Problem with Thermodynamics. Period dependence on temparature.by AudriusR Tags: cylinder, dependence, hydrogen, period, piston 

#1
Dec1211, 02:14 PM

P: 11

1. The problem statement, all variables and given/known data
http://img267.imageshack.us/img267/5830/80849533.jpg 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 m_{1}=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 questionmarks. Why? 



#2
Dec1211, 07:23 PM

P: 1,361

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 ? 



#3
Dec1311, 09:50 AM

P: 11

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
Dec1311, 10:35 AM

P: 1,361

Problem with Thermodynamics. Period dependence on temparature.
Just let dx = x.




#5
Dec1511, 07:36 AM

P: 11

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
Dec1511, 05:35 PM

P: 1,361

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! 



#7
Dec1511, 05:49 PM

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AM 



#8
Dec1511, 08:19 PM

P: 1,361

I not too sure about my value for k though. 



#9
Dec1511, 09:58 PM

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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(VdV)^γ = P2(V+dV)^γ = K Since the net pressure on the piston is the difference between the two pressures: P1P2 = F/A = K/(VdV)^γ  K/(V+dV)^γ = K/(VAx)^γ  K/(V+Ax)^γ Letting A be the crosssectional area of the cylinder and L the length of the space on each side at equilibrium the net force is: [itex]F = K(A/(ALAx)^γ  A/(AL+Ax)^γ) = KA^{(1γ)}(1/(Lx)  1/(L+x))^γ[/itex] For x << L, (1/(Lx)  1/(L+x))^γ ≈ (2x/L^2)^γ So it seems to me that: [tex]F =  KA^{(1\gamma)}\left(\frac{2}{L^{2}}\right)^\gamma x^\gamma = kx^\gamma [/tex] where [itex]k = KA^{(1\gamma)}\left(\frac{2}{L^{2}}\right)^\gamma[/itex] AM 



#10
Dec1611, 02:05 AM

P: 11





#11
Dec1611, 07:32 AM

P: 1,361

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
Dec1711, 08:00 AM

P: 11

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 ( VdV)^? Consideringt that: p1(V+dV) + p2 (VdV) = 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
Dec1711, 09:21 AM

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From my earlier post:
[itex]F = kx^\gamma [/itex] where [itex]k = KA^{(1\gamma)}\left(\frac{2}{L^{2}}\right)^\gamma[/itex] and [itex]K = PV^\gamma[/itex] 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 [itex]K = P_0V_0^\gamma = nRT_0V_0^{\gamma 1}[/itex] So the force equation becomes: [itex]F = nRT_0V_0^{(\gamma 1)}A^{(1\gamma)}\left(\frac{2}{L^{2}}\right)^\gamma x^\gamma[/itex] Since [itex]A^{(1\gamma)} = A^{(\gamma1)}[/itex] we can reduce this to: [itex]F = nRT_0\left(\frac{V_0}{A}\right)^{(\gamma 1)}\left(\frac{2}{L^{2}}\right)^\gamma x^\gamma[/itex] But V0/A is just the length of the chamber on each side of the piston, ie. L. So this reduces nicely to: [itex]F = 2^\gamma nRT_0L^{(\gamma 1)}L^{2\gamma} x^\gamma = 2^\gamma nRT_0L^{(\gamma + 1)}x^\gamma[/itex] AM 



#14
Dec1711, 09:30 AM

P: 11

I'm very grateful for your answer



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