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

During the fast process the input energy to the gas is well defined as mgh. But during the reversible, slow process the energy given to the gas is the aformentioned area under the (mg - F(t) ) curve.
The two end states are not the same. The two states involved in the fast process cannot be joined by a single isentropic process. If you choose to control F(t) so as to end up at the same final gas pressure as Pf of the fast process, you will end up at a different final volume. You will have to add a second reversible process (NOT isentropic) to get to the final state of the fast process.

Bob Sciamanda
Physics, Edinboro Univ of PA (Em)
treborsci@verizon.net
http://mysite.verizon.net/res12merh/

--------------------------------------------------
From: "LaMontagne, Bob" <RLAMONT@providence.edu>
Sent: Sunday, January 17, 2010 5:22 PM
To: "Forum for Physics Educators" <phys-l@carnot.physics.buffalo.edu>
Subject: Re: [Phys-l] T dS versus dQ

If the energy input is well defined, isn't the final equilibrium increase in the temperature of the gas well defined? After waves have dissipated, there doesn't seem to be any other way for the energy input, mgh, to manifest itself - leaving S = k lnT the same.

Bob at PC
I know you are all being patient with me, but I really am trying to get to that "aha" moment where it all comes together.

________________________________________
From: phys-l-bounces@carnot.physics.buffalo.edu [phys-l-bounces@carnot.physics.buffalo.edu] On Behalf Of Bob Sciamanda [treborsci@verizon.net]
Sent: Sunday, January 17, 2010 4:51 PM
To: Forum for Physics Educators
Subject: Re: [Phys-l] T dS versus dQ

Bob LaMontagne wrote:
. . the oscillation follows the same P-V adiabatic curve during the
oscillation as it does in the slow adiabat . . .

I'll try one final time:
During the "fast" (irreversible) process there is no defined "P-V adiabatic
curve" to be followed. P of the gas isn't even defined during this process.

Bob Sciamanda
Physics, Edinboro Univ of PA (Em)
treborsci@verizon.net
http://mysite.verizon.net/res12merh/

--------------------------------------------------
From: "LaMontagne, Bob" <RLAMONT@providence.edu>
Sent: Sunday, January 17, 2010 11:44 AM
To: "Forum for Physics Educators" <phys-l@carnot.physics.buffalo.edu>
Subject: Re: [Phys-l] T dS versus dQ

Sorry - I was trying to follow the thread - someone had proposed letting a
weight fall through mgh to a stop. If there is no stop then the mass
certainly oscillates. However, again following the thread of adiabatic,
the oscillation follows the same P-V adiabatic curve during the
oscillation as it does in the slow adiabat (neglecting for the moment the
sound wave, etc., points that have been raised on this thread). It
overshoots the final stopping point (h) along that adiabat and then
retraces itself - the net result being the area under the reversible
quasi-equilibrium process - mgh.

Bob at PC
________________________________________
From: phys-l-bounces@carnot.physics.buffalo.edu
[phys-l-bounces@carnot.physics.buffalo.edu] On Behalf Of Bob Sciamanda
[treborsci@verizon.net]
Sent: Sunday, January 17, 2010 1:51 AM
To: Forum for Physics Educators
Subject: Re: [Phys-l] T dS versus dQ

Bob LaMontagne wrote:
Yes it does the work mgh, but it produces significant KE of the piston.
If
mg is >> P A, then almost all of mgh goes into the piston KE.

True, when you just drop the weighted piston, it will fall and oscillate
momentarily. But when it has finally comes to rest, where has all of the
mgh energy gone? Under the stated constraints it can only have gone to
the
gas.

Bob L wrote:
If you want to compress the gas, then (F + mg)/ A must be > = P at any
instant. (Here F is an applied force to the piston . . .

In the reversible case, you slowly lower the weight by pulling up with the
variable force F so that at every instant the net external downward
pressure
(mg-F)/A only infinitesimally exceeds the gas pressure P. The gas
pressure
P slowly rises in value and you continuously reduce the strength of F
until
finally F=0 and mg/A=P. The total work done on the gas is then the area
under the (mg-F) curve. This will be less than mgh.

Bob Sciamanda
Physics, Edinboro Univ of PA (Em)
treborsci@verizon.net
http://mysite.verizon.net/res12merh/

--------------------------------------------------
From: "LaMontagne, Bob" <RLAMONT@providence.edu>
Sent: Saturday, January 16, 2010 10:27 PM
To: "Forum for Physics Educators" <phys-l@carnot.physics.buffalo.edu>
Subject: Re: [Phys-l] T dS versus dQ

Yes it does the work mgh, but it produces significant KE of the piston.
If
mg is >> P A, then almost all of mgh goes into the piston KE. If you want
to compress the gas, then (F + mg)/ A must be > = P at any instant. (Here
F is an applied force to the piston and, again, we are assuming no energy
passing through the cylinder wall.) The force of the piston acting on the
gas cannot be greater than the force of the gas back on the piston.

It's like a car hitting a brick wall and driving through it. The wall
didn't break because the force of the car on the wall was greater than
the
wall on the car. Their mutual equal and opposite force at some point
exceeds the breaking strength of the wall.

Bob at PC

________________________________________
From: phys-l-bounces@carnot.physics.buffalo.edu
[phys-l-bounces@carnot.physics.buffalo.edu] On Behalf Of Bob Sciamanda
[treborsci@verizon.net]
Sent: Friday, January 15, 2010 2:03 PM
To: Forum for Physics Educators
Subject: Re: [Phys-l] T dS versus dQ

Bob LaMontagne wrote:
My claim was that the final temperatures would be the same if the process
proposed would be done quickly or slowly.

Bob, let me try to make more explicit what others have already said:

If you simply drop the weighted piston, it does the work mgh, all of
which
goes to increase the gas temperature.
But if you want to very slowly lower the weight, you will have to hold
back
with a varying upward force against the weight of the "falling" piston
throughout the process.
The downward force of the piston upon the gas is now a varying force,
increasing slowly to the final value mg, during the compression process.
The area under this variable F(x) curve is less than mgh, so that less
work
is done on the gas to go towards a temperature increase.

Hope this helps.

Bob Sciamanda
Physics, Edinboro Univ of PA (Em)
treborsci@verizon.net
http://mysite.verizon.net/res12merh/

_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
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_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
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_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l
_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l

_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l
_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l