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Heat produced on shaking |
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| Mar7-05, 08:48 AM | #1 |
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Heat produced on shaking
OK, here is a question I found in a book some time ago :-
An insulated box containing a monatomic gas of molar mass M moving with a velocity v is suddenly stopped. Find the increment in the gas temperature as a result of stopping the box. I thought on this and what came to my mind was this - I should first know the the exact definition of temperature to tackle this question. As I know, temperature is defined very precisely by the zeroth law of thermodynamics. But here, I need a definition based on the energy of the molecules. Of course, temperature is directly proportional to the total internal energy of the gas. And what is internal energy? It is the sum of several forms of mechanical energies one of them being the kinetic energy. Now, we know that the KE depends on the frame of reference, it comes that the temperature should also depend on the frame of reference. But does it? This is one question that comes from the main question. The problem is not as simple as it seems. I asked several people the same question but all did some baseless energy conservation calculations and actually reached the answer. No one seems to be realising that there is something serious here. If you reach the answer, then you may like to think on the validity of your arguments by applying the same method to a slightly modified question - instead of the container stopping to zero velocity, its velocity suddenly changes to -v. Now find the change in temperature. Help please! |
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| Mar7-05, 09:49 AM | #2 |
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If the container is moving at velocity [tex] v[/tex] towards the right, it will have two components of energy: -One measured in the reference frame I said above, and it represents the local thermodynamic state: [tex] U_1=mc_vT_1[/tex] is the internal energy which contains the microscopic kinetic energy of the molecules. Under the assumption of local thermodynamic equilibrium there exists a mean velocity of fluctuation of the molecules, which is given by the Kinetic Theory. -Another measured in the reference frame you want, always if you are coherent with the rest of the calculation. I have chosen the reference frame attached to ground. It represents the macroscopic kinetic energy: [tex] E_1=1/2 m v^2[/tex] Once the container have been stopped, its total energy is only the thermal energy based on the local thermodynamic properties: [tex] U_2=mc_vT_2[/tex] Under the assumption of adiabatic walls, the container doesn't exhange any heat with the external environment. And suppossing the effects of internal (viscous dissipation) are negligible, the desacceleration process can be approximated as an isentropic one: [tex] U_1+E_1=U_2[/tex] which gives us an important result: [tex] T_1+\frac{v^2}{2c_v}=T_2[/tex] This can be reshaped into a similar form: [tex] \frac{T_2}{T_1}=1+\frac{v^2}{2c_vT_1} [/tex] This last result is similar to one found in Gas Dynamics Theory. The last term is of the order of the Mach Number. If you study Gas Dynamics some time, you will realise that the behavior of a compressible fluid particle inside a flow when being accelerated/desaccelerated, can be described with the whole approximation of this container problem. The temperature [tex] T_2[/tex] represents the stagnation temperature of the flow, and it depends by this proper definition on the reference frame chosen. On the other hand, the temperature [tex] T_1[/tex] is the so called Static temperature, and it is only a function of the local thermodynamic state as I said. I hope this helps you. Otherwise post your doubts. |
| Mar7-05, 09:53 AM | #3 |
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But yes, in case of an ideal gas the temperature is proportional to the average kinetic energy of the gas. But in an ideal gas the particle-motion is isotropic, the same number of particles moving to the left as moving to the right. So only is the frame of reference in which the average velocity is zero the temperature is proportional to the average kinetic energy. So the answer is no: temperature doesn't depend on the frame of reference. |
| Mar8-05, 01:39 AM | #4 |
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Heat produced on shaking
Thanks Clausius for your reply. But does your equation apply to my second problem? Suppose the container, instead of stopping completely, changes its velocity to -v. In that case, your [tex]E_{1}[/tex] term does not change. Will the temperature not change in that case?
Secondly, I would prefer a much rigorous treatment taking care of the definition of temperature. First, what's the definition? Is it - The avg. KE of the molecules as seen from the centre of mass reference frame? If yes, then I would prefer to think like :- The initial temperature = avg. of half * m * (v-v0)^2 where v0 is the velocity of the centre of mass from the ground frame Similarly, find the expression for the final temperature and then see what types of relations they exhibit. |
| Mar8-05, 01:42 AM | #5 |
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So do you mean, that (in the classical sense) if you have been told the KE's of all the molecules of a gas, and their masses also, then you can't find the temperature? |
| Mar8-05, 04:47 AM | #6 |
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Leave me some time to see if I am able to explain this mathematically. |
| Jul15-05, 12:32 AM | #7 |
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This is a JEE 2 marks question hehe :D and yes i did conservation of energy without thinking too much. I always had this small niggling doubt that Kinetic Energy of a system depends on your frame of reference. I cannot give an alternate definition of temperature or anything of that sort but i can say that if the velocity changes from v to -v there will be no temperature change. Firstly let this be clear...we are talking abt temperature change of box. now we have decreased the velocity from v to 0 and the temperature of box can be calculated from conservation of energy(friction, sound dissipation absent). i do not know anything abt temperature's relevance to frame...i just know that if conservation of energy holds...i can calculate temperature using conservation principle. immediately heat transfers from the box to the surrounding to maintain thermal equilibrium. now when box is again given a velocity of -v ... it gains KE and logically...and of course mathematically...box shud attain same temp as it had when it had velocity of v.
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| Jul15-05, 01:47 AM | #8 |
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You're forgetting that work needs to be done by some external body to bring the box to rest. How do you use energy conservation without including this term ?
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| Jul15-05, 03:35 AM | #9 |
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| Jul15-05, 08:58 AM | #10 |
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How is the system isolated if an external body is interacting with it ? And since when did JEE start asking "7th class stuff" ?
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| Jul15-05, 09:32 AM | #11 |
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Hey are you all Indians here like me? If not, how do you know about JEE. In case you want to tell that it's a very prestigious exam and is famous throughout the world then do tell me because I just qualified in the exam one month ago ;-)
And yes, applying energy conservation is not that simple here, toocoolsashi. Indeed, the system is isolated and you can use conservation of energy, i.e., the total energy = constant. Now, some part of this total energy is related to the temperature and some to miscelleneous stuff like KE. So, you have, energy related to temperature + miscelleneous stuff = constant. To get anything out of this equation, you need to know too much about the first term. You need to be aware of the relation between temp. and energy, for example. And this is not simple, since you may say that temperature is just a proportionality constant times the avg. KE of the molecules, but, KE from which reference frame? Ground reference frame? Or the frame of the container? Or that of the Zem mattresses of the planet Squornshellous Beta? |
| Jul15-05, 10:47 AM | #12 |
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It's not hard to calculate the work done by the external body - just use momentum conservation. Of course, this may lead to surprising results, but such is physics. |
| Jul15-05, 11:19 AM | #13 |
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| Jul15-05, 02:42 PM | #14 |
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Recognitions:
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OTOH, if the deceleration takes a long time, the kinetic energy of the gas is allowed to do work on the decelerator, and in the limit, there is no temperature rise. |
| Jul15-05, 04:06 PM | #15 |
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| Jul15-05, 05:25 PM | #16 |
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Recognitions:
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It's like throwing a ball of putty against a wall. The momentum is transferred to the earth. But the energy heats the putty.
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| Jul15-05, 05:44 PM | #17 |
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