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Thermodynamics (piston)

  1. Jul 24, 2015 #1
    By the formula PV/T = PV/T,
    If there is a piston, and it compresses, so the volume gets smaller, the temperature will go down too..

    But, I think if the volume gets smaller, the space is smaller, so the gas particle will collide more frequently, which makes the temperature goes up.
    And, my logic breaks what formula says.
    I really have no idea what actually V/T = V/T says.

    And, I want to know what pressure really is on gas particle.
    Is it the force given by a gas particle to another per the gas particle area? Or, is it the pressure given by gas particles to the piston ??
    If the piston compresses, will the pressure also get bigger?
    Last edited: Jul 24, 2015
  2. jcsd
  3. Jul 24, 2015 #2
    The actual formula is PV = nRT, don't mind nR it's just a constant, so PV/T doesn't change too and you already know that, but for just re-arrange the Eq P/T = nR/V, as the volume gets smaller the P/T ratio does up, this know depend on the kind of the piston and the process (isothermal process thus no temperature change or isobaric which means to pressure change), but I'll assume that you have an everyday piston and an everyday use, then It's temperature goes up and so does the pressure, Pressure is force per unit m*m, if you take an element of the surface of your piston and calculate the force acting (Impulses) you'll end up getting the pressure, as the volume compress collision happens quite frequent thus pressure goes up, the kinetic energy of the gas particles rises too, Good lucl !
  4. Jul 24, 2015 #3


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    Focusing first on this: really? Suppose T is held constant via equilibrium with a large body of fluid, while the piston is slowly pressed. What does that imply if V is cut in half?
  5. Jul 25, 2015 #4
    But, I'm a bit confused now.
    PV = nRT
    The piston is pushed down -> volume of gas gets smaller
    So, of course that the pressure will get bigger.
    But, the temperature??? It'll stay the same, right ?
    Because volume gets smaller, but the pressure gets bigger, and I think the moles stay the same too.

    Of course the pressure will be doubled.
  6. Jul 25, 2015 #5


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    Ok, good so far. I think you understand the case of constant temperature. The other extreme is no heat transfer at all. Then, temperature increases and pressure increases more than the constant temperature case. The idealized form of this is isentropic (reversible) compression. See the discussion here:

  7. Jul 25, 2015 #6
    Just wanted to add this is known as an adiabatic process (no heat transfer).
  8. Jul 25, 2015 #7
    Adiabatic = no heat transfer
    Isotermic = the temperature stays the same

    What's the difference ?
  9. Jul 25, 2015 #8


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    They are very different. To maintain temperature while a piston is pushed, heat must be transferred out of the system. With an adiabetic process, there is no heat transfer, and temperature increases as the piston is pushed (due to work done on the gas by whatever is pushing the piston). The link I gave before shows how to compute the temperature increase (under idealized assumptions) for a given volume reduction for an adiabetic=isentropic process.
  10. Jul 25, 2015 #9


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    Better not drive a car, motorcycle, or use a gasoline powered mower until you figure this out. It might fool you and refuse to run until you know what you want to have happen inside the engine.
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