A New Thermodynamics

Blog: Negative Work:New Conceptualization

By Kent W. Mayhew

www.newthermodynamics.com

In my previous blogs I have discussed that lost work (Wlost) involved the displacement of Earth’s atmosphere against gravity. And that it is equal to:

 

Wlost=PatmdV                                    (1)

 

Where Patm is the atmosphere’s pressure and dV is the atmosphere’s displacement volume or if you prefer the volume increase of the expanding system. See blog on "Lost Work"

 

The above explains why useful processes are not reversible. It also implies that the second law should not be applied the way it is traditionally applied to useful processes. In other words both entropy and the second law need reconsideration and thermodynamics as a science is in dire need of an overhaul.

 

 

Negative work

 

Although “negative work” conceptually feels counterintuitive, it is mathematically expressed by simply placing a negative sign in front of the work term. To visualize negative work, all one needs to do is to take a syringe, and place the plunger all the way to the bottom as is illustrated in the top of Fig. 1.7.9.  Now hermetically seal the syringe’s opening with your finger and finally apply a volume expanding force (pull out on the syringe’s plunger). The volume inside the syringe expands, thus we are performing negative work onto the volume of space that is now being created the inside of the syringe.

 

  If the initial volume (Vi) of gas within the expanding volume was zero, then our expansion of the hermetically sealed syringe to some final volume (Vf), means that our final pressure (Pf) is that of a vacuum (zero pressure). Therefore, ignoring any blackbody radiation within the syringe, then our final energy state within that created volume is zero. Accordingly, the work done onto the volume within the syringe (Wvol) in creating the vacuum is:

          

          Wvol  =  -Patm(Vf-Vi) = -PatmVf = -Watm                        (2)

 

Where Patm is the atmosphere’s pressure, Watm is the work done onto the atmosphere

 

Now ask, has the energy of the atmosphere changed? Since the atmosphere’s upward displacement was due to the introduction of a volume of nothingness (ignoring the blackbody radiation within), then the total energy of the atmosphere has not changed. 

 

Importantly negative work tends to create unstable volumes. In the case of the expanded syringe, once the plunger is released, the plunger will come crashing down into the syringe removing the recently created volume within the syringe. One could rightfully argue that since the pressure inside of the syringe was less then atmospheric, then the atmosphere’s pressure will drive the plunger into the syringe, and this would be the mechanical answer.

 

From an energy perspective, we could say that that the plunger went crashing down into the syringe because the interior of the syringe signified negative work, when compared to its surroundings hence was an inherently unstable volume. Ultimately, when a volume of nothingness displaces a mass in a gravitational field, then this is negative work to the volume that was previously occupied by that mass.

Again how we consider the work maybe a matter of perspective. From a thermodynamics perspective, the expansion of the syringe does not mean much. We could take the syringe full of nothing, bring it into outer space, then release the plunger and nothing will happen. However, claiming that nothing happens is from the perspective of inside the syringe. The reality is that that a volume of something replaces the volume that was once occupied by the syringe, as we bring the syringe out into Earth’s atmosphere’s surroundings, i.e. outer space. Okay are now verging upon metaphysics.

Consider what man on the moon sees. Although it was not noticed on Earth, the man on the moon clearly saw the Earth’s atmosphere’s volume increase, as the syringe was pulled. He then looks at his extraordinary accurate instrumentation and notes the Earth’s temperature remained constant. Did the actual energy of the Earth and its atmosphere isothermally increase? He ponders, perhaps it was due to cosmic rays, adding energy onto the system, Earth. So the man on the moon checks and no cosmic rays struck the Earth at that instant of time. He then recalibrates his equipment, and no errors are found. Facing a conundrum, the man on the moon turns blue.

If only the man on the moon knew all the intricacies going on in the system Earth, he would then understand that there were no real changes, due to the syringe’s expansion. We must emphasize that the created vacuum is not exactly a volume of nothing because it does contain blackbody radiation, but the point is in terms of mass in a gravitational field, the vacuum is a volume of nothingness. Adding or subtracting a volume of nothingness to a system does not necessarily change the system’s total energy.

Now ponder that the syringe starts at a specified non-zero value, and then negative work is done onto that volume of gas within the syringe. Now the pressure inside of the syringe decreases as its volume increases. Herein we shall consider that the gas inside of the syringe is an ideal gas hence the ideal gas law: PV=NkT=C’ = constant, applies. Therefore the negative work done onto the volume of gas within the syringe becomes:        

 

           Wvol = -(NkT)In(Pi/Pf) = -(NkT)In(Vf/Vi)                          (3)

 

    From the perspective of Earth and its surrounding atmosphere as a system, again the expansion of the syringe did not represent a change to the system’s energy. Although Earth’s atmosphere was upwardly displaced, the magnitude of the potential energy increase equals the magnitude for the negative work associated with the expanded volume. 

 

To see work done on a vacuum [eqn (2)] vs gas [eqn (3)]: see Graph 1.7.1 vs 1.7.2 

 

We have discussed that negative work is unstable. If one wanted you could equally argue that negative work is reversible. If anyone is interested I could expand this into cavitations and how they correlate to negative work. Just let me know

 

 

Understanding eqn 3) : Ideal Gas and Work

A mathematical analogy for this work is obtained by defining the process as isothermal (dT=0), thus the ideal gas law [PV=NkT] tells us that the pressure multiplied by volume equals a constant (C’), i.e.:

PV=NkT=C’                                                              (4)

 

Rewriting  eqn 4:

 

P=C’/V                                                                       (5)                                         

 

When the work required varies with the location at which the force is applied, the line integral is often used to calculate the work involved. Now work is the integration of infinitesimal changes. Therefore, the total work (W) is defined by the integration of infinitesimal work (dw), as follows:

 

W=integral (dw) = integral(Pdv)                               (6)                                                       

Notice that we transformed infinitesimal work (dw) into infinitesimal volume change (dv). Work now takes the more general form that being the summation of all the work associated with each infinitesimal volume change. Substituting eqn 5 into eqn 6, gives:  

 

W=integral (dw) = C’integral (dv/V)             (7)

 

Performing the integration gives:

 

W=C’In(Vf/Vi)                                              (8)

 

which can be rewritten for the isothermal work as:

 

W=(NkT)In(Vf/Vi)                                        (9)

 

Eqn 9 adheres to the traditional conceptualization for work. However, it possesses the following inherent ambiguity that is not traditionally acknowledged. That being that it can equally be written in terms of pressure change,

 

             W=(NkT)In(Pi/Pf)                                       (10)

 

  Although it is a moot point ,  a possible  reason that traditional thermodynamics is always written in eqn 9 rather than eqn 10 is the preference of volume over pressure as a parameter of relevance. In my previous blogs I gave a reason for this oversight that being that volume increases are wrongly attributed to entropy increases, when entropy (S) is considered in terms of Boltzmanns consideration [S=kIn(@)], where @ is the number of accessible states

 

 

 Thanks for reading what I write.

 

I am still looking for help in rewriting the science. It is mostly done but I need to throw ideas of some people to see if they understand and more brains involved the better the final product and hopefully outcome will be.  If it interest anyone with an open mind scientific mind who also has thick skin and a want to actually accomplish something.

 

 

Copyright  Kent Mayhew

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