PV diagrams for ideal gas

  • Thread starter slaw155
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  • #1
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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?
 

Answers and Replies

  • #2
ehild
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See the picture . You need the blue area from the really zero pressure.

ehild
 

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  • #3
Andrew Mason
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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?
Use the formula for area of a trapezoid: (L1 + L2)/2 x W

AM
 
Last edited:
  • #4
ehild
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Use the formula for area of a parallelogram: (L1 + L2)/2 x W

AM
You meant trapezoid, I think :smile:

ehild
 
  • #5
Andrew Mason
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You meant trapezoid, I think :smile:

ehild
Right. A trapezoid.

AM
 

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