Finding work for a gas in a piston cylinder

[jdawg]

Homework Statement

A gas in a piston-cylinder assembly undergoes a process for which the relationship between pressure and volume is pV²=Constant. The initial pressure is 1 bar, the initial volume is 0.1 m³, and the final pressure is 9 bar. Determine the final volume and the work for the process.

Homework Equations

The Attempt at a Solution

V₂=sqrt((p₁)(V₁²)/(p₂))
V₂= 0.03333 m³

W=∫ pdV
W=∫ C/(V²) dV
=C[-1/V]
=C[-1/(V₂-V₁)]
=[(p₂V₂²)-(p₁V₁²)][-1/(V₂-V₁)]

=[((900000Pa)(0.0333m³)²)-((100000Pa)(0.1m³)²))][-1/(0.0333-0.1)]

I think the answer is supposed to be W= -19.97KJ, I'm not sure what I'm doing wrong.

 

[Chestermiller]

Homework Statement

A gas in a piston-cylinder assembly undergoes a process for which the relationship between pressure and volume is pV²=Constant. The initial pressure is 1 bar, the initial volume is 0.1 m³, and the final pressure is 9 bar. Determine the final volume and the work for the process.

Homework Equations

The Attempt at a Solution

V₂=sqrt((p₁)(V₁²)/(p₂))
V₂= 0.03333 m³

W=∫ pdV
W=∫ C/(V²) dV
=C[-1/V]
=C[-1/(V₂-V₁)]
=[(p₂V₂²)-(p₁V₁²)][-1/(V₂-V₁)]

=[((900000Pa)(0.0333m³)²)-((100000Pa)(0.1m³)²))][-1/(0.0333-0.1)]

I think the answer is supposed to be W= -19.97KJ, I'm not sure what I'm doing wrong.

Your integration is incorrect, and your determination of the constant C is incorrect. $C=p_1V_1^2=p_2V_2^2$

 

[jdawg]

Oh ok, the constant part makes sense. How is the integration wrong though?
I thought ∫1/x^2 dx = -1/x ?

 

[Chestermiller]

Oh ok, the constant part makes sense. How is the integration wrong though?
I thought ∫1/x^2 dx = -1/x ?

Yes, that's correct, but you substituted the integration limits incorrectly.

Chet

 

[jdawg]

Thanks for your help!