How Is Net Work Calculated in a Cyclic Ideal Gas Process?

[eriadoc]

Homework Statement

An ideal gas follows the three-part process shown in the figure.

At the completion of one full cycle, find the net work done by the system.

Homework Equations

W=P*deltaV; Total Work = Wab+Wbc+Wca
A=1/2bh

The Attempt at a Solution

Work done from C to A is zero, because the volume is not changing. Work done from B to C is (50kPa)(-3) = -150kJ.

Work done from A to B is where I'm lost. I know it's supposed to be the area, so I did .5(100kPa)(3) = 150kJ.

Therefore, total work would be zero. The answer is 150kJ, but I can't figure out why. I've read the answer on Cramster and it honestly confused me even more. I feel like I'm close, but I just need an explanation more so than the math. TIA.

 

[da_nang]

It's not the area of the triangle you're looking for, it's the area under the graph to the V-axis you want. See it as a basic integral W = _a∫^bpdV. To put it more simply, the work done by one process is that integral. Since you're looking for the total work done by the system, all you really need to do is calculate the area of the triangle as it already has subtracted W_bc.

 

[Andrew Mason]

da nang is right. The work done by the gas from A to B is the whole area under AB not just the purple triangle - ie. the purple triangle PLUS the area under BC = 300 KJ.

AM