Ideal Gas Law with spring, no numbers, closed container

In summary, the conversation discusses a closed cylinder with a tight-fitting but frictionless piston in equilibrium. The left chamber has pressure p0 and length L0, while the spring on the right is compressed by ΔL. The questions ask for an expression for ΔL in terms of p0, L0, A, M, and k, an expression for the net force on the piston when it is moved a small distance to the right, and an expression for the oscillation period assuming the piston is released and oscillates around the equilibrium position. The ideal gas law and simple harmonic motion equations are used to solve the problems. The final solution for the oscillation period is T = 2pi/√k/M.
  • #1
MrMaterial
33
0

Homework Statement


The closed cylinder of the figure has a tight-fitting but frictionless piston of mass M. the piston is in equilibrium when the left chamber has pressure p0 and length L0 while the spring on the right is compressed by ΔL.

a. What is ΔL in terms of p0, L0, A, M, and k?

b. Suppose the piston is moved a small distance x to the right. Find an expression for the net force on the piston. Assume all motions are slow enough for the gas to remain at the same temperature as its surroundings

c. If released, the piston will oscillate around the equilibrium position. Assuming x << L0 find an expression for the oscillation period T.

qdtHj.jpg



Homework Equations


for a.
Not quite sure, All this chapter in the book really taught me was the Ideal Gas Law pV = nRT

Given that the question is concerned that I include M I suppose they want me to use kinematics? I don't know how M (mass of the piston) affects the gasses, or ΔL.

I know pressure = Force/Area, and Force = Mass X Acceleration -k X Δx since it's a spring

for b.
force of spring = -kΔx
p1 = nRT/V1
V=A*L0
p1*v1 = p2*v2

for c.
not sure

The Attempt at a Solution



Note: spring is being compressed, p=0 on other side of piston

first off, V1 = A*L0

then p1 = (-kΔL)/A (I'm lead to believe p = 0 when ΔL = 0

I need to fit in M somewhere to fulfill a.

p1*A = Fsp = -kΔL or, since it's a Force, M*acceleration.

need k, cross out F=Ma approach

what does M effect?
  • not volume
  • not temperature
  • movement
  • pressure?
  • Fsp?

I wonder... Potential spring energy is Usp = 1/2k * Δx^2...

Kinetic energy is 1/2m * v^2

could I turn the isothermal (p1v1 = p2v2) into kinematics? energy?

forces to the right are sourced from p1 and forces to the left are sourced from the spring

p1 = 1/V1

V1 = A*L0... but plus ΔL if it ever gets larger or smaller

p1 = 1/(A*(L0+ΔL))

equilibrium let's p1 also equal Fsp/A

-k*ΔL/A = 1/(A*(L0+ΔL))

I think I am stuck here

I need some clues, or hints as to how I should approach this. I feel that once I find out how M effects ΔL the rest won't be so hard!
 
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  • #2
For part A, I don't think you need all the properties. For example the mass will probably be a component of part C but for part A just do a free body diagram on the mass. The mass isn't moving so the force from the gas = the force from the spring
 
  • #3
That's what's so confusing to me that there's really not too much to say about the system when it's in that state! Yet a. is asking for... an equation by the sound of it. That seems to be impossible so I will assume you are right about that.

I might as well move on to b.
 
  • #4
There is an equation. If you do a free body diagram on he mass M the force pushing to the right is P0*A. The force pushing on the mass by the spring is k*ΔL. Set them equal to each other because the forces are equal and solve for ΔL.

Part b the mass is moved to the right a very small distance x. This is where you use the ideal gas law but to=tf so Po*Vo=Pf*Vf

You know Po, V0 is lo*A, the new Vf is (Lo+x) * A solve for Pf. Force f is Pf*A

Part c. Since x is small this is simple harmonic motion for a spring
 
  • #5
ok so I did up part a.

decided I should find the force of pressure Fp

p=F/A
so Fp = p*A
p=nRT/V
v=A*(L0+ΔL)

combine them both into

nRT/A*(L0+ΔL) * A = Fp
simplified
Fp = nRT/(L0+ΔL)

<-----------*----------->
Fsp(-kΔL) Fp(nRT/(L0+ΔL)

and that's all I could do for a.



For b. I made a Fnet equation:

Fnet = Fsp + Fp

Left is negative and Right is positive

Fnet = -kΔL + nRT/(L0+ΔL)

T stays the same so obviously the Fnet is going to go left, and real quick unless there is a substantial gain in T (which there isn't)

Attempting c. need to find out how to do oscillations... and what "x << L0" even means!
 
  • #6
note, on a. when it said to show ΔL in terms of p0, L0, A, M, and k, I interpreted that as something like "find an equation to solve ΔL using these variables" I must have been mistaken
 
  • #7
What's the definition for the period of a spring/mass system in simple harmonic motion?
 
  • #8
i slightly remember doing stuff like that two quarters ago. Going to have to look through some notes I guess!
 
  • #9
So close to answering c.

5 hours of straight physics, and I can't refresh myself with kinematics. Gonna try this again tomorrow morning and see if I can answer it.
 
  • #10
c.


ω = √k/M

so 2pi/T = √k/M

2pi = T*√k/M

2pi/√k/M = T

I guess that's all they ask for!
 

What is the Ideal Gas Law?

The Ideal Gas Law is a principle that describes the behavior of an ideal gas in terms of its pressure, volume, temperature, and number of moles. It states that the product of pressure and volume is directly proportional to the product of the number of moles and the temperature of the gas.

How does the Ideal Gas Law apply to a system with a spring?

In a closed container with a spring, the Ideal Gas Law still applies. The spring acts as a constraint on the volume of the gas, and the pressure of the gas will increase or decrease accordingly. However, the number of moles and temperature must remain constant for the Ideal Gas Law to hold.

Why is the container considered "closed" in this scenario?

A closed container in the context of the Ideal Gas Law means that there is no exchange of matter with the surroundings. This means that the number of moles of gas inside the container remains constant, and the system is isolated from any external influences.

How does the Ideal Gas Law relate to the spring constant of the system?

The spring constant does not directly affect the Ideal Gas Law in this scenario. However, it can indirectly impact the pressure of the gas by changing the volume of the container. As the spring is compressed or expanded, the volume of the container changes, resulting in a corresponding change in pressure.

Can the Ideal Gas Law be applied to real gases with non-ideal behavior?

While the Ideal Gas Law is a useful approximation for many gases, it is not entirely accurate for real gases with non-ideal behavior. This is because real gases have intermolecular forces and occupy a finite volume, unlike ideal gases. However, the Ideal Gas Law can still be a useful tool for understanding and predicting the behavior of real gases in many scenarios.

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