We introduced the idea of work in Chapter 9. Work is the energy transferred between a system and the environment when a net force acts on the system over a distance. The process itself is a mechanical interaction, meaning that the system and the environment interact via macroscopic pushes and pulls. Loosely speaking, we say that the environment (or a particular force from the environment) “does work” on the system. A system is in mechanical equilibrium if there is no net force on the system.

FIGURE 19.3 reminds you that work can be either positive or negative. The sign of the work is not just an arbitrary convention, nor does it have anything to do with the choice of coordinate system. The sign of the work tells us which way energy is being transferred.

FIGURE 19.3 The sign of work.

In contrast to the mechanical energy or the thermal energy, work is not a state variable. That is, work is not a number characterizing the system. Instead, work is the amount of energy that moves between the system and the environment during a mechanical interaction. We can measure the change in a state variable, such as a temperature change $\text{Δ}T=T_{\text{f}}-T_{\text{i}}$, but it would make no sense to talk about a “change of work.” Consequently, work always appears as $W$, never as $\text{Δ}W$.

You learned in 《 SECTION 9.3 how to calculate work. The small amount of work dW done by force $\vec{F}$ as a system moves through the small displacement $d\vec{s}$ is $dW=\vec{F}\cdot d\vec{s}$. If we restrict ourselves to situations where $\vec{F}$ is either parallel or opposite to $d\vec{s}$, then the total work done on the system as it moves from $s_{\text{i}}$ to $s_{\text{f}}$ is

Let’s apply this definition to calculate the work done when the volume of a gas changes. FIGURE 19.4a shows a gas of pressure $p$ whose volume can be changed by a moving piston. The system is the gas in the cylinder, which exerts a force $\vec{F}{}_{\text{gas}}$ of magnitude $pA$ on the left side of the piston. To prevent the pressure from blowing the piston out of the cylinder, there must be an equal but opposite external force $\vec{F}{}_{\text{ext}}$ pushing on the right side of the piston. In practice, this would likely be the force exerted by a piston rod. Using the coordinate system of Figure 19.4a, we have

FIGURE 19.4 The external force does work on the gas as the piston moves.

Full Alternative Text

Suppose the piston moves a small distance $dx$, as shown in FIGURE 19.4b. As it does so, the external force (i.e., the environment) does work

If $dx$ is positive (expanding gas), then $dW$ is negative. This is to be expected because the force is opposite the displacement. $dW$ is positive if the gas is slightly compressed $(dx<0)$.

If the piston moves $dx$, the gas volume changes by $dV=Adx$. Thus the work is

If we allow the gas to change in a slow, quasi-static process from an initial volume $V_{\text{i}}$ to a final volume $V_{\text{f}}$, the total work done by the environment is found by integrating Equation 19.7:

Equation 19.8 is a key result of thermodynamics. Although we used a cylinder to derive it, it turns out to be true for a container of any shape.

What is this work done on? Not, as you might think, the piston. Work is an energy transfer, but the piston’s energy is not changing. It is in equilibrium, with equal forces exerted on both faces. Further, the motion of the piston is very slow—it’s a quasi-static process—so the piston has negligible kinetic energy. The piston is merely a movable boundary of the gas, so the work is done by the environment on the gas.

• When a gas is compressed, $\text{Δ}V<0$, $W>0$, and energy is transferred from the environment to the gas (the system).

• When a gas expands, $\text{Δ}V>0$, $W<0$, and energy is transferred from the gas to the environment.

When a gas expands—for example, when the burning fuel in a cylinder of your car’s engine expands and, through a series of mechanical linkages, transfers energy to the wheels—it’s often said that “the gas does work on the environment.” But be careful! In any volume change, both the environment and the gas do work; it’s not one or the other. In fact, because $\vec{F}{}_{\text{gas}}=-\vec{F}{}_{\text{env}}$, the work done by the gas is simply the negative of the work done by the environment: $W_{\text{gas}}=-W_{\text{ext}}$. In an expansion, where energy is transferred from the gas to the environment, $W_{\text{ext}}<0$ means that $W_{\text{gas}}>0$. It thus makes sense to say “the gas does work.” The important point is that the work $W$ that appears in the energy principle, and now in the laws of thermodynamics, is—by definition—the work done by the environment, not the work done by the system. To say that “the gas does work” in an expansion simply means that we need to use a negative value of $W$ in the energy equations to show that energy is being transferred out of the system.

We can give the work done on a gas a nice geometric interpretation. You learned in Chapter 18 how to represent an ideal-gas process as a curve in the pV diagram. FIGURE 19.5 shows that the work done on a gas is the negative of the area under the pV curve as the volume changes from $V_{\text{i}}$ to $V_{\text{f}}$. That is,

FIGURE 19.5 The work done on a gas is the negative of the area under the curve.

Full Alternative Text

Figure 19.5a shows a process in which a gas expands from $V_{\text{i}}$ to a larger volume $V_{\text{f}}$. The area under the curve is positive, so the environment does a negative amount of work on an expanding gas. Figure 19.5b shows a process in which a gas is compressed to a smaller volume. This one is a little trickier because we have to integrate “backward” along the $V\text{-}\text{axis}$. You learned in calculus that integrating from a larger limit to a smaller limit gives a negative result, so the area in Figure 19.5b is a negative area. Consequently, as the minus sign in Equation 19.8 indicates, the environment does positive work on a gas to compress it.

EXAMPLE 19.1 The work done on an expanding gas

How much work is done in the ideal-gas process of FIGURE 19.6?

FIGURE 19.6 The ideal-gas process of Example 19.1.

FIGURE 19.6 Full Alternative Text

MODEL The work done on a gas is the negative of the area under the pV curve. The gas is expanding, so we expect the work to be negative.

SOLVE As Figure 19.6 shows, the area under the curve can be divided into two rectangles and a triangle. Volumes must be converted to SI units of ${\text{m}}^3$. The total area under the curve is 250 J, so the work done on the gas as it expands is

$$W=-(\text{area under the }pV\text{curve})=-250\text{J}$$

REVIEW We noted previously that the product $\text{Pa}{\text{m}}^3$ is equivalent to joules. The work is negative, as expected, because the external force pushing on the piston is opposite the direction of the piston’s displacement.

Equation 19.8 is the basis for a problem-solving strategy.