I am a little confused, from the first law of thermodynamics (energy conservation)

$$\Delta E = \delta Q - \delta W $$

If the amount of work done is a volume expansion of a gas in, say a piston cylinder instrument at constant pressure,

$$\Delta E = \delta Q - p \, dv$$

Here $p$ is the constant pressure and $dv$ is the change in (specific) volume.

So, when do I take into account

$$\delta W = d(pv) = p \, dv + v \, dp$$

I am assuming that for cases of boundary work, at constant pressure, the $v \, dp$ term is zero.

So under what conditions should I consider the $v \, dp$ term?

  • 1
    $\begingroup$ Where are you getting $\delta W = d(pv)$ from? I'm skeptical that it's true since it would seem to imply that if you were to increase the temperature of an ideal gas sample in a closed, rigid box (and thus increase its pressure while keeping volume constant), then the gas would do work, but a gas does no work in such a situation. $\endgroup$ Commented Jan 23, 2013 at 20:04
  • $\begingroup$ @joshphysics why don't I see what you see? So when do we use vdp? I was always under the impression that $$\delta W = d(pv) = p dv + vdp$$. Generaly for constant pressure processes we would neglect $$vdp$$... $\endgroup$
    – dearN
    Commented Jan 23, 2013 at 20:08
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    $\begingroup$ I am also skeptical about $\delta W=d(pv)$, since work is not an exact differential form (that's why you write $\delta$ instead of $d$), while $d(pv)$ is exact: it's the differential of $pv$, which is a state function. $\endgroup$
    – Martino
    Commented Jan 23, 2013 at 20:14
  • $\begingroup$ @Bzazz you are right. so, $$\delta W = p dv$$ with an integral from state 1 to 2 is correct? $\endgroup$
    – dearN
    Commented Jan 23, 2013 at 20:17
  • $\begingroup$ Usually it is in systems like the one you mentioned. I don't have exceptions in mind at the moment, but I'm not an expert. $\endgroup$
    – Martino
    Commented Jan 23, 2013 at 20:53

8 Answers 8


$PdV$ is boundary work. $VdP$ is isentropic shaft work in pumps (as you have identified above), gas turbines, etc. Now you must realize that even in a pump or turbine the mechanism of work is still $Pdv$, i.e., the gas pushing on the blade out of its way. But, then there is work required to maintain the flow in and out of the device/control volume, which requires flow work $PV$ so the net reversible work from a steady-flow device turns out to be shaft $vdP$.

Why flow work $PV$? To push a packet of fluid with volume $V$ forward into a device you have to do work against the pressure of the fluid already in the device, i.e., overcome the back force of that fluid. This implies the work you do in pushing your new packet of length $L$ and cross-section area $A$ into the device is: \begin{align*} \int Fdx = \int_{0}^{L}PAdx = PV \end{align*} It must be noted that in a steady-flow device (unlike in a piston) the back pressure $P$ is constant.

Now consider the device (e.g., turbine to be a control volume). The energy of the fluid going in is its internal energy and the work invested into the fluid to enter the device: $U_{entry}+P_{entry}V_{entry}=H_{entry}$. Similarly for exit from the device. The net change across the device is $\Delta H$. For a differential device (or across a small change) this is $dH$. The work output from the shaft of then device is the $\delta W= dH$.

Now if the device is isentropic, i.e., adiabatic-reversible. The Gibbs equation provides: \begin{align*} &dH=TdS+VdP=VdP\\ &\delta W =dH=VdP \qquad (\text{isentropic}\; dS=0) \end{align*}

Therefore $VdP$ is isentropic shaft work from a flowing device.

Important points: 1) Both internal energy and enthalpy are state variables, therefore can be measured for a system static or flowing. This is why sometimes there is a tendency to use $U$ and $H$ incorrectly. The true purpose of $H$ is to capture the work required to push/maintain a flow against a back pressure, i.e., it incorporates the $PV$ part. Therefore when you write an energy balance with flows coming in and out, the energy crossing boundary is not just $U$ but $H$ and this distinction must be kept in mind.

2) $VdP$ is isentropic steady-flow shaft work. The isentropic is key here.

  • $\begingroup$ Nice, thanks. I think the point of the answer by Bryson S. is that $H_{exit}=H_{entry}-W$ for an adiabatic process, where $W$ is the (positive) work output of the shaft, so $H$ decreases as shaft work is performed. $\endgroup$
    – Art Brown
    Commented May 10, 2014 at 1:07

$\delta W=\mathrm{d}(PV)$ is wrong. We always have $\delta W=P\mathrm{d}V$ (unless there are other interactions like magnetic field). In fact, the reason that many books opt to denote infinitesimal work as $\delta W$ instead of $\mathrm{d}W$ is to emphasize that work is not an exact differential.

$V\mathrm{d}P$ manifests in these formulas $$\mathrm{d}H=T\mathrm{d}S+V\mathrm{d}P$$ $$\mathrm{d}G=-S\mathrm{d}T+V\mathrm{d}P$$ where $H$ and $G$ are not internal energy, but enthalpy and Gibbs free energy. These two concepts are more useful than internal energy in isobaric processes.


When one system does mechanical work on another, the work done (i.e. the amount of energy transferred) is equal to the product of the force acting and the displacement in the direction of the force. When the force acts on a boundary around a system at pressure $p$ and the volume of the system changes by $dV$ then the work done on that system is $$ \delta W = - p dV $$ where on the left I am using the $\delta$ symbol to refer to an improper differential (one which is not a change in a any function of state). Notice that it was not necessary to mention the conditions. This result can be applied to a flow process, for example, as I will show in a moment.

Next, the quantity $pV$ (the product of pressure and volume) is a perfectly well-defined physical quantity, and it has dimensions of energy, but it is not work. Do not call it "$W$" or you will confuse yourself or your readers. This is not to deny that there may be some situation in which the work done is equal to $pV$ (for example, one could imagine expanding a system from zero volume at constant pressure), but the quantity $pV$ is simply the product of pressure and volume and does not earn a name because it is not sufficiently important to earn a name.

The quantity $H = U + pV$ does earn a name. It is called enthalpy, and it is useful because for a closed simple mechanical system, $$ dH = dU + d(pV) = dU + p dV + V dp = T dS + V dp. $$ Because the right hand side has just two terms and they are independent, we have a useful quantity. It is useful in processes at constant pressure, for example.

Let's now consider a flow process in which work is done, often called "shaft work" because typically it would be causing a turbine to rotate. Take a fixed mass of fluid, and push it into one end of a turbine, at constant pressure. Let $V_i$ be the initial volume of the whole turbine, including all the fluid and pipes etc. Let $V_1$ be the volume of the mass of fluid which gets pushed into the turbine, let $p_1$ be the pressure on the input side of the turbine. Then the work you have to do $on$ the system, to push this fluid into it, is $$ W_i = \int_{V_i}^{V_i - V_1} - p_1 dV = p_1 (V_i - (V_i - V_1)) = p_1 V_1 . $$ Let $V_2$ be the volume occupied by this same mass of fluid as it arrives at the output side (it might have expanded or contracted, depending on what kind of device we have), and let $p_2$ be the output pressure. The work done $on$ the system by this part of the overall process is $$ W_o = \int_{V_i}^{V_i + V_2} - p_2 dV = p_2 (V_i - (V_i + V_1)) = - p_2 V_2 $$ So the total amount of work done $on$ the system is $$ W_i + W_o = p_1 V_1 - p_2 V_2 . $$ We normally define the "shaft work" to be the work done $by$ the system; this is $$ W_{\rm shaft} = p_2 V_2 - p_1 V_1. $$ In the case of an incompressible fluid, one would have $V_2 = V_1$ and then $$ W_{\rm shaft} = (p_2 - p_1) V_1 $$ which, for a small amount of work, could be written $V \Delta p$, but notice that the $\Delta p$ is not a differential; it is a finite change in pressure between input and output of a turbine. It could be (and usually is) a large quantity. But of course you can always also consider two close regions in a continuous gradient of pressure with an incompressible fluid, and then you would get $V dp$.

The main thing to notice in all the above is that the work done is always given by $-p dV$ in every case. You just have to understand how to apply that formula to various physical processes. Even in the final case where we got $V dp$, it really came from two instances of $-p dV$ in certain specific conditions (applied at two sides of a pressure gradient in the flow of an incompressible fluid).

  • $\begingroup$ Thanks a lot, this clears my confusion. in all the PV state diagrams people describe work done as the area under the curve projected to X-axis and not the Y-axis, Hence dw=Pdv always. $\endgroup$
    – Diza
    Commented Feb 20, 2022 at 8:17

Sankaran is correct in that the magnitude of the net reversible shaft work $\delta w$ is $vdP$, but he is incorrect in ascribing to this quantity a positive sign. In reality, the shaft work done by the system must be $-vdP $, here's why...

If, by definition, a differential change in enthalpy $dh = TdS + vdP$, then for an adiabatic process where $\delta q = TdS = 0$, the differential change in enthalpy must be

$dh = (0) + vdP = vdP$.

Now, if the work done by the system is positive ($\delta w > 0)$ its change in enthalpy must be negative $(dh < 0)$. These two facts in conjunction must mean that the net reversible shaft work done by the system is:

$\delta w = -vdP = -dh$

I ran into this problem myself in trying to derive the specific work produced by an isentropic turbine, and you will definitely get the wrong answer if you start with the premise that the differential shaft work is simply $vdP$.


I hate answering my own question but I'd like to share this with you as it is definitive:

$$\mathrm{d}h = \mathrm{d}u + \mathrm{d} (p v)$$ $$h_2 - h_1 = u_2 - u_1 + p(v_2 - v_1) + v (p_2 - p_1)$$ Now, for a pump working in the compressed liquid (subcooled liquid zone), it is noted that the change in specific volume $v$ is minimal when the pump adds pressure energy to the liquid. It is minimal because liquid is incompressible.

This leaves us with the following terms:

$$h_2 - h_1 = u_2 - u_1 + v (p_2 - p_1)$$ $$\Delta h = C (T_2 - T_1) + v(p_2 - p_1)$$

An other assumption that is made that the fluid flow through the pump does not raise the temperature much which would allow us to drop the $C (T_2 - T_1)$ term where $C$ is the specific heat of the liquid.

  • This leaves us with the equation that was troubling me:

$$h_2 - h_1 = v (p_2 -p_1)$$

So this is valid for a pump working in the compressed liquid region of the $P-v$ or $T-v$ diagram.

If you'd like to comment further, please do so. This has been quite enlightening for me already! :)

  • $\begingroup$ Is $h$ here enthalpy? $\endgroup$
    – Siyuan Ren
    Commented Jan 24, 2013 at 15:26
  • $\begingroup$ @C.R. yes, $h$ is the enthalpy? Would you like to edit the answer as you noticed that it wasn't defined? $\endgroup$
    – dearN
    Commented Jan 24, 2013 at 15:38

Generally we call p dV as the displacemnt work done by a piston expanding in a cylinder containing some amount of gas. This is applicable to only NON FLOW PROCESSES. But the work v dp is applied to only flow processes or control volume processes. you can simply use the steady flow energy equation in place of this to calculate the work done in a flow process.


First of all, this is called flow work and it is calculated in this way when we have

  • when the system is irreversible in nature
  • when work crosses the boundary
  • when the thermodynamic system is an open system e.g. turbines, compressors, pumps etc etc then we use integral -vdp to calculate the flow Work

We can write δW=d(pv)=pdv+vdp when p is not a function of v, i.e. when p & v are independent variables.


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