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7.7: Measuring Pressure-Volume Work

  • Page ID
    152029
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    By definition, the energy of a system can be exploited to produce a mechanical change in the surroundings. The energy of the surroundings increases; the energy of the system decreases. Raising a weight against the earth’s gravitational force is the classical example of a mechanical change in the surroundings. When we say that work is done on a system, we mean that the energy of the system increases because of some non-thermal interaction between the system and its surroundings. The amount of work done on a system is determined by the non-thermal energy change in its surroundings. We define work as the scalar product of a vector representing an applied force, \({\mathop{F}\limits^{\rightharpoonup}}_{applied}\), and a second vector, \(\mathop{r}\limits^{\rightharpoonup}\), representing the displacement of the object to which the force is applied. The definition is independent of whether the process is reversible or not. If the force is a function of the displacement, we have

    \[dw={\mathop{F}\limits^{\rightharpoonup}\left(\mathop{r}\limits^{\rightharpoonup}\right)}_{applied}d\mathop{r}\limits^{\rightharpoonup} \nonumber \]

    Pressure–volume work is done whenever a force in the surroundings applies pressure on the system while the volume of the system changes. Because chemical changes typically do involve volume changes, pressure–volume work often plays a significant role. Perhaps the most typical chemical experiment is one in which we carry out a chemical reaction at the constant pressure imposed by the earth’s atmosphere. When the volume of such a system increases, the system pushes aside the surrounding atmosphere and thereby does work on the surroundings.

    When a pressure, \(P_{applied}\), is applied to a surface of area \(A\), the force normal to the area is \(F=P_{applied}A\). For a displacement, \(dx\), normal to the area, the work is \(Fdx=dw=P_{applied}A\ dx\). We can find the general relationship between work and the change in the volume of a system by supposing that the system is confined within a cylinder closed by a piston. (See Figure 4.) The surroundings apply pressure to the system by applying force to the piston. We suppose that the motion of the piston is frictionless.

    Screen Shot 2019-10-02 at 5.03.40 PM.png
    Figure 4. Pressure–volume work.

    The system occupies the volume enclosed by the piston. If the cross-sectional area of the cylinder is \(A\), and the system occupies a length \(x,\) the magnitude of the system’s volume is \(V=Ax\). If an applied pressure moves the piston a distance \(dx\), the volume of the system changes by \(dV_{system}=A\ dx\). The magnitude of the work done in this process is therefore

    \[ \begin{align*} \left|dw_{system}\right| &=\left|P_{applied}A\,dx\right| \\[4pt] &=\left|P_{applied}dV_{system}\right| \end{align*} \nonumber \]

    work is positive if it is done on the system

    We are using the convention that work is positive if it is done on the system. This means that a compression of the system, for which \(dx<0\) and \({dV}_{system}<0\), does a positive quantity of work on the system. Therefore, the work done on the system is \(dw_{system}=-P_{applied}dV_{system}\) or, using our convention that unlabeled variables always characterize the system,

    \[dw=-P_{applied}dV \nonumber \]


    This page titled 7.7: Measuring Pressure-Volume Work is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Paul Ellgen via source content that was edited to the style and standards of the LibreTexts platform.