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### Topic: Reversible compression of solid material  (Read 6237 times)

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#### Hand15

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##### Reversible compression of solid material
« on: March 20, 2008, 07:25:50 AM »
Well... honestly I have a problem with my homework.

So what is the work done when a solid is compress "isothermally and reversibly" with Coefficient of thermal expansion and isothermal compressibility value are given and assumed to be constant?

The problem is compress isothermally and reversibly. I never heard a solid is compress reversibly, so the work would not be simply w = P * delta V

Strange because usually it is gas... but in this homework the material given is solid....

Could someone help me?

#### eschewthecashew

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##### Re: Reversible compression of solid material
« Reply #1 on: March 21, 2008, 07:33:13 PM »
Hmmm, this is my idea...
reversibility just implies that at any given moment in the transformation, the system and surroundings are in equilibrium, in which case, pressure is not constant. So, you would have to find an expression for pressure (based on isothermal compressibility coefficient, assumed to be unchanging throughout the transformation) and integrate over the change in volume (based on on coefficient of thermal expansion, also assumed to be unchanging throughout transformation). Presumably, for a solid, the amount of work in this case would be small.

#### Arkcon

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##### Re: Reversible compression of solid material
« Reply #2 on: March 21, 2008, 08:25:51 PM »
I can't help but wonder, when they asked this question about the compression of a solid, could they possibly mean an elastomer?  or a spring?  But I haven't had to think about physical chemistry in so log, I wouldn't know the formulas to use, anyway.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

#### Hand15

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##### Re: Reversible compression of solid material
« Reply #3 on: March 22, 2008, 01:46:06 PM »
Hmmm, this is my idea...
reversibility just implies that at any given moment in the transformation, the system and surroundings are in equilibrium, in which case, pressure is not constant. So, you would have to find an expression for pressure (based on isothermal compressibility coefficient, assumed to be unchanging throughout the transformation) and integrate over the change in volume (based on on coefficient of thermal expansion, also assumed to be unchanging throughout transformation). Presumably, for a solid, the amount of work in this case would be small.

Thanks for the advice.... finally I can find the answer, never thought it need such a difficult calculus technique