# Problem with Thermodynamics. Period dependence on temparature.

1. Dec 12, 2011

### AudriusR

1. The problem statement, all variables and given/known data
http://img267.imageshack.us/img267/5830/80849533.jpg [Broken]
Picture. The volume of that cylinder is V = 1l. In that cylinder we have 1g of hydrogen m=0,001 kg. The cylinder is separated in two equal parts with m1=0,005 kg and d=0.006m length piston. Imagine that we pushed that cylinder and piston in there began to swing. We need to calculate swings period dependence on temperature of the gases.

2. Relevant equations

We need to find k in this formula. F= -k x ( it's the force, which trying to return piston back to his place) And place it in this formula T = 2 Pi sqrt(m1/k)

3. The attempt at a solution
In my opinion that F is the difference of pressures in that cylinder.
I would calculate both pressures with this formula:
p V = m/M R T

and then I would deprive.

F would be : F = (delta)p S

But when I calculate, my V and m depends on the x, and he disappears.

In my head there are no more thoughts, so I'm asking you for help.

P.S I tried to put symbols of Pi and delta, but instead of them in the post there were question-marks. Why?

Last edited by a moderator: May 5, 2017
2. Dec 12, 2011

### Spinnor

Below I keep temperature constant, for small displacements I think that is OK. Hope the following helps.

For more accuracy you may want to expand V^2 = [V_o + dV]^2 about V_o ?

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Last edited: Dec 12, 2011
3. Dec 13, 2011

### AudriusR

Sorry, but I didn't get it, how to get f(x) function, when we have f(dx) ?
Or maybe it's possible to calculate period with this function?

4. Dec 13, 2011

### Spinnor

Just let dx = x.

5. Dec 15, 2011

### AudriusR

I'm really sorry for my questions, but I didn't get, how to get exact answer for Periods depence on temperature ( T(T) ). Maybe, if I don't ask you too much, you could give me another hint.

6. Dec 15, 2011

### Spinnor

On my sketch I wrote

F = [-nRTA^2dx]/v^2

let dx = x

F = [-nRTA^2x]/v^2

This is of the form,

F = kx

which you should know from the harmonic oscillator. Now F is a function of several numbers, one of them is T (all but x are roughly constant for small displacements x). The corrections go as x^2?

The problem gets harder if we need to be more accurate. I don't know how far you need to go.

Good luck!

Last edited: Dec 15, 2011
7. Dec 15, 2011

### Andrew Mason

Assume the cylinder is insulated and that the compressions and expansions are adiabatic. You have to use the adiabatic condition PV^γ = K in order to solve this. You cannot just assume that the restoring force is linear.

AM

8. Dec 15, 2011

### Spinnor

I'm confused then. If my sketch of the pressure verses position of the cylinder above is close then for small displacements the sum of the two pressures will be some linear function of x (for small x)? For small displacements F = kx.

I not too sure about my value for k though.

9. Dec 15, 2011

### Andrew Mason

I am saying that the restoring force is not linear. It is proportional to -x^γ

Since the initial volumes are the same (say, V) and the pressures and temperatures are initially the same in the equilibrium position, and since the total volume is always the same:

P1(V-dV)^γ = P2(V+dV)^γ = K

Since the net pressure on the piston is the difference between the two pressures:

P1-P2 = F/A = K/(V-dV)^γ - K/(V+dV)^γ = K/(V-Ax)^γ - K/(V+Ax)^γ

Letting A be the cross-sectional area of the cylinder and L the length of the space on each side at equilibrium the net force is:

$F = K(A/(AL-Ax)^γ - A/(AL+Ax)^γ) = KA^{(1-γ)}(1/(L-x) - 1/(L+x))^γ$

For x << L, (1/(L-x) - 1/(L+x))^γ ≈ -(2x/L^2)^γ

So it seems to me that:

$$F = - KA^{(1-\gamma)}\left(\frac{2}{L^{2}}\right)^\gamma x^\gamma = -kx^\gamma$$

where $k = KA^{(1-\gamma)}\left(\frac{2}{L^{2}}\right)^\gamma$

AM

10. Dec 16, 2011

### AudriusR

Thank you for a detailed answer, but since given data include V and lenght of piston d I'm not so sure, if I can get exact result in those data. Maybe, you could suggest something?

11. Dec 16, 2011

### Spinnor

If you graph pressure on one side verses piston displacement and then expand about small displacements no matter what the function for pressure we get,

P(dx) = c1 + c2dx + c3dx^2 + ...

For small displacements pressure difference will go as dx? If we add the pressure on both sides the constants c1 will cancel and to lowest order in dx we will have for the force on the piston,

F = 2Ac2dx + higher order terms.

The constant c2 will be found via Andrews help,

"Assume the cylinder is insulated and that the compressions and expansions are adiabatic. You have to use the adiabatic condition PV^γ = K in order to solve this. ... ."

12. Dec 17, 2011

### AudriusR

So making simple mathematics, what result I should get ?

Maybe, you could look at my calculations:

F= 2Ac2dx

c2 = KA^(1-?) (2/L^2)^?

F= 2K A^(2-?) (2/L^2)^? dx

So period formula looks like this:

T= 2 ? sqrt( m1/k)

k= 2K^(2-?) (2/L^2)^?

T = 2 ? sqrt ( A^? m1 (L^?)^2/ 2^(?+1) A^2 K )

K = p1 (V + dV)^? = p2 ( V-dV)^?

Consideringt that:

p1(V+dV) + p2 (V-dV) = m/M R T ( Am I right? )

K = 2 (mRT/M)^?

And the result is

T(T) = 2 ? sqrt( V^? m1 L^? M^? / 2^(?+2) A^2 (mRT)^? )

Also there is a problem, I couldn't get result by given data ( L and A hadn't been given)

Any ideas, how to remove those and is this is a good result ?

13. Dec 17, 2011

### Andrew Mason

From my earlier post:

$F = -kx^\gamma$

where $k = KA^{(1-\gamma)}\left(\frac{2}{L^{2}}\right)^\gamma$ and $K = PV^\gamma$ where P and V are the pressure and volume at any time eg. P0 and V0, the initial pressure and volume.

To find the relationship to temperature you would have to express the adiabatic condition in terms of T0, the initial temperature and V0, the initial volume. You do this by substituting for P0: P = nRT/V

$K = P_0V_0^\gamma = nRT_0V_0^{\gamma -1}$

So the force equation becomes:

$F = -nRT_0V_0^{(\gamma -1)}A^{(1-\gamma)}\left(\frac{2}{L^{2}}\right)^\gamma x^\gamma$

Since $A^{(1-\gamma)} = A^{-(\gamma-1)}$ we can reduce this to:

$F = -nRT_0\left(\frac{V_0}{A}\right)^{(\gamma -1)}\left(\frac{2}{L^{2}}\right)^\gamma x^\gamma$

But V0/A is just the length of the chamber on each side of the piston, ie. L.

So this reduces nicely to:

$F = -2^\gamma nRT_0L^{(\gamma -1)}L^{-2\gamma} x^\gamma = -2^\gamma nRT_0L^{-(\gamma + 1)}x^\gamma$

AM

Last edited: Dec 17, 2011
14. Dec 17, 2011