Question

how to come, dw=pdv in isobaric process.

Answer 1

Here's my GUESS.
if you start with work = force times distance, and then see that pressure = force/area, and that volume is area times length, and then "doodle" a little you may find that force times distance and pressure times volume are the same idea...
force/area times area times length = force times length = force times distance which is work.
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Answer 2

This is commonly derived in the text books in the specialised case of a cylinder which leads to the conclusion that \(dW = F * ds = pA * ds = P~ dV\)

But I think you can generalise out fairly easily.

If we start with the definition of work: \(\delta W = \mathbf F \bullet \delta \mathbf r \)

....and consider a small surface area of the overall generalised containing surface, \(\delta \mathbf S ( = \delta S ~ \mathbf n)\), then the pressure is acting perpendicular to that area and so \(\delta W = \mathbf F \bullet \delta \mathbf r \) becomes \(\delta W = p ~ \delta S ~ \delta r ~ \mathbf n \bullet \mathbf n = p ~ \delta V ~ \mathbf n \bullet \mathbf n \)



This is because \(\delta S ~ \delta r \approx \delta V\) as \(\delta S\) is effectively extruded along \(\mathbf n\) by amount \(\delta r\).

ie: For element \(\delta S\), displacement \(\delta \mathbf r\) has to occur in the direction of the force, so \(\delta \mathbf S\) and \(\delta \mathbf r\) point in the same direction.