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FYI: This is not a homework q; I am just curious... You can assume that...

  • the syringe tip cap creates a "perfect" seal, i.e., keeps air from flowing in/out!
  • the syringe is completely depressed before starting the experiment.
  • the outside barrel has metric ruler.
  • we have access to accurate temperature & atmospheric pressure readings.
  • the syringe is hung in the following way...

enter image description here

Is the following equation correct for determining the mass of the weight, based on how far it causes the plunger to depress...

$$ m = \frac{A\cdot B\cdot x}{a\cdot L} $$

where...

  • A = Surface area of the (black) stopper face.
  • B = Bulk modulus of air given temperature & atmospheric pressure readings.
  • x = Plunger displacement (due to suspending weight) after waiting for a "long" time.
  • a = Acceleration due to gravity.
  • L = Length of syringe barrel.
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Yes you are basically right, and I will add a historical note. When Robert Boyle (1627-1691) did his experiments which led to what we now call Boyle's law, he compared air to a spring and wrote of "the spring of the air", very much in the same spirit as you are doing here. He titled his book "New Experiments Physico-Mechanicall, Touching the Spring of the Air, and its Effects"; it was an important contribution to the development of experimental physics. His approach was along similar lines to yours---thinking through in precise mechanical terms the consequence of a change in volume of a given amount of air. There is one slight drawback in the use of an ordinary syringe, in that they normally have a lot of friction. The experiment would work better with a metal piston and cylinder, with o-rings and grease a bit like those used in a car engine.

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  • $\begingroup$ Yes, I thought about the friction... The syringe manufacturers could VERY accurately measure the force ($= F_{\text{tc off}}$) required to overcome the friction between stopper & barrel (i.e., w/ the tip cap off) engraving the outcome; then my equation would look like this... $$ 0 = F_{\text{tc on}} + F_m \Longrightarrow 0 = \left(\frac{A\cdot B\cdot x}{L} + F_{\text{tc off}}\right) + \left(-m\cdot a\right) \Longrightarrow m = \frac{A\cdot B\cdot x}{a\cdot L} + \frac{F_{\text{tc off}}}{a} $$ Does it still look correct, after I've made the update? $\endgroup$
    – Landon
    Commented Jun 2, 2019 at 23:42

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