# Thermodynamics- work in a cylinder piston device with spring

1. Sep 16, 2008

### oneosix

1. The problem statement, all variables and given/known data
Horizontal frictionless cylinder piston device- piston is forced againtst a spring which exerts a force directly propotional to the gas volume. Patm (101.13KPa) also acts on the outer face. Considering the gas as a system, calculate the work for the process from an initial state of 200Kpa, 0.1m^3 to a final volume of 0.3m^2. If the spring is taken as a system find the work for the same process.

2. Relevant equations
Work on spring=1/2kx^2
F=PA
W= integral(PdV) (sorry dont know proper equation tools)
PV/T= constant

3. The attempt at a solution
I must be missing something easy, because it doesnt seem like enough info is given.
Anyway
let x=distance the piston moves
V1= Ah = .1
V2=A(h+x)= .3
rearrange gives x=2h

sum of forces=0 at each state (can i do this?)
P1.A= Patm.A+kh
P2.A= Patm.A+k(h+x)
these 2 eqns gives: kx=P2.A-P1.A

use Work on spring:
W= .5*kx*x
W= .5*(P2.A-P1.A)*2h
use Ah=V1
W= P2.V1-P1.V1

this is as far as i have been able to get and i dont even know if im on the right track. It appears i need to calculate final pressure, but i'm unsure how. the answers are given which are 59.74 and -39.48 so have thought about working backwards, but dont know how to go about it

Any help is greatly appreciated

(edited: typos)

2. Sep 16, 2008

### Topher925

Welcome to the forums oneosix.

I think this is where you are getting stuck. The sum of the forces is obviously not 0 because the piston is moving. In order for them to be 0 you would have to consider the inertia of the piston which you did not. I think you started off on the right foot though.

I would approach this problem by first assuming that the gas in the piston is ideal and at constant temp. That is p1v1 = p2v2 along with no frictional affects. Then do a force balance approach, or Fgas = F spring. That relation will allow you to solve for x and you can use either of the equations for the gas force or spring force to determine the force as the piston moves, this will be a function of x. I think you know how to handle it from there.

I derived the final equations of this problem but wont be able to verify my answer unless you post the spring constant.

Last edited: Sep 16, 2008
3. Sep 16, 2008

### oneosix

This is the problem- the spring constant is not given. I think it is based on it being directly proportional to the gas volume, however i do not know how to go from there. Thanks for your help, I'll see what i can get from your input
cheers

4. Sep 17, 2008

### Topher925

Sorry for my brief laps in stupidity but, yes, you don't need the spring constant you only need to know that the force of the spring is linear. The first question of the problem is rather simple and it is only evaluating the work performed by the gas, however I can't seem to get the same answer that you posted. I found my answer by simply evaluating the integral

$$\textbf{W = }$$$$\int{p}$$$$\textbf{dV}$$

But the answer I get is a little less than half of what you posted. Perhaps I am doing something wrong and someone can correct me.

As for the second answer, it is obviously based off the first. The work done by the spring should be the negative of the work done by the gas and should be less due to the atmospheric pressure that is against the piston.

5. Sep 21, 2008

### corilat

To solve this problem, you basically need an equation for the pressure vs volume and the easiest way to get it is with a graph. Since the forces acting on the gas are always:
Ftot = kV + Fatm ,
the graph will look like a straight line, intersecting the P axis at Patm and passing through the known point (V1, P1). Thus your equation ends up as:
P (MJ) = 0.987V + 0.1013.
This leads to the correct result when you integrate wrt volume. (W = 59.74 J)
And then as was stated, to find the work on the spring you can subtract the work due to atmospheric pressure, or just integrate the equation: P = 0.987V.
I think the idea of this problem was that the system was always in equilibrium so the forces are actually always balanced.