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P~9a~~--------------------------

PHYSICAL
CHEMISTRY

Dr. J.N. Gurtu

Dr. H.C. Khera

M.Sc., Ph.D.

M.Sc., Ph.D.

Former Principal

Reader & Head, Deptt. of Chemistry,

Meerut College, Meerut

loP. College, Bulandshahr.

»

PRAGATI PRAKASHAN


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© Authors-Physical Chemistry Vol. I

PRAGATI PRAKASHAN
Head Office: Educational Publishers
PRAGATI BHAWAN
240, W. K. Road, Meerut-250 001
SMS/Ph. : (0121) 6544642, 6451644
Tele/Fax: (0121) 2640642, 2643636
Visit us at : www.pragatiprakashan.in
e-mail:
Regd. Office:
New Market, Begum Bridge, Meerut-250 001

Edition 2010

ISBN-978-81-8398-496-6

Published by : K.K. Mittal for Pragati Prakashan, Laser Typesetting : Pragati Laser Type
Setters Pvt. Ltd., (Phone: 2661657) Meerut. Printed at: Arihant Electric Presss, Meerut.


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CONTENTS

IIUlij'
MATHEMATICAL CONCEPTS AND COMPUTER
Logarithmic and Antilogarithmic Relations
Find out the values of the following
Differentiation with Examples

Numerical Problems
Integration and give important formulae
Numerical Problems
Terms Permutation and Combination with Examples
24-41. Numerical Problems
42. Probability Definition
43-50. Numerical Problems
51. Logarithmic, Trigonometric Series
52. Maxima and Minima
53-57. Numerical Problem
58. Functioning, Characteristics, Limitations
Computer Programing Flow Charts
Fortan, Cobol, Basic, Pascal
Operating System
Exercise
Multiple Choice Questions
Fill in the Blanks
True or False
1.
2.
3.
4-12.
13.
14-23.
24.

1-55
1
5
7

11-13
13
16

22
. 23

29
32
32
33
34
38
49

50
51
52
52
54
54

o
o
o

111~lij"
GASEOUS STATE

5~102


1.

Nature of R and its value in different units

56

2.

Short account of kinetic theory of gases and derivation of
kinetic equation

57

3.

(a)

59

4.

Short account of kinetic theory of gases. Derivation of
PV = RT and show how the various gas laws are
consistent with it?
(b) Expression for kinetic energy of one mole of gas
Values of Cv and Cp from kinetic equation and variation of
CplCv with molecular complexity of the gas

62



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(viii)

Distribution of molecular velocities of Maxwell's law
65
(a) Average velocity, root me~n square velocity and most 67
probable velocity and relation among them
(b) Calculation of RMS velocity from kinetic theory of gases 67
7. Kinetic equation of gases
68
8. Ideal gas and its difference with a real gas
70
9. (a) Limitations of PV = RT and improvements suggested by 70
vander Waals. Derivation of vander Waals equation.
(b) Units of vander Waals constants
70
(c) Show that effective volume of gas molecules is four 70
times greater than actual volume of molecules
10. (a) Critical phenomenon, calculation and determination of 75
critical constants, short note on continuity of state
79
11. Some short questions on vander Waals equation
81
12. Short notes on :
(a) Various equation of state
(b) Law of corresponding states
(c) Mean free path

(d) Critical phenomenon and its utility
(e) Collision frequency
(0 Law of equipartition of energy
(g) Specific heat ratio
(h) Boyle temperature
(i) Continuity of state
13. Methods for producing cold and liquefaction of gases, 86
inversion temperature
o Numerical Problems
98
o Multiple Choice Questions
100
o Fill in the Blanks
101
o True or False
102

5.
6.

IIn"j'"
CHEMICAL AND PHASE EQUILmRIUM
Chemical Equilibrium

I

I

103-171


1.

Law of mass action and eqUilibrium constant

103

2.

Short notes on the following :
(i) Work function
(ii) Free energy
Thermodynamic derivation of law of mass action

104

3.

108


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4.
5.
6.
7.

Thermodynamic derivation of van't Hoff isotherm
Thermodynamlc··-derivation of van't Hoff isochore or van't
Hoff equation

Thermodynamic derivation of Clapeyron equation and
Clausius-Clapeyron equation
Le-Chatelier-Braun principle and applications to different
equilibria
o Numerical Problems

109
111

Explanation and illustration of phase, component and degree
of freedom
Phase rule and its thermodynamic derivation
Explain: Can all four phases in a one component system
co-exist in equilibrium?
Application of phase rule to water system
Application of phase rule to sulphur system
Short notes on :
(a) Non-variant system in phase rule studies
(b) Triple point
(c) Transition point
Two component system, graphical representation, reduced
phase rule equation, condensed state
Application of phase rule to lead-silver system
Application of phase rul~ to potassium iodide and water
system
Determination of number of phases and components of
different systems
Calculation of degree of freedom
Determination of number of phases, components and degree
of freedom of different systems

Ideal solutions, vapour pressure of such solutions
Non-ideal or real solutions, vapour pressure curves of
completely miscible binary solutions
Theory of fractional distillation of binary solutions
(a) Theory of partially miscible liquid pairs, e.g.,
(i) Phenol-water system
Oi) Triethyl-amine water system

126

IPhase Equilibrium I

1.

2.
3.
4.
5.
6.

7.
8.
9.
10.
11.
12.
13.
14.
15.
16.


114
118
122

-129
131
131
134
137

138
140
142
143
143
144
144
147
150
154


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(x)

(b)

(iii) Nicotine-water system
Influence of impurities on critical solution temperature


154

IDistribution Law I
1.

2.

Nemst's distribution law. limitations. applications
157
Nemst's distribution law. modification when the solute 162
undergoes dissociation or association
Numerical Problems
163
o Multiple Choice Questions
169
Fill in the Blanks
170
True or False
170

o
o
o

COLLOIDAL STATE
1.

2.


3.

4.

172-210

Explain the terms : colloidal state and colloidal solution, 172
methods for preparation and purification of colloidal solutions
(a) Difference between true solution. colloidal solution and 177
suspension
(b) Types of colloidal systems.
177
Preparation of colloidal solutions of AS2S3. Fe(OH)3 gold. 178
sulphur. silicic acid. carbon. mastic. iodine
Short notes on :
180
(i) Lyophilic and lyophobic colloids
(ii) Peptisation
(iii) Dialysis
(iv) Ultramicroscope
(v) Tyndall effect
(vi) Brownian motion
(vii) Electrophoresis
(viii) Electro-osmosis
(ix) Coagulation
(x) Hardy-Schulze law
(xi) Protection
(xii) Gold number
(xiii) Stability of lyophilic colloids
(xiv) Iso-electric point

(xv) Emulsion
(xvi) Gel
(xvii)Electrical double layer or Zera potential


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(xi)

5.

(a)
(b)

6.

Explain the following facts:
199
(a) A sulphur sol is coagulated by adding a little electrolyte,
whereas a gelatin sol is apparently unaffected.
(b) What happens when a colloidal solution of gold is
brought under the influence of electric field?
(c) What happens when an electrolyte is added to colloidal
solution of gold?
(d) What happens when a beam of light is passed through a
colloidal solution of gold?
(e) A colloidal solution is stabilised by addition of gelatin.
(f) Presence of H2S is essential in AS2S3 sol though H2S
ionises and should precipitate the sol.
(g) Why ferric chloride or alum is used for stoppage of
bleeding?


7.
8.
9.

Applications of colloids in chemistry
Sol-gel transformation
Note on thixotropy
Multiple Choice Questions
Fill in the Blanks
True or False

Origin and significance of charge on a colloidal particle
Classification of the sols:
Gold, Fe(OH)3, gelatin, blood, sulphur, AS2S3

198
198

200

2C4
206
208
209
209

o
o
o


III~I'M
CHEMICAL KINETICS AND CATALYSIS
Chemical Kinetics

I

I

211-276

1. (a)

Explain the terms: rate of chemical reaction, velocity 211
coefficient, molecularity and order of reaction
(b) Difference between molecularity and order of reaction
(c) Why reactions of higher orders are rare?
(d) Factors which affect reaction rates?
2. Zero order reaction, rate expression, characteristics.
215
3. Half order reaction, rate expression, characteristic
216
4. First order reaction, rate expression, characteristics, examples 217
5. (a)
(b)

Pseudo-unimolecular reactions
Study of kinetics of hydrolysis of methyl acetate

6. Half life period for a first order reaction


219
220


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7. Second order reaction, rate expression, characteristics,
examples and study of kinetics
8. Half life period for a second order reaction
9. Third order reaction, rate expressions, characteristics and
examples
10. nth order reactions, rate equation and characteristic
11. Methods employed in determining the order of reaction
12. Energy of activation and temperature coefficient
13. Activation energy, potential energy barrier and Arrhenius law
14. Collision theory for unimolecular reactions
15. Mathematical treatment of transition state theory, comparison
with collision theory.
Numerical Problems

o

221
224
225
227
228
230
233

235
236
242

ICatalysis I
1.

2.

3.
4.
5.

Catalyst, catalysis, types and classification of catalysis. 251
characteristics of catalytic reactions
Notes on the following:
258
(a) Catalytic promoters
(b) Catalytic poisons
Theories of catalysis, industrial applications of catalysts
261
Enzyme catalysis, characteristics and examples of enzyme 265
catalysis, kinetics of enzyme catalysis
Note on acid-base catalysis
269
Multiple Choiae Questions
270
Fill in the Blanks
274
True or False

275
o Log and Antilog Tables
(i)-(iv)

o
o
o


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MATHEMATICAL CONCEPTS AND COMPUTER
MATHEMATICAL CONCEPTS
Problem 1: Expillin the logarithmic and antilogarithmic reilltions with
suitable examples.

[A] Index
Multiplication of equal terms:
then multiplication will be x n , i.e.,

If the term x is multIplIed

Il

tImes,

x x x x x x x x ... n times = x"
Here x is called the base and n is called index.

[8] Laws of Index

m

and ~----tn-II

x"

3. xo= 1

4. (Xlll)" = Xlllll

5. (xyt =x"l

6. (xlyr = x"ly"

1
d x -n =1
7.xn =-an
-n
n
X

8.

xlln

X

= n{;

X


[C] Logarithms
Definition: If ab = c; then exponent 'b' is called the logarithm of number
'c' to the base 'a' and is written as log.. c = b, e.g., J~ = 81 ~ logarithm of
81 to the base 3 is 4, i.e., log3 81 = 4.
Note: a b = c, is called the exponential form and loga c = b is called the
logarithmic form, i.e.,
T3 = 0.125
(i)
(Exponential form)
(Logarithmic form)
log2 0.125 =-3
1
(Logarithmic form)
(ii)
log64 8 ="2
(64)112 = 8

(Exponential form)

Laws of Logarithms
[I] First Law (product law) :
The logarithm of a product is equal to the slim of logarithms of its
factors.
logo (m X 11) = log" III + log" /I


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2


PHYSICAL CHEMISTRY-I

loga (m X n xp) = loga m + logan + lo~p
Remember:
10& (m

+ n);/:. logam + logan

[II] Second Law (Quotient law) :
The logarithm of a fractl~n is equal to the difference between the
logarithm of numerator and (he logarithm of denominator.

Remember:
loga m

-1-- ;/:. loga m - loga n
oga Il

[III] Third Law (Power law) :
The logarithm of a power of a number is equal to the logarithm of
the number multiplied by the power.
loga (mt = n loga m

Corollary: Since
.,

Jlr-

-vm = m


l/n

1Jrl/n
1
lo~ -vm =lo&m
=-Io~m

n
Note:
(a) Logarithms to the base 10 are known as common logarithms.
(b) If no base is given, the base is always taken as 10.
(c) Logarithm of a number to the same base is always one, i.e.,
loga a = 1; 10glO 10 = 1 and so on.
(d) The logarithm of 1 to any base is zero, i.e.,
loga 1 = 0; logs 1 = 0; 10glO 1 =0 and so on.
(e) 10gJO 1 = 0; 10gJO 10 = 1;
10gIO 100 = 2
[.: 10gIO 100 = 10gJO

102 = 2 log 10 10 = 2 xl = 2]

Similarly, 10g]O 1000 = 3; 10gIO 10000 = 4 and so on.

Example: If log 2 = 0.3010 and log 3 = 0.4771; find the value of:
(i) log 6
(ii) log 5
(iii) log -V24
Solution:
(i)


log 6 = log (2 x 3) = log 2 + log 3 = 0.3010 + 0.4771 = 0.7781

(ii)

10
log 5 = log 2" = log 10 -log 2 = 1 - 0.3010 = 0.6990
(': log 10= 1)


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MATHEMATICAL CONCEPTS AND COMPUTE:,:R-'--_ _ _ _ _ _ _ _ _---=::3

(iii) log ..J24 = log (24)1/2 = ~ log (23 X 3)

="21 [3 log 2 + log 3] ="21 [3 X 0.3010 + 0.4771]
=0.69005

[0]

Common Logarithms and Use of Four Figure Log
Tables

[I] Common Logarithms : Logarithms to the base 10 are known as
common Logarithms. If no base is given, the base is always taken as 10.
[II] Characteristics and Mantissa: The logarithm of anum ber consists of two parts :
(i) Characteristic-It is the integral part of the logarithm.
(ii) Mantissa-It is the fractional or decimal part of the logarithm.
For exampLe, in log 273 = 2.4362, the integral part is 2 and the decimal
part is .4362.

Therefore, characteristic = 2 and mantissa = .4362.
[III] How to Find Characteristic?
(i) The characteristic of the logarithm ofa number greater than olle
is positive and is numerically one less than the number of digits before the
decimal point.
In number 475.8~ the number of digits before the decimal point is three.
..
Characteristic of log 475.8 = 2, i.e., (3 - 1 = 2)
Similarly, Characteristic oflog 4758 = 3, i.e., (4 - 1 = 3)
Characteristic of log 47.58 = 1, i.e., (2 - 1 = 1)
Characteristic of log 4.758 =0, i.e., (l - 1 =0)
(il) The characteristic of the logarithm of a number less than one in
negative and is numerically one more than the number of zeros immediately
after decimal point.
The number 0.004758 is less t;lan one and the number of zeros immediately after decimal point in it are two.
~ :. Characteristic of log 0.004758 = - (2 + 1) = -3, which is also written
as 3.
Note: To find the characteristic of the logarithm of a number less than
one, count the number of zeros immediately after the decimal point and add
one to it. The number so obtained with negati~ sign gives the characteristic.
..
Characteristic of log 0.3257 = - 1 = 1
[Since, the number of ~eros after decimal point = 0]
Characteristic of log 0.03257 = - 2 = 2
[Since, the number of zeros aftet; decimal point = 1]
Characteristic of log 0.0003257 = -4 = 4 and so on.


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4


PHYSICAL CHEMISTRY-I

[IV] How to Find Mantissa?
The mantissa of the logarithm of a number can be obtained from the
logarithmic table.
A logarithmic table consists of three parts:
(1) A column at the extreme left contains two digit numbers starting
from 10 to 99.
(2) Ten columns headed by the digits O. 1,2, 3,4,5,6, 7, 8,9.
(3) Nine more columns headed by digits 1,2,3,4,5,6.7,8,9.
A part of the logarithmic table is given below: (Difference to be added)
0

1

2

3

4

5

6

7

8


9

1 23 456

789

30 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 134 678 JO 11 13
31 4914 4928 4942 4955 4969 4983 4997 5011 5024 5038 1 34 678 JO 11 12
32 5051 5065 5079 5092 5105 5119 5132 5145 5159 5172 134 678 9 11 12

(a) To find the mantissa of the logarithm of one digit number.
Let the number be 3.
:. Mantissa of log 3 =Value of the number 30 under O.
= 0.477l.
(b) To find the mantissa of the logarithm of two digit number.
Let the number be 32.
:. Mantissa of log 32 = Value of the number 32 under O.
= 0.5051.
(c) To find the mantissa of the logarithm of three digit number.
Let the number be 325.
:. Mantissa of log 325 = Value of 32 under 5 = 0.5119
(d) To find the mantissa of the logarithm of a four digit number.
Let the number be 3257.
Mantissa of log 3257 = Value of 32 under 5 plus the difference
under 7
=0.5128
[5119+9=5128]

1. How to find the logarithm of a number from the logarithm table?
First. find the characteristic and then mantissa.

Suppose the number is 3257.
Characteristic of log 3257 =3
Mantissa of log 3257 = 0.5128
Therefore,

log 3257 =3.5128

[Note: The mantissa of the logarithms of all the numbers having the
same significant digits is the same. While tinding the mantissa, ignore the
decimal point.]


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5

MATHEMATICAL CONCEPTS AND COMPUTER

For example:

log 3257 = 3.5128
log 325.7 = 2.5128
log 3.257 = 0.5128
log 0.3257 = 1.5128
log 0.003257 = 3.5128. and so on]

[Note ..:.
3.4682 is equivalent to (-3 + 0.4682) and -3.4682
= -(3 + 0.4682) =-3 - 0.4682 i.e., in 3.4682, the mantissa is positive, while in -3.4682, the mantissa is negative.
(2) Remember, the mantissa should always be written positive.

(3) To make the mantissa positive, subtract 1 from the integral part
and add 1 to the decimal paJ;t, Thus,
-3.4682 =-3 - 0.4682 =(-3 - 1) + (l - 04682)
(1)

=- 4 + 0.5318 =4.5318]
Problem ~ : Find out the values of the following:
(i) ~.8321 + 1.4307
(ii) !.9256 - 4.5044
(iii) 1.7544 x 2
(iv) 2.3206 + 3
Solution:
(i)

3.8321
1.4307

Since, after adding decimal parts, we have 1 to carry.

1.2628

:.

-

-

1 + 3 + 1 = 1 - 3 + 1 =-1 == 1.

(ii) 1.9256


-4.5044
3.4212

Since,

-

1 - 4 =-3 =3

(iii) 1.7544

x2

-

1.5088 Since,

--

lX2+1=2+1=1

(iv) 2.3206 + 3
2.3206
3
3 + 1.3206
3

To make the integral part 2 divisible by3.


3 + 1.3206
3

Take

-

-

= 1 + 0.4402 = 1.4402

2= 3+1


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6

PHYSICAL CHEMISTRY-I

[E]

Antilogarithms
If log 5274 = 3.7221, then 5274 is called the antilogarithm of 3.722l.
We write, antilog 3.7221 = 5274.
To find an antilogarithm, the antilogarithm tables are used.
The antilogarithm tables are used in the same way as the logarithm

tables. The only difference between the two tables is that the column at the


extreme left of the log table contains all two digit numbers starting from 10
to 99; whereas an antilog table contains numbers from .00 to .99 (i.e., all
fractional numbers with only two digits after decimal) in the extreme left
column of it.
x

0

1

2

3

4

5

6

7

8

9

123 456 789

0.35 2239 2244 2249 2254 2259 2265 2270 2275 2280 2286 1 1 2 233 445
0.36 2291 2296 2301 2307 2312 2317 2323 2328 2333 2339 1 12 233 445

037 2344 2350 2355 2360 2366 2371 2377 2382 2388 2393 I 12 234 455
0.38 2399 2404 2410 2415 2421 2427 2432 2438 2443 2449 1 1 2 234 455

[Note :
(i)
Antilog tables are used only to find the antilogarithm of decimal
part.
(ii) To find the antilog of 2.368 means to find the number whose log
is 2.368].

Ex. 1 : If log x = 2.368,jind x.
Solution :
log x = 2.368
x = antilog 2.368
Antilog 2.368 = the number characteristic of whose log is 2 and mantissa
is 368.
From antilog table, the value of .36 under 8 is 2333.
Since the characteristic of log of the number is 2.
:. The number has 2 + 1 =3 digits in its integral part (i.e., 3 digits before
the decimal point).
Antilog 2.368 = 233.3.
Ex. 2 : Find the antilog of 2.3536
Solution: From the antilog table, find the value of .35 under 3 and add
to it the mean difference under 6. The number thus obtained is 2257.
Now place the decimal point so that the characteristic of its log is 2.
..
Antilog 2.3536 = 0.02257
[Note:
(i) Antilog 0.5362 = 3.438
(ii) Antilog 2.5362 = 343.8

(iii) Antilog 4.5362 = 34380

Antilog 1.5362 = 0.3438
Antilog 1.5362 = 0.003438
Antilog 5.5362 = 0.00003438 and so on].


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MATHEMATICAL CONCEPTS AND COMPUTER

7

Problem 3: Explain differentiation with suitable examples.

[I] Differentiation
If y = f(x) , x is the function of x, then there will be a change inf(x) with
respect to every change in x. If the increment in x is denoted by cSx , increment
inf(x) will be
f(x + ax) - f(x)
.
The ratio of increment in function f(x) and increment in variable x,
i.e.,
f(x + cSx) - f(x)

Ox
is called the difference quotient.
But, if cSx -+ 0, then this ratio tends to a definite quantity. This definite
quantity is called the differential coefficient of f(x) with respect to x and it is
denoted by


d1;)

or

~

or f'(x) or y'

Therefore, differential coefficient of f(x) with respect to x,

!!l = !:lfJE =
dx

dx

lim f(x
x~o

+ cSx) - fix)
cSx

The process of finding differential coefficient of f(x) with respect to x
is called 'differentiation' If Ox = It, then

!!l =

lim f(x + It) - f(x)
It


x~o

dx

[II] Some Standard Derivatives
(1)

!

(xn) = nxn -

1,

where n is a real number

For example, differential coefficient of ;?= 5x5 - 1 = 5x4
(2)

.!£(..Jx) = _1_

2Vx

dx

(3) _d (eX) = eX

dx
d mx
mx
(4) .dx (e ) = me

For example, differential coefficient of e3x = 3e 3x

d

<

X

(5 ) dx (a') = a loge a
For example, differential coefficient of 5x is

.!£
(5") =5x log 5
dx
e


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8

PHYSICAL CHEMISTRY-I

d
1
(6) - (log x) =dx
e
x
For example, differential coefficient of log (x + 1)
d [log (x


+ 1)]

_1_. 1 = _1_
x+1
x+l

dx

d
dx

1

(7) --- (log x) = - log e

For
d

-

d'(

(log x)
2

a

a


X

example,
1

differential

coefficient

of

log2 x,

i.e.,

=-log~ e
X

-

[III] Differentials of Trigonometric Functions
d ( .

(l) dx
(3)

SIn

(2)


x) = cos x

2

(4) dx (cot x) = cosec x

d

(6) dx (cosec x) = - cosec x cotx

x

~ (sin mx) = m cos mx
d

sinx

d

d
~
dx (tan x) = sec- x

d
(5) -d (sec x) = sec x tan x

(7)

~ (cos x) = -


(8)

~ (cos I1lx) = -

m

sin mx

2

(9) -d (tan IIlx) = 111 sec mx

x

[IV] Differentials of Inverse Functions
(1) dd (sin-I x) =

x

bI-x

(2) -d (cos-I)
x = - -1 -

{l=7

dx

(3) dd (tan-I x) = _1_ry


x
1+x1
d (-I
(5) -d
sec x) = _~
x
x'lx- - 1

(4 ) -d (cot-I)
x = - - -1- ,
dx
1 +x~
d (
_)
1
(6) -d cosec x =- _~
x 'IX- - 1

x

[V] Differentiation of a Product
The differential coefficient of product of two functions is the sum of
the product of one function and the differential coefficient of the other.
d
du
du
-- (ll . u)
dx

=u . - + u . dx


dx

For example, to find the differential coefficient of xe-' with respect to
x, we have
d
t
d r
x
d
dx (xe-) =x . d; (e') + e . d-; (x)
=x.e-'+ex.l
l

=e (x+1)


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9

MATHEMATICAL CONCEPTS AND COMPUTER

[VI] Differentiation of a Division
If u and v are any two functions of x, then

d

d


~(~)== v .~(u)-u.~(v)
dx v

(v/

For example, to find the differential coefficient of xm Iloge x with respect
to x, we have

!!.- (~) == loge (x) . d;d(m
X ) dx loge x

m d
X

•

d; (loge x

)

(loge x)2
loge x (mxm -I) -

==

(loge x)2
m-I

==


x'" . (lIx)

m. 1ogeX. X

m-I

-x

(loge xi
m

x -

I

•

(m loge x - 1)
ADS.

(loge x)2

[VII] Partial Differentiation
We know that the differential coefficient off (x) with respect to x is
. f(x + ox) - f(x)
11m
Iix~O
x
provided that limit exceeds and is expressed as,


a

/' (x) or

'

~ [f(x)]

If u == fix, y) be a continuous function of two independent variables
x and y, then the differential cofficient w.r.t. x (taking y as constant) is called
the partial derivative or partial differential coefficient of u w.r.t. x and is
represented by different symbols such as

au
ax ' E1
ax ,Ix (x, y) ./x·
Symbolically, if u == f(x, y) then
· f(x + Ox, y) - f(x, y)
11m
Iix ..... O
Ox
If it exists then. it is called the partial derivative or partial differential
coefficient of u w.r.t. x. It is denoted by,

au
ax

or

af

ax

or

Ix or

Ux


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10

PHYSICAL CHEMISTRY-I

Similarly, by keeping x constant and allowing y alone to vary, we can
define the partial derivative or partial differential coefficient of u w.r.t. y. It
is represented by anyone of the following symbols.

au af
oy' Oy,/y (x, y),/y
ou = lim f(x, y + By) - fix, y)
oy dy-+O
By

Symbolically,

provided this limit exists.

then


u = ~ + 2hx2y + bi
ou
2·
ox = 3ax + 4hxy

and

-=2hx +2by

For example, if

ou
oy

2

[I] Rules of Partial Differentiation
Rule (1) : (a) If u is a function of x, y and we are to differentiate partially
w.r.t. x, then y is treated as constant.
(b) Similarly, if we are to differentiate u partially w.r.t. y, then x is treated
as constant.
(c) If u is a function of x, y, z, and we are to differentiate u w.r.t. x, then
y and z are treated as constants.
Rule (2) : If z = u ± v, where u and v are functions of x and y, then

oz = ou + ov

ax ax - ax


and

oz = ou + ov

ay ay - ay

Rule (3) : If z = uv, where u and v are functions of x and y, then

oz 0
ov
ou
-=-(uv)=u-+vox ox
ox
ox
oz 0
ov
ou
-=-(uv)=u-+vand
oy oy
oy
oy
Rule (4) : If z =~, where u and v are functions of x and y, then
v
oz
ox
and

=~ (~ ) = v ( ~ ) ox

v2


V

ou.)

u(

f)

(OV)

~; =;y (~)= v ay ~ u ay
(

Rule (5) : If z = f(u) , where u is a function of x and y, then

OZ = oz OU and OZ = oz ou
ox ou· ox
oy ou· oy


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11

MATHEMATICAL CONCEPTS AND COMPUTER

Problem 4. Determine thejirst order partial derivatives ofyx.
Solution. Let z = yx
Taking logs.


log z =x log y

... (i)

az

. . w.r.t. x. -1 -;- = Iog y
O I·ftierenttatmg
Z aX

az = z Iog y= yx Iog y
ax
Now differentiate equation (i), w.r.t y, we get

;~~=x(~)=~
az (X)
ay=Z
y =yx(X)
y =xy

x-I

or

Problem S. Find the differential coefficients ofthejollowing:
(i) x S
(U)

M


(iii) X-SIl

Solution,
(i)

!!:.-. (x5 ) = 5~ - I = 5x4

(ii)

!!:.-. U =.!!:.- (x)3/2 =1 x~-l = 1 xl/2-

'1'1')
(1

dx

dx
d
dx

2

dx

(-5/2)
X

-5


=T X

-7/2 - I
2

2

-5

=T x

-7/2

Problem 6. Find the differential coefficients of the following:
(i) ax4
(ii) 7 loglo x
Solution.
(i)
(ii)

:.x (ax = a :.x (x4) = 4ax3
:.x (7 loglO x) = 7 :.x (log lO x) = 7
4

)

~ loglo (e)

Problem 7. Find the differential coefficients oj the following:
2

(i) 4x + ~ + 10
(ii) tan x 1- sin x
2
(iii) x sin x
Solution .
. d
2
d
2
d
d
(I) dx(4x +6x+1O)= dx(4x)+ dx"(6x)

+ dx(1O)=8x+6

..) dx
d (tan x sm
.)
d (tan x ) + dx
d (-sm
.)
(II
x = dx
x = sec2 x + cos x
d (.
d (2).
...) dx
d (2.
(III
x sm x ) -::: X .z dx

sm x ) + di
x . sm x

= "'='
~ SIn-X


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12

PHYSICAL CHEMISTRY-I

Problem 8. Find the differential coefficients of the following:
(i)

sin x

x

(ii)

(x+lf

xl

Solution.
2d(.

(i)


)

.

d(2)

~ (Sin x) = x ;t;; sm x - sm x.;t;; x
(X 2)2

x2

dx

=
=

i

cos x - sin x . 2.x
X4

X

cos X

2 sin x

-


x

2

d
d
2
(x+l) ;t;;(x)-x.;t;;(x+l)
2

••) - d [
X
(II
dx (x + 1)2

(x+ 1)4
_ (x+ 1)2. I-x. 2. (x+ 1)
(x+l)4
_ (x+ 1) [x+ 1-2.x] _ (I-x)
(x+l)4
-(x+l)3

Problem 9. Differentiate the following with respect to x :
(i) sin 5x
(ii) log sin x
(iii) (3x 2 + Il
Solution.
(i) Let y = sin 5x => y = sin t, where t =5x

(Meerut 2007)


~=cost

..

dt =5
dx
dv dv

and

dt
d; = di .dx
= cos t . 5 = 5 cos 5x.
(ii) Let y = log sin x => y = log t, where t = sin x

4r_l

dt - t
dt
dx =cosx

and

4r = 4r.
dx

4r =_._1_ . cos x = cot x

or


(iii)

dt = 1 . cos x
dtdx t

1

(3x

dx

2

smx

2

+ 1)2 = 2 (3x + 1)2-1 . (6x)
= 2 (3x2 + 1) + (6x)
=12x (3x2 + 1)


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13

MATHEMATICAL CONCEPTS AND COMPUTER

Problem 10. Calculate the value of ~ from the following:

(i) x=acost,y=asint
Solution.
(i) x = a cos t, y = a sin t
dx
.
dv
- = - a sm t ::::.L = a cos t
dt
' dt
cos t
!:!l. =d dt dt = a cos t
y
. dx - a sin t = - si~
dx

!:!l. = -

or

dx

1

cott

Problem ll.lfy.= tan- [1

~x2}

then calculate the value of


Solution. Let x = tan 8, then
.•
Y = tan -I [ 2 tan 82
I-tan 8
..

Problem 12. If y

J= tan-

1 ( tan

~.

28) = 28 = 2 tan - 1 x

dy _ '2
dx - 1 +x2"
=:

d .
tan-1 (Sin x + C?S x) then calculate the value of f!:1...dx
cos x - SID X

SoIu tion. Let y = tan -I (sin x + cos
. x~

cosx- smx


:

Dividing ilie :::::~(f;;::n~)~ ~; r:~(: ~Jl~: + ~
...

!:!l.
dx --

1.

Problem 13. Explain integration and give important integral formulae.

[I] Integration
The reverse process of differentiation is called the integration and it is
represented by a symbol

f.

If differential coefficient of f(x) is <I>(x), i.e.,
d
- [((x)] = <I>
dx

f

then

Therefore,

<I>


f

dx =f(x)

<I>(x) dx


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14

PHYSICAL CHEMISTRY-I

is called the integration of c1>(x) with respect to x.

[II] Tables 01 Integral Formulae
(1) J al(x) dx = a JI(X) dx + C

± c1>(x)} dx = JI(X) dx ± J c1>(x) dx

(2) J If(x)

(3) Jl. dx=x, JOdx=C

n :n;:, (n;t - 1)

(4) Jx dx:%

For example, integration of x5 is given by,

5+1
1 6
5
x dx = ; + 1 =6" x
(6) J

f
~dx=IOgeX

(7) f C dx=e

x

(8)faXdx=~
loge a
(9) ISin x dx = - cos x + C

(10) Jcos x dx = sinx + C
(11)

f

tan x dx = loge sec x + C = - 10& cos u + C

(12) fcotx dx= 10& sin x + C = -loge cosec x + C

(13) fcosec x dx = loge (cosec x - cot u) + C = loge tan
2

(14) f sin xdx=

2

(15) J cos x dx =
2

t
t +t

x-t sin x cos x + C
x

sin x cos x + C

(16) Jsec x dx = tan x + C

~+C


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MATHEMATICAL CONCEPTS AND COMPUTER

(17)

15

f

cosec 2 x dx = - cot x + C
2


(18) ftan x dx = tan x - x + C
2

(19) fcot xdx=-cotx-x+C
(20) f sec x tan x dx = sec x

(21) Icosec xcotx dx= - cosec x
(22)
(23)

f ~1 ~X2

l

l

dx=sin- x=-cos- x

f~
dx = tanl+x

l

l

x = - coe x

(24) ff(y) dx = ff(y) dy :


(25)

f~=.!tan-l~+
C
2
2

(26)

J

x +a

a

dx
x- -a

a

1

~=-2

a

x-a
2
2
loge--+C,x >a

x+a

1
a-x
2
2
=-2 log--+C,x a
a+x

(27)

Ih=
a -x

sin-

1

~a + C, a > 0

[III] Definite Integral
When any functionf(x) is integrated between the lower limit and upper

limit of x, then it is called the definite integral.
For f(x), if the lower and upper limits of x are a and b, respectively,
then

f

b

f(x) dx =

[F(x)]~ =F(b) - F(a)

a

where

ff(X) dx = F(x).

For example,J(x) =x 5 and a = 2, b =3, then


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16

PHYSICAL CHEMISTRY-I

1 [(3)6 -

(2)6] = 729 - 64 = 665
6
Problem 14. Evaluate the following integrals:
=

(i)

f

X7

dx

S~lution.
(ii)

f

f

(li)

(i)

f

eX dx

7+ 1

7

x dx

8

=; + 1 =~

eX dx = eX

Problem 15. Find the value of the following integrals:
(i)

f

3

(x + 2x + 7) dx(ii)

Solution. (i)

f

f

(x

3

2)2 dx

-

f f f
3

3

(x + 2x + 7) dx =

x dx +

4

(ii)

f

2x dx +

2

4

7 dx

=£+~+7x=£+i+7x
4

(x

3

-

2)2 dx =

=

2

f
f f f

4

3

6

(x + 4 - 4x ) dx

6

x dx +

x7

4 dx -

3

4x dx

4x3+ 1

=-+4x--7
3+1

x7
4X4
=-+4x--

7

4

7

= ~+4x-x4
7

f
f

Problem 16. Evaluate the integral
·
(.)
So Iubon.
I

f~-!1-

-

.
SIO X

1

(1

-

d~

-smx

.

(l. + sin
x) dx
. X )sm x). (1 + Sin