PV diagrams for ideal gas: Finding Work Done

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Homework Help Overview

The discussion revolves around calculating the work done by an ideal gas as represented on a pressure-volume (PV) diagram. The graph features a straight line from point A at 300 kPa, 0.5 L to point B at 500 kPa, 1 L, with a false origin at 200 kPa, 0 L.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the area under the graph and the work done, with one participant calculating the area as a triangle plus a trapezium. There is a noted discrepancy between their calculation of 100 J and the textbook's 200 J, prompting questions about potential errors in reasoning.

Discussion Status

Some participants have offered guidance regarding the calculation methods, suggesting the use of specific area formulas for trapezoids and parallelograms. Multiple interpretations of the area calculation are being explored, but there is no explicit consensus on the correct approach yet.

Contextual Notes

Participants are working under the assumption that the area under the graph directly correlates to the work done, and there is a discussion about the implications of the false origin on the graph.

slaw155
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Homework Statement


Imagine a graph of pressure in kPa (y-axis) against volume in L (x-axis). There is a straight line moving from point A at 300kPa, 0.5L to point B at 500kPa, 1L. The false origin on this graph is 200kPa, 0L. What is the work done by the gas?


Homework Equations



W=pV

The Attempt at a Solution


I know that work is area under graph so its basically the area of a triangle plus area of trapezium. But this gives work = 100J but textbook says 200J. Where have I gone wrong?
 
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See the picture . You need the blue area from the really zero pressure.

ehild
 

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slaw155 said:

Homework Statement


Imagine a graph of pressure in kPa (y-axis) against volume in L (x-axis). There is a straight line moving from point A at 300kPa, 0.5L to point B at 500kPa, 1L. The false origin on this graph is 200kPa, 0L. What is the work done by the gas?

Homework Equations



W=pV

The Attempt at a Solution


I know that work is area under graph so its basically the area of a triangle plus area of trapezium. But this gives work = 100J but textbook says 200J. Where have I gone wrong?
Use the formula for area of a trapezoid: (L1 + L2)/2 x W

AM
 
Last edited:
Andrew Mason said:
Use the formula for area of a parallelogram: (L1 + L2)/2 x W

AM

You meant trapezoid, I think :smile:

ehild
 
ehild said:
You meant trapezoid, I think :smile:

ehild
Right. A trapezoid.

AM
 

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