How Is Work Calculated in a Gas PV Diagram for Path a-d?

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Homework Help Overview

The discussion revolves around calculating the work done by a gas in a PV diagram, specifically along the path a-d. The problem involves understanding how work is determined when both pressure and volume change, and the original poster seeks clarification on the appropriate method to use.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of the equation W=P(delta)V and question its applicability when pressure is not constant. There is mention of integration and calculating the area under the curve as a method for determining work.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of how to calculate work in this scenario. Some guidance has been offered regarding the use of integration and the geometric interpretation of work as the area under the curve, but no consensus has been reached.

Contextual Notes

Participants are considering the implications of changing pressure and volume, and there is an acknowledgment of the complexity involved in integrating non-constant pressure. The original poster's question highlights a potential gap in understanding the relationship between pressure changes and work calculation.

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Problem.
http://img294.imageshack.us/img294/8068/pvdiagram2fo.gif
Find Work done by each path.

My Solution.
a-b, c-d >> W=P(delta)V
a-c, b-d >> W=0

My Question.
What is the work done by the gas (path a-d)?

My Thoughts.
There is a change in both pressure and volume. However, the equation to find the work in this situation is "W=P(delta)V". There is no (delta)P.
 
Last edited by a moderator:
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could anyone help him?I'm kinda interested in the answer to this as well!
 
Last edited:
Are you familiar with integration? The answer is always found by calculating the area under the curve, which is exactly what you did for all the other ones. In this case, the area is a trapezoid
 
so the work done from a to d is W = deltaP * deltaV
very good, P is no longer constant so integrate that as well
as if it was that simple
 
Last edited:

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