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a question about position of piston after achieving thermal equilibrium |
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| Aug18-12, 10:14 AM | #1 |
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a question about position of piston after achieving thermal equilibrium
A vertical cylindrical container contains some helium gas that is in thermodynamic equilibrium with the surroundings. The gas is confined by a movable heavy piston. the piston is slowly elevated a distance H from its equilibrium position and then kept in the elevated position long enough for the thermodynamic equilibrium to be reestablished. after that,the container is insulated and then the piston is released. after the piston comes to rest,what is the new equilibrium position of the piston?
This question is quite complex to me and i am totally at a loss. i know the second process is an adiabatic process but not sure about the first,the answer is the piston dropped by 0.6H to 0.4H at the new equilibrium,but who can show me how to arrive at it? |
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| Aug18-12, 10:46 AM | #2 |
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Assume an initial temperature T0. So what is the temperature of the gas after the first part? (Hint: easy). How do you determine the state of the gas (P, V, T) after the second part? Hint: this involves the [itex]\gamma[/itex] for Helium. AM |
| Aug18-12, 11:01 AM | #3 |
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during the first part,is the pressure constant? if it is,T1 would be T0(h+H)/h.
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| Aug18-12, 11:06 AM | #4 |
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a question about position of piston after achieving thermal equilibrium
After the adiabatic process,P1V1γ=P2V2γ,for monoatomic helium,γ is 1.67,but still not sure what process the first is,i think i am stuck at this point
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| Aug18-12, 11:19 PM | #5 |
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| Aug19-12, 01:14 AM | #6 |
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Once you figure that out, all you have to do is apply [tex]P_1V_1^\gamma = P_2V_2^\gamma[/tex] to determine what the new volume will be in terms of H. AM |
| Aug19-12, 02:04 AM | #7 |
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by p0v0=p1v1,i can get p1=p0h/h+H as p0=p2=patm, i have p0h/(h+H)*(h+H)1.67=p0*(h+kH)1.67,assuming new equilibrium is kH away from the old equilibrium position simplify the above eqn i have h3(h+H)2=(h+kH)5 i think i would be able to get k=0.4 from this equation,but actually i am not very sure about how to solve it,could you give me hints on this final step? thank you |
| Aug19-12, 09:19 PM | #8 |
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In looking at this problem again, the adiabatic condition does not apply here as this is not a quasi-static adiabatic compression.
Since the cylinder is insulated, no energy is lost from the system. The work done by the atmosphere and the piston ( by the change in gravitational potential energy of the piston at height H to its final position ) is converted into heat flow into the gas. I'll take a closer look at this and see if this is solvable. AM |
| Aug20-12, 12:29 AM | #9 |
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The final pressure is equal to the original one: pi=po+mg/A, the atmospheric pressure + the pressure due to the weight of the piston. (A is the cross section area of the piston).
The force F=Apo+mg does the work when the piston is released from the elevated position. The work done by the external force when the piston moves from height H to height h is W=(Apo+mg)(H-h)=Api(H-h). This work is converted entirely into the internal energy of the helium gas. ΔU=W. The internal energy of n mol He is 3/2 nRT at temperature T. The change of internal energy is ΔU=3/2nR(Tf-Ti) Initially the gas had volume Vi, temperature Ti and pressure pi: Ti=PiVi/(nR). In the final equilibrium state, the volume is Vf=Vi+Ah, the pressure is equal to pi, so Tf=pi(Vi+Ah)/(nR). Write up the equation ΔU=W in terms of H, h, Vi, and pi and solve for h. ehild |
| Aug20-12, 03:19 AM | #10 |
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| Aug20-12, 03:48 AM | #11 |
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What have you done? It is very simple. ΔU=3/2nR(Tf-Ti), W=Api(H-h)
Ti=PiVi/(nR), Tf=pi(Vi+Ah)/(nR). Plug in and all unknowns cancel, except H and h. ehild |
| Aug20-12, 04:07 AM | #12 |
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so actually the information about the first isothermal expansion is not useful except telling me pressure is not changed? why for the adiabatic process pi=pf,i thought pi should be lower than pf? |
| Aug20-12, 04:20 AM | #13 |
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The pressure changes in the first isothermal process, but it does not matter.
The initial volume of the cylinder is assumed to be Vi, but it cancels at the end. ehild |
| Aug20-12, 04:26 AM | #14 |
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| Aug20-12, 04:39 AM | #15 |
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The pressure is different before and after the final compression step, but the process is not quasi-static. The gas is not in equilibrium when the piston moves downward by its own weight, and as pressure is defined only for equilibrium, you can not speak about pressure of the gas during the process. The work on the gas is done by external forces, those caused by the atmospheric pressure and gravity.
pi means the pressure before anything happened to the gas. pf is the final pressure after the piston has been released. ehild |
| Aug20-12, 05:02 AM | #16 |
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Thank you for answering so many questions, i think i learned a lot. |
| Aug20-12, 06:07 AM | #17 |
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You are welcome.
Remember the change of internal energy is equal to the heat added and the work done on the gas: ΔU=Q+W. The work W is done by the external force, and it is -∫PdV only for quasi-static processes, when the pressure of the gas balances the external force exerted on the piston. ehild |
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