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I need to balance forces on the "plate" of a piston situated in a closed container of a given spring constant k and area A with a compression x. It is also given that the initial pressure inside the container is P°

Thus PA = kx

Now here is my question, would P be the value of the net final pressure inside the container or would it be the change in pressure relative to initial pressure ? I think it should be the former but i've been told otherwise and can't wrap my brain around as to why is it so.

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Initially, the spring is uncompressed. The force on the piston due to the internal pressure of the container is balanced by the force on the piston due to the external pressure of the atmosphere.

Once the spring is compressed, there are three forces acting on the piston:

$F_1=$ The outward force $kx$ due to the spring

$F_2=$ The inward force $P_0A$ due to atmospheric pressure

$F_3=$ The outward force $P'A$ due to pressure inside the container

The force balance gives us

$$kx = (P_0-P')A$$

Since $P_0$ was the initial pressure inside the container, the pressure $P=P_0-P'$ is the decrease in pressure inside the container relative to the initial pressure, not the final pressure inside the container.

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  • $\begingroup$ what if the restoring force is pushing up the gas initially rather than the gas pushing the spring down then the net force we give will be +ve and we would get P+P' to be pressure and hence net pressure is what prevails , if thats not the case "the net pressure is P-P'" which leads to the answer being concluded as change in pressure(i first thought this was the answer) but falls down as net in the other case $\endgroup$
    – Naveen V
    Mar 5 at 12:01
  • $\begingroup$ also the outward force wouldnt make the plate to move it just makes the air molecules to move outwards ( due to nlm 3rd law) hence it is not taken into consideration , the external force we give is what is taken into consideration $\endgroup$
    – Naveen V
    Mar 5 at 12:05
  • $\begingroup$ What does x represent on the OP's diagram? $\endgroup$
    – Chet Miller
    Mar 5 at 13:43
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I finally understood your question , initially i was wrong in the comments so i deleted them.

$PA = kx$

We have a piston which is movable already experiencing an initial pressure $P^{\circ}$ , if the system is in equilibrium then we can use this pressure to be the $P$ in the equation

If it is not in equilibrium then we give an external force for making it in equilibrium.

If the piston is moving up(in non equilibrium state) and then we are giving a force $PA$ in the direction of the original pressure force by gas

Else its moving down we give the force -PA to make it equal so that it reaches original position enter image description here

Hence it would be $$PA+P^{\circ}A=kx$$ or $$P^{\circ}A-PA=kx$$ (if the gas is pushing down we give force opposite to it hence- sign) depending on the situation

So I believe it must be the net pressure on the plate that should matter

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This assumes that there is vacuum on the spring side of the piston. Let x represent the distance of the piston from the closed end of the cylinder (x=V/A) and let $x_0$ represent the value of x when the spring is not extended. Then the force balance on the piston (assuming negligible piston mass) reads: $$PA=k(x-x_0)=k\left(\frac{V}{A}-x_0\right)$$

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