LOGARITHMS AND EXPONENTIAL FUNCTIONS 35
A
- The work done in an isothermal expansion of
a gas from pressurep 1 top 2 is given by:
w=w 0 ln
(
p 1
p 2
)
If the initial pressurep 1 = 7 .0 kPa, calculate
the final pressurep 2 if w=3w 0
[p 2 =348.5 Pa]
4.9 Laws of growth and decay
The laws of exponential growth and decay are of the
formy=Ae−kxandy=A(1−e−kx), whereAandk
are constants. When plotted, the form of each of these
equations is as shown in Fig. 4.7. The laws occur
frequently in engineering and science and examples
of quantities related by a natural law include:
A
y
0 x
y = A(1−e−kx)
(b)
0 x
(a)
A
y
y = Ae−kx
Figure 4.7
(i) Linear expansion l=l 0 eαθ
(ii) Change in electrical resistance
with temperature Rθ=R 0 eαθ
(iii) Tension in belts T 1 =T 0 eμθ
(iv) Newton’s law of cooling θ=θ 0 e−kt
(v) Biological growth y=y 0 ekt
(vi) Discharge of a capacitor q=Qe−t/CR
(vii) Atmospheric pressure p=p 0 e−h/c
(viii) Radioactive decay N=N 0 e−λt
(ix) Decay of current in an
inductive circuit i=Ie−Rt/L
(x) Growth of current in a
capacitive circuit i=I(1−e−t/CR)
Problem 24. The resistance R of an elec-
trical conductor at temperatureθ◦Cisgiven
by R=R 0 eαθ, where α is a constant and
R 0 = 5 × 103 ohms. Determine the value of
α, correct to 4 significant figures, when
R= 6 × 103 ohms andθ= 1500 ◦C. Also, find
the temperature, correct to the nearest degree,
when the resistanceRis 5. 4 × 103 ohms.
TransposingR=R 0 eαθgives
R
R 0
=eαθ.
Taking Napierian logarithms of both sides gives:
ln
R
R 0
=ln eαθ=αθ
Hence α=
1
θ
ln
R
R 0
=
1
1500
ln
(
6 × 103
5 × 103
)
=
1
1500
(0. 1823215 ...)
= 1. 215477 ···× 10 −^4
Hence α=1.215× 10 −^4 ,
correct to 4 significant figures
From above, ln
R
R 0
=αθ
hence θ=
1
α
ln
R
R 0
WhenR= 5. 4 × 103 ,α= 1. 215477 ...× 10 −^4 and
R 0 = 5 × 103
θ=
1
1. 215477 ...× 10 −^4
ln
(
5. 4 × 103
5 × 103
)
=
104
1. 215477 ...
(7. 696104 ...× 10 −^2 )
= 633 ◦C, correct to the nearest degree