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First law of thermodynamics

by vasudevan349
Tags: thermodynamics
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vasudevan349
#1
Mar3-13, 07:51 AM
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I began applying first law of thermodynamics (delQ=dE+delW) to a freely falling body.
We see that for this case delQ=0, so dE=-delW.
As dE consists of both potential and kinetic energies we see that neglecting any viscous losses, dE=0. (At max. height, KE=0,PE=mgh and at bottom most point PE=0,KE=.5*m*v^2=mgh).
This concludes that delW=0.
But what happens to the work done by gravity?
Is there any restriction placed on delW to be used in first law?
If you say that work done by the gravity is compensated by change in potential energy, how to convey this using first law of thermodynamics?
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Doc Al
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Mar3-13, 08:01 AM
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Quote Quote by vasudevan349 View Post
If you say that work done by the gravity is compensated by change in potential energy, how to convey this using first law of thermodynamics?
When you use a gravitational PE term, you are already accounting for gravity and any work it does. To also include a dW term for gravity would be to count it twice.

So choose one or the other, but not both.
Radhakrishnam
#3
Mar4-13, 12:32 AM
P: 35
It is an interesting question. It helps clarify many interesting aspects of thermodynamics.

First law of thermodynamics concerns with internal energy changes of the system as a result of interactions with the surroundings.

In the question under consideration, the system (assuming it to be a rigid solid mass that suffers no change of shape as it falls) suffers no internal energy change. Consequently the algebraic sum of heat interaction and work interaction with the surroundings must be zero. Since heat interaction is zero work interaction is necessarily zero.

Having said that, I may ask you to ponder over what your system is, what the surroundings is and how to account for work interaction between system and surroundings in any problem you tackle in thermodynamics. That helps, I hope, clarify the concepts in thermodynamics to a large extent.


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