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

AI Thread Summary
The net work done by an ideal gas in a cyclic process is calculated by summing the work done in each segment of the cycle. The work from point C to A is zero due to no volume change, while the work from B to C is -150 kJ. The work from A to B requires calculating the area under the curve, which includes both the triangle and the area under BC, resulting in a total of 300 kJ. The confusion arises from misinterpreting the area needed for the calculation, emphasizing the importance of considering the entire area under the curve for accurate results. Understanding this integral approach clarifies how to determine the total work done in the cycle.
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Homework Statement



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

Walker4e.ch18.Pr025.jpg


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.
 
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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 = abpdV. 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 Wbc.
 
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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
 
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