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Simple (edit: Thermal) Physics Question

  
 
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kingnat
Old 10-24-2007, 12:41 PM     Post subject: Simple (edit: Thermal) Physics Question #1 (permalink)  
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 Given the level of discussion Boost's post received I'd thought I'd throw another interesting question out there. It should be most interesting if people attempted to answer the question immediately after reading before checking out everyone else's reply... but I doubt many will do that.

This is not meant to have tricky language, so feel free to ask questions if you are confused. This is a physics question, so assume everything is ideal and such...

An ideal gas of temperature T, volume V, and pressure P, is inside a piston which is surrounded by an insulating jacket, and placed vertically on a table, (such that the piston head could move vertically). The frictionless piston head of mass M is free to move, but initially is at rest at a height H above the table. Slowly, a large number of small masses (considerably smaller than M) are added to the piston head.

Describe what happens to the gas after the masses are added, specifically whether the temperature, volume, and pressure increase, decrease, remain the same, or if there is not enough information given to determine the answer.
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a500lbgorilla
Old 10-24-2007, 01:04 PM #2 (permalink)  
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This is a thermodynamics problem, not a simple physics problem.

Smithers, use the amnesia ray.
You mean the revolver, sir?
Precisely.
 
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kingnat
Old 10-24-2007, 01:07 PM #3 (permalink)  
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Quote:
Originally Posted by a500lbgorilla
This is a thermodynamics problem, not a simple physics problem.
What's not simple about thermo?
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Old 10-24-2007, 01:11 PM #4 (permalink)  
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Quote:
Originally Posted by kingnat
Quote:
Originally Posted by a500lbgorilla
This is a thermodynamics problem, not a simple physics problem.
What's not simple about thermo?
Trying to call me out? I'm not very good at thermo unless it's applied to a compressible flow

Smithers, use the amnesia ray.
You mean the revolver, sir?
Precisely.
 
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kingnat
Old 10-24-2007, 01:20 PM #5 (permalink)  
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Quote:
Originally Posted by a500lbgorilla
Quote:
Originally Posted by kingnat
Quote:
Originally Posted by a500lbgorilla
This is a thermodynamics problem, not a simple physics problem.
What's not simple about thermo?
Trying to call me out? I'm not very good at thermo unless it's applied to a compressible flow
Not in the least... I don't know squat about compressible flow... if anyone ever asks me a question about a real system I'm lost. Most physics folk treat thermo like the red-headed step-child of statistical mechanics, so much so that many schools are shifting away from teaching thermodynamics at the upper-level (or at least highly limiting it) in favor of more stat mech. I don't agree with that, but that's what's happening.... does this mean you aren't going to answer my question?
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a500lbgorilla
Old 10-24-2007, 01:29 PM #6 (permalink)  
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Anyway, here's my guess.

So balancing the forces you get (Internal - Ambient)*Area = M*g and ambient is considered constant. To increase M means you must increase P.

According to PV=RT, this is done either through decreasing volume or increasing T. To decide which it is, we must consult some laws.

The 1st law involves enthalpy which involves temperature. I don't know if I remember it exactly, but here goes.

I believe its Win - Qout = h2 - h1 = something.

The Qout is zero because it is an insulated piston. However, there is boundary pressure work being done. Work is a N*m, pressure being a N/m^2 says I probably need to multiply (Internal - Ambient)*Area*Distance Compressed. Which indicates a volume change will probably occur. Also, the work into the system will be positive so the volume will decrease. I guess Area*dheight is a dVolume? So the boundary work is PdV.

So PdV = h2 - h1 = cp(T2 - T1)? So for a dV to occur, a temperature change must occur. The left hand side is positive for this equation so T2-T1 > 0, suggesting temperature increases.

So as you add mass, the volume decreases, pressure increases and temperature increases.

EDIT maybe I should check the 2nd law to see if one of those remains constant. It's an isentropic process so ds = 0. Lemme think.

EDIT EDIT and where is this temperature increase coming from? Energy of the system remains constant. Well, I guess the energy of the system will change. The piston has some potential energy which is altered for a kinetic energy which will be passed into the fluid before equilibrium is reached again.

I'm gonna check out my book to remind me the 2nd law.

Smithers, use the amnesia ray.
You mean the revolver, sir?
Precisely.
 
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spoonitnow
Old 10-24-2007, 01:34 PM #7 (permalink)  
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a500lbgorilla
Old 10-24-2007, 01:42 PM #8 (permalink)  
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yah, I dont see anything about the 2nd law that would help. I'm satisfied with my answer.

Smithers, use the amnesia ray.
You mean the revolver, sir?
Precisely.
 
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Jibalob
Old 10-24-2007, 01:58 PM #9 (permalink)  
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Quote:
Originally Posted by a500lbgorilla
So as you add mass, the volume decreases, pressure increases and temperature increases
This is pretty basic knowledge, why all the formulae and laws?
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Old 10-24-2007, 02:13 PM #10 (permalink)  
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Quote:
Originally Posted by Jibalob
Quote:
Originally Posted by a500lbgorilla
So as you add mass, the volume decreases, pressure increases and temperature increases
This is pretty basic knowledge, why all the formulae and laws?
So that I know why it will happen. Understand the process and everything will fall in line.

Also I don't have a very good internal model for temperatures, so anytime temps are added I have a little trouble working them through in my head.

EDIT and my way, you could predict the magnitudes of change for each of the variables.

Smithers, use the amnesia ray.
You mean the revolver, sir?
Precisely.
 
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Warpe
Old 10-24-2007, 03:24 PM #11 (permalink)  
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 Volume decreases and pressure increases, and both will remain at their new level. Temperature increases but will decrease again to reach equilibrium with its surroundings unless the insulating jacket is a perfect insulator (not enough information given).



 
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