4
$\begingroup$

This concept was a bit hard to grasp, because I'm bad at seeing the real-world implications of multiplication. Division makes sense, but multiplication is harder for me. Pressure-volume work is measured in $1$ Pa $\times$ m$^3$. It says that $1$ Pa $\times$ m$^3 = 1$ J.

Does that mean that no matter what the volume of the thermodynamic system is, if it exerts $1$ Pa onto its surroundings, its doing $1$ Joule of work? Again, I'm not very learned in science, math and thermodynamics, but this looks like $1$ Pa $\times$ $1$ m$^3 = 1$ J. Is this another way of saying:

$x$ Pa $\times$ $1$ m$^3 = x$ J

OR

$1$ Pa $\times$ $x$ m$^3 = x$ J

Basically saying that either the volumetric quantity or the pressure quantity has a $1/1$ relation with the energy of the work (J), meaning the other quantity is equal to $1$.

$\endgroup$
6
$\begingroup$

Note that the volume is not the system volume but the displaced volume, or the volume exchanged by moving an interface against a pressure resistance.

With this in mind, yes, if the resistance is a constant 1 pascal, then the work in joules numerically equals the volume exchange in cubic meters. If the volume exchange is in total a cubic meter, then the work in joules numerically equals the constant resistance in pascals, as you write.

Mathematically, we'd define a small amount of work $w$ as the instantaneous pressure $P$ times a small amount of volume shift $dV$: $$w=P\,dV;$$ integrating, we have $$W=\int_{V_1}^{V_2}P\,dV,$$ where $W$ is the total work done, and if $P$ remains constant, we can bring it outside the integral to give $$W=P\Delta V=P(V_2-V_1).$$

Although this is commonly called $P-V$ work, to avoid confusion, never forget that that's a shorthand for $P-dV$ or $P-\Delta V$ work. This follows the general pattern of a generalized force (here, the pressure) acting over a generalized displacement (here, the volume exchange).

$\endgroup$
1
  • $\begingroup$ I see. This is not what I believed. I thought the volume being referred to was the total volume of a system; the pressure being the static pressure exerted onto the surroundings, like for example the walls of a container. But its the displaced volume. So if a pressure of $2$ Pa moves $3$ m$^3$ of water, then the work of this will be: $2$ Pa $\times$ $3$m$^3 = 6$ J. $\endgroup$
    – A. Kvåle
    Apr 9 at 19:28
3
$\begingroup$

Does that mean that no matter what the volume of the thermodynamic system is, if it exerts $1$ Pa onto its surroundings, its doing $1$ Joule of work?

No. For a closed system, pressure-volume work is the work done on or by the system resulting in contracting or expanding the boundary of the system (decreasing or increasing its volume). The general expression for boundary work is

$$W=\int_{V1}^{V2}PdV$$

In order to calculate the work you need to know pressure as a function of volume, i.e.,you need to know the process connecting the initial and final volume. For example, for a constant pressure process, the work is simply

$$W=P(V_{2}-V_{1})$$

Otherwise, $PV$ is simply the product of two system properties, pressure and volume, that happens to have units of energy.

For example, for an ideal gas, the ideal gas law gives the relationship between the product of pressure and volume and temperature at equilibrium.

$$PV=mRT$$

Where $m$ is the constant mass of the gas and $R$ is the gas constant.

Hope this helps.

$\endgroup$
3
$\begingroup$

Basically saying that either the volumetric quantity or the pressure quantity has a 1/1 relation with the energy of the work (J), meaning the other quantity is equal to 1.

The definition of mechanical pressure-volume work is the integral of the external pressure times the differential change in volume of the system $w = -\int p_{ext} dV$. In SI, the units can be Pa m$^3$. The units can also be lbf ft from (lbf/ft$^2$)(ft$^3$) or Torr L or N m from (N/m$^2$)(m$^3$). These are all unit designations for mechanical work.

The expression 1 J = 1 Pa $\times$ 1 m$^3$ is a unit conversion expression. The unit on the left is a derived unit for energy. The units on the right consist of a derived unit for pressure (Pa = 1 N/m$^2$ = kg/m s$^2$) and a fundamental unit for length m$^3$.

The relationship is true regardless of whether it is used to convert the units of mechanical expansion/compression work to energy or whether it is used to convert the $pV$ component in the ideal gas law to an energy to compare with the $nRT$ term (that has the same units) or even whether it is used to convert $pV$ in the definition of enthalpy $H \equiv U + pV$ from pressure times volume to energy units.

The relationship is not saying that one or the other quantity is equal to 1. How can you appreciate this? Consider that all of these conversions give the same value of 1 J.

$$ 0.5 Pa \times 2 m^3$$ $$ 1 Pa \times 1 m^3$$ $$ 2 Pa \times 0.5 m^3$$

The mistake is perhaps to believe that multiplication of units in a unit conversion factor can be "inverted" to get a numerical value of one or the other factors. You cannot and must not go down that path. The better approach in calculations is to take three steps. First, calculate the number using the equation. Second, determine and validate the units on the number by unit analysis through the equation. Finally, convert the calculated units as needed to other forms.

$\endgroup$
2
  • $\begingroup$ Ah, I understand. If we somehow measure the amount of joules exerted by a pressure, then we can just either figure out the pressure, or the volume, and we will know the other quantity. If the amount of joules are 2, and the pressure is 4 Pa, then the system must be of $0.5$m$^3$ volume, and vice verca. However, this energy, e.g. 2 joules; is it the energy of work, or the energy of the system? Since its called pressure-volume work, I would assume the energies mentioned are those exerted on the surroundings. Is it so? $\endgroup$
    – A. Kvåle
    Apr 9 at 19:22
  • $\begingroup$ We do not "measure the amount of Joules exerted by a pressure". We measure the pressure. We multiply that pressure by the volume displacement to obtain mechanical work. Or we multiply the pressure by the system volume to obtain the term $pV$ that is added to internal energy to obtain the enthalpy. The units of Pa-m$^3$ are converted to J when we do the multiplication. Both units represent ENERGY. The latter (J) is the accepted derived unit in the SI system. $\endgroup$ Apr 9 at 22:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.