PV diagrams for ideal gas: Finding Work Done

[slaw155]

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?

 

[ehild]

See the picture . You need the blue area from the really zero pressure.

ehild

 

[Andrew Mason]

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

 

[ehild]

Use the formula for area of a parallelogram: (L1 + L2)/2 x W

AM

You meant trapezoid, I think

ehild

 

[Andrew Mason]

You meant trapezoid, I think

ehild

Right. A trapezoid.

AM