Change in potential energy of air in a piston-cylinder system

[Pripyat]

1. Homework Statement
This is my first question and I'll try to stick to the forum rules as much as possible.

imgur.com/a/2JyZC
Imagine a cylinder filled with air which is uniformly distributed and a piston above it, so the air can expand vertically. Let the system be the air itself, so the piston is not included. Due to a resistor in the cylinder, the volume of the air increases with 0.045m^3. The area of the piston surface is 0.09m^2. The mass of the air in the cylinder is 0.27kg, what is the change in potential energy of the air after the process?

ΔV=0.045m³
A_piston=0.09m²
m_air=0.27kg
g=9.81m/s²
ΔPE=...

Homework Equations

ΔPE=m_air*g*Δz
And I assume Δz being the change in height of the center of mass of the air in the cylinder.

The Attempt at a Solution

1) The change in volume and the area of the piston is given, so the air expands with ΔV/A_piston=(0.045/0.09)m=0.5m

2) The center of mass of the air moves with half of this calculated length, so 0.25m.

3) ΔPE=(0.27*9.81*0.25)J=0.662J

According to my book:
What is the change in potential energy of the air, in kJ?
Ans. ≈1.055*10^-3kJ

Not sure where I went wrong on this one. It's almost twice my answer. Is it wrong to take the change in height of the center of mass?

Thanks in advance.

 

[scottdave]

Welcome to PhysicsForums. We will provide input or ask questions to guide you to arrive at a solution to the problem.

The piston has a mass, and it moved up too. It seems that you are thinking the piston does not play a role. What if it had a 1 kg piston, instead of 45 kg piston? Would the volume change be the same for the same amount of heating? I am guessing the resistor heats the air, which then expands (from looking at the picture). Also, they give you Δu of air. Do you have any formulas to incorporate that? Or maybe I'm missing what they are asking for.

 

[Pripyat]

Hello scottdave and thanks for replying,

the mass of the piston plays a role in the main question: What is the heat transfer Q of the resistor to the air for a volume change of 0.045m^3? The piston is at rest in the initial and final state. The pressure of the air, p_air, stays constant during the entire process.

It is assumed there is no friction between piston and cylinder. Also, the cylinder and the piston are both good insulators so heat transfer between these two objects and the air can be neglected. Further, it's a closed system so the following energy balance could be applied:

(ΔPE+ΔKE+ΔU)_air=Q-W

PE=gravitational potential energy
KE=kinetic energy
U=internal energy
Q=net energy transfer by heat to the system
W=net energy transfer by work from the system

1) I neglected ΔKE and ΔPE for the main question

2) ΔU=Δu*m, u being the specific internal energy

3) For the work, we need to know p_air and here is where the mass played a role:
Force balance for when the piston is at rest
p_air*A_piston=m_piston*g+p_atm*A_piston

4) W= ∫p_air*dV → pressure of air stays constant → W=p_air*ΔV

5) Correction:
So Q_resistor=Q=Δu*m+W=42*0.27+1.049*4.5=16.1kJ
Side question: What is ΔPE_air (which is usually neglected because it's relatively small)?
I thought I should use the height difference of the center of mass of the contained air for my calculation, but apparently the book says otherwise.