James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

228 Chapter 3 Differentiation Rules
Since the concentration of the product increases as the reaction proceeds, the derivative
dfCgydt will be positive, and so the rate of reaction of C is positive. The concentrations
of the reactants, however, decrease during the reaction, so, to make the rates of reaction
of A and B positive numbers, we put minus signs in front of the derivatives dfAgydt
and dfBgydt. Since fAg and fBg each decrease at the same rate that fCg increases, we
have
rate of reaction − dfCg
dt
− 2 dfAg
dt
More generally, it turns out that for a reaction of the form
− 2 dfBg
dt
aA 1 bB l cC 1 dD
we have
2 1 a
dfAg
dt
− 2 1 b
dfBg
dt
− 1 c
dfCg
dt
− 1 d
dfDg
dt
The rate of reaction can be determined from data and graphical methods. In some cases
there are explicit formulas for the concentrations as functions of time, which enable us
to compute the rate of reaction (see Exercise 24).
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ExamplE 5 One of the quantities of interest in thermodynamics is compressibility.
If a given substance is kept at a constant temperature, then its volume V depends on its
pressure P. We can consider the rate of change of volume with respect to pressure—
namely, the derivative dVydP. As P increases, V decreases, so dVydP , 0. The compressibility
is defined by introducing a minus sign and dividing this derivative by the
volume V:
isothermal compressibility − − 2 1 V
Thus measures how fast, per unit volume, the volume of a substance decreases as the
pressure on it increases at constant temperature.
For instance, the volume V (in cubic meters) of a sample of air at 258C was found to
be related to the pressure P (in kilopascals) by the equation
V − 5.3
P
The rate of change of V with respect to P when P − 50 kPa is
dV
dP
Z − 2 5.3
P−50 P 2
The compressibility at that pressure is
Z
P−50
dV
dP
− 2 5.3
2500 − 20.00212 m3 ykPa
− 2 1 V
dV
dP
Z − 0.00212 − 0.02 sm 3 ykPadym 3
P−50 5.3
50
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