Hello I had a question, related to a recent experiment we've done, that I have been thinking about for a while now. Hopefully someone can help as it has been bugging me! Consider two situations: A: a syringe is filled with 5ml of air and then sealed B: the same syringe but now filled with 10ml of air and then sealed. Why is it much harder to pull the plunger in case A? And if we were to pull each of the plungers down with the same force, why would plunger B move more?
Thanks!
Answers and Replies
Hello I had a question, related to a recent experiment we've done, that I have been thinking about for a while now. Hopefully someone can help as it has been bugging me! Consider two situations: A: a syringe is filled with 5ml of air and then sealed B: the same syringe but now filled with 10ml of air and then sealed. Why is it much harder to pull the plunger in case A? And if we were to pull each of the plungers down with the same force, why would plunger B move more?
Thanks!
What are your thoughts? Can you use the ideal gas law and a little math to try to figure it out?
Hi berkeman Thanks for your reply. Ok, not sure if this is right but... pV = NkT ⇒ p = NkT/V assuming NkT is constant, ⇒ δp = NkT ln(V)δV (where N is the total number of particles and k is Boltzmann's constant) ⇒ δV ∝ δp / Nln(V) Is this right?
so if we pull down the plunger with a certain force (cause a certain change in pressure), the resulting change in V should be inversely proportional to the number of particles and the log of the initial volume (which is an increasing function). If the initial volume of the syringe was higher, the number of particles would also be higher so Nln(V) would be higher. So, this shows that the bigger the initial volume, the less the plunger will move when it is pulled with the same force. But the opposite actually happens!
jbriggs444
It is not clear what is motivating the differential or why one would think that the first derivative of 1/x is ln x. Even that skips over the question of why the force required to withdraw the syringe would be proportional to the absolute pressure inside the syringe. Possibly a few steps back are in order.
If you draw a free body diagram for the piston in the syringe, what is the force required to withdraw the piston in terms of the pressures inside and outside the syringe?
Likes berkeman
The ideal gas law is PV=nrt. P is pressure in pascals. V is volume in cubic meters I believe. N is number of moles. R is gas constant, which is the boltzmanns constant times Avogadro's number. T is temp in kelvin. R is 8.314 roughly. Not sure why you are using log function.
Each unit you pull the plunger down is a smaller percentage of the larger volume so it requires less force to pull it down 1 unit for the larger volume than the smaller volume.
If you draw a free body diagram for the piston in the syringe, what is the force required to withdraw the piston in terms of the pressures inside and outside the syringe?
I wanted to understand why less force is required when pulling down the plunger if the initial volume of air inside the syringe is greater. That's why I wanted to work out what the incremental change in volume depends on. Can I ask if the following makes more sense: δ(pV) = δ(nRT) ⇒Vδp + pδV = 0 (assuming withdrawing syringe is an isothermal change) ⇒δV = - (V/p)δp = -(V^2/nRT)δp ⇒δV ∝ -V^2
It is not clear what is motivating the differential or why one would think that the first derivative of 1/x is ln x. Even that skips over the question of why the force required to withdraw the syringe would be proportional to the absolute pressure inside the syringe. Possibly a few steps back are in order.
Which shows that the change in volume when there is a change in pressure, depends on the initial volume, V - so the bigger the intial volume, the bigger the volume change, along the lines Austin Z W mentioned above (thanks Austin).
Nidum
It is often useful to consider limit states when exploring a problem that you don't initially understand . What if there was : (a) No air in the syringe ?
(b) A very large volume of air in the syringe ?
It is often useful to consider limit states when exploring a problem that you don't initially understand . What if there was : (a) No air in the syringe ?
(b) A very large volume of air in the syringe ?
(b)When there is a large volume of air, experience tells me that the plunger moves down effortlessly. I am not sure why though?
Because the gas inside has more molecules it will take up a larger space easier than a smaller number of molecules will.
Make like Einstein and do a thought experiment.
CWatters
I wanted to understand why less force is required when pulling down the plunger if the initial volume of air inside the syringe is greater.
You should really define what you mean by "pulling down the plunger". For example a syringe that is only a quarter full has more travel remaining than one that is half full. Are you comparing the force required to pull out the plunger? Increase the volume by a fixed amount (eg 2ml) ? or perhaps increase the volume by the same percentage increase?