Problem with ideal gas law and a spring Help please

[ml_lulu]

Problem with ideal gas law and a spring! Help please!

Homework Statement

We have a box divided in two parts by a piston without friction, and in one part there are n moles of an ideal gas and a Spring orf constant K and natural longitude L which keeps the piston in equilibrium. According to this, and knowing that the temperature of the gas is T0, find the amount for K.

Homework Equations

PV=nRT
F=-KX
E=(KX^2)/2

The Attempt at a Solution

I know its (nRTo)/6L^2 because the book has the answers, but I don't know where it came from! Please help!

 

[denverdoc]

what do you know besides eqn. How do we link the two systems?

The common part to both is the piston, so Pressure of the gas multiplied by its area exerts a force on it.

This force must be the same as that exerted by the spring in order for system to be in equilibrium, that is piston is not moving.

Does this help?

 

[ml_lulu]

I forgot to tell that the longitude of the spring in equilibrium is 3L (which means that x=2L, I think?) So, what I did is:

P= F/A
F=-kX
PV=nRT

kX/A=nRT/V

And volume is one, because its an ideal gas, so

K=nRTA/X

K=nRTA/2L

And that's all I know

 

[ml_lulu]

Forget it! I did it

F/A*V=nRT
F*L=nRT
kX*L=nRT
K=nRT/X.L
K=nRT/2L*3L
K=nRT/6L^2

Thanks anyways! :)

 

[denverdoc]

Thanks, I was wondering where the 6 came from

Thats getting real close,

the importat links are being made:

There is nothing that says ideal gas has volume of 1,
in fact it is 22.4 L for n=1.

what we can say P=nRT/(A*L) where L here is the Length at equilibrium under final condition.

we know that force is P*A=nRT/L

we also know that F=K*L where L is above.

then k=nRT/L^2 now relate the L I used vs that you were given at equilibrium as I had some trouble understanding problem description.