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Jun 25, 2013 at 23:11 comment added joshphysics @DepeHb Using the sign conventions I used in the response, one would have $\delta W = +PdV$. This can be used with ideal gases for example. And yea, for isothermal processes you would use that expression along with the ideal gas law to write $P$ as a function of $V$, and then you would explicitly perform the integral.
Jun 25, 2013 at 21:02 comment added Spine Feast Thanks for the answer. BTW, when is it true that $\delta W=-pdV$? I recall this being a useful tool for quick calculations, but surely its applicability is limited? For example, in path (3), the definition $\Delta T = 0$ implies $\Delta E =0$, so $\delta Q = \delta W$. But I know of no way to integrate $\delta Q$, and since it's not a state variable, I can't just write $Q_b - Q_a$...
Jun 25, 2013 at 20:49 vote accept Spine Feast
Jun 25, 2013 at 20:33 history answered joshphysics CC BY-SA 3.0