Re: [Phys-l] T dS versus dQ

[Carl Mungan <mungan@usna.edu>]

I wrote:

> > I have an ideal gas in a thermally insulated cylinder and piston. I 
> > suddenly compress the piston. To be specific, suppose the piston's 
> > speed profile starts from zero, rapidly (practically stepwise) 
> > rises up to the speed of sound, then drops rapidly back to zero 
> > once the gas is compressed by dV (which is negative). In practice I 
> > might accomplish this by having a huge weight sitting on the piston 
> > which is at the top end of the cylinder and held in place by a pin. 
> > I pull out the pin and let the piston fall a distance dx until it 
> > slams into a stop.
> >
> > I think we have T dS > 0 (because the process is certainly 
> > irreversible), dQ = 0, dE > 0 (because work -P dV was done on the 
> > gas) and so it doesn't look like dQ can be equal to either T dS or 
> > dE. I "computed" dQ by noting that the cylinder (including the 
> > piston) is thermally insulated (and I'm further helped that the 
> > process is so fast there isn't time for heat transfer even if it 
> > weren't insulated, noting that no thermal insulation is perfect in 
> > the real world).

John M wrote:

> I agree with Carl that his scenario seems to me to be a case in 
> which  dq equals zero while dE and TdS don't.  ("TdS" doesn't even 
> really  make sense for this or any other irreversible 
> noninfinitesimal  process.)  Because processes like this are 
> essentially the norm, I'm  pretty sure I (and perhaps Carl as well) 
> must be misunderstanding  John Denker's point.  I wasn't following 
> the thread closely up until  now, so maybe I missed some critical 
> caveats along the way.
>
> In general it seems to me that for any specified change one can write:
>
>   dE = dq + dw = TdS + dw_rev
>
> where all quantities are understood as integrals, dw_rev is the work 
> that would have been done along a reversible path, and TdS is an 
> integral over the same reversible path, so that
>
>   dq = dE - dw
>         = TdS + dw_rev - dw
>
> Thus,
>
> a) dq = dE only when no work is done
>
> and
>
> b) dq = TdS only when dw_rev = dw
>
> This all seems pretty standard to me so I still can't help but think 
> I'm missing something.
>
> It occurs to me as I finish this that the difference may have to do 
> with restricting one's attention to infinitesimal processes which, I 
> guess, are always reversible.  But in that case dq is always equal 
> to  TdS, isn't it?

On the one hand, John M is correct that I was sloppy in stating that 
-P dV is the work. The work is positive but since I'm moving the 
piston rapidly it cannot be easily calculated as an integral over the 
actual process but instead should be calculated from W = E_final - 
E_initial which we can do because we have well-defined equilibrium 
initial and final states. I believe that is also the main point that 
Bob S was trying to make. I agree with John M's summary points (a) 
and (b) which seem nicely stated.

However, I'm not sure I agree with his final paragraph. Look at 
Schroeder page 112. I think we *can* have irreversible infinitesimal 
processes. It's just that dQ doesn't equal T dS *and* dW doesn't 
equal -P dV for them. Carl
--
Carl E Mungan, Assoc Prof of Physics  410-293-6680 (O) -3729 (F)
Naval Academy Stop 9c, 572C Holloway Rd, Annapolis MD 21402-1363
mailto:mungan@usna.edu     http://usna.edu/Users/physics/mungan/