Re: CONSERVATION OF ENERGY

[John Mallinckrodt <ajmallinckro@CSUPomona.Edu>]

Emilio writes:

>  I wonder how people would explain the following two processes:
>
>    a) I have two containers, A and B, joined by a tube with a valve.
> A is empty (vacuum), while B contains air at normal pressure. I open
> the valve, so that the air also gets into A without producing any work.

The system energy (E) gets redistributed among its internal potential 
energy (Pi, due to molecular interactions), internal kinetic energy 
(Ki, calculated with respect to the CM of the system), and bulk 
translational kinetic energy (Kb, due to motion of the CM).  
Initially Kb = 0, and E (= Pi + Ki + Kb) is less than Ki due to the 
fact that Pi is negative at "normal pressure."  Just after the valve 
is opened, the primary effect is that Ki begins to reduce in favor of 
Kb as the molecular velocities become somewhat "organized" in the 
direction of container A.  After a short period of "sloshing around," 
Kb is once again eliminated but now, since the average intermolecular 
distance is smaller Pi is larger (less negative) than before.  As a 
result Ki is smaller and this is reflected in a moderate "cooling" 
(decrease in temperature) of the gas.

I have always particularly liked this example for illustrating the 
differences between the many different useful ways of defining work 
and the system energy changes that are related to them.  One can 
calculate work using all forces (A), using external forces only (X), 
or using internal forces only (I). One can use displacements of the 
points of application relative to an inertial frame (G) or relative 
to the CM frame (C) or one can "pretend" that the net force on the 
system is applied at the center of mass and use its displacement (C).

 Work         Name                   Associated energy change
(F * x)
 A   G     Total work                       Ki      + Kb
 X   G     External work                    Ki + Pi + Kb (= E)
 I   G     Internal work                       - Pi
 A   C     Pseudowork                                 Kb
 A   R     Relative total work              Ki
 X   R     Relative external work           Ki + Pi
 I   R     Relative internal work              - Pi

Consider what happens after the valve is opened:  There is an 
unbalanced external force from the rear wall of container B that does 
no "external work" (in the "lab frame") so it does not change E.  
However, it does do positive "pseudowork" (in that same frame) and an 
equal but opposite amount of "relative external work" (which is 
frame-independent.)  The positive "pseudowork" increases Kb and the 
negative "relative external work" decreases the sum of the internal 
energies Ki + Pi.  (In this case, most of that decrease happens to 
show up in Ki.)  

Later on, the rear wall of container A begins to exert a force which 
eventually exceeds that of the rear wall of container B.  This force 
again does no "external work" but it does negative "pseudowork" 
(decreasing Kb) and an equal but opposite "relative external work" 
(increasing Ki + Pi).  During the "sloshing around" period, the 
"pseudowork" and the "relative external work" reverse signs several 
times, but, in the end, the total "pseudowork" and the total 
"relative external work" done are both identically zero.  This leaves 
the total of the internal energies (Ki + Pi) unchanged.  

Finally, during the process the molecules have also on average 
receded from each other against an attractive intermolecular force.  
Thus, a negative "internal work" (or "relative internal work") has 
been done increasing Pi and, necessarily, decreasing Ki.

>    b) In a container (usually a cylinder with a piston) I have a
> mixture of air and gasoline vapor. I ignite it via a spark plug and
> the temperature rises.

I'll keep this one short!  Positive "internal work" is done 
decreasing Pi.  However no "relative external work" is done so Ki + 
Pi is unchanged.  Thus Ki increases.

(Is anyone still here?)

John
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 A. John Mallinckrodt      http://www.intranet.csupomona.edu/~ajm
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