- #1

Markel

- 84

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For example, for an Isothermal expansion, both pressure and volume are changing. For some reason we call V=dv and relate P to V by the ideal gas law and integrate to get work:

W= nRT ln (V

_{F}/ V

_{i})

Now, could we do the same thing, and call work the integral of:

W= [tex]\int[/tex]Vdp

Which would lead to

W= nRT ln (P

_{f}/ P

_{i})

If both quantities are changing? How do we know which to set as the differential, and which to relate in terms of the other variable?

or why couldn't I do a double integral? first over the volume, then over the pressure?

Also: Why does the LAtex code show up all messed up when I hit 'preview post'...?thanks