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Chapter 2 because the force and the displacement are in opposite directions. The total work
The First Law of Thermodynamics
done on the object by all forces is zero. The work–energy theorem (2.16) gives
w K 0, in agreement with the fact that the object started at rest and ended
at rest. (We derived the work–energy theorem for a single particle, but it also
applies to a perfectly rigid body.)

Exercise
A sphere of mass m is attached to a spring, which exerts a force F kx on
the sphere, where k (called the force constant) is a constant characteristic of the
spring and x is the displacement of the sphere from its equilibrium position (the
position where the spring exerts no force on the sphere). The sphere is initially at
rest at its equilibrium position. Find the expression for the work w done by some-
one who slowly displaces the sphere to a final distance d from its equilibrium
1
2
position. Calculate w if k 10 N/m and d 6.0 cm. (Answer: kd , 0.018 J.)
2

2.2 P-V WORK
Work in thermodynamics is defined as in classical mechanics. When part of the sur-
roundings exerts a macroscopically measurable force F on matter in the system while
this matter moves a distance dx at the point of application of F, then the surroundings
has done work dw F dx [Eq. (2.8)] on the system, where F is the component of F
x
x
in the direction of the displacement. F may be a mechanical, electrical, or magnetic
force and may act on and displace the entire system or only a part of the system. When
F and the displacement dx are in the same direction, positive work is done on the
x
system: dw 0. When F and dx are in opposite directions, dw is negative.
x
Reversible P-V Work
l The most common way work is done on a thermodynamic system is by a change in
the system’s volume. Consider the system of Fig. 2.2. The system consists of the mat-
ter contained within the piston and cylinder walls and has pressure P. Let the external
pressure on the frictionless piston also be P. Equal opposing forces act on the piston,
System
and it is in mechanical equilibrium. Let x denote the piston’s location. If the external
pressure on the piston is now increased by an infinitesimal amount, this increase will
produce an infinitesimal imbalance in forces on the piston. The piston will move
b
inward by an infinitesimal distance dx, thereby decreasing the system’s volume and
x increasing its pressure until the system pressure again balances the external pressure.
During this infinitesimal process, which occurs at an infinitesimal rate, the system will
Figure 2.2
be infinitesimally close to equilibrium.
A system confined by a piston. The piston, which is part of the surroundings, exerted a force, which we denote by
F , on matter in the system at the system–piston boundary while this matter moved a
x
distance dx. The surroundings therefore did work dw F dx on the system. Let F be
x
the magnitude of the force exerted by the system on the piston. Newton’s third law
(action reaction) gives F F . The definition P F/A of the system’s pressure P
x
gives F F PA, where A is the piston’s cross-sectional area. Therefore the work
x
dw F dx done on the system in Fig. 2.2 is
x
dw PA dx (2.25)
The system has cross-sectional area A and length l b x (Fig. 2.2), where x is the
piston’s position and b is the position of the fixed end of the system. The volume of