Why $VdP$ term omitted in isothermal Work?

Context: I'm asking about classical thermodynamics, that is "ideal gas", closed system, reversible processes etc.

Why is the $$VdP$$ term omitted in calculation of work during isothermal process in a closed system? Why only the PdV term considered? Clearly, during isothermal process not only V changes but also the pressure P drops/grows too, so $$VdP$$ is not zero as would have been during isobaric process.

• Why do you think VdP represents work for a closed system? Work is Fds, not d(Fs). Apr 17 at 13:08
• @Chet Miller, Well it's hard for me to tell why... I guess, I look at the enthlapy and how it's treated in books dH = dU + d(PV) and all that. but I don't see d(PV) anywhere close to the closed classical ideal gas problems (the first part of any book on the subject). I can't formulate a better question... :( Apr 17 at 13:14
• Some people say "Vdp is NOT work, hence does not appear" but others call Vdp shaft or open system work.... so... I can't get the idea that "VdP never was in the first law of W for a closed system" Apr 17 at 13:24
• Vdw is not equal to the shaft work unless the flow is isentropic. Also, in a flow system, the shaft work is not the total work. Apr 17 at 14:00
• Have you studied the derivation of the open system version of the first law of thermodynamics? Apr 17 at 16:12

The terse reason is that it was never there in the first place. The total differential for the internal energy is defined to be, $$\begin{gather*} dU = \delta w + \delta q \end{gather*}$$ where $$\delta q$$ is the inexact differential of heat and $$\delta w$$ is the inexact differential for work. Often, the only meaningful work being done is $$P-V$$ work, which, following on the line integral definition of linear work, $$\begin{gather*} w = \int_\gamma \vec{F} \cdot d\vec{r} = \int_\gamma P dV \end{gather*}$$ The differential for $$P-V$$ work is thus in the form, $$\begin{gather*} \delta w_{rev} = - P dV \end{gather*}$$ where the subscript refers to the fact that this only applies for reversible processes. For a non-reversible process we would just use the external pressure to the system $$P_{ex}$$ rather than the constantly equilibrated pressure of the system.

As an aside, based on a Clausius analysis of entropy, we can use the differential, $$\begin{gather*} \delta q_{rev} = T dS \end{gather*}$$ Then, combining the two relations, we get the total differential for the internal energy for reversible processes, $$\begin{gather*} dU = TdS - PdV \end{gather*}$$ This is called the fundamental thermodynamic relation because it reveals that the internal energy is a natural function of $$S$$ and $$V$$. Because the internal energy is a state function, this will also hold for arbitrary processes. All we lose is the ability to identify each term with either heat or work specifically. Using a Legendre transform to define the enthalpy, $$\begin{gather*} H = U + PV \\ \implies dH = dU + VdP + P dV \end{gather*}$$ actually does have a $$VdP$$ term, though this is not due to work but the product rule of calculus.

• I almost get the idea that "The terse reason is that it was never there in the first place. " But only almost. Why is it not there? What if it were? ... Sorry for unclear comments and questions, I can't make it clearer. Apr 17 at 13:20
• There not much sense asking what if since it simply isn’t. If it were included your equations would be flat out incorrect and would give results that do not align with experimental reality. The grounding reason is simply that the differential we write follows directly from the definition of work, so your question might better be phrased as “why is work the way that it is.” Apr 17 at 13:28
• Alright, can I ask this: "Let's have a d(PV) and solve it, we will have PdV + VdP. Ok? So why is PdV considered work and VdP isn't? (By some people, I'm not sure it's universal) Apr 17 at 13:34
• Because work has nothing to do with $d(PV)$. I cannot state it more plainly. Apr 17 at 13:35
• Is PdV in (PdV + VdP) just a coincidence then? Apr 17 at 13:37

The thermodynamic work is closed system/reversible process is $$dW = -P dV$$ There is no $$V dP$$ term in this formula, the why is answered in any undergraduate textbook (if your question is where this formula comes from, then you should revise your post to indicate that). The result has nothing to do with ideal gas, it is the general expression for reversible work in closed system.

• "the why is answered in any undergraduate textbook" Can you help me and answer in couple of lines? I mean, the pressure clearly changes yet dP part is omitted... Maybe I'm confusing something between differentiation and ... well, variable changing... i don't know... Apr 17 at 13:12

It depends on what you mean by "work". When people say $$\delta W = p_0dV$$ is the work, where $$p_0$$ is the pressure of the environment and $$dV$$ is the increase of volume of the system and decrease of volume of the environment, then depending on its sign they mean the work done on or absorbed by the environment (or by the system). In a reversible process $$p=p_0$$ so $$\delta W_{rev}=pdV$$. But there is another way of looking at the balance of work.

Let us start with simplest case. The energy equation $$dU=TdS-pdV+\mu dN$$ holds for the simplest body. Assume that it can exchange $$S,V,N$$ at potential levels $$T_0,p_0,\mu_0$$ representing thermal, pressure and chemical reservoirs having combined energy change $$dU_0=T_0dS_0-p_0dV_0+\mu_0 dN_0$$, then the total energy variation is $$dU+dU_0=0$$ with $$dV+dV_0=0$$ and $$dN+dN_0=0$$ but $$dS+dS_0=\sigma \ge 0$$. Here $$\sigma >0$$ is the irreversibly produced entropy in the system at temperature $$T$$; if and only if the process is reversible $$\sigma =0.$$

From energy conservation, therefore $$0=dU+dU_0\\=T_0dS_0 +T(\sigma - dS_0)-(p_0-p)dV_0+(\mu_0-\mu)dN_0$$ and after rearrangement $$-T\sigma =(T_0-T)dS_0-(p_0-p)dV_0+(\mu_0-\mu)dN_0 \tag{1}$$ You could write this Eq.1 as $$-T\sigma =\Delta T dS_0-\Delta p dV_0+\Delta \mu dN_0 \tag{2}$$ where $$\Delta T,\Delta p, \Delta \mu$$ denote the energetic potential drop to which the transported work quantities $$dS_0,dV_0,dN_0$$ are subjected, respectively. Each term is a separate work term:

1. $$\Delta T dS_0$$ is the thermal work in which the thermal reservoir supplies entropy $$dS_0$$ and this entropy is moved through $$\Delta T$$ temperature drop;
2. $$-\Delta p dV_0$$ is a spatial work term in which $$dV_0$$ volume is moved through $$-\Delta p$$ pressure drop;
3. $$\Delta \mu dN_0$$ is a chemical work term in which $$dN_0$$ amount of chemical species is moved through $$\Delta \mu$$ chemical potential drop.

If and only the process is reversible, ie, $$\sigma = 0$$, these work terms are in perfect balance with each other.

Now in the special case that the process is adiabatic $$dS_0=0$$ and closed $$dN_0=0$$ then $$-T\sigma = -(p_0-p)dV_0$$. Since the process cannot now be balanced with another work term it must, per force, be irreversible and all of the $$\Delta p dV_0$$ work is dissipated and turns to heat. On the other hand if it is still closed but diathermal then we have $$-T\sigma =\Delta T dS_0-\Delta p dV_0$$ and this can now be reversibly so that the thermal and spatial works be balanced, $$\Delta T dS_0=\Delta p dV_0$$, and that would be a Carnot engine operating at temperatures $$T_0$$ and $$T$$ while absorbing thermal energy (heat) $$T_0S_0$$ and rejecting absorbing thermal energy (heat) $$TS_0$$, resp.

So the punchline: the $$\Delta p dV_0$$ work term represents all the pressure-volume work that is done during the energetic exchange between the system and its environment.

• Ok, It looks to me that you have written a good answer for "not my question" but I like it. It gives me smth to think about. Really appreciate. Thanks. Apr 17 at 14:47
• you are right, just forgot to add the punchline but now I have, see above. Apr 17 at 15:16
• and for further details regarding what happens when the process is isothermal and reversible, which is specifically your question, see my answer here physics.stackexchange.com/questions/779195/… Apr 17 at 15:25

From the comments, it looks like you’re still wondering why $$V\,dP$$ doesn’t appear in the definition of work or in the First Law when it appears in the differential form of the enthalpy $$H\equiv U+PV$$ (or $$dH=dU+d(PV)=dH+P\,dV+V\,dP$$).

First, what is thermodynamic work anyway? It’s the raising of many particle energies in concert. (In contrast, heating is the broadening of the distribution of particle energies.) This requires a so-called generalized force—which is sometimes just literally a mechanical force—and a so-called generalized displacement. Not just a position, a displacement. So right away we have an asymmetry: We're talking about the value of one parameter and a difference in another parameter. The generalized force $$F$$ drives the change at any moment, and the generalized small displacement $$dx$$ describes the extent of particle energy elevation. The infinitesimal work is then in the form of $$F\,dx$$, integrated to give path-dependent work $$\int F\,dx$$.

For instance, pushing up on a body locked in place does nothing to raise its gravitational potential energy; a table doesn’t do work on the objects sitting on it. It’s the displacement that embodies the energy transfer process.

(Other examples are a mechanical force and displacement, a pressure and a volume shift, a surface tension and an area change, a stress and a volumetric strain, an electric field and extent of polarization, a magnetic field and extent of magnetization, and a chemical potential and amount of shifted matter. These pairs are called conjugate thermodynamic variables; their product gives units of energy.)

This summarizes why $$P\,dV$$ looks the way it does when defining expansion–compression work. Why, then, is enthalpy defined using the complete term $$\boldsymbol{PV}$$, with $$\boldsymbol{d(PV)}$$ appearing in the differential form? Doesn’t this give $$\boldsymbol{P\,dV+V\,dP}$$? Yes, but $$PV$$ here is a unique term. It conceptually describes the work done on the atmosphere—at surrounding pressure $$P$$—to make room for the entire system at volume $$V$$. (In other words, it's the result of evaluating the expansion work in the special case of $$\int_0^V P\,dV$$ when the surrounding pressure $$P$$ is constant.) This lets us account for the fact that Nature prefers denser systems when the surrounding pressure is higher. Again, this is a special case; systems don’t generally wink in and out of existence. When do they effectively wink in and out of existence? When a chemical reaction trades some compounds for others—and we use the enthalpy to characterize the so-called heat of reaction. When matter is pushed in and out of a control volume—and we use the so-called shaft work $$V\,dP$$ to analyze the energy required/collected when the inlet and outlet pressures differ.

Note carefully the difference between these processes and the process of simple expansion and compression, in which the work done by the system is again simply $$P\,dV$$ (where $$P$$ is the external pressure).

Now, isn't this just splitting hairs regarding "making room for the system"? Can't we just model expansion as removing the entire system and inserting it again at the larger volume? Yes, and the enthalpy change would be $$\Delta H=-\int P\,dV+\Delta (PV)$$. Note that we've accounted for the loss in internal energy of the system from doing the expansion work. If the surrounding pressure remains constant, this reduces to $$\Delta H=0$$.

To summarize: It would be a mistake to conflate $$P\,dV$$, or infinitesimal expansion work, with one term of the differential enthalpy, which looks similar but has an additional $$V\,dP$$. Work is not equivalent to enthalpy. Terms such as "shaft work" and "flow work" arose in part as a convenient shorthand to refer to enthalpy changes, but they aren't a part of the generalized framework of thermodynamic work discussed above—and it's this thermodynamic work that alters the internal energy, as described by the First Law $$\Delta U=Q-W$$ (for work done by the system).

• Your take on "generalized work" and treating the $PdV$ as a product of "intensive х d(extensive)" quantitites is understood. Thus $VdP$ is a ... well.. an odd thing since it is of the form: "extensive х d(intensive)" which is quite odd to interpret in this generalized framework... Apr 17 at 14:56
• Agreed. That’s why it doesn’t belong in the definition of work. This was your original question, correct? Please let me know what remains unclear. Apr 17 at 14:58
• Well... i got this argument but... confusion is still there :) You say: PdV+VdP is a unique term. It describes the work done on the atmosphere to make room for the entire system at volume V. So... you call it work. But then again you didn't just a moment ago, right? Or other people don't call it that. Is it work or what? Apr 17 at 15:02
• Or rather: why making room for the system takes PdV+VdP as work, and only PdV in closed classical isothermal expansion, when clearly in this expansion it does the same thing as "making room for a system"? Apr 17 at 15:03
• Please see my edits addressing this point. Apr 17 at 16:18

There is no $$VdP$$ work because mass isn't moved into and out of a closed system. The work in a closed system is boundary work, the work required to expand or contract the boundaries of the system, i.e., change its volume.

I know that VdP is a "flow work" of some sorts... but I wondered why it is never interpreted as the "changing pressure" part of the isothermal work... I'm stuck in somekind of a dumb idea about isothermal work and I can't get out :

You need to go back to the basic definition of work, which is force times the displacement of material in contact with the force. For the closed system the force is the pressure and the displacement is the change in volume.

So for a closed system a change in volume must always be associated with the change in pressure in order for work to be done. Without a displacement of volume (and the mass contained there in), a change in pressure does no work. For example, if you heat a rigid vessel containing a gas its pressure will increase but since the volume doesn't change no work is done. There is no displacement of mass. With regard to the open system $$\Delta(PV)$$ is the work required to move mass into and out of system.

Finally, with regard to a reversible isothermal process for an ideal gas there is a change in volume along with the change in pressure where PV is a constant, o work is done.

Hope this helps.

• yeah, I know that VdP is a "flow work" of some sorts... but I wondered why it is never interpreted as the "changing pressure" part of the isothermal work... I'm stuck in somekind of a dumb idea about isothermal work and I can't get out :) Apr 17 at 13:18
• @coobit I’ll update my answer to explain Apr 17 at 13:27
• so you name VdP as "work", right? Why then some people do not consider it to be work but just have the units of work... Is it because VdP and Fds are not derivable from eachother? Apr 17 at 13:36
• @coobit See update to my answer Apr 17 at 14:41
• VdP is not the work to move mass into and out of an open system. Apr 17 at 14:59