Please Register or Log in to view the hidden image! I have a question - how is energy conserved in the above scenario?? It is simply a diagram of a weight being put onto a piston, and then taken off when that piston compresses to its maximum depth. Very simply, the piston starts with a set pressure, volume and temperature in illustration (1), descends to illustration (2), and when the weight is taken off the piston expands again to illustration (3). Between (1) and (2) the gas in the piston heats up. Given enough time it will reach maximum compression where the temperature of the piston equals the surroundings. Then, when the weight is taken off the piston rises between (2) and (3), absorbing heat and eventually reaching the same expansion level as in (1). The pistons in illustration (1) and (3) are identical - same pressure, volume and thus the same temperature. IE equal energy. But, the weight has in fact descended! It's gravitational* energy should be converted to heat somewhere in this system. So, energy of the piston is equal at the beginning and the end of the scenario - it seems like the piston absorbs the same amount of heat going from (2) to (3) as it emits going from (1) to (2). After all, the only resistance which is offered by the piston between (1) and (2) is because of compression pressure. To hammer the point in - ignoring the weight and just looking at the piston, (1) to (2) is just the reverse of (2) to (3), or same as (3) to (2). Piston (1) has the same amount of energy as piston (3), which is more than piston (2). Of course, more than piston (2) because going from (1) to (2) energy is released, and going from (2) to (3) energy is absorbed. I could go on and elaborate, but this is really all there is to it. How is energy conserved? Please Register or Log in to view the hidden image! :shrug: * it does not have to be gravitational potential, it could also be electrostatic potential
that can't be it because the rock starts stationary, and ends stationary - its whole potential energy is used to compress the piston, no kinetic energy. I think my mistake may be considering just the thermal energy of the piston rather than the energy of the entire system.. I'm just a novice when it comes to thermodynamics.
I can't see the picture but I think your description is good enough. There is a piston at equilibrium and you then you put a weight on the piston and it compresses the air in the cylinder. You have converted gravitational energy to the 'pressure' potential energy. As you noted the gas heats up due to the increase in pressure. That means as the gas cools it will contract and the piston will lower a bit more. This cooling means that the heat is lost. If the weight is removed the piston will raise back up. You will be converting the 'pressure' potential energy back to gravitational potential energy. The important point is that the piston will not attain the same height because of the energy loss in the form of the heat loss in the cyclinder. This delta height will occur even if the piston and cyclider are friction free. Does this answer your question?
Ignoring the temperature of the piston due to compression and decompression, it seems that the return to step 3 from step 2 would generate the same amount of heat that simply dropping the rock on the ground would. Think of the piston like a dampened, compressed spring...that resistance goes "somewhere".
As the gas expands it cools down. If the temperture is allowed to equalize with room temperature, the cylinder will again reach its beginning position. Air conditioners work on the principal that as a compressed gas expands it absorbs heat. It is then compressed and the heat generated is radiated, usually outside. The compressed gas returns to be expanded again and cooling process continues. This is the same for refrigerators and air conditioning.
Well OnlyMe it sounds like you are equating "equalizing with room temperature" with the cylinder returning to its starting position, while Origin was referring to the heat lost to the surroundings while the piston is compressed. I was speaking along the same lines as Origin, except my conclusion was not that the piston wouldn't reach its original height, but rather the heat added to the surroundings would be the same as if the rock were dropped directly on the ground.
Believe it or not, a gas can expand (or be compressed) without changing temperature. Temperature is just the average kinetic energy per particle, so what's to stop all the molecules having the same velocity when a gas expands, given that the number of collisions between particles should decrease? Why would that happen when a gas compresses, given that the number of collisions should increase?
The HEAT would remain the same, but the temperature would decrease as the gas expands. Temperature is a function of heat over volume.
If the system is isolated: 1) the internal potential energy remains constant over time. 2) the internal kinetic energy remains constant over time. therefore, the temperature remains constant over changes in volume (as long as this occurs 'adiabatically'). Temperature will change if the system exchanges energy with its environment. This is a direct consequence of Boyle's Law.
Not exactly what I was saying. Put the weight on a cylinder and the gas is compressed. As the gas is compressed it gets hot and given time and opportunity that heat may dissipate into the room. Remove the weight and the gas will expand, cooling as it does so. The cylinder will rise not quite as high as it had been initially, as the cool gas will take up less space than it did initially. Now if you wait, the excess heat which moved from the compressed gas into the room, will have an opportunity to move back into the now uncompressed gas. As the temperature of the gas in the cylinder again equalizes with the room air it will again take up more space than it did when cool.., the cylinder will return to its original height. All assumes that the room represents a closed system and no additional heat is added or lost from outside sources.
Using utterly technical terms, heat is the flow of thermal energy. IE a glass of water has no heat. Temperature is a function of average kinetic energy over volume. Yes, this is an isothermal process. However, it is difficult/impossible to achieve in real heat engines. That is a good question! I am wondering why an expanding gas has to cool, and why a compressed gas has to heat up. It seems to me that the amount of energy should stay equal - but it may have to do with the thermodynamics of the process. I think even the ideal gas laws predict this, when there is an infinitely small particle size. Yes, I think this has to be the case, I mean the compressed piston certainly has potential energy which can be used for work ... what is bothering me though is that the piston cannot lift the rock back up to its original position. Granted though, it cannot be a 100% efficient process.
One way to think about it is in terms of reversible and irreversible physical changes in an isolated system (of many particles). If the gas expands 'suddenly', chaotic motion ensues and Boyle's Law no longer holds. The statistics change 'unpredictably' and the system will have new values for temperature and pressure when the system returns to an equilibrium state. That is, when the particles once again have a Maxwell-Boltzmann distribution of energies.
It's not clear from the OP what "system" is supposed to conserve energy. Is the piston insulated so that no heat can escape or enter? If not, then what, exactly, is the closed system in this example? Is the air around the piston and mass included? Exactly which energy do you think should be conserved?
You're confusing heat with temperature. Adiabatic is no loss in heat, NOT temperature. If you have 1 cubic cm of gas at 50 degrees C, do you really expect it to remain at 50 degrees C after it expands to 1 cubic km?? That shouldn't pass the common-sense test!
I don't think so. Heat is a form of energy, temperature is a statistic. If the gas is a closed system, and it expands 'uniformly', then yes, it should remain at the same temperature. That is, if you do a statistical measurement of n particles before and after the (adiabatic) expansion, they will have the same distribution of energies. The details of how to measure the energies (i.e. take the temperature) of a collection of widely separated or closely spaced particles, are not relevant, even though that's what you need to do.
Arfa, what you're saying is that a pocket-full of warm air could heat the entire universe. It should strike you as absurd without much thought.
Now you appear to be confusing heat energy with temperature. Think of it as a distribution of n particles. If the volume increases and the particles become more widely spaced, the energy density decreases. But that only means it will take longer to measure enough particles to find a statistic, or 'take the temperature' of the n particles. If however the n particles expand rapidly, they should end up with widely separated energies, and it will take time for the system to reach a new equilibrium; however the total energy won't change (unless the system isn't closed). The same argument applies the other way too: if you rapidly compress n particles their energy distribution will not be uniform--part of the new volume might have a different temperature than another part. But the total energy is always conserved, even if it gets distributed non-uniformly by chaotic processes (like rapid expansion or compression in a closed system). This suggests that expanding or compressing a real gas cannot be a closed system, because it can affect the energy distribution (of n particles in equilibrium).
Arfa...temperature is not energy. Heat is energy, and it is heat which does not change in an adiabatic process. It's late on a Friday night and I don't have the motivation to explain this right now. Please spend 30 seconds Googling this before you respond...
That's correct. Have I said temperature is energy, somewhere? The total energy of a closed system (of n particles) doesn't change, that's also correct. The temperature doesn't change either, IF and only IF the expansion/compression (i.e. change in volume) is adiabatic. That means you will measure the SAME temperature in an expanded gas, even if the expansion is nearly infinite. It will just take a nearly infinite amount of time. But the time it takes to measure a temperature is not physically relevant to the statistics of energy distribution. Think about how long it might take to measure the kinetic energies of all the galaxies in the visible universe, as if they are n particles in an adiabatically expanded gas.
