1

For a reversible process, it is assumed that the external pressure $P_{ext}$ is infinitesimally different from internal pressure $P_{int}$.

So in reversible process, I can have $~P_{int}=P(V,T)~$ but also $~V=V(P_{int},T)~$.

Then what is the obvious reason/advantage for defining $~\delta W=-P(V,T)dV~$ instead of just define $~\delta W=-V(P,T)dP~?~$
it seems that I can always convert one to the other

Qmechanic
  • 201,751
P'bD_KU7B2
  • 243
  • 1
    If the volume is constant, what kind of work have been done? Work is force times displacement, and changing the volume is a form of displacement. How does change in pressure with constant volume do work? – Ofek Gillon Feb 05 '22 at 15:28
  • @OfekGillon that is what I am confused about, if we can just "define" work to be the change in pressure with constant volume (displacement times force), calculation can still be carried, so what is the reason for insisting on defining force over displacement? – P'bD_KU7B2 Feb 05 '22 at 15:42
  • If the process is not at constant temperature, PdV+VdP is not zero. And, for a constant temperature, there is also a sign change. – Chet Miller Feb 05 '22 at 15:56
  • @ChetMiller I understand PdV and VdP have different values, but in a reversible process, they seem to carry the same information (except the value difference). I can measure how volume changes with the change of pressure, but I can also do it in the opposite order. I can use V-P diagram instead of P-V. It seems to me (and correct me if I am wrong) it is just a matter of choice about which number I decide to give the name "work". So I don't understand why, in the case of reversible process, PdV is more "privilege" than VdP, I know its better in irreversible since I can only measure $P_{ext}$. – P'bD_KU7B2 Feb 05 '22 at 16:08
  • You are aware that work is the vector product of force and displacement, right? Just because the units of Fdx is the same as xdF does not mean that they are anything like the same. – Chet Miller Feb 05 '22 at 16:53
  • @ChetMiller I understand they are both dot products and give different scalar result, but they are both scalars that represents the property of the system. Would you mind elaborating more on the difference between these two terms? – P'bD_KU7B2 Feb 05 '22 at 17:28
  • 1
    Fdx is what we define as work, and xdF is not.. – Chet Miller Feb 05 '22 at 17:34
  • @ChetMiller Yes, that is my question there, the reason you can say PdV is work just seem to be "its just how we define it", but it doesnt give the reason of why we define it in this way – P'bD_KU7B2 Feb 05 '22 at 17:48
  • 1
    You realize that PdV is the same as Fdx, right? – Chet Miller Feb 05 '22 at 17:49
  • @Chet Miller, Yes, I do, what I don't understand is in a reversible process, I think (correct me if I'm wrong) you can have an invertible function $M$ such that $M(xdF)=Fdx$, so I can either choose to say the work is $Fdx$ or redefine it as $xdF$, then why pick one over the other? – P'bD_KU7B2 Feb 05 '22 at 17:57
  • 1
    Essentially a duplicate of https://physics.stackexchange.com/q/382726/2451 – Qmechanic Feb 05 '22 at 19:01

3 Answers3

3

Every system is defined by a variety of extensive variables: entropy, volume, mass, charge, surface area, magnetization, etc. The internal energy $U$ is a function of these variables—$U=U(S,V,N, Q,A,M,\dots)$—and we seek to know how $U$ changes when the variables change:

$$dU=\left(\frac{\partial U}{\partial S}\right)_{V,N,Q,A,M\dots}dS+ \left(\frac{\partial U}{\partial V}\right)_{S,N,Q,A,M\dots}dV+ \left(\frac{\partial U}{\partial N}\right)_{S,V,Q,A,M\dots}dN+\cdots.$$

We happen to call one of these coefficients, $\left(\frac{\partial U}{\partial V}\right)_{S,N,Q,A,M\dots}$, the negative pressure $-P$. We can increase the energy of a system for positive values of $-P\,dV$. That’s why the parameters are ordered the way they are.

Chemomechanics
  • 25,546
1

in reversible process, why define $\delta W=-PdV$ instead of $\delta W=-VdP$?

You need to differentiate between work done in a closed system and an open system.

For a closed system, where there is no flow of mass into or out of the system, work is referred to as "boundary work". This is the work only associated with expanding or contracting the boundaries of the system, i.e., changing the volume of the system. The often used example is the expansion or compression of a gas in a cylinder fitted with a movable piston. For a closed system, where expansion work is considered positive,

$$\delta W=PdV$$

It applies to an irreversible as well as a reversible process. For the irreversible process, $P$ is the external pressure.

For an open system, where work is required to push or pull mass into and out of the system because of pressure differences at the boundary, work is then called flow work and

$$\delta W=VdP$$

Hope this helps.

Bob D
  • 71,527
  • Thank you for the answer. What I am wondering is, in the reversible case, assume I am doing measurements on $P$ and $V$ during an expanding process, I can plot P-V diagram, or V-P diagram. If I assign the area under V-P diagram (VdP) instead of P-V (PdV), it gives the information about the reversible process just as good as the P-V diagram. So why can't I assign the name "Work" to the area under V-P diagram (and then further define the internal energy to be $U=Q+VdP$), but rather choose the area of P-V diagram? – P'bD_KU7B2 Feb 05 '22 at 16:21
  • @rookie Consider a reversible isobaric (constant pressure) expansion process. The area under the PV diagram (expansion work) is $P\Delta V$. The area under the VP diagram is zero. – Bob D Feb 05 '22 at 17:10
  • But if I choose the define "Work" as VdP, then I can just say in this case, the "Work" done is zero. What I am thinking is if I have two variables, $W_1=VdP$ and $W_2=PdV$, then in a reversible process, there should be a bijective mapping from $W_1$ to $W_2$, where the $W_1=0$ represents a different value in $W_2$, which represents the reversible isobaric work. – P'bD_KU7B2 Feb 05 '22 at 17:24
  • "But if I choose the define "Work" as VdP, then I can just say in this case, the "Work" done is zero." How can you arbitrarily dismiss the expansion work so that the system expands without using any energy. I really don't understand what you are trying to do here. But that's OK. It's probably just me. – Bob D Feb 05 '22 at 17:36
  • I am not dismissing the work PdV in this case, what I am trying to say is if there is a function $F(VdP)=PdV$, then $F(0)$ is the isobaric work that you are talking about, in this case, I can redefine "Work" to be VdP instead of PdV – P'bD_KU7B2 Feb 05 '22 at 17:43
  • I still don't get it. But more importantly, why bother "redefining" work to be VdP instead of PdV. To what purpose? It seems to me it will only obfuscate rather than enlighten people on the concept of work in a closed system. – Bob D Feb 05 '22 at 17:54
1

Maybe I misunderstand your question, but pdV-work follows naturally from the definition of work $W$ from classical mechanics (in one dimension): $$ dW = F dx = pA dx = p dV $$ where $p$ is pressure, $A$ is the cross sectional area, $F$ is the applied force, $dx$ is the displacement, and $dV(=Adx)$ is the displaced volume.

Furthermore, the thermodynamic identity takes the form $$ dU = TdS - pdV $$ for internal energy. If you formulate the thermodynamic identity in terms of enthalpy you obtain $$ dH = TdS + V dp. $$ So the two terms ($pdV$ and $Vdp$) are related through the Legendre transformation, and their use depends on which potential is relevant to the problem at hand.

MOOSE
  • 441