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I was watching a video, and a question came to my mind. Suppose I have a cylindrical piston. I have a rock that rests on a roof, which keeps the gas underneath at a constant pressure & volume. That is, the force exerted on the roof by the gas is equivalent to the force exerted by the rock. There is no friction on the roof, and we're in a vacuum. Suppose I now cut a piece of that rock, being careful not to apply an additional force to the roof. Physics says the volume will increase until the force exerted on the roof by the gas = that force exerted by the rock. My question is, how quickly will the volume rise? Or, how quickly will the pressure fall?

Newton tells us F=ma. Acceleration of the roof will lead to an increase in the gas's volume, which leads to a decrease in the pressure, which should then lead to a decrease in the force exerted on the roof. So, the net force on the roof will go to zero, but there ought to be a net downward force of a magnitude > 0 to decelerate the roof and this force must be applied just long enough to stop the gas.

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Interesting question. As you state we need to apply Newton's law and find the equation of motion of the system. For simplicity we first assume that we only have gravity and the force due to the gas (we'll see later that we need friction to get the expected behavior of the system):

$F_\text{tot}=F_\text{grav}+F_\text{gas}=(m-\Delta m)a$

$(m-\Delta m)g+\Delta PA=(m-\Delta m)a$

where $g=-9.81$ m/s$^2$ is the acceleration due to gravity, $m$ the mass of your initial rock and piston, $\Delta m$ the part you cut of the rock, $\Delta P$ the difference in pressure in your cylinder and outside, $A$ the area of your piston, and $a$ the acceleration of the piston and rock. Since we assume that we work in vacuum $\Delta P=P=nRT/V$, where $P$ and $V=Ah$ are the pressure and volume of your cylinder and $h$ is the height of your cylinder. We can now write the equation of motion as

$\frac{d^2}{dt^2}h(t)=\frac{nRT}{(m-\Delta m)h(t)}+g$,

You can solve this equation numerically using Mathematica for instance. For $m=10$ kg, $V=1$ L, $h=1$ and $\Delta m=1$ kg you find

enter image description here

So as you can see the system behaves like an oscillator. But we would expect that the volume is increased by a factor of 10/9 so that $h=1.11$ m. The oscillations occur because there is no friction to stop the piston. We can introduce friction by adding a term to the equation of motion that is proportional to the velocity of the system, but in the opposite direction

$\frac{d^2}{dt^2}h(t)=\frac{nRT}{(m-\Delta m)h(t)}+g-c\frac{d}{dt}h(t)$,

where $c$ is the friction coeffcient. Taking $c=1$ kg/s and the same values as before we get after solving in mathematica

enter image description here

Which shows the expected behavior.

EDIT: after the comment of fibonatic I decided to include the effect of the adiabatic expansion. Although I left out the decrease in temperature initially for simplicity, it turns out that it acts as a friction force. When assuming an adiabatic behavior we rewrite the pressure as

$P'=P\left (\frac{V}{V'} \right )^\gamma=\frac{nRT}{Ah(t)}\left (\frac{h(t)}{h(t)+\tfrac{d}{dt}h(t)} \right )^\gamma$

Using $\gamma=1.4$ for air, $c=0$ kg/s (no damping force) and the other values as before, I get:

enter image description here

Here I have assumed that the cylinder is thermally isolated so that the temperature (and hence pressure) do not increase after the adiabatic expansion.

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  • $\begingroup$ So you are assuming that the temperature $T$ remains constant? I would think that a better approximation would be adiabatic expansion, such that the temperature does change, which makes the problem non-linear. $\endgroup$
    – fibonatic
    Dec 1, 2015 at 18:37
  • $\begingroup$ I did assume constant T, as I did not want to cmplicate matters too much. I should have mentioned it though. $\endgroup$
    – Paul
    Dec 1, 2015 at 19:00

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