Thermodynamics piston–cylinder assembly Question

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Discussion Overview

The discussion revolves around a thermodynamics problem involving a piston–cylinder assembly containing air. Participants explore the relationships between pressure, volume, mass, work, and heat transfer during a process where the pressure-volume relationship is described as pV = constant. The conversation includes attempts to solve for mass, work, and heat transfer, while addressing unit consistency and the implications of the process on temperature.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant calculates the mass of air using the ideal gas law, arriving at approximately 4.66 kg, but expresses uncertainty about the correctness of this value.
  • Another participant points out that the specific gas constant for air should be 287 J kg-1 K-1, questioning the use of 0.286 in the calculations.
  • There is a debate about whether work done during the process is zero, with some arguing that since pV is constant, work should be zero, while others assert that since pressure changes, work must be done.
  • One participant suggests that a temperature change occurs during the expansion, which contradicts another participant's assertion that temperature remains constant if pV is constant.
  • After clarifying the problem statement, it is acknowledged that pV = constant, leading to a consensus that temperature does remain constant.
  • One participant claims to have calculated work as 277.2 kJ, contingent on the correctness of the mass calculation.

Areas of Agreement / Disagreement

Participants express disagreement regarding the implications of the pressure-volume relationship on temperature and work. While there is some consensus on the mass calculation, the discussion remains unresolved on the nature of work done and the temperature change during the process.

Contextual Notes

There are unresolved issues regarding unit consistency, particularly the transition between different unit systems (e.g., bar to Pascal). Additionally, the initial confusion over the problem statement led to differing interpretations of the thermodynamic relationships involved.

Who May Find This Useful

This discussion may be useful for students studying thermodynamics, particularly those working on problems involving ideal gas behavior and piston-cylinder assemblies.

derrickb
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Homework Statement


A piston–cylinder assembly contains air, initially at 2 bar, 300 K, and a volume of 2 m3. The air undergoes a process to a state where the pressure is 1 bar, during which the pressure–volume relationship is pV 5 constant. Assuming ideal gas behavior for the air, determine the mass of the air, in kg, and the work and heat transfer, each in kJ.


Homework Equations


Pv=RT, v=specific volume
W=\intPdV
Q-W=ΔE


The Attempt at a Solution


Pv=RT
(2 bar)(100 kPa/bar)v=(.286)(300 K)
v=0.429 m3/kg

v=V/m
m=V/v
m=2 m3/0.429 m3/kg
m=4.66kg

I feel like this may be incorrect.

W=\intPdV
Would W=0 because PV is constant?

I'm kind of lost at this point. If anyone could point me in the right direction, I would really appreciate it
 
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derrickb said:

Homework Statement


A piston–cylinder assembly contains air, initially at 2 bar, 300 K, and a volume of 2 m3. The air undergoes a process to a state where the pressure is 1 bar, during which the pressure–volume relationship is pV 5 constant. Assuming ideal gas behavior for the air, determine the mass of the air, in kg, and the work and heat transfer, each in kJ.


Homework Equations


Pv=RT, v=specific volume

You're not necessarily dealing with 1 kg of air. So your formula should be
pV =mRspecificT, V = actual volume, m= actual mass, p in Pascals. 1 Pascal ~ 1e-5 bar.

Rspecific for air is 287 J kg-1 K-1 so where did you get 0.286? Are you using SI units thruout I hope?
 
rude man said:
You're not necessarily dealing with 1 kg of air. So your formula should be
pV =mRspecificT, V = actual volume, m= actual mass, p in Pascals. 1 Pascal ~ 1e-5 bar.

Technically, the formula you put is the same as mine, I just divided V by m to get a separate variable v. That formula shouldn't only apply to 1 kg of air. In the steps after, I substituted V/m in for v and solved for m. There isn't anything wrong with this; I guess it's just the longer way of doing it. As for units, I thought that I had them correct. Maybe the .286 is kJ/kgK? I'm not really sure
 
derrickb said:
As for units, I thought that I had them correct. Maybe the .286 is kJ/kgK? I'm not really sure

Yes, that would account for your being off by a factor of 1000.

You need to stick to SI units. The SI unit is the Joule, not kilojoule.
 
Ok I'll make sure to watch units. The answer I got for mass wouldn't change, the decimal place will just move. Can anyone help with the work part of the problem? Would work be 0 because pV remains constant?
 
derrickb said:
Ok I'll make sure to watch units. The answer I got for mass wouldn't change, the decimal place will just move. Can anyone help with the work part of the problem? Would work be 0 because pV remains constant?

Work = ∫pdV.
pV0.5 = cosntant.

Since you're told p changes, then so must V. And if V changes, work will be done either on or by the system.
 
To expand on what rude man said, there is also a temperature change during the expansion. That is how the pressure and volume are controlled to satisfy pv0.5 = const. Use rude man's equations of the previous post to calculate the amount of work done. Tell us what you get.

Chet
 
I don't understand why there would be a temperature change in this problem. It says pV as a whole remains constant, but the values of p and V change. If pV is constant, mRT remains constant and, assuming there isn't a change in mass, the temperature remains the same. I finished the problem and got a mass=4.65kg, W=277.2kJ, and Q=277.2kJ. If T remains constant, Q must equal W because ΔE=0.
 
derrickb said:
I don't understand why there would be a temperature change in this problem. It says pV as a whole remains constant, but the values of p and V change. If pV is constant, mRT remains constant and, assuming there isn't a change in mass, the temperature remains the same. I finished the problem and got a mass=4.65kg, W=277.2kJ, and Q=277.2kJ. If T remains constant, Q must equal W because ΔE=0.

It does not say pV remains constant. I think it says pV0.5 remains constant. So T changes.

Your answers are correspondingly wrong.

I have to admit that the statement as you posted it is questionable. Could you dig up the original for us?
 
Last edited:
  • #10
Yeah I just reread the question statement that I posted and it must have changed when I copy and pasted it from the pdf of the book. Instead of pV 5 constant, it says pV=constant. Sorry for all the confusion.
 
  • #11
derrickb said:
Yeah I just reread the question statement that I posted and it must have changed when I copy and pasted it from the pdf of the book. Instead of pV 5 constant, it says pV=constant. Sorry for all the confusion.

OK, so temperature does remain constant.

But work is still not zero.
 
  • #12
YEah I figured out how to solve for work and my answer was 277.2 kJ I believe.
 
  • #13
derrickb said:
YEah I figured out how to solve for work and my answer was 277.2 kJ I believe.

That answer is correct if you computed the mass right, which I did not check.
 

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