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

Question

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 $dU=-nC_v\,dT=\delta W=P\,dV$?

Answer 1

The short answer to the title of your post is no. A simple example is a closed system isochoric (constant volume) process involving an ideal gas. Heat transfer increases or decreases the pressure (and temperature) and no boundary work is done.

Hope this helps.

Answer 2

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.