By Kent W. Mayhew

**Illusionary Reversible Work: A Traditional Big Mistake**

It should be stated that often isothermal work is wrongly defined in terms of the expanding system parameters i.e

W= [(NkT)In(Vf/Vi)]sys (1)

Eqn (1) is too often mistakenly considered as being reversible work in thermodynamics.

for example see:

1)http://www.Socratic.org/questions/what-is-reversible-isothermal-expansion.html

2) http://www.chemistry-desk.blogspot.com/2011/051/isothermal-reversible-expansion-work-of.html

3) http://www.hyperphysics.phy-astr.gsu.edu/hbase/thermo/isothermal.html

4) http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node33.html

The above perversion of logic clearly demonstrates how traditional thermodynamics continues to put the cart ahead of the horse. This is discussed in the Parameters Blog. It is simply wrong because the work done by an expanding system is lost work onto the surrounding atmosphere i.e. work is done external to the expanding system, as defined by the exact differential: W=(PdV)atm = PatmdVsys

There are those who realize that work is lost onto the surroundings but they fail to understand that this is Lost work.

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Another example of perverse logic can be witnessed in link below:

A) http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/isoth.html

In the above they realize that for an isothermal expanding gas that the gas's total energy does not change i.e montaomic ideal gas Egas = 3PV/2 i.e. dEgas=0 Note: This does not consider the blackbody/thermal radiation residing in freespace.

Since dE=0, then they correctly write that based upon the first law: Q=W (2)

However they then wrongly consider the work in eqn (2) in terms of eqn (1). I.e.

Q=W= [(NkT)In(Vf/Vi)]sys (3)

The correct solution is in terms of irreversible lost work. I.e.

Q=W= (PdV)atm (4)

For more info on incorrect eqn (3) vs correct eqn (4) please see parameters.

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