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Physics 8 Online
OpenStudy (anonymous):

For the process shown in the p-V diagram in the following figure, the total work in going from a to d along the path shown is

OpenStudy (anonymous):

OpenStudy (anonymous):

(a) 15 × 10^5 J (b) 9 × 10^5 J (c) 6 × 10^5 J (d) 1 × 10^5 J

OpenStudy (anonymous):

@cormacpayne this is really hard I have no clue

OpenStudy (anonymous):

@campbell_st @hartnn I need your help please

OpenStudy (anonymous):

@cormacpayne please check this for me!

OpenStudy (anonymous):

If I remember right, in a PV diagram, the amount of work done along a path is the area under the curve (or path).

OpenStudy (anonymous):

can you explain more please, I know I asked you alot

OpenStudy (anonymous):

It's no problem. A key feature of the diagram is that the amount of energy expended or received by the system as work can be estimated as the area under the curve on the chart. For a cyclic diagram, the net work is that enclosed by the curve.

OpenStudy (anonymous):

and how do I find the area!

OpenStudy (anonymous):

You just have to find the area under the thick blue line, specifically the horizontal line at the top since the area under vertical lines is zero. So find the area under the blue horizontal line that is 3 x 10^5 pascals "tall" and 3 m^3 "long"

OpenStudy (anonymous):

oh yeah so it would be 9*10^5

OpenStudy (anonymous):

Correct!

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