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# Re: [Phys-l] T dS versus dQ

• From: Carl Mungan <mungan@usna.edu>
• Date: Wed, 13 Jan 2010 10:19:44 -0500

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/