A Closer Look at Work
A system doesn’t contain work. Rather, we can calculate the amount of work that
a system does in moving an object through a distance. Mathematically, we can de-
fine work as the product of the applied force and the distance the object moved.
Work (w) = force (F) × distance (d)
To relate this equation to a chemical system, as is done by the rocket scientist, the
force exerted by an expanding gas is directly related to the product of the pressure
of a system and the area in which it is expanding. Pressure, as we’ll discover in
Chapter 11, is the force exerted by a gas per unit area on the walls of its container.
Force (F) = pressure (P) × area (A)
Substituting this equation into that for the calculation of the amount of work,
we get
Work (w) = P × A × d
This equation can be simplified by considering the product of
the area (A) multiplied by the height (h), a measure of distance. As
shown in Figure 5.12, the base of the cylinder in the initial figure
has a known area (A). Moreover, the cylinder has a known height
(h), the distance from the bottom to the top of the cylinder. There-
fore, the gas inside the cylinder has a known volume (V), because
V
= A × h. In the final state in Figure 5.12, the piston has been dis-
placed by a known distance (h). We use the Greek letter (delta)
to represent the change in some quantity. Therefore, h means “the
change in height.” The resulting three-dimensional measurement
is the change in the volume of the system (V). Specifically,for this
case, V is the final volume of the system minus the initial volume
of the system: V = V
final
− V
initial
. In other words, A × hfora
reaction is best represented by V.
Work (w) = pressure (P) × change in volume (V)
Our final modification to the equation that defines work is an
adjustment to account for the loss of energy when a system does
work and the gain of energy when a system has work done on it.
We add a negative sign to the equation to illustrate this fact:
w =−PV
This equation indicates that when the volume of a system expands (i.e., V is pos-
itive), work is done by the system, and, as a result, energy is lost from the system
(w
system
= “−”). Conversely, when the volume of a system contracts (i.e., V is
negative), the system gains energy, and work is done on the system (w
system
= “+”).
Because pressure is recorded in atmospheres and volume in liters, the units for
work (energy) are in L·atm. This is mathematically related to the SI unit of energy
known as the joule (J), in honor of the English physicist James Prescott Joule
(1818–1889). We will derive this relationship later, but for now we can assume
that 1 L·atm = 101.3 J.
EXERCISE 5.2 Work, Work, and More Work
Calculate, in L·atm and joules, the work associated with the contraction of a gas
from 75.0 L to 30.0 L at a constant external pressure of 6.20 atm. (1 L·atm = 101.3 J)
First Thoughts
The road signs concerning work can be confusing. When a system expands, work
is done by the system (w
system
= “−”) on the surroundings (w
surroundings
= “+”).
176 Chapter 5 Energy
∆h
(a) Initial state (b) Final state
Area = A
∆V
FIGURE 5.12
Gas expansion does work. The initial volume of the gas in-
creases by an amount ∆V. When the pressure remains
constant, this becomes P∆V, or work.