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Adiabatic piston-cylinder problems always constant temperature?

  1. Sep 5, 2011 #1
    Okay I think I get this but I just want to make sure:

    In a frictionless, fully insulated piston-cylinder arrangement, when an external force is applied to the piston and the gas inside the cylinder is compressed, the temperature of the gas remains constant, and none of the work done to the system (system being defined as the gas inside the cylinder) is converted to heat. Therefore the pressure of the gas can be calculated using the ideal gas law PV=nRT assuming T does not change, or by PV^k=constant for some k which is specific to the gas in the cylinder, (and is equal to the specific heat ratio cv/cp).

    The part about the temperature of the gas not changing had me hung up for a bit, because in the refrigeration cycle you can change the temperature of the refrigerant by altering the pressure. However, this is due entirely to the heat transfer with the surroundings: if the refrigerant is compressed in a cylinder, there are more molecular collisions with the cylinder walls, transferring heat out of the system, and lowering the temperature of the refrigerant. If the cylinder were completely insulated, heat transfer would not occur and the refrigerant would not change temperature.

    Right?
     
  2. jcsd
  3. Sep 6, 2011 #2

    Drakkith

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    As far as I understand, compressing a gas causes an increase in temperature, and expanding a gas causes a drop in temperature.
     
  4. Sep 6, 2011 #3
    Excellent! Let's get the discussion going.

    According to the ideal gas law PV=nRT, where R is always a constant and n is a constant in this case (we have a fixed quantity of gas), when V goes down, T also goes down and/or P goes up. For example if we let P,V, and T represent the initial state and p,v, and t represent the final state, and we reduce the volume by 1/2 so that V=2v, we have
    V = nRT/P = 2v = 2nRt/p --> nRT/P = nRt/p (nR cancels) -->
    T/P = 2t/p

    But there are an infinite number of values that fit this equation:
    T=2t, P=p
    T=3t, 2P=3p
    T=t, P=(1/2)p
    and so on...

    My question is: which of these is the case? I think the answer is that, if the cylinder is fully insulated and no heat escapes, T remains constant while P varies. If some heat does escape that would increase the temperature of the surroundings and decrease the temperature of the gas. I don't fully understand what happens though, which is why I'm asking the question.
     
  5. Sep 7, 2011 #4

    cjl

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    Unfortunately, you are wrong. In the case that no heat escapes, the situation is such that PVgamma is constant (where gamma is dependent on the gas species). It's not exactly intuitive, but it just so happens that the temperature will increase as it is compressed, and similarly, the temperature will decrease as it is expanded.
     
  6. Sep 7, 2011 #5

    Drakkith

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    I don't think you are using the ideal gas law correctly. The gas in your example WILL increase in temperature. If you were to compress the gas very slowly and not have it perfectly insulated then the heat would be absorbed by the rest of the cylinder and engine over time and the gas wouldn't heat up very much.

    Check here: http://en.wikipedia.org/wiki/Adiabatic_process
     
  7. Sep 7, 2011 #6
    Thanks for helping me with this guys. I think I'm starting to get it.

    There are two equations that explain what happens, not just one as I had assumed:

    (1) PVK = C
    (2) PV=nRT

    So as the volume deceases, the pressure increases according to equation (1). That value for P is then used in equation (2) to solve for T.

    If you set the 2 equations equal to each other by solving both for P you get

    C/VK = nRT/V

    C = VK-1nRT

    T = C/(VK-1nR)

    The special case is in the ideal gas model where k=1, so
    VK-1 = V0 = 1 and we have

    T = C/(nR) = constant

    In all other cases (all real processes) the temperature of the gas increases by

    TF - TI = (C/R)[(1/VFK-1) - (1/VIK-1)]

    Since VF < VI, it follows that 1/VFK-1 > 1/VIK-1, so TF - TI > 0, so the temperature increases.
     
  8. Sep 7, 2011 #7
    At the molecular level, I guess this could be explained by collisions between molecules within the gas: when this happens some of the translational energy (kinetic energy of the molecule moving through space) is converted to vibrational or rotational energy (kinetic energy of how the molecule vibrates and rotates; affects temperature). As volume decreases, the frequency of these collisions increases. In the ideal gas model gas molecules are point masses and thus do not collide, therefore this does not happen so temperature remains constant. In real gasses it does happen, which causes an increase in temperature.
     
  9. Sep 7, 2011 #8

    Andrew Mason

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    It has nothing to do with intermolecular collisions. The temperature of an ideal gas increases as it is compressed. This is just conservation of energy - the first law of thermodynamics. If you do work on the gas but no heat flows into or out of the gas, the internal energy must increase, which means temperature increases.

    As the walls of the container move in to confine the gas to a smaller space, the kinetic energy of the gas increases. The molecules of the gas rebound with greater energy from a moving wall than from a stationary wall. That happens regardless of how fast the walls move because the slower it moves the more molecules there are that hit the moving wall. The same total energy is imparted to the molecules regardless of the speed of compression.

    AM
     
  10. Sep 7, 2011 #9
    So for an ideal gas, how do you know what the temperature is after compression?
     
  11. Sep 7, 2011 #10
    Okay so I think what I had screwed up was thinking that K=1 for an ideal gas in PVK=C. I was thinking of the compressibility factor Z, (Z=1 for an ideal gas). K never equals 1. So now I think that what I said before (quoted here) is right except for the part about the "special case" of an ideal gas. Anyone agree?
     
  12. Sep 7, 2011 #11

    sophiecentaur

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    Why should the temperature increase?
    Imagine a piston in a cylinder. As the piston moves upwards, the molecules that hit it will bounce off with more energy than they arrived. That corresponds to an increase in temperature.
     
  13. Sep 7, 2011 #12
    That makes sense, and helps me understand it on a molecular level, thanks. Do you think I am on the right track with how to calculate the increase in temperature?
     
  14. Sep 7, 2011 #13

    Andrew Mason

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    I agree. Your expression of temperature is correct. [itex]K = \gamma = C_p/C_v > 1[/itex]. From the first law the heat capacity at constant pressure must always greater than the heat capacity at constant volume because no work is being done at constant volume.

    AM
     
  15. Sep 7, 2011 #14
    I see. If X amount of energy enters a gas through heat, under constant pressure the gas will expand and some of that energy will be converted to work. If instead you are under constant volume, it must all be going to internal energy and therefore the temperature will increase more in the constant volume case. Therefore the specific heat (Q/dT) is lower. So k = Cp/Cv is always greater than one.

    Thanks, AM
     
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