A cylinder with a piston to seal it starts empty. The are of the piston give a force of about 100 pounds. Lift the piston till there is 100 liters of vacuum. Now 2 cases: 1) release the piston, it slams down almost instantly. Since 1 atmosphere is about 10 newtons per centimeter, and 1 liter is 10 centimeter liters, 1 atmosphere-liter is 100 newton meters or joules. 100 liters makes for 10,000 joules of energy, as it hits the bottom. So loud bang, lots of heat. 2) Here is the real question. What happens if instead a hole is opened in the piston, letting air rush in. Ideal gas law says if volume increase/decreases work happens, but opening volume to connect, should create no work. In fact, opening a compressed bottle of air it cools of, as far as that air goes, it is expanding against a wall of other gas. Not precise i know. Either way, the atmosphere is lifted 100 liters, then dropped. Do that with a giant can of sand, drop it, bang. Open it, sand trickles down. Energy does not go the same place. Initially, the air expanding to fill the vacuum should cool slightly, then more air from outside the room, 100 liters of it, comes in to the room to equalize pressure. So, where does the 10,000 newtons go? Does the air in the room gain 10,000 joules of heat? or is the 10,000 joules distributed through the atmosphere, as would happen with the sand example?
In all cases the energy goes into increased heat. Whether that happens slowly or quickly is not relevant.
yes, but where does the heat go? With the closed piston, the "heat" hits the floor, and so is concentrated. with opening the cylinder, the floor clearly does not get heated up. The question is, is the heat concentrated in the room, or distributed among all the air moved, including outside the room?
heat is not the only form of energy in the problem during expansion, the atmosphere is lifted, not heated. Just like lifting a hundred pound weight. That lift is distributed however imperceptibly all the way to the very fuzzy top of the fluid. In the first case, pressure remains uniform as the atmosphere falls. In the second case, a low pressure wave propagates outward indefinitely, as far away as you can detect the "pop" of opening the vacuum. Is that where the heat goes? Any experiments like this out there?
Welcome to PF! You should be more precise and consistent with your units: it is pretty hard to follow your post. Regardless of whether you really mean pounds or Newtons, 100 L (0.1 cubic meter) of vacuum (at sea level) is all you need to know for the energy. For simplicity, we'll say we have a .1 square meter column, 1m high. The force is 1010 N and the potential energy is 10,100 J. More complicated, but Conservation of Energy tells us it has to work out the same. The correct way to approach any thermodynamics problem is to assume conservation of energy holds and then look for the energy. So if the gas cools as it enters, what else happens to it that could affect the energy released? It is actually up to you, since you didn't specify an important issue: is the container insulated? Yes! So cooler means less energy, but more mass/denser means more energy, right? What do you mean? Where do you think the energy goes?
What do i mean by energy does not go the same place? Drop a mile high can of sand, it all stays in contact, entire force is transmitted to the bottom as it hits. Energy concentrated at the bottom. Force on the ground initially the weight plus the force due to momentum. Really big bang. Open a hole in the bottom of a mile high can of sand, the sand on the bottom falls first, and that fall travels up the can. Energy of the fall of the top grain of sand impacts on where it lands, at the top of the pile. Energy distributed. Force on the ground builds gradually to reach the total weight. Force on ground never exceeds total weight by much. No big bang. So my question regarding air, is energy distributed in an analogous manner to filling grains of sand? Or is there some mechanism i'm overlooking that would concentrate heat at the location of the vacuum? Having worked with pressurized air, i can say if you allow air to enter a lower pressure container, it does not appear to heat up, but is that because it cooled off first, then heated up as pressure equalized? I am not sure, and can find no formulas covering this type of air movement. Unlike the ideal gas law implies, air does act somewhat as a container of sorts for itself. so one more times, not asking how much energy or if it ends up as heat. Where does the heat end up after air is done moving? Like the mile high sand can, pressure on the "floor" of the vacuum chamber never exceeds atmospheric pressure, and builds "slowly" if you consider the speed of sound slow.
I am not sure what you are saying here. Potential energy of the sand is the same from the start, regardless of whether you drop it all at once or drop it grain by grain. And also, the kinetic energy of each grain of sand when it reaches the ground is the same whether all grains are dropped at once or one by one. The impulse upon hitting the ground is also the same in both cases. The difference is that all the tiny little forces of each grain of sand add up at once to give the larger force and the bang when dropped all at once. By the way, the gas ( air ) that has entered the container will heat up and have a higher temperature in your hole experiment. That increase in temperature has an equivalent to the loud bang energy. Just thought I would mention that for you to ponder the question, "why does that happen."
Heat is just the random movement of molecules. Air molecules above the piston are colliding with it and thereby pushing it down. When you lift it against the force of the air you are accelerating air molecules. Imagine a ball bouncing of a wall. If the wall moves toward the ball it will accelerate the ball and give it more kinetic energy. So when you lift the piston you are giving the air molecules more energy and that means the air gets warmer. When the air pushes the piston down that process is reversed. The molecules slow down after the collision and therefore some of the heat of the air is transformed into kinetic energy of the piston. If the piston suddenly disappears air molecules will move into the cylinder. The average kinetic energy of the molecules will not change since they do not collide with a moving object, only with the stationary cylinder walls. While the air rushes into the cylinder their movement will not be completely random anymore but partly ordered. So one could say part of the heat energy of the air got transformed into energy of motion. However that does not mean that the kinetic energy of the molecules changed. It just means that ordered motion doesn't count as heat. After the cylinder has filled with air the motion of the molecules will be completely random again and the temperature of the air will be the same as it was before the piston disappeared. So in short, you didn't loose or gained any energy here. The amount of energy in the system is the same before and after the experiment. Heat energy just got turned into motion temporarily and then immediatly back into heat.
There is a difference depending if the air performs work on an external object, such as the piston, or if the air experiences free expansion, without performing any work on an external object. Wiki article: http://en.wikipedia.org/wiki/Joule-Thomson_effect
That describes is a constant enthalpy process, where the gas does an expansion. This scenario of this post is not an expansion, but the filling up of a container with gas to an equal pressure as outside the container. Enthalpy of gas when outside container = Internal energy of gas when inside container. See http://web.mit.edu/16.unified/www/SPRING/thermodynamics/notes/node16.html
thanks for everyone's answers. I posted this because it has a direct bearing on a project i am undertaking, and the direction of the project depends on the answer. I have my own theory but I am not sure of it, hence post. Regarding any heating occurring locally, From experience, filling containers with compressed air, from a large tank of compressed air does not heat the container noticably. And if it did, to fill a container to 150 psi or 10 atmospheres, putting 1000 liters in a 100 liter tank would certainly make the container very hot. Why? 10,000 joules is a lot. Specifically, 100 joules will heat 1 liter of air 120 C, which added to say 30 ambient is 50 C above the boiling point. If all that energy went into the air, it would be scalding hot. Consider doing this under a pool of water. Expanding raises the pool, and does work equal to the weight risen. Pool does not get hotter, it gets higher, even if that's practically immeasurable, as in the ocean. Step 2, must behave almost the same with or without the piston, because water is incompressible. The weight drops, impact is at bottom of cylinder, sound and heat result from release kinetic energy, focused at the bottom where the impact occurs. Air (gas) is not a true fluid, by which i mean liquids behave like fluids all the way down to the molecular scale. (ok chunky fluids there). Air is a cloud of particles bouncing around a total vacuum on ballistic trajectories at the speed of sound. There is no mechanism to focus energy from a distance, as there is in water, via incomprehensibility. A big person is also about 100 liters. Every step you take displaces 100 liters of air, but does not take 10,000 joules of work. The front of your body doesn't do "work" on the air, and air does not do work on the back of your body, because your front doesn't boil and your back freeze, every step. Something else is happening, enthalpy. Flow happens around you with negligible work. So, what I think happens in case B, is that upon opening, the air immediately by the opening expands, cooling, creates a net momentum into the container (momentum of not more than 100 liters of air or 140 grams) and that that net momentum is halted upon equalization, bringing the temperature of that air back to ...? What? its starting point? Or the amount of heat generated by 140 grams dropping? Not sure. 140 grams of air must also enter the room, the house, the neighborhood, the city. The weight still falls in case A and case B. So where did the energy go in case B, if the container does not become scalding hot? I believe the 100 pound weight drop in the second case speeds up (heats) the atmosphere all the way to the top of the atmosphere "pool", an imperceptible amount. If so, why does not case A also heat the air? case A allows for no expansion/re-compression. Case B requires localized expansion/re-compression, in a wave propagating outward. Case A forces air to behave as a constant volume fluid, case B allows it to behave as a cloud of independent particles. And I'm not sure I'm right, as i said. So I wanted other opinions on where the heat went (goes). Thank you all for your thoughtful answers. It gives me more to think about. I did not want to bias answer by posting my theory, but there it is and further comments welcome.
for everyone's information, the mit link predicts a 200 C rise. 1 liter of Air rises 30 C for each 25 Joules, so mit thinks 166 joules per liter rise. Its worth noting that's more than the energy it took to evacuate the tank, 100 joules per liter, so respectfully, not plausible. I don't have a vacuum pump, but do have a 6 gal (23 liter) 7 kilogram steel tank. Measured temperature with infrared thermometer. Added about 6.7 atmospheres, a few seconds, was surprised to see a rise of almost 2 C. 7000 grams of steel times .45 specific heat yields 6300 Joules. Divided by 23 liters and by 6.7 atmospheres yields 40 joules per unit air liter (at 1 atmosphere). Input pressure was regulated, so valid model of constant pressure. Back calculating that would be a 54 C rise. All measurements within say 20%, so lot of room for error. Also, the steel was losing heat to the atmosphere, but not much i think, as it was less than 2 C average difference on the room air. Worst case measurement errors would not get to 200 C rise or 166 Joules per liter. Also, there were 23 liters of air to start with, so thermal mass 23 * 1.4 grams is lost in thermal mass of tank. I claim that if anything, having air already in the tank would increase work of adding more air due to back pressure. Certainly not exactly the same experiment, but I would expect very similar results from adding 8 units of air or adding 7 units of air to 1 already there. Feel free to explain why that is wrong. So, I was surprised at 40 joules per liter, but its much less than either the full atmospheric pressure x 1 liter prediction and less by a lot than mit's. From first law, the heat in the air before entering the void is all the initial heat energy in the system. The only other energy in the system is the momentum of 1.4 grams per liter entering, which is not insignificant, since its travelling a sizable fraction of the speed of sound. Rather all the molecules are effectively traveling at the speed of sound, but spread out over a half sphere in direction, so net motion in, at average speed of ? half the speed of sound? I'd love to hear from anyone able to run an accurate experiment, or anyone from mit who can defend there calculations in the face of experimental data.
Mit link above points to a page which at the bottom, 2.3.3, describes exactly my question. Not sure 1 point was clear re: Mit (link above) problem labeled 2.3.3 - One can create a liter of vacuum inside the atmosphere with 100 joules of work pulling. If opening that up created 166 joules (per mit, 200 C) you could do that with 100 liters ten times a second and create 66000 or 66 kilowatts net surplus energy, and run an electric generator on that, sucking power from nowhere. Enough to power your home and car. Good work MIT! :-) really, they probably just used the constant for helium instead of air, but to be fair they do say air. Also, they ignore kinetic energy saying "initial and final" kinetic energy are zero, so its negligible to the end result. Initial and final energy of a stopped car, which speeds up, then crashes into a cliff, are also 0. So I'm not following that logic. Crashed cars are still hotter than stopped cars. Air definitely turns to really fast wind in the middle here. -------- The statement of the problem says work is done on the system, but the system is initially vacuum. If the expansion were done against constant pressure, the given formula Po*Vo would hold. You cant do work on a vacuum, nothing to collect the energy. Whatever the right formula is, the initial work is 0, Po*Vo represents a rectangle of work on a pressure volume plot. Does not fit the problem. Also, as stated in the problem, the gas is doing work on the system (vacuum), so would be getting colder not hotter. By this problem solution, gas expanding into the vacuum of space still grows 200 C, as the solution is not dependent on dimensions, just flow.
calculating an upper bound on heat inside the volume. To summarize, the problem is what will happen to the heat generated by allowing the atmosphere to fill a container initially holding vacuum. The proposed solution by MIT is that the answer is that work equal to Po*Vo is added to the container. There are 2 errors in the solution and a more general error in the sense that treating this like a closed system suspended in mid air with no container is at best approximate. From this they arrive at a rise of 200 C, computed by "gamma". However, they apparently do not use the correct value of Gamma for air, which is 1.4. Assuming ambient is 300 Kelvin, the final value of the temperature would be 420 kelvin, or a 120 C rise. The same as you get from adding 100 Joules to a liter of air, 120 C rise. They draw an imaginary line or balloon around a volume of air equal to the volume of the container, and assume adiabatic compression of this air into the container. This is not correct, since both heat and air can freely pass the imaginary line (meaning it is not adiabatic compression), but its not a bad upper bound. They also assume the volume of air outside will equal the volume of air inside. That is, once the container opens, the initial volume of the system is 2*Vo, initial pressure 1/2Po, and final volume is v0, final temperature X*To where X is greater than 1. This is very incorrect. As the problem is stated, the initial volume of the imaginary balloon is unknown. Although since compression heating is happening, it must be smaller than Vo, because the temperature will increase. So taking po as 1, the final density must be 1/X. And since initially the temperature was also 1 on a relative scale, the initial unknown volume was also 1/X. Let initial temperature, volume, density, and pressure all equal 1 (relative scale) for clarity. Stating the problem in terms of Vu for Volume unknown we know: Initial volume = 1+Vu Final volume = 1 Volume Ratio = 1/(1+Vu) Beta = gamma -1 = 1.4-1 = .4, the exponent of the energy curve, from integrating the pressure curve. Heat Ratio Hr= (1/1+Vu)^-0.4, simplifying = (1+Vu)^0.4 Initial density = Vu/(1+Vu) Final density = Vu Initial Temperature = 1 Final temperature = 1/Vu finally, ending temperature is both 1/Vu and 1*Hr or (1+Vu)^0.4: 1/Vu = (1+Vu)^0.4 See graph for solution, Vu=.792, Temp final = 1.26*T initial or 78 C rise for 300 Kelvin initial. I mentioned this was an approximation, essentially it assumes there is a weightless, perfectly insulating balloon surrounding the starting point. But since its imaginary, this depends on an imaginary balloon with 1 atmospheric pressure on one side and 0.6-ish atmosphere on the other. Also, while the total volume is collapsing at the speed of sound, heat is also escaping at the speed of sound. Also, realistically initial pressure outside the imaginary line must be less than 1,and so some of the work done is done on air outside the imaginary line. The "solution" system cannot be accurately analyzed as a real system, as it is closed to neither heat nor mass. The 78 C rise corresponds to a 65 joule per liter energy. So this model says 35 joules of every 100 does not reenter the volume. The very crude experiment "measured" 40 Joules, or a 54 C rise. So I'm actually pleased at that. A non-adiabatic compression would yield lower temperatures. The experiment shows your mileage may vary, which I expect is also true if the opening were varied from a pinhole to a full square meter. Still not sure how to accurately calculate this, but I finally have a model and measurements that are predicting similar quantities of heat inside the volume. And for all the carping about MIT's inaccurate model, I must thank them for helping me get a handle on a reasonably accurate method to calculate at least an upper bound.