Piston Pressure: How to Calculate Forces and Work in a Gas Cylinder System

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Homework Help Overview

The problem involves calculating forces and work in a gas cylinder system, specifically focusing on a piston with a given diameter and gas pressure. The original poster seeks assistance with multiple parts of the problem, including the forces exerted by the gas and the environment, as well as the work done during the gas expansion.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the conversion of pressure units from atmospheres to pascals and question the accuracy of their calculations. The original poster expresses confusion regarding the forces calculated and the implications of environmental pressure.

Discussion Status

Some participants have provided guidance on unit conversion, while others express frustration over their calculations. There is a lack of consensus on the correct approach to determining the forces and work, with multiple interpretations being explored.

Contextual Notes

The original poster notes a potential misunderstanding regarding the pressure of the environment and its role in the calculations. There is also mention of the need for accurate unit conversions, which may be affecting the results.

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Homework Statement


A cylinder with a 16-cm-diameter piston contains gas at a pressure of 3.0 atm.

A.How much force does the gas exert on the piston?

B.How much force does the environment exert on the piston?

C.The gas expands at constant pressure and pushes the piston out 10 cm. How much work is done by the environment?

D.How much work is done by the gas?

Homework Equations


P=F/A=N/m^2
W=Fd

The Attempt at a Solution


A. F=(.08^2)(pi)(3)(1.013e5)=6110 N [Correct]
B. F=(.08^2)(pi)(1)(1.013e5)=2040 N[Incorrect]
C. [Unable to calculate due to part B]
D. W=(6110)(.1)=611 J[Correct]

Isn't 1 atm the pressure of the environment? Any help would be appreciated.
 
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im pretty sure you have to convert atms to Pa. 1 x 10^5 Pa = 1atm
 
alicemunro said:
im pretty sure you have to convert atms to Pa. 1 x 10^5 Pa = 1atm

I did convert it into Pa 1 atm = 1.013e5

F = (1.013e5 Pa)(.08 m)^2(pi)= 2037 N, I submitted this but it says incorrect.
 
I can't seem to get this right, am I doing something wrong? Please help, its getting frustrating.
 
Anyone?
 

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