# Pressure and spring constant problem

1. Apr 23, 2006

### kellyneedshelp

hi, i am having trouble with this problem:

A cylinder with a diameter of 5.04 cm has a movable piston attached to a horizontal spring. The cylinder contains 1.40 liters of an ideal gas at 20.9°C and 0.99 atm pressure. Under these conditions, the spring is unstretched. The temperature of the gas is increased to 95°C and it is found that the spring compresses 3.41 cm. Determine the spring constant k.

I am not sure how to relate the change in pressure with a spring constant. So far, I tried finding the new pressure using (P1*V1)/(T1) = (P2*V2)/(T2) like this:
(.99atm*1.4L)/(273.15+20.9) = (P2*1.4L)/(273.15+95)
and got P2 = 1.239atm

How can I use this to find the spring constant? I tried using the eq. P=F/A but then I didn't know the area of the cylinder to use.

Thanks!!

2. Apr 23, 2006

### Hootenanny

Staff Emeritus
You are correct in using $F = PA$.You are given the diameter of the cylinder and hence you can work out the area.

~H

3. Apr 23, 2006

### kellyneedshelp

isn't the area of a cylinder = 2*pi*r*h ?
how do i find the area of the cylinder without knowing its height/length?
thanks!

4. Apr 23, 2006

### lightgrav

The Force applied by Pressure is perpendicular to the Area.
That is, the Pressure Force is the Normal Force from long ago...
... in F = P.A , the Force pierces (acts *through*) the Area,
so you want to use the "cross-sectional" Area, perp. to the motion
(not the area of the *side* of the cylinder).

5. Apr 23, 2006

### kellyneedshelp

do you mean use the area of one end of the cylinder, like just pi*r^2?

6. Apr 23, 2006

### lightgrav

Yes.
A Force perp to that end Area would push the cylinder, correct?
A Force perp to the sides would just try to enlarge the chamber diameter

7. Apr 23, 2006

### kellyneedshelp

ok, well i tried using A=pi*(.0252m)^2 = .001995 m^2
and P = 1.239atm
to get F = (1.239)*(.001995) = .0007868 N
and then use F=-kx
where x=0.0341m
to get k=.07249 N/m
but this is not the right answer.

am i not following what you are saying correctly?

thanks=)

8. Apr 23, 2006

### lightgrav

Apparently, there's 0.99 atm of Pressure on the OUTside of the piston
(because F_spring = 0 when INside Pressure = 0.99 atm).
One presses leftward, one presses rightward ...

9. Apr 21, 2008

### as11

Hi, I'm trying to figure out the same problem.

What does it mean for there to be .99 atm pressure outside vs. inside, in terms of solving a problem? Conceptually, is it that the pressure outside is pushing against the top of the cylinder, but the pressure inside is greater than the outside, thus pushing outward and compressing the spring?

So would that mean F$$spring\rmfamily$$ + F$$outside pressure\rmfamily$$ = F$$inside pressure\rmfamily$$? So, -kX + P1A = P2A?

10. Apr 19, 2009

### begbeg42

yes some help on this would be nice...
using the ideal gas law, since nr=constant=PiVi/Ti=PfVf/Tf, thus
PfVf=a constant

F=PA
Fspring=kx

then -kx-P(outside)A=P(inside)A
where A=pi(r^2)

but now I am stuck...I understand that initially P(outside)A=P(inside)A and thus kx=F(spring)=0 but Im not sure how to go from here
thanks

11. Apr 19, 2009

### begbeg42

how do you solve for the final pressure & and the final volume? and is what I posted above correct? or is there some flaw i'm missing
again, thank you

12. Apr 19, 2009

### begbeg42

is this problem/ piston isochoric or isobaric?

13. Apr 19, 2009

### begbeg42

i did what kelly did and also got the wrong answer...I do not know how to apply the thing about the outside pressure