Work from a pressure-volume graph

[theonlytycrane]

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Using W = P * deltaV, I thought that I would find the work done for each leg of the triangle and sum the results for the total Work. For the green leg, I wasn't sure how this would be calculated as Pressure and Volume are both changing.

The solution, however, is to interpret the Work done by the gas in the cycle as the area of the triangle. I'm trying to better understand why this interpretation is used. Is my initial approach invalid?

Thanks!

 

[bee17]

Whenever you have a pressure vs. volume graph, work is the area under the curve (or in this case, the area enclosed by the curve) . I'm not particularly sure why... probably some calculus mumbo-jumbo we didn't learn in my physics for non-majors class 🙂

For the MCAT, just know work is the area under the curve, or, if it is a cycle, the area of the shape made by the curve.

 

[Axes]

yup

 

[theonlytycrane]

Updating my own (old) question 🙂

W = PV comes from W = Fdcos(theta) where theta = 0.

Using Pressure = Force / Area, W = Pressure * Area * d = PV.

 

[aldol16]

More concretely, dW = P*dV. The variables are independent, so separation of variables and integration are valid and you can just integrate the left with respect to W and the right with respect to V, given you W = integral(P*dV).