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The Physics of the Kinetic Theory of Gases: Energy, Momentum, Speeds, Balloons, and Blowing on Your Arm

This blog post will discuss some of the physics behind the things molecules do – as gases, liquids, or solids – and also get you thinking about the concepts momentum, kinetic energy, rate (of a physical molecular-scale process), and equilibrium constant (of a physical molecular-scale process). These may all sound like difficult, high-level ideas – but if molecules, which don’t have much in the line of brains, can act according to these ideas, you, with brains, can follow them too!
What are molecules? First, at their simplest, they’re just clusters of atoms stuck together in such a way that they don’t come apart on their own. (If they did, they would do so immediately; there’s no such thing as delayed disintegration, the way that isotopes of some elements are radioactive.) Wait a minute, you say, what if I heat a pure substance and it breaks down? That happens when one molecule of the substance crashes into another molecule of the substance with enough energy to break at least one bond between atoms, on at least one molecule. That’s a chemical reaction! That’s not “on their own”, there has to be a collision (or absorption of energy from some other source) with something other than just that single molecule.
Next, the behavior of molecules: at the simplest, acting as a gas, molecules bounce around, hitting each other and the walls of any container they’re in (anything in their way, actually), endlessly! They don’t: 1) stick to each other – that would start to make them into a liquid or a solid, and 2) they don’t care much what else they’re running into – either they happen to have enough “oomph” (energy) to trigger a reaction with or in what they ran into, or they don’t. If they do, they (may) react, and if they don’t, they bounce off!
KINETIC ENERGY
Now, what about this bouncing off process could possibly be interesting? Consider yourself, bouncing off a wall, for example. You run at the wall (or get pushed!), start to squish against it, squish more against it, reach a position of maximum squish (when some of your energy has been stored as squished muscles, compressed bones, and so forth), start to unsquish from the wall (but are still partly pressed against it), unsquish more (almost free of the wall), and unsquish more so that you’re totally free of the wall and moving away from it. All of this happens pretty fast, but those are the stages of what happens to you physically. Surprise – that’s exactly what molecules do when they hit a wall, or hit each other! The only thing that separates your behavior from the molecules’, in fact, is that you generally lose some of your energy of motion (kinetic energy) to heat, sound (that splat sound!), tearing of tissues (with eventual bruising), and so forth. Molecules don’t. They just either bounce, or react chemically with each other (we’re not getting into things like rotation, vibrations, or electronic transitions here).
Now, in the collision with the wall, I said, you may lose some of your kinetic energy, which is the energy of motion; alternatively, think of it as the energy that was used to get you going to the speed with which you hit the wall, and also, it’s the energy that could be recovered in stopping you to a standstill (think regenerative braking in an electric car). That kinetic energy lost could appear ultimately as heat in either you or the wall. If NO energy appears to have been lost, we call that an “elastic” collision; either you bounced off with exactly the speed you came in with, or else you set the wall in motion, and together, you and the wall added up to the same kinetic energy as you came in with. (Pretty shoddy wall, but take it from me, never shove drywall hard with your foot!)
With molecules, it’s the case that, if you sum up the kinetic energy of the two molecules (for each, its kinetic energy = (mv^2)/2, where m is the mass of the molecule, v is its velocity (a speed, with a particular direction), and “^2” means the velocity is squared) *before* they bounce, and then also *after* they bounce, those two sums are equal. You might say, there was a total kinetic energy held by the two molecules before the collision, and that’s the same as the total kinetic energy held by the two molecules after the collision. (By the way, the “particular direction” that a velocity has, disappears when you square it.) No total kinetic energy was lost; we say the energy was “conserved”. Notice, I did NOT say that each molecule kept its own kinetic energy! In fact, usually some is passed from one molecule to the other; it’s just not “lost”. Even a big object like you never “loses” energy, but sometimes it is passed from one form (such as kinetic energy of a moving mass) into other forms – heat (which is still kinetic energy, but added up as a total for all the molecules individually), gravitational potential energy (perhaps you used some of your impact to climb the wall – but remember what I said about drywall!), electrical potential energy (you punched a generator as you hit the wall, and charged a battery a little), and so on.
There’s another little complication to this, which is that the kinetic energy of an object, relative to you, depends on how fast you are moving relative to the object. So, if an object is moving by on a train, and has a kinetic energy relative to me (we say, “in my frame of reference”), and I hop onto the train, now that object has no kinetic energy relative to me – even though I didn’t do anything to it! Still, the idea of total energy not being changed in a collision, doesn’t change, no matter what your (constant) motion may be as an observer (note, you can’t be changing direction or speed, as you observe the collision – you must be in an “inertial frame of reference”).
So, kinetic energy is something that is *conserved* in collisions between gas molecules – and, since the gas in a closed container (let’s say a balloon) is just a lot of molecules colliding continually, the total kinetic energy that they have is conserved – that is, it remains constant as time passes (we say, “constant over time”). We have a name for the idea of all this kinetic energy present in this bunch of gas in a container (which we call a “system”): it is heat energy. Molecules moving around “have” heat, or rather we should say, heat simply IS molecules moving around. There’s one other term we apply, used when we *measure* heat: temperature. Temperature is nothing more than the *average* kinetic energy of the moving molecules – some of them are moving slowly, some faster, a few even faster – so they each have a different individual kinetic energy – but sampled as an average, which is what a thermometer does, there is a definite, specific value for temperature.
It’s kind of curious, what the typical (bulb-type) thermometer does in detail. It allows the gas you’re measuring (with moving molecules) to hit the bulb of the thermometer for a while, which makes the molecules in the wall of the bulb wiggle around, too – and the wiggling is passed through into the liquid inside the bulb, which then starts wiggling (as molecules) proportionally to the wiggling of the wall – which liquid then expands to a volume proportional to the amount of wiggling, and you read the liquid level in the stem of the thermometer. All the molecules concerned – whether solid, liquid, or gas -- wiggle (move, either freely in space, or in smaller volumes as determined by their neighboring molecules) with exactly the same average kinetic energy (that’s what we’d say if we were looking with microscopic-size vision) – they all have the same temperature (what we say, when we look with normal-size vision). That’s an important equivalence,
“IMPORTANT LESSON #1”
so let me stress it again: there is just one process that’s governing, the kinetic energy exchange among molecules, no matter what – but you can think about it either at the *microscopic* scale, which looks like molecules bouncing, or at the *normal everyday* scale, which looks like gases, liquids, and solids having *temperatures*. You need to be able to appreciate the underlying connection here, that the behavior of individual molecules, has observable, predictable effects on the behavior (things we normally measure) of large groups of molecules.
MOMENTUM
We use the term “kinetic energy” for both small items like molecules, and for large objects (like you bouncing off a wall); the equation above for kinetic energy ( kinetic energy = (mv^2)/2 ) applies equally well for each one. There is one other term that also applies to both large and small objects, that’s important for both, and that’s the property called *momentum*. Momentum is also related to objects and their motion, and it’s even simpler than kinetic energy. First, think of kinetic energy as “splat” – everyday-size things with a lot of kinetic energy tend to go “splat” when they hit something else (think, a bullet hitting a wooden wall). Then momentum is “bump” – the ability to bump something else, to get it moving. Obviously you can’t have just one of these without the other. But, you can have them in different ratios. You will see what I mean in a moment, and why it’s important. But first, what would make you (on rollerblades, say) able to give someone else (on rollerblades) a bump?
Obviously, if you’re not intending to mess with them, but hit them accidentally, you give them a bigger bump if: 1) you’re heavier, and 2) if you’re moving into them faster. As equations go, that means p (we call momentum by the letter p) = mv (or, momentum = mass times velocity). That looks quite a bit like the formula for kinetic energy, doesn’t it? Ah, but the differences!
First, the formula for momentum just has one “v” in it, not squared as kinetic energy had. That means the *direction* of the motion is always important for momentum. You have a particular amount of it *in a particular direction*; we call a property that has this directionality to it a *vector*. (A thing that doesn’t have a direction, because it never had, or because it lost it due to something like the squaring operation as did kinetic energy, is called a *scalar*. Temperature is a scalar, for example, you can’t have a temperature in a particular direction!). Because of that, every time we work with momenta (the plural of momentum), we have to keep track of directions. Frequently we reapportion a momentum (decompose it into pieces) as pieces in the x-direction and in the y-direction; you’ll cover that (lots!) in physics, but it’s not important here.
What is important about momentum (which we also call by names such as “inertia” as a general property, or “impulse” when we’re calculating it as an amount transferred from one object to another in a collision, or by a force acting for a time on an object) is two things: 1) it is ALWAYS conserved – when you add it up for a pair of objects before a collision, and compare it to the sum afterwards, it’s always the same. There’s never any transfer to another form, as there sometimes is with kinetic energy – although, there are some hidden aspects that you will learn about in physics. Also, 2) there’s no name for the “average” momentum of a group of molecules, for a couple of reasons. First of all, the average momentum of the gas in a closed container is zero! Considered as a whole, the mass of the gas isn’t travelling anywhere (unless you are moving the container!), so the average velocity is zero (although the mass is still whatever the mass of the gas in the container happens to be), and the momentum of the system (the large group of molecules) must thus be zero. Secondly, the average speed (the momentum, without direction) of the molecules, considered by adding their speeds up individually and dividing by their number, is different depending on what molecule they are!
This is odd, isn’t it: for kinetic energy, there was an average for all the molecules in a system of gas, and it didn’t depend on what kind of molecules they were. So, it could have been a mixture of hydrogen and carbon dioxide, for example, and if you considered the hydrogen separately and the carbon dioxide separately, each set of molecules would have had the same *average* kinetic energy each (and hence, the same temperature). Not so for molecular speed. In order to see this, imagine one molecule with a mass of “1” (never mind the units for now) and another with a mass of “100”. For the average kinetic energy (mv^2)/2 to be equal for them, would mean that the heavier should be moving at 1/10 the speed of the lighter one; then the v^2 term becomes 1/100, which when multiplied by the mass ratio 100, equals unity. But for momentum, the heavier one has the calculation 100 * 1/10 = 10 vs. the lighter one; the more massive molecule carries more individual “momentum” (disregarding direction for a moment).
The heavier molecules still have the same temperature as the lighter ones, and that’s not all: because the *pressure* measured for a gas arises from all the gas molecules bouncing off the walls of their container (and off your pressure gauge!), and bouncing off is a reversal of their momentum (to some extent, also depends on the angles of the bounce), you might think that heavier molecules might therefore have a different pressure than light ones. But they don’t; although they’re moving with higher momentum, because they’re moving more slowly, they arrive less frequently at the walls, and these two effects exactly cancel each other out, mathematically. So, every gas molecule behaves the same – makes the same contribution to the properties temperature, pressure, volume, and so forth that we measure in the laboratory – no matter what mass it has.
BEHAVIOR OF GASES AND BALLOONS
Now we get to some interesting problems in which we’ll be able to use some of the ideas above. The first is the behavior of balloons. Consider a balloon blown up with helium (you could use hydrogen, but then you might get blown up as well). As you’re well aware, it has buoyancy – it tries to ascend in the air, you have to hold it down or store it on the ceiling. Why does it do this? The stock explanation is that the displaced fluid (surrounding air) exerts a buoyant force, and that force is upwards because it’s a buoyant force. That’s circular reasoning, people!
Let’s consider the matter in a little more detail. Inside the balloon is the helium; it’s pushing in all directions (as fluids do; in physics we call anything that doesn’t retain its shape a fluid, so that includes gases and liquids) outwards on the walls of the balloon, and the outside air is pushing back. The walls of the balloon aren’t moving anyplace; the forces are balanced. The forces, incidentally, aren’t trivial – 1 atmosphere = 760 torr = 101 kPa (kilopascals, don’t worry about the units) is about 15 pounds per square inch; over a typical balloon surface of 4” radius with ~2000 in^2 area that’s 3000 pounds. The only thing that saves the balloon from annihilation is that 3000 pounds of force directed inward are balanced by 3000 pounds of force directed outward! But if that were all, that *wouldn’t* give us any buoyant force! What *does* give us a buoyant force, is the variation of air pressure with altitude (outside the balloon) and the variation of helium pressure with altitude (inside the balloon).
You may be aware, that the atmosphere gets thinner (less dense) as you go upwards. There’s a molecular-level reason for this – gas molecules, as they bounce around at sea level, say, could travel upwards (and would, if they didn’t keep hitting other molecules above them). As they do this, they “lose steam” due to the pull of gravity. The average nitrogen molecule, for example, would never be able to escape the Earth’s gravitational field – it would simply hop in a parabolic trajectory and come back down. Only the few highest-kinetic-energy molecules are able to travel higher; the sluggards spend all their time down low. So, an atmosphere composed of mostly nitrogen, as Earth’s is, has more molecules near the planet surface, and less and less at higher and higher altitudes, because fewer and fewer molecules have the energy to get up higher. Less molecules present at higher altitude translates into a lower pressure (all other factors being equal). Therefore, the atmosphere surrounding the *top* of a balloon is putting less pressure force on it than is the atmosphere at the *bottom* of the balloon. You might think, “Oh, how could just a tiny difference of pressure make that much difference?” But if you work it out, you’ll see that it does. Now, if the balloon were completely rigid-skinned, and empty of gas, the difference of these two external pressures would sum to the average density of the external atmosphere times the volume of the balloon, as a buoyant force directed upwards on the balloon. But, there is also the helium present inside the balloon. Doesn’t this cancel out the external (“buoyant”) forces? Alas, not – a gas with lighter molecules has a less steep pressure gradient with altitude – the molecules are already moving faster, remember? – and therefore they tend to distribute out to a greater altitude, compared with nitrogen. That means a smaller gradient of pressure with altitude for helium – so while it might balance the force of the external air at the bottom of the balloon, it then has much more than enough force to balance the force of the external air at the top of the balloon – and the balloon finds itself pushed upwards. But just for kicks, consider an imaginary balloon that’s: 1) flat on the bottom, 2) has straight cylindrical walls upwards into space indefinitely, 3) filled inside with helium, and 4) itself is weightless. Would it have buoyancy (assume that it’s filled with enough helium to exactly balance the pressure of the external air on the bottom of the balloon)?
If you answered “NO”, give yourself a star! There can be no buoyant forces upward from the walls of the imaginary balloon – they don’t have any sideways slant, and gases only exert pressure perpendicular to surfaces – and the bottom forces are balanced, so no buoyancy! It’s only the condition of the column of the inner gas (helium) being capped off in the real balloon, as opposed to extending indefinitely into space, that permits buoyancy!
DIFFUSION OF GASES AND BALLOONS
Balloons (real or imagined) are one of my favorite toys with which to talk about physics.
Did you ever wonder what happens to balloons, if you lose them to the ceiling, and then they come floating down again in a couple of days? They don’t seem that much smaller, and yet your teacher (especially physics teachers!) might say, they’re lost helium. So what’s with losing helium, but not deflating completely?
What’s going on here, is a phenomenon called diffusion. What is it? Everything that has moving molecules (by now you should know that that’s gases, but it’s true also for liquids, and in certain circumstances solids), has those molecules continuously moving. If the molecules are free to move (i.e., not tied down by their size and entanglement with neighboring molecules, or tacked down by electrical forces, etc.), they do move. That means, each of them may follow a path that meanders here and there, as it is bumped by other molecules. There’s no reason to stay in one place. So, if you looked at the movement, over time, of many molecules, some would (by chance) not move far, others would get further, a few make it even further, a rare few make it even further, and so on. The movement is incidentally termed a “random walk” (you can and should look it up, it has neat mathematical properties); the process and the result (transport of molecules away from their origin) are both called “diffusion”.
How is this different than the movement of water when you pour it, say (which we call a “bulk flow”)? In diffusion, each of the molecules is always moving independently, limited only by collisions with neighbors, and the neighbors have, on the average, zero velocity (remember from above? velocity = speed AND *direction* of movement). If the neighbors *do* have a net average velocity, then the molecule you’re looking at is dragged right along into the flow. This will move it *much* faster than it would ordinarily move by diffusion. But there’s another speed to consider yet – the speed of the individual molecule. That is ordinarily even faster than either diffusion or bulk flow! So, why is diffusion so slow? It’s slow because the molecule keeps bumping into other molecules and changing directions. Imagine trying to push your way through a dense crowd of people. It doesn’t matter if you can run fast, you just can’t push your way through fast! That’s what it’s like for a molecule (although, gas molecules travel a lot further between collisions relative to their actual size than you would through the crowd). Now, how about bulk flow and diffusion? Ah analogy here would be a room packed with blindfolded people. If you suddenly dropped the floor out, they’ll all fall out in a hurry! That’s bulk flow. But, if you open a door such that only one person can get out at a time, and shout “fire”, they’ll all get out eventually, but it will take quite a while (and when they catch you, you will end up in a room ….) – this is diffusion (you blindfolded them so that they had to bump around, like molecules, until they found the door by accident).
So what happens with the balloon? The walls of the balloon are thin, but (if latex rubber) somewhat porous – they have spaces between the molecules of rubber that molecules can gradually bump their way through until they pass out of (or into) the balloon. What’s important here is that these passageways are not large with respect to the distance the molecules travel between bumps; therefore, if one molecule happens to enter one of these passageways (called a “pore”), that doesn’t “clue in” the surrounding molecules that there’s a large hole there by a big pressure decrease. Each molecule is free to enter a pore if it encounters one; it may eventually emerge on the other side, or it may eventually pop back into where it started.
Now, here’s what’s important about this. On the inside of the balloon, there’s a lot of helium (initially, anyway), and on the outside, essentially zero. If a molecule enters a pore from the inside of the balloon, and pops back into the balloon, it (or another molecule) is free to try again. There hasn’t been any change on the number of molecules inside. But if it does make it out, chances are good that it will diffuse away from the balloon, never to be seen again! The balloon has essentially irreversibly lost helium – even though there’s nothing (microscopically) preventing that molecule from diffusing back in through the pore, you might say that the average rate of movement away from the balloon surface is much greater than the average rate of finding the pore again and diffusing back in. This is:
IMPORTANT LESSON #2
The actions a molecule can take may be reversible – it can go one way or the other – but the *net movement* may be governed by other factors, such as probability. If doing one thing has a higher *probability* than doing a second thing, molecules will do more of the first and less of the second, which will result in more product of the first thing than of the second thing (all other factors being equal). This is a “rate thing”, but it will have effects on things you may study later in chemistry, such as equilibrium.
How does this work for the balloon? Immediately after it is inflated (even before you pay for it, just imagine) the helium starts working its way out of the balloon, molecule by molecule (well, atom by atom, in the case of helium). At the same time, however, the molecules of air – 78% nitrogen (N2), 21% oxygen (O2) and 1% argon (Ar) are starting to work their way in, through the same pores. How can things pass different directions at the same time? Seems weird – but molecules do it all the time – they move independently, except when they happen to collide with each other. It’s useful to realize – this is not a phenomenon governed in any way by the inflation pressure of the balloon (which is not much compared with atmospheric pressure); it takes place because of, and proportional to, the absolute pressures of the gases at each side of the balloon surface, the composition of gas on each side of the surface, and lastly, to the molecular weight of each kind of gas. For this last, gases with more massive molecules diffuse more slowly. You could perhaps figure this out from the discussion of “average kinetic energy” above; the speed of diffusion is proportional to (m)^(-1/2), in other words the 100x heavier gas molecule diffuses at 1/10 the speed of the 1x molecule.
Now, put this all together: the helium start diffusing out; the nitrogen starts diffusing in, at a rate of (4/28)^(-1/2) that of the helium, or about 30% as fast, and oxygen diffuses in at about 7%. So for each mL of helium lost, about 0.37 mL of other gases leak in. Eventually the rate of helium loss tapers, as it becomes depleted inside the balloon – but by this time, the balloon is partially inflated with air, which it will not easily lose (it diffuses in as fast as it diffuses out). If you assume that the balloon shrinks to 37% of its original volume, this is still 72% of its original diameter; no wonder you didn’t see much difference!
SO WHY DOES YOUR BREATH FEEL HOT OR COLD DEPENDING ON HOW HARD YOU BLOW AGAINST YOUR ARM?
Whoa, that’s quite a jump of topic! Are you saying that my arm is some kind of balloon, or something?
Not exactly. But it’s a surface, with properties, and various molecules here and there inside and outside of it. And these molecules are subject to things like diffusion, and bulk flow, and average kinetic energy (i.e. temperature) effects, and so on. And a few other things besides!
First, what is the role of water loss from the skin? If you aren’t exercising and having to shed heat by evaporating sweat, you still have insensible water loss, estimated (http://www.anaesthesiamcq.com/FluidBook/fl3_2.php) as 400 mL day^-1 through the skin and 400 mL day^-1 in exhaled vapor. When the passive heat flow from the body by vasodilation alone is sufficient to disperse all heat produced by metabolism, the body does not sweat (ref. slide 100/103 of a cardiac physiology course, http://www.people.vcu.edu/~mikuleck/PHIS502CVf.ppt). Suppose that you are standing outside, on a not too hot day, not sweating, and you either breathe or blow hard on your arm. These two actions feel as if the impinging breath is respectively warm or cool. But why?
The first thing that has to be considered is the possible role of insensible water loss through your arm skin. This is limited by such things as the physiological state of your skin, in particular the thickness of the stratum corneum, the dry, waxy outermost layer of consolidated epidermal cells. The thicker it is, the slower the diffusion of water molecules through it; so if you’ve recently showered and rubbed it down with a loofa sponge, you may be losing somewhat more water than otherwise. Another way of losing it, it to press cellophane tape onto it (hard), then to strip the tape off. It will pull off a single layer of stratum corneum. Repeat this about 20 times (depending on skin site), and the layer of skin will start appearing shiny, even moist – you have now removed the stratum corneum, and you are losing water at a significant rate right there, similar to the water loss that burn victims experience on their burned skin. But ordinarily, the stratum corneum is the rate-limiting barrier to water loss: once a water molecule makes it through, it’s easily gone into the surrounding air. There is no additional “burden” of having to get away from the skin, before being irreversibly lost to the body.
There is always a heat loss associated with transition of a water molecule from a liquid (as it is in the stratum corneum) to a gaseous form. It’s not inconsiderable: 540 cal/g vaporized – but, considered per day for the whole body, 400 mL lost from the skin amounts to only ~200 Calories ( or 400 Cal including respiratory loss); that’s quite small compared with the adult human basal metabolism rate of 1000 – 2500 Cal day^-1, implying that the majority of basal metabolically produced heat is lost in other ways. These other ways include radiative, conductive, and convective routes; per site http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/coobod.html and http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/bodcon.html#c2 the evaporative, radiative, and conductive plus convective losses are on the order of 380, 2740, 225 Cal day^-1 respectively, thus radiative loss dominates.
So, what does breathing gently, or blowing, on your arm do? The radiative loss rate isn’t much affected by whether you lean over your arm or not (try it! You won’t feel a difference, unless your body is blocking a path to a very cold (winter night clear sky) or very hot (a blazing fire) distant location). The conductive route is into whatever you’re sitting on, typically. But convective! Here, there are only three factors which dictate heat loss (or gain): 1) the temperature difference between skin surface and surrounding air, 2) the thickness of the boundary layer (the unstirred air layer right next to your arm), and 3) the heat conductivity of the gas in the boundary layer. Let’s consider these in order.
1) The temperature difference skin-to-air. Typically, arm skin temperatures range from perhaps 32 down to 28°C (http://www.people.vcu.edu/~mikuleck/PHIS502CVf.ppt slide 95/103); thus it is easy to see why warm gently exhaled air (at 37 °C) would warm it. However, if one blows vigorously, exhaled breath turbulently mixes with ambient air, and the mix, if below skin temperatures, will then be perceived as cooling, and the more so as higher wind speeds cause progressive decreases in the boundary layer thickness. One can test this by pumping a thermostatted air supply at the particular skin patch: below the skin temperature, the breeze is the more cooling the stronger it is; at the skin temperature, it generates no net temperature feel, at any speed; and above the skin temperature, it is the more warming the stronger it is (how a convection oven works!)
This disposes of two of the three factors which might influence heat flow, and hence perceived heating or cooling. The third factor, heat conductivity, may be broken into factors of the density of the boundary layer gas and heat capacity per gas molecule. Ordinarily the density is essentially that of air, since the surrounding gas *is* air, at atmospheric pressure; but both the density and the heat capacity per molecule heavily depend on the molecular weight of the gas. Hence, a helium stream on the skin (even at 25°C) would be perceived as cooling, as it conducts heat away much better than does air; and a cloud of carbon dioxide (such as in the fog above a pail of water with a large chunk of dry ice at the bottom), especially if unstirred, is perceived as heating, because it retards the conduction of heat away from the skin, both effects referenced to the feel of ordinary air comparably applied.

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