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Find the change in temperature during adiabatic compression

  1. Jan 18, 2016 #1

    Titan97

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    1. The problem statement, all variables and given/known data
    WP_20160118_18_15_15_Pro (2).jpg
    Untitled.png

    2. Relevant equations
    $$PV^{\gamma}=\mathrm{constant}$$

    3. The attempt at a solution
    Initially, the piston will get a velocity ##v##.

    But the chamber will move too because of the forces exerted by the gas.
    Since total external force is zero, velocity of centre of mass is constant.

    Initial velocity of centre of mass: $$\frac{mv+4m\times 0}{m+4m}=\frac{v}{5}$$

    Then I used energy conservation.
    with respect to earth, final velocity of both piston and chamber will be zero. (right?)
    $$\frac{1}{2}m{v}^2=nC_{\mathrm{V}}\Delta T$$
    I got $$\Delta T=\frac{mv^2}{3nR}$$.
    But the answer given is $$\Delta T=\frac{4mv^2}{15nR}$$.
     
  2. jcsd
  3. Jan 18, 2016 #2

    BvU

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    Two remarks (I haven't got a clue about the solution :frown:):
    Cylinder plus piston are ##5m## as I read it.
    Particle collides elastically - so it bounces back. Cylinder gets momentum and there is no friction, so in the final state it still is moving
     
  4. Jan 18, 2016 #3
    I got the wanted solution by calculating the velocity of the piston after the elastic collision, taking into account only the mass of the piston for the first contact - then you can calculate the velocity of piston, chamber and gas (index pcg) at the point of maximal compression (at the point, when piston and chamber proceed with the same speed).

    The final velocity of piston, chamber and gas won't be zero with respect to the ground. You have to take into account the kinetic energy of piston, chamber and gas (index pcg) in your formula.

    $$\frac{1}{2}m{v}^2=nC_{\mathrm{V}}\Delta T+\frac{1}{2}m_{pcg}{v_{pcg}}^2$$
     
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