# Does the $Vdp$ term in thermodynamics always imply some kind of work?

I found an answer here: What is $v \, dp$ work and when do I use it? It says that $$V\,dP$$ term is isentropic shaft work.

Let us consider an adiabatic piston gas system which has some sand mass on top of the piston. The mass is slowly decreasing such that the process is reversible. So for the gas in the system $$PV=nRT$$. Now let us differentiate on both sides $$PdV+VdP=nRdT$$, (Assuming that gas does not dissociate). For adiabatic process $$-dU=-nC_v\,dT=\delta W=P\,dV$$ so putting $$P\,dV$$ as $$-nC_v\,dT$$ we see that $$Vdp=nR\,dT+nC_v\,dT$$.What does the $$V\,dP$$ term signify for non flow systems? Are variable external pressure systems even considered closed systems? And if it implies some kind of work, why isn't is considered in $$dU=-nC_v\,dT=\delta W=P\,dV$$?

• The term "shaft work" implies an open system with gas flowing through the control volume, not a closed system like the one you are considering. Have you not learned about the open system version of the 1st law of thermodynamics yet? The shaft work is the result of turning a turbine or other open system work device. – Chet Miller May 2 at 11:57
• @Chet Miller that is what i am asking if VdP means something in closed systems too? – Sarvesh Maheshwari May 2 at 16:27
• It guess its the differential change in enthalpy at fixed entropy for a constant composition single phase system. Otherwise, I can't think of anything. – Chet Miller May 2 at 16:52

The $$V\,dP$$ expression is not limited to thermodynamics—and so not limited to thermodynamically large, closed, equilibrium, steady state, etc. systems. It is in fact just the work-energy theorem $$dW=F\,dx$$ rewritten with a factor of area moved from the force to to distance. (That is, $$F=PA$$ and $$dx=dV/A$$.) So in any case where it possible to calculate the kinetic pressure on the boundary of a changing volume, $$V\,dP$$ gives the correct expression for the work done.