Page 43 - Physical chemistry eng

20 CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics

TABLE 2.1 Types of Work
Types of Work Variables Equation for Work Conventional Units
V f
3
Volume expansion Pressure (P), volume (V) w =- P external dV Pa m = J
L V i
x f #
Stretching Force (F), length (l) w = F dl N m = J
x
L i
s f #
2
-1
Surface expansion Surface tension (g) , area (s) w = G dS (N m )(m ) = J
Ls i
Q
Electrical Electrical potential (f) , electrical charge (Q) w = fdQ¿ V C = J
L 0

EXAMPLE PROBLEM 2.1
a. Calculate the work involved in expanding 20.0 L of an ideal gas to a final volume
of 85.0 L against a constant external pressure of 2.50 bar.
b. An air bubble in liquid water expands from a radius of 1.00 cm to a radius of
–1
3.25 cm. The surface tension of water is 71.99 N m . How much work is done in
increasing the area of the bubble? Assume that the system is the contents of the bubble.
c. A current of 3.20 A is passed through a heating coil for 30.0 s. The electrical
potential across the resistor is 14.5 V. Calculate the work done on the coil.
d. If the force to stretch a fiber a distance x is given by F =-kx with
k = 100. N cm -1 , how much work is done to stretch the fiber 0.15 cm?

Solution
V f
a. w =- P external dV =-P external (V - V )
i
f
LV i
-3
5
10 Pa 10 m 3
=-2.50 bar * * (85.0 L - 20.0 L) * =-16.3 kJ
bar L
b. A factor of 2 is included in the following calculation because a bubble has an
inner and an outer surface. We consider the bubble and its contents to be the sys-
point in opposite directions, giving rise to the negative
tem. The vectors and S
G
sign in the second integral.
#
s f s f
2
2
w = G dS =- gds = 2g4p(r - r )
i
f
Ls i Ls i
10 -4 m 2
2
2
-1
2
2
=-4p * 71.99 Nm (3.25 cm - 1.00 cm ) * 2
cm
=-0.865 J
Q
c. w = fdQ¿= fQ = Ift = 14.5 V * 3.20 A * 30.0 s = 1.39 kJ
L 0
d. We must distinguish between F, the restoring force on the fiber, and F œ , the force
œ #
exerted by the person stretching the fiber. They are related by F =-F œ . If we
calculate the work done on the fiber, F dl = F¿dl because the vectors F œ and
dl point in the same direction
œ # x f kx 2 x f 100. N m -1 * x 2 0.15
w = F dl = kxdx = B R = B R = 1.1 J
L 3 2 x 0 2 0
x 0
If we calculate the work done by the fiber, the sign of w is reversed because F and
dl point in opposite directions.