutes across. I've therefore cropped out the central region for easier viewing.
The Ring Nebula is about 2,300 light-years from Earth, and is currently about two and a half light-years across. Measurements of its expansion rate suggest that it has been expanding for about \(1,610\pm240\) years.
The processes forming the Ring Nebula have to do with the life cycles of stars. When stars about the mass of the Sun exhaust the hydrogen in their cores, they go through a complex process of fusing the helium produced by hydrogen fusion into heavier elements, then those into heavier elements, up the periodic table till they get stuck at carbon, having insufficient mass to fuse it to anything higher. During this time, due to other concurrent processes, their atmospheres swell up to become hundreds of times larger than before. As the star runs out of fusible material in its interior it gradually loses its grip on its outer atmosphere which puffs off into space, and which would have been observed starting sometime between A.D. 250 and A.D. 670.
This escaped atmosphere is what we're actually seeing when we look at the Ring Nebula. The core of the progenitor star has contracted down to a small white dwarf of mostly carbon about the size of Earth, but containing about the mass of the Sun. It currently has a temperature of about 125,000 K (~225,000 \(^\circ\)F) and lights up the surrounding atmosphere like a beacon as it blows away. The white dwarf at the center of the Ring Nebula is too faint to be seen in this picture, but is estimated to weigh about 20% more than the Sun currently does.
One of the reasons that I haven't had a picture of this famous (and not un-photo-genic) nebula up before, is because I'd already taken a picture of it...sorta. Sometime during the summer of 2010, I think, I tried imaging it using the narrow-band filters on the imager. Unfortunately, the night I chose had some very thin, high clouds, and I quickly learned that just because a star is bright enough over the entire visible light spectrum to serve as a guide star, does not mean it will be bright enough when you are only looking at the minuscule fraction of its light that comes through a narrow-band filter. Basically, it lost tracking during the exposure, the resulting picture was ruined, and I just never got around to imaging it again, there being plenty of other objects in the summer and autumn sky to keep me busy. This September I finally got around to imaging it and I'm glad I did, for completeness' sake if nothing else.
Perhaps it's appropriate that I have this picture less than a week after The Hobbit came out, as this object, Messier 57, is popularly known as the Ring Nebula. It's a small planetary nebula (small on the sky, not physically) found in the constellation Lyra, the Lyre, best seen during the summer and autumn. When I say small, it's only about 1.5 by 1 arc-minutes in diameter; compare that with Messier 55 from my last post, at 19 arc-minutes across. I've therefore cropped out the central region for easier viewing.
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| Messier 57, the Ring Nebula, in Lyra, at 100% resolution from the camera. |
The processes forming the Ring Nebula have to do with the life cycles of stars. When stars about the mass of the Sun exhaust the hydrogen in their cores, they go through a complex process of fusing the helium produced by hydrogen fusion into heavier elements, then those into heavier elements, up the periodic table till they get stuck at carbon, having insufficient mass to fuse it to anything higher. During this time, due to other concurrent processes, their atmospheres swell up to become hundreds of times larger than before. As the star runs out of fusible material in its interior it gradually loses its grip on its outer atmosphere which puffs off into space, and which would have been observed starting sometime between A.D. 250 and A.D. 670.
This escaped atmosphere is what we're actually seeing when we look at the Ring Nebula. The core of the progenitor star has contracted down to a small white dwarf of mostly carbon about the size of Earth, but containing about the mass of the Sun. It currently has a temperature of about 125,000 K (~225,000 \(^\circ\)F) and lights up the surrounding atmosphere like a beacon as it blows away. The white dwarf at the center of the Ring Nebula is too faint to be seen in this picture, but is estimated to weigh about 20% more than the Sun currently does.
One of the reasons that I haven't had a picture of this famous (and not un-photo-genic) nebula up before, is because I'd already taken a picture of it...sorta. Sometime during the summer of 2010, I think, I tried imaging it using the narrow-band filters on the imager. Unfortunately, the night I chose had some very thin, high clouds, and I quickly learned that just because a star is bright enough over the entire visible light spectrum to serve as a guide star, does not mean it will be bright enough when you are only looking at the minuscule fraction of its light that comes through a narrow-band filter. Basically, it lost tracking during the exposure, the resulting picture was ruined, and I just never got around to imaging it again, there being plenty of other objects in the summer and autumn sky to keep me busy. This September I finally got around to imaging it and I'm glad I did, for completeness' sake if nothing else.
Tuesday, December 11, 2012
Globular Cluster Photo Series (Part 28): M55
Today I have another globular cluster picture for you, and this one just happens to be the next in the Messier catalog: Messier 55, in Sagittarius. This globular cluster is much closer than M54, at a moderately distant 17,600 light-years. It appears almost twice as large on the sky at 19.0 arc-minutes, but is a mere third its actual size at 96 light-years in diameter. It's also a lot less compact than M54 (class XI out of XII), and really looks quite nice.
Not every object in Charles Messier's catalog was discovered by him (and he gave credit where it was due), and M54 is one such object. It was discovered by an astronomer named Nicholas Louis de Lacaille from an observatory in South Africa in 1752. Messier, having heard of this discovery, tried several times to locate the cluster starting in 1764, but was stymied by its low apparent height from his location in Paris (it is located 30 degrees south of the celestial equator, which makes it rather difficult to see from mid-northerly latitudes). In fact, it wasn't until 1778 – 14 years later – that Messier was actually able to find it, after which he included it in his famous catalog of objects.
All in all, M55 is a rather nice looking cluster, if I say so myself.
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| Messier 55 in Sagittarius. |
All in all, M55 is a rather nice looking cluster, if I say so myself.
Saturday, December 8, 2012
Globular Cluster Photo Series (Part 27): M54
It's been a while since I had any astronomical images to show, hasn't it? I haven't been able to use the imager for a while now, due to a combination of poor weather and being busy, but I do have a few images from September lying around that I never got around to reducing. Today I have the first of those, a picture of the globular cluster Messier 54 in Sagittarius.
Messier 54 is an interesting globular in several ways. For starters, it doesn't actually belong to our galaxy – or at least is a relatively recent acquisition. It appears to originate from the Sagittarius Dwarf Elliptical Galaxy (or SagDEG), a small nearby satellite galaxy of the Milky Way currently residing opposite the galactic core from us. SagDEG has four known globular clusters of its own, of which Messier 54 is the largest and main one.
Because it's on the other side of the core, M54 is the most distant cluster I've yet photographed, at a whopping 87,400 light-years away, easily surpassing the next most distance cluster I've shown here (M53, 58,000 light-years). For comparison, the Milky Way Galaxy itself is only about 100,000 light-years across. Despite its great distance, M54 still appears a relatively large 12.0 arc-minutes across on the sky, fully one-third the diameter of the full Moon. At its distance, that translates into the incredible diameter of about 306 light-years, making M54 larger than nearly every other globular cluster in the Milky Way (and certainly all the ones I've shown so far). It is also very luminous, shining with the light of 850,000 Suns, being outshone only by the brilliant cluster Omega Centauri (which is also a lot closer).
M54 is also one of the denser globular cluster, being a class III on the density scale (with class I being the densest and XII the least dense). It's also possible, according to a 2009 paper, that there may be a black hole with a mass 10,000 times that of the Sun at the center of the cluster, which is unusual for a globular cluster. All in all, it's a fascinating cluster.
Messier 54 is an interesting globular in several ways. For starters, it doesn't actually belong to our galaxy – or at least is a relatively recent acquisition. It appears to originate from the Sagittarius Dwarf Elliptical Galaxy (or SagDEG), a small nearby satellite galaxy of the Milky Way currently residing opposite the galactic core from us. SagDEG has four known globular clusters of its own, of which Messier 54 is the largest and main one.
Because it's on the other side of the core, M54 is the most distant cluster I've yet photographed, at a whopping 87,400 light-years away, easily surpassing the next most distance cluster I've shown here (M53, 58,000 light-years). For comparison, the Milky Way Galaxy itself is only about 100,000 light-years across. Despite its great distance, M54 still appears a relatively large 12.0 arc-minutes across on the sky, fully one-third the diameter of the full Moon. At its distance, that translates into the incredible diameter of about 306 light-years, making M54 larger than nearly every other globular cluster in the Milky Way (and certainly all the ones I've shown so far). It is also very luminous, shining with the light of 850,000 Suns, being outshone only by the brilliant cluster Omega Centauri (which is also a lot closer).
M54 is also one of the denser globular cluster, being a class III on the density scale (with class I being the densest and XII the least dense). It's also possible, according to a 2009 paper, that there may be a black hole with a mass 10,000 times that of the Sun at the center of the cluster, which is unusual for a globular cluster. All in all, it's a fascinating cluster.
Saturday, December 1, 2012
A New Room
Falling squarely in the category of "not particularly important, and only mildly life-altering", I've decided to move from my current room on the ground floor in the house where I live up to the empty room on the second floor.
This is exciting for me, as I don't think I've ever had a room above the ground floor (given the number of houses I've lived in across my lifespan, I actually had to pause to give that thought). Being on the second floor for this house means it'll be a bit warmer in general (good in the winter, not so nice in the summer), and also that I shouldn't have to worry about encountering centipedes anymore. The experience of "moving" has also pushed me to do some cleaning and tossing of stuff I no longer need. I don't think of myself as someone who spends much money on material items, so it's been a bit of a shock to see just how much stuff I've nevertheless managed to accumulate in my three years here.
I just started moving stuff upstairs today, but I'm hoping to be done or almost done by tomorrow night and actually be sleeping up there. Need to get some sleep now, as moving heavy stuff upstairs turned out to be more tiring than I expected. A hui hou!
This is exciting for me, as I don't think I've ever had a room above the ground floor (given the number of houses I've lived in across my lifespan, I actually had to pause to give that thought). Being on the second floor for this house means it'll be a bit warmer in general (good in the winter, not so nice in the summer), and also that I shouldn't have to worry about encountering centipedes anymore. The experience of "moving" has also pushed me to do some cleaning and tossing of stuff I no longer need. I don't think of myself as someone who spends much money on material items, so it's been a bit of a shock to see just how much stuff I've nevertheless managed to accumulate in my three years here.
I just started moving stuff upstairs today, but I'm hoping to be done or almost done by tomorrow night and actually be sleeping up there. Need to get some sleep now, as moving heavy stuff upstairs turned out to be more tiring than I expected. A hui hou!
Friday, November 30, 2012
A New Job
In the category of "fairly important and relatively life-altering events", as of two days ago I have accepted a position as a Data Quality Assistant at the James Clerk Maxwell Telescope, starting early next year. As this is a full-time position, I'll be leaving my job at the Mauna Kea Visitor Information Station in three weeks' time.
Having written that, I'm having trouble thinking of anything else to add to it. I've learned a lot while working at my current job, and will definitely miss my coworkers when I leave (though I intend to resume volunteering, just as I did before I was hired there). I'm also very excited (and a bit trepidatious) to be working in a job that, I feel, fits well with my problem-solving and computer abilities. And working for one of the best astronomical observatories in the world is both stimulating and intimidating! However things go, there will certainly be some changes for me in the coming weeks. A hui hou!
Having written that, I'm having trouble thinking of anything else to add to it. I've learned a lot while working at my current job, and will definitely miss my coworkers when I leave (though I intend to resume volunteering, just as I did before I was hired there). I'm also very excited (and a bit trepidatious) to be working in a job that, I feel, fits well with my problem-solving and computer abilities. And working for one of the best astronomical observatories in the world is both stimulating and intimidating! However things go, there will certainly be some changes for me in the coming weeks. A hui hou!
Tuesday, November 27, 2012
Relativistic Gaming Fun
How many games can you name off the top of your head that involve simulating the effects of relativity? Before Sunday I'd have had a hard time naming even one, but since then I've discovered not just one, but two.
The first one is a simple Flash game called Velocity Raptor, a game that takes place in two dimensions of space and one of time, right from the comfort of your own web browser! It features a whimsical art style and story suitable for all ages, and calming, ambient music. This game simulates the effects of special relativity by slowing the speed of light down to 3 miles per hour. Naturally, this changes things in ways that we are not normally equipped to think about, and the game is mind-bending while still managing to be fun. It builds up, introducing you first to a Newtonian world, then the world of relativity as measured (without taking into account light travel time), then finally the world as seen, where light takes a noticeable amount of time to reach you and things begin to appear to deform in wild and amazing ways. I found myself smiling quite a bit while playing this game as I watched the world around my character (the eponymous Velocity Raptor) warping and stretching . The progression is done well, introducing new concepts (such as the Doppler shift, or the relativity of simultaneity) in simple cases before tasking you with using your new-found knowledge to solve a puzzle to advance. (It actually reminds me a bit of Portal and Portal 2's approach to teaching new concepts, and I think that's a great thing. More games should be like that.) Be warned, the last levels are very difficult.
The second game I came across is called A Slower Speed of Light, produced by the MIT Game Lab (no, I didn't know MIT had a game lab before either). This game is an actual stand-alone program that you have to download (it's free) and run on your computer. Unlike Velocity Raptor, it's a full three-dimensional (well, four-dimensional, since it involves relativity) first-person view game. Similar to Velocity Raptor, it involves slowing down the speed of light rather than you moving close to the measured speed of light. However, it goes about it differently: in the game, your goal is collect 100 orbs, each of which, when collected, slows down the speed of light by a little bit. This has the effect of starting you in a basically Newtonian world that gradually gets more and more relativistic as you collect more orbs. The final ones can prove challenging to collect, not because of any obstacles, but because it can be difficult to judge position accurately when turning at near light-speed. The simulation mostly involves the Doppler shift and the Searchlight Effect, but upon collecting all 100 orbs those effects are turned off and speed of light is dropped to just above your walking speed, allowing you to see the Lorentz transformations that take place at significant fractions of light-speed. It's a lot easier than Velocity Raptor in that there is no way to actually lose, and watching spacetime warp and deform around you in first-person view is incredibly cool. The game authors are working on cementing the underlying game engine and planning to release it sometime next year as open-source software, so hopefully we'll start seeing more games that are truly relativistic.
I can definitely recommend these games to any aspiring physicists out there. It's incredibly cool to see these relativistic concepts come alive. But even if you're not a physicist, you can play them and experience some of the fun we get to have discovering these ideas!
The first one is a simple Flash game called Velocity Raptor, a game that takes place in two dimensions of space and one of time, right from the comfort of your own web browser! It features a whimsical art style and story suitable for all ages, and calming, ambient music. This game simulates the effects of special relativity by slowing the speed of light down to 3 miles per hour. Naturally, this changes things in ways that we are not normally equipped to think about, and the game is mind-bending while still managing to be fun. It builds up, introducing you first to a Newtonian world, then the world of relativity as measured (without taking into account light travel time), then finally the world as seen, where light takes a noticeable amount of time to reach you and things begin to appear to deform in wild and amazing ways. I found myself smiling quite a bit while playing this game as I watched the world around my character (the eponymous Velocity Raptor) warping and stretching . The progression is done well, introducing new concepts (such as the Doppler shift, or the relativity of simultaneity) in simple cases before tasking you with using your new-found knowledge to solve a puzzle to advance. (It actually reminds me a bit of Portal and Portal 2's approach to teaching new concepts, and I think that's a great thing. More games should be like that.) Be warned, the last levels are very difficult.
The second game I came across is called A Slower Speed of Light, produced by the MIT Game Lab (no, I didn't know MIT had a game lab before either). This game is an actual stand-alone program that you have to download (it's free) and run on your computer. Unlike Velocity Raptor, it's a full three-dimensional (well, four-dimensional, since it involves relativity) first-person view game. Similar to Velocity Raptor, it involves slowing down the speed of light rather than you moving close to the measured speed of light. However, it goes about it differently: in the game, your goal is collect 100 orbs, each of which, when collected, slows down the speed of light by a little bit. This has the effect of starting you in a basically Newtonian world that gradually gets more and more relativistic as you collect more orbs. The final ones can prove challenging to collect, not because of any obstacles, but because it can be difficult to judge position accurately when turning at near light-speed. The simulation mostly involves the Doppler shift and the Searchlight Effect, but upon collecting all 100 orbs those effects are turned off and speed of light is dropped to just above your walking speed, allowing you to see the Lorentz transformations that take place at significant fractions of light-speed. It's a lot easier than Velocity Raptor in that there is no way to actually lose, and watching spacetime warp and deform around you in first-person view is incredibly cool. The game authors are working on cementing the underlying game engine and planning to release it sometime next year as open-source software, so hopefully we'll start seeing more games that are truly relativistic.
I can definitely recommend these games to any aspiring physicists out there. It's incredibly cool to see these relativistic concepts come alive. But even if you're not a physicist, you can play them and experience some of the fun we get to have discovering these ideas!
Thursday, November 22, 2012
Hauʻoli Lā Hoʻomaikaʻi!
Happy Thanksgiving everyone!
Once again it's that time of year where I am reminded how thankful I am to have loving family and friends, a steady job (even if I did have to work this evening), and a good living situation.
Speaking of working, I was able to see what kind of Thanksgiving dinner the cooks made up at Hale Pōhaku:
I will note that this is actually not my plate, but that of a friend of mine. I only thought to take a picture of our Thanksgiving dinner after I saw him doing so, and since I'd already cleared half my plate I asked to use his. Also, I hadn't noticed the cranberry sauce on the salad bar. That minor detail was easily fixed, and all in all I had a pretty good Thanksgiving dinner. The weather was completely overcast the entire evening as well, so it was nice and quiet at the VIS, for which I was thankful. Hope your Thanksgiving was as pleasant as mine!
Once again it's that time of year where I am reminded how thankful I am to have loving family and friends, a steady job (even if I did have to work this evening), and a good living situation.
Speaking of working, I was able to see what kind of Thanksgiving dinner the cooks made up at Hale Pōhaku:
I will note that this is actually not my plate, but that of a friend of mine. I only thought to take a picture of our Thanksgiving dinner after I saw him doing so, and since I'd already cleared half my plate I asked to use his. Also, I hadn't noticed the cranberry sauce on the salad bar. That minor detail was easily fixed, and all in all I had a pretty good Thanksgiving dinner. The weather was completely overcast the entire evening as well, so it was nice and quiet at the VIS, for which I was thankful. Hope your Thanksgiving was as pleasant as mine!
Tuesday, November 20, 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.
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.
Tuesday, November 13, 2012
Making Fudge.
Today I made fudge for the first time, and am currently munching on the first-fruits of my labors. Well, labors may be too strong a word – the whole process took fifteen minutes from start to finish, tops. It turned out to be easier than I was expecting, though it was still a good learning process. Some things I learned:
- If a recipe includes phrases like “stirring constantly” in it, it's probably a good idea to get any ingredients that come later in the recipe ready before becoming trapped in a cycle of time-critical stirring from which you can't break away.
- Pure vanilla extract? Very strong flavor. Very, very strong. Discovered this after spilling a bit (due to the hasty way I was rushing to open the bottle) and idly licking it off my fingers. I'd heard that before, of course, but wasn't quite expecting that particular burst of flavor.
- Molten fudge has a consistency close to that of pāhoehoe lava, if the shapes it formed as it cooled and congealed in the pan are any indication. Especially so soon after my trips through the Kaumana lava tubes, I was struck by the many similarities between the rock formations there and the fudge formations that formed in my pan. Fudge: chocolate lava. Or is it lava: rock fudge? Fudge:chocolate::lava:rock? (If you'll pardon the logical formalism.)
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