PV graph question (trying to determine the work)

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To determine the work done by the system from points A to C on the PV diagram, one must calculate the area under the curve between these points. The initial attempt at calculating work resulted in an incorrect value of 1596J. The correct approach involves using the formula for work, which is pressure multiplied by the change in volume (W = P * ΔV). It's essential to accurately identify the areas corresponding to the processes A to B and B to C on the graph. Properly applying these principles will yield the correct work done by the system.
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Homework Statement


Look at the PV diagram of a system above. (attached) The units for P and V are atmospheric pressure atm and liter L. Process A to B and B to C are straight lines on it.
How much is the work done by the system from A to C in J?



Homework Equations


i know that work is the area under the graph, but when i add up areas, answer is wrong for some reason??


The Attempt at a Solution


W = .5(4*3) + .5(4*1.5) * 152J
= 1596J

this is wrong though
?

 

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>>>>Work Done is...[ p*delta V] i.e. pressure multiplied by CHANGE IN VOLUME.

This is enough for a hint...^^^
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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