October 04, 2022, 08:28:58 PM
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Topic: Conceptual problems with Irreversable Isothermal Expansion  (Read 4568 times)

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Offline Abedeus

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Hey, recently I've been doing Reversable and irreversable isothermal expansions in my thermodynamics module, I'm having conceptual problems with were the equations come from, by all means I could always parrot fashion remember the equations but I always find it extremely benneficial to remember where they came from, should somone ask me "Why".

As for the problems, the topic seems funnily underexplained, as far as i can tell the work done (energy wise) is the force * distance (quite an oldskool equation yes) and that Pressure*Area*Distance moved = The work done by the gas in an irreversable isothermal expansion.

The problem isnt the equations used, but the reasoning, surely the force wouldnt be equal to the external pressure, but:

(Pinternal*A-Pexternal*A) i.e the net force for the expansion being the force exherted by the internal gas minus the force being exherted by the external gas against the internal gas'es force (yeah, my english sucks :( ). But most textbooks and explinations just seem to imply that the force required is equal to the force required to push against the external pressure.

I sort of get it for expansions, but for compressions many referances seem to use the same equation but with a -dv (resulting in a negative work done by the gas), but wouldnt the equation for compression be Pinternal*dV.

Sorry its confusingly written, but I'm having a brain dead moment on this issue :(

Offline Yggdrasil

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Re: Conceptual problems with Irreversable Isothermal Expansion
« Reply #1 on: May 28, 2007, 01:33:28 PM »
(Pinternal*A-Pexternal*A) i.e the net force for the expansion being the force exherted by the internal gas minus the force being exherted by the external gas against the internal gas'es force (yeah, my english sucks :( ). But most textbooks and explinations just seem to imply that the force required is equal to the force required to push against the external pressure.

Here's a good analogy.  Let's say you are trying to push the same block against two different surfaces.  In the first case, you push the block on a surface with very low friction.  In the second case, you push the block on a surface with very high friction.  Given that you push with the same net force in both situations, you are still doing more work in the second case.

Quote
I sort of get it for expansions, but for compressions many referances seem to use the same equation but with a -dv (resulting in a negative work done by the gas), but wouldnt the equation for compression be Pinternal*dV.

The equation for work is always dw = -integral(PextdV).  For expansions, dV > 0, so dw < 0, meaning that the gas does work on the surrounding and loses energy.  For compressions, dV < 0, so dw > 0, meaning that the surrounding do work on your gas and the gas gains energy.

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