I've seen work often represented by the formula $$W = PΔV$$ But there are also other formulas, which represent different types of work. For example,

Non Flow Work $$ W =\int_{1}^{2} PdV $$

Steady Flow Work $$ W =\int_{1}^{2} -VdP $$

Polytropic Work $$ W = \frac{P_2V_2-P_1V_1}{1-n} $$

I just want to ask, what is the general formula or at least representation of a formula for work in terms of pressure and volume that encompasses everything? One where I can kind of derive or at least have an intuition where the specific formulas for work are coming from.

If it is just work times pressure and volume, how do you determine which one becomes the differential variable and which one stays constant in an integral? When is the delta sign used in place of differential? When are both not constant, just like in the polytropic process?

I've studied about types of work in thermodynamics but am not that familiar so I decided to ask here to have a more clear comparison and contrast on the different concepts.


2 Answers 2


Here I will define $W$ to be the work done on a gas. In all cases, the formula $$W_{12}=\int_1^2 P \; dV$$ will correctly determine the work done on the gas as it moves from state $1$ to state $2$ (assuming that the pressure and volume are always well defined). Notice that if we were to define $W$ as the work done by a gas instead, then the sign would flip (as is the case in the equation for steady-flow work).

Sometimes though, there are more convenient ways to find the work, depending on what kind of problem you are trying to solve. For example, if you are finding the work done by an adiabatic process, then you know that $P_iV_i^\gamma=P_fV_f^\gamma$, so the work done in this case is $$W=\frac{P_fV_f-P_iV_i}{1-\gamma},$$ which can be found by differentiating both sides of the adiabatic condition, solving for the differential and plugging it into $W=\int_i^f P dV$. The same is true if the pressure is not changing (isobaric process); then we can remove pressure from the integral to conclude $$W=P\Delta V.$$ In conclusion, the formula you decide to use will be a function of the problem you are trying to solve. Of course, you can always use $W=\int_i^f PdV$ if you really want to, but it's often more convenient to directly apply equations that work specifically for whatever problem you are solving.

  • $\begingroup$ Wow, I loved how you weaved in my formulas to your explanation. But I just want to ask, isn't the formula for steady state work equals to volume times differential of pressure? Your explanation accounts for the sign change but how about the variable switch between the two? $\endgroup$
    – AndroidV11
    Dec 21, 2021 at 0:24
  • $\begingroup$ Integrating the volume over the pressure will give exactly the same result as integrating the pressure over the volume (assuming the path taken in the PV space is the same). For steady-state work, this switch in differentials becomes advantageous because it is usually easier to integrate volume in steady flow systems (the volume is either constant or changing at a constant rate by definition, which makes it easier to integrate). This is similar to why moment of inertia is computed by integrating over mass instead of over radius -- even though both give the same answer, its easier to have dm. $\endgroup$
    – Roger Yang
    Dec 21, 2021 at 3:58


In the IUPAC/Kelvin paradigm with $dU = \delta q + \delta w$, work done by a mechanical piston is defined as below

$$ \delta w = -p_{ext}\ dV $$

where $p_{ext}$ is the external pressure on the piston. When the volume expands against constant external pressure, $dV$ is positive and work done by the piston is negative. Work energy exists the system.

We recognize a these different cases for a closed system.

Constant Pressure Expansion $p_{ext} = p$

$$ w = -p\Delta V $$

Free Expansion $p_{ext} = 0$

$$ w = 0 $$

Reversible Work $p_{ext} = p_{int}$

$$ w = - \int\ p_{int}\ dV $$

Ideal Gas - Reversible Polytropic $p_{int}V^n = n R T$

  • Isobaric $n = 0$
  • Isochoric $n = \inf$
  • Isothermal $n = 1$
  • Adiabatic $n = \gamma$ with $\gamma = $ ratio of specific heat capacities

Example: Ideal Gas - Reversible Isothermal Work $p_{int} = n R T / V$

$$ w = - n R T \ln(p_f/p_i) $$

Reversible Flow - Enthalpy Work

$$ dH \equiv dU + d(pV) = \delta q + V dp $$

$$ w = \int\ V\ dp $$

Alternative Perspective

In the Clausius/mechanical engineering paradigm, $dU = \delta q - \delta w$. We correspondingly remove the negative sign on the definition for work done by the system and the enthalpy (flow) work recovers a negative sign.


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