Thermodynamics piston–cylinder assembly Question

In summary: As long as you're dealing with Pascals and meters and Kelvin, your mass will come out in kilograms.In summary, the problem involves a piston-cylinder assembly with 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, with the pressure-volume relationship being pV=constant. Assuming ideal gas behavior, the mass of the air is 4.66 kg, and the work and heat transfer are both 277.2 kJ. The temperature remains constant throughout the process.
  • #1
derrickb
22
0

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=[itex]\int[/itex]PdV
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=[itex]\int[/itex]PdV
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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  • #2
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?
 
  • #3
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
 
  • #4
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.
 
  • #5
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?
 
  • #6
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.
 
  • #7
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
 
  • #8
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.
 
  • #9
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.
 

What is a thermodynamics piston-cylinder assembly?

A thermodynamics piston-cylinder assembly is a system used to study the principles of thermodynamics. It consists of a piston that can move freely in a cylinder, creating a closed space where gas or fluid can be compressed or expanded.

What is the purpose of a thermodynamics piston-cylinder assembly?

The main purpose of a thermodynamics piston-cylinder assembly is to demonstrate and study the principles of thermodynamics, such as heat transfer, work, and energy conservation. It is also used to investigate the properties of different gases and fluids under varying conditions.

How does a thermodynamics piston-cylinder assembly work?

A thermodynamics piston-cylinder assembly works by using a piston to compress or expand a gas or fluid inside the cylinder. This change in volume and pressure allows for the study of the gas or fluid's thermodynamic properties, such as temperature, pressure, and volume.

What are the applications of a thermodynamics piston-cylinder assembly?

A thermodynamics piston-cylinder assembly has many applications, including in research and development of engines, refrigeration systems, and power generation. It is also used in educational settings to teach the principles of thermodynamics.

What are the limitations of a thermodynamics piston-cylinder assembly?

One of the main limitations of a thermodynamics piston-cylinder assembly is that it is a simplified model and does not fully represent real-world systems. It also has limitations in terms of the range of temperatures and pressures that can be studied. Additionally, it does not take into account external factors such as friction and heat loss.

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