- #1

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That is, I know that [itex] dW=P \text{ }dV[/itex], but why is that true physically? I realize that [itex] W=f \cdot d \text{ and } F=P \cdot A [/itex] .Thanks.

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- #1

- 6

- 0

That is, I know that [itex] dW=P \text{ }dV[/itex], but why is that true physically? I realize that [itex] W=f \cdot d \text{ and } F=P \cdot A [/itex] .Thanks.

- #2

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W = P x V...P = F/A and V = A x d

W = F/A x A/d ...= F x d

- #3

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- #4

- 350

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- #5

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Oh...wow I'm silly. Thanks!

- #6

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you are in good company!

- #7

Gold Member

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- #8

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think about a situation where the internal pressure is much higher than the external presssure. how do you calculate work then?

thafeera

- #9

- 350

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P x V units are Pa x m^3 = (N/m^2) x m^3 = Nm = energy units

v x t units are ms^-1 x s = m = distance units

F x e units N x m = energy units

V x Q units are volts x charge = V x C = (J/C) x C = J = energy units

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