Re: [Phys-l] App. for Was: Re: T dS versus dQ

[Bernard Cleyet <bernardcleyet@redshift.com>]

Is this what youall are envisioning?


Poperating = Pgas(1 ± u/vx)2; vx2 = RT/Mwhere u is the absolute value of the velocity of the piston, vx is the average velocity of the gas molecules moving toward the piston, and M is the molar mass of the gas. The positive sign applies to compressions and the negative sign to expansions.

and:


The idea that thermodynamic work depends on the speed of the process is not new. The concept of a reversible process carried out in a series of infinitesimal steps suggests an extremely slow process. It is generally recognized that the effective voltage of a battery decreases with increasing cur- rent, with the reversible voltage represented by the value mea- sured (or extrapolated) at zero current. The same principle can be applied to mechanical processes. This study suggests that the definition of mechanical work should be
w = −∫foperatingdl
with foperating representing the force that is exerted on the sys- tem at its boundary while the movement is occurring. The application of this definition to a system confined by a pis- ton gives
w = −∫PoperatingdV
with Poperating representing the pressure exerted on the face of the piston as the piston moves.
For processes involving the compression or expansion of perfect gases, these considerations lead to the approxima- tion
u2 2 RT Poperating = Pgas 1 ± v	; vx	= M
x
The positive sign applies to compressions and the negative sign to expansions, and M represents the molar mass of the gas and u is the absolute value of the velocity of the piston.


and:


The adiabatic gas law apparatus of Pasco Scientific al- lows a gas to be rapidly compressed to approximately half of its initial volume while simultaneously recording tempera- ture, pressure, and volume. Because of the speed of the com- pression, the process is assumed to be adiabatic. The equipment manual describes the calculations as if this were a reversible process. The manual also describes the correspond- ing expansion experiment, but cites complications such that “the expansion data do not give good quantitative results”.
McNairy (14) describes experiments with this appara- tus, “The adiabatic process is simulated by rapidly compress- ing gas in a cylinder by forcefully pushing down on a lever that is attached to a piston in the cylinder, ...”. He refers to the “approximately 100-ms period of the stroke” with which the gas is compressed from approximately 240 cm3 to 100 cm3. While the author does not report temperatures for this experiment and there is no measurement of the applied pres- sure, the experiment is clearly performed in an irreversible manner. There is no indication that the applied pressure is constant, but the nature of the experiment leads one to ex- pect that the applied pressure is probably much closer to the final pressure than to the pressure of the gas throughout the compression. 

14:  TPT 34 178


http://pubs.acs.org.oca.ucsc.edu/doi/pdfplus/10.1021/ed082p874

{J. Chem. Educ.  82 (6) 874 [2005]}

bc continues googling


On 2010, Feb 09, , at 23:41, John Mallinckrodt wrote:

> Bob LaMontagne wrote:
>
> > it would be nice for the smiling professor to be able to give a  
> > ballpark estimate of how large the temperature difference is  
> > between slow and rapid adiabatic compression for specific instances  
> > (say mach 0.0001 versus mach 0.9).
>
> Okay.  I'll give it a very rough, back-of-the-envelope go, but no  
> guarantees!
>
> I figure that if you compress a monatomic gas to half its volume in a  
> rigid container using a rigid piston moving at the sound speed,  
> you'll hit something like half of the gas molecules one time and give  
> them something like two to three times their original speed on  
> average.  (I have in mind an average molecule that is moving  
> generally toward the piston but at, say, a 45 degree angle in the lab  
> frame and having an initial velocity equal to the rms speed which is  
> 34% faster than the piston is moving.  If that molecule makes an  
> elastic collision with the piston, I calculate that it will leave  
> with about 2.3 times its initial speed.)
>
> If half the molecules have their speeds increased by a factor of 2.3  
> then the gas would increase its total energy--and, therefore, its  
> temperature by a factor of about 3.  This is to be compared with an  
> isentropic compression that increases the temperature by 30%.
>
> While I don't have a lot of confidence in this calculation, it seems  
> to me that it might not be a bad estimate for a monatomic gas having  
> a mean free path on the order of the cylinder length.  I'm not sure  
> how it might need to be modified for a shorter mean free path (which  
> would lead to a partial thermalization of the gas during the motion  
> of the piston) or for a diatomic gas (which might have its  
> vibrational and rotational modes excited by the collisions as well.)   
> In any event, I'm certain that you would never actually see such a  
> large temperature rise simply as a result of the substantial heat  
> flow to the (actually non-rigid) walls that would take place with the  
> first round of energetic gas collisions.
>
> John Mallinckrodt
> Cal Poly Pomona 
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