Should we not be using Pdv for calculating the work done during a reversible adiabatic process ? Why are we using Vdp ? Can some one please explain this ? (P.S. I'm taking an introductory course on Thermodynamics , so please keep it simple )

(Edit: The process in question is a steady state process. Does this fact change the work transfer formula ?)

  • $\begingroup$ Can you give more information on your process? $\endgroup$ – rmhleo Aug 17 '15 at 21:34
  • $\begingroup$ I'm sorry,but I've provided all the information I have . $\endgroup$ – Kishore Pattabhiraman Aug 18 '15 at 3:19
  • 1
    $\begingroup$ See this. $\endgroup$ – Ant Aug 18 '15 at 3:28

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