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Thermodynamics problem

  1. Nov 8, 2012 #1
    - A cylinder with a gas inside moves with a velocity V. What would be the decrease in temperature caused by the sudden stop of the cylinder.

    My solution:
    Δ(kinetic energy) = Δε = mV²/2 = (3/2)nRΔT -> ΔT = MV²/3R

    But the book solution is different
    Δ(internal energy) = ΔU = mV²/2 = nℂvΔT -> ΔT = MV²/2ℂv

    He considered mV²/2 to be the variation of internal energy istead of kinetic energy. As far as I know, the formula mv²/2 is for ε, not U. Am I wrong?

    Which one is the correct solution?

    []'s
    John
     
  2. jcsd
  3. Nov 8, 2012 #2

    ehild

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    Hi John,


    3/2 nRT is the kinetic energy of the random translational motion of the molecules in the gas (n is the number of moles). In this problem, the whole cylinder moves with velocity V. When the cylinder stops, the energy of its translational motion (mV2/2 transfers to the random motion of the molecules, that is, to the internal energy of the gas. The internal energy of an ideal gas is nCvT.

    ehild
     
    Last edited: Nov 9, 2012
  4. Nov 9, 2012 #3

    Andrew Mason

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    Is this a monatomic ideal gas? If so, both solutions appear to be the same.

    It is not clear what you mean by Δε. This is just a first law problem: Q = ΔU + W (W = work done by the gas). In this case Q = 0 so the work done ON the gas = -W = ΔU.

    AM
     
    Last edited: Nov 9, 2012
  5. Nov 9, 2012 #4
    Δε is the variation of the kinetic energy
    The gas has not to be monoatomic, it is a random gas (mono, di or poliatomic).
     
  6. Nov 9, 2012 #5
    I did't understand yet. Why the energy is added to the internal and not kinetic energy?
     
  7. Nov 9, 2012 #6

    Andrew Mason

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    It is added to the internal energy because of the first law. Internal energy of the gas may consist of kinetic energies of translation, rotation, vibration plus potential energy. If it is an ideal gas, U will be entirely kinetic.

    In this case, there is no heat flow into or out of the gas. There is only work done on the gas (assuming the cylinder was to stop very suddenly) in stopping the translational motion of the centre of mass of the gas. According to the first law, this results in a change in internal energy of the gas that is equal in magnitude to the work done on the gas.

    The work done on the gas is equal to the kinetic energy of the centre of mass of the gas immediately before the cylinder stopped (mv^2/2). If the cylinder stops very quickly, the centre of mass of the gas will keep moving relative to the cylinder with the same kinetic energy. It will keep moving until the gas in the forward end of the cylinder compresses to a maximum. Then the gas will expand in the other direction until pressure builds up on the other end and the process continues as a damped oscillation until equilibrium is reached, at which time that mechanical energy (mv^2/2) dissipates and ends up as additional internal energy of the gas. So Δε = mv^2/2 = ΔU = nCvΔT

    AM
     
    Last edited: Nov 9, 2012
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