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Compression due to temperature rise vs volume reduction

  1. Oct 20, 2015 #1
    If I have 2 sealed tanks A and B of 1 cu ft volume that contain air at atmospheric pressure and are at 300 deg K. temp absolute.

    For tank A I raise the temperature from 300K to 600K and I suppose the pressure will double due to this and would be 14.7 * 2 = 29.4psi absolute

    For tank B I use a piston and compress the air in it to 29.4 psi. I don't really know how to calculate the heat of compression in this case.

    Now If I allow both tanks to cool to ambient.
    Would the pressure in both tanks reduce to atmospheric pressure?

    What is the fundamental difference between the pressure increase due to temperature and and pressure increase due to mechanical compression?
     
  2. jcsd
  3. Oct 20, 2015 #2

    BvU

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    Tank A does return to same pressure.
    Tank B has double pressure.
    See ideal gas law.

    In tank B the compression process "can" be done reversibly and then it follows the ideal gas law. The gas gets twice as hot from the work you do on the piston.
     
  4. Oct 20, 2015 #3
    1. Why would Tank B NOT return to the original pressure of 14.7 psi?
    2. By reversible compression did you mean isothermal compression?
     
  5. Oct 20, 2015 #4
    This is not correct. For adiabatic reversible 2x compression of an ideal gas, the temperature ratio is ##2^{\gamma -1}##, where ##\gamma## is the ratio of the constant pressure- and constant volume heat capacities.
     
  6. Oct 20, 2015 #5
    Are you currently taking a course in physical chemistry or thermodynamics? If so, you are probably getting ahead of your current course material. Material like this will be covered later. For now, you are probably learning about the ideal gas law, correct? My advice: just use that for the present to answer the questions that it can address on its own, and wait until later to see how it is applied to more complicated problems.

    Chet
     
  7. Oct 20, 2015 #6
    Thanks for your explanations
    But I really do need to understand this topic, could you kindly explain the true difference between pressure rise due to temperature and pressure rise to do mechanical compression. Even a clue or a reference would do. I would refer to it and then get back if necessary.
    Many thanks.
     
  8. Oct 20, 2015 #7
    BvU gave you a good answer at one level, but apparently that was beyond where you are now. Please tell us what level you are approaching this from. For example, are you familiar with the first law of thermodynamics? Do you know what internal energy is? You apparently have never heard the term reversible process. Are you familiar with the term adiabatic process? We need to know this so that we can properly answer you at a level you can understand.

    Chet
     
  9. Oct 20, 2015 #8
    what does mechanical compression do to volume?
     
  10. Oct 20, 2015 #9
    it reduces it?
     
  11. Oct 20, 2015 #10
    In a container the rapid motion and collisions of molecules with the walls of the container causes pressure (force on a unit area).Where the motion of molecules depend on temperature.So, the more you rise temperature the more pressure rises.
    Mechanical comparison is more simple than that comparison of temperature.Its just an equalization process. When you blow air into a balloon, the balloon expands because the pressure of air molecules is greater on the inside of the balloon than the outside. Pressure is a property which determines the direction in which mass flows. If the balloon is released, the air moves from a region of high pressure to a region of low pressure and the balloon deflates.
     
  12. Oct 20, 2015 #11

    BvU

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    My mistake: I did indeed mean reversible isothermal (it's easier on the math) and then the work you do is heat given off to the surroundings. So the gas doesn't get twice as hot.

    To be honest, I completely oversaw what Chet points out.
     
  13. Oct 20, 2015 #12
    Right. So, do you know the ideal gas law, and what it says about volume, pressure, and temperature?
     
  14. Oct 20, 2015 #13

    russ_watters

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    So how can a vessel with a permanently reduced volume return to its starting conditions?
     
  15. Oct 20, 2015 #14
    I am familier with these terms. You could freely use them in your answers.
     
  16. Oct 20, 2015 #15
    PV = nRT
    In mechanical compression
    V would reduce
    P would go up
    T would go up
     
  17. Oct 20, 2015 #16
    If a vessel is permanently reduced volume that is if it is a rigid container. I suppose it would never return to starting conditions.
    Are you referring to Tank A the one that we are giving heat to?
     
  18. Oct 20, 2015 #17
    OK. So, if you are familiar with all that, please describe in words (qualitatively) what happens in the adiabatic compression of a gas in terms of the heat added to the system, the change in internal energy, and the work done by the system on the surroundings. Then write for us an equation that applies the first law to the adiabatic compression of a gas.

    Chet
     
  19. Oct 20, 2015 #18

    ogg

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    Escape. In school we usually start with the various gas laws PV=k, P/T = k, P=nk, etc. and come to the Ideal Gas Law PV=nRT. Why? because it is so useful and general.
    Its often good for REAL gasses to within a couple of percent (at STP or higher temps and/or lower pressures) and even the worst cases are generally only "off" by 5 or 10% (away from the material's critical points). But it assumes (if you derive it from "fundamental" principles (ie statistical mechanics)) that the gas is composed of point particles which do not interact. Well, how many Angels can dance on the head of a pin? I mean, how many moles, n, of a gas composed of infinitely small point-particles can you add to a tank? Yeah, as many as you want. And according to a beginner's application of the IGL, what you'll change is P (although it doesn't forbid a change in T or in both P & T). But changing T is trickier. There is NOTHING in the IGL which tells you anything about what happens to the other variables when you change one of the FOUR variables. If the problem is expressed carefully, the variables which are independent are declared and the variables which are held constant are also mentioned. In your cases, in tank A, by assumption n and V are held constant and T is increased by 300°. In your tank B, by assumption, n is held constant but now you have two dependent variables T and P and one independent variable, V. It takes more assumptions to conclude that T doesn't drop to 150°K! (which would keep P constant as V is halved). Right? In thermodynamics, which this question is about (mostly), one of the first things you learn about is systems. Open, closed and Isolated are the three most often taught at the elementary level. So in tank A, you add heat and nothing else and then allow heat to escape and then ask what the difference will be. In tank B, you change the volume of the system, allow heat to escape (T to return to 300°) and ask what the difference will be. Its an easy IGL question. for nRT to be constant, then if you have halved V you must double P otherwise PV will not equal P'V'. If you think about it in more pragmatic terms, stepping on a 12 inch piston so that it compresses to 6 inches will NOT!! heat the contents by 300°. The actual energy used depends on how the molecules (or atoms if its full of He or something similar)
    will heat up as they are crowded...NOT something point-like particles need be concerned with.) (In other words, the fact that a gas changes T as it expands or is compressed is PROOF that atoms (molecules, actually) are not point-like and/or do interact.) Elementary thermodynamics deals with situations where gravity, the weak force, strong force, and even most electromagnetic radiation can be (and is) ignored. (Just picture a tank 1000km high, think pressure will be uniform?? (and how about Temperature? Will T EVER become uniform????!?!(hint: velocity is a measure of kinetic energy = molecular heat, and escape velocity is the velocity at which gravity is "overcome"...). Further information on how T of a gas increases with energy (work, heat) can be found under Heat Capacity in Wikipedia, but reading through it requires a bit of partial differentiation (of simple multivariate equations, f(x,y,z)) which may be something you've not learned yet...(check out the section on diatomic gasses).
     
  20. Oct 20, 2015 #19

    ogg

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    So, in Tank A your only change from start to finish is ....nothing (since heat escaped and T goes back to 300°K). In tank B that's not true, you've done work (not handled at all with the IGL!) gotten a T rise then allowed heat (energy) to escape and are left with ½V. Thermodynamics assumes you don't need to know about HOW a process is carried out, that the thermodynamic quantities (these are only a subset of all the variables of a real world system, but a very very useful subset) are ONLY dependent on the initial (starting) state and the final state of the system...(the path to get from initial to final state can vary infinitely). P,V,T and n (by assumption) are thermodynamic variables (when system is "at equilibrium", what ever that means). So, to answer your question about the state of the final system in tank B, if you have the same temperature as initially, the same number of molecules, n, and ½V then P' must be 2P. Worrying about how hot it gets WHILE the system is changing is another story - one involving heat, energy conservation, heat capacities, and doing work on the system. All this goes away, if we just worry about the initial and final state (and don't worry about how we got there). The critical question is: do we have to worry about how we got there? The answer is: if we're worried about the final pressure, no. But if we're worried about how much heat is given off, or how much work was done on the gas, then yes.
     
  21. Oct 20, 2015 #20
    The is not correct. Even the temperature of an ideal gas (comprised of point particles) will increase when the gas is compressed adiabatically. Collision of the point particles with the moving piston causes the average kinetic energy of the particles to increase. The piston is doing work on the gas. When we calculate the work done on an ideal gas, we do not take into account the effect of the "crowding" of the molecules, yet we still obtain an increase in the internal energy and temperature of the gas.
     
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