BAB 2. MAT KIM INTEGRAL CALCULUS

(1)

INTEGRAL CALCULUS

SUWARDI BY

(2)

2.1 INTRODUCTION

• _{TWO MAJOR APPROACHES}

– _{ANTIDERIVATIVES – to mathematically}

generate integrals

– _{INTEGRATION – to assign a physical meaning}

to the integral

• _{The other approach:}

– _{To consider the integral as the sum of many}

similar, infinitesimal elements

• _{The process of taking integration, as the}

inverse of differentiation

• _{TWO MAJOR APPROACHES}

– _{ANTIDERIVATIVES – to mathematically}

generate integrals

– _{INTEGRATION – to assign a physical meaning}

to the integral

• _{The other approach:}

– _{To consider the integral as the sum of many}

similar, infinitesimal elements

• _{The process of taking integration, as the}

(3)

• _{In this chapter we shall consider the}

reverse process. Knowing the effect of individual changes, we wish to

determine the overall effect of adding together these changes such that sum equals a finite change

• _{Before considering the physical}

significance and the application of

integral calculus, let us briefly review the general and special methods of integration

• _{In this chapter we shall consider the}

reverse process. Knowing the effect of individual changes, we wish to

determine the overall effect of adding together these changes such that sum equals a finite change

• _{Before considering the physical}

significance and the application of

integral calculus, let us briefly review the general and special methods of integration

(4)

2.2 INTEGRAL AS

ANTIDERIVATIVE

2.2 INTEGRAL AS

ANTIDERIVATIVE

• _{Function y = f(x), differentiation of}

this function, symbolized by the equation

• _{Function y = f(x), differentiation of}

this function, symbolized by the equation

)

(

'

)

(

x

f

dx

x

df

dx

dy



dy

_

f

'

(

x

)

dx

Where f’(x) denotes the first

derivative of the function f(x) with respect to x

Where f’(x) denotes the first

derivative of the function f(x) with respect to x

(5)

• _{In this chapter we shall pose the following}

question: What function f(x), when

differentiated, yields the function f’(x)?

• _{For example, what function f(x) when}

differentiated, yields the function f’(x) = 2x?

• _{Substituting f’(x) = 2x into equation gives} • _{In this chapter we shall pose the following}

question: What function f(x), when

differentiated, yields the function f’(x)?

• _{For example,}_{what function f(x) when}

differentiated, yields the function f’(x) = 2x?

• _{Substituting f’(x) = 2x into equation gives}

x dx

dy dx

(6)

• _{This function,}_f(x)_{, for which we are}

looking is called the integral of the differential and symbolized by the equation

• _{This function,}_f(x)_{, for which we are}

looking is called the integral of the differential and symbolized by the equation





f

x

dx

x

f

(

)

'

(

)



:

the integral sign

; f’(x):

the

integrand



:

the integral sign

; f’(x):

the

integrand

(7)

• _{If f’(x) = 2x, then f(x) = x}2. f(x) = x2

is not the complete solution but f(x) = x2 + C is the complete solution.

• _{C is the constant of integration, and}

this constant always is included as

part of the answer to any integration. Thus,

• _{If f’(x) = 2x, then f(x) = x}2. f(x) = x2

is not the complete solution but f(x) = x2 + C is the complete solution.

• _{C is the constant of integration, and}

this constant always is included as

part of the answer to any integration. Thus,









x

dx

x

C

(8)

2.3 GENERAL METHODS OF

INTEGRATION

2.3 GENERAL METHODS OF

INTEGRATION

• _{Let us consider several general methods of}

integration

• _{Let us consider several general methods of}

integration



du

(

x

)



u

(

x

)



C

.

1 

a du a



du au  C

. 2

C

n

u

du

u

n n









1 .

3

(9)



x dx  x  C a

4 )

(

4 3

Examples: Examples:

             C

RT H C

T R

H dT

T R

H dT

T R

H dT

RT H

b) ₂ 1₂ 2 1

(10)







d

u



u



C

u

du

ln

.

4 





f

(

x

)



g

(

x

)

dx





f

(

x

)

dx





g

(

x

)

dx

.

5 



e



C

m

dx

e

1 .

6 





kx



C

k

dx

kx

1 cos

sin

.

7 



x



C

k

dx

x

1 sin

cos

.

(11)

2.4 SPECIAL METHODS OF

INTEGRATION

2.4 SPECIAL METHODS OF

INTEGRATION

• _{Many of the functions encountered in}

physical chemistry are not in one of the general forms given above

• _{Special methods of integration:}

– _{Algebraic Substitution}

– _{Trigonometric Transformation}

– _{Partial Fractions}

• _{Many of the functions encountered in}

physical chemistry are not in one of the general forms given above

• _{Special methods of integration:}

– _{Algebraic Substitution}

– _{Trigonometric Transformation}

(12)

Algebraic Substitution

• _{The mathematical functions can be}

transformed into one of the general forms in section 2.3 or into one of the forms found in table of integral by

some form of algebraic substitution

• _Examples

• _{The mathematical functions can be}

transformed into one of the general forms in section 2.3 or into one of the forms found in table of integral by

some form of algebraic substitution

• _Examples



1 

.

2 Evaluate

)

(13)

• _{Let us attempt to transform this}

integral into the form

• _{Let u = (1 – x}2), Then du = -2x dx.

Hence,

• _{Let us attempt to transform this}

integral into the form

• _{Let u = (1 – x}2), Then du = -2x dx.

Hence,



u n du











x  x2 5 dx 



u5 du  u6  C  1 x2 6  C

6 1 6

1 1

(14)



   

  



 _dT

kT E e

b) Evaluate E/kT ₂ (

kT

E

u







Let

Then ₂ dT. Hence

kT E du  



_ 



   

  



   



 _dT _e _du _e _C _e _C

kT E

(15)



_

.

Evaluate

)

(

nb

V

dV

c

. form

the into

integral the

transform to

attempt us

Let

_

u du

Let u = V – nb. Then du = dv. Hence,







    

 u u C V nb C

du nb

) ln(

(16)



x

dx

d

)

Evaluate

sin

cos

(

Let u = sin x. Then du = cos x dx. Hence



2 x x dx  u2 du  u3  C  sin 3 x  C

3 1 3

1 cos

(17)

Trigonometric Transformation

• _{Many trigonometric integrals can be}

transformed into a proper form for integration by making some form of trigonometric transformation using

trigonometric identities.

• _{For example, to evaluate the integral}

we must make use of the identity

• _{Many trigonometric integrals can be}

transformed into a proper form for integration by making some form of trigonometric transformation using trigonometric identities.

• _{For example, to evaluate the integral}

we must make use of the identity



sin2 xdx





x 1 cos 2 2

(18)

• _Thus, • _Thus,







 x 



dx 



cos 2x dx

2 1 2

1 2

cos 1

2 1

• _{Integrating each term separatly gives,} • _{Integrating each term separatly gives,}



x dx x  sin 2x  C

4 1 2

(19)

• _{Again, integration of integral of this}

type are more parctically done by using the Table of Integrals

• _Example:

• _{Again, integration of integral of this}

type are more parctically done by using the Table of Integrals

• _Example: Evaluate

_

cos3 2x dx

• _{Integration of this function can be}

accomplished using integral from the Table of Integral. Here, a = 2 and b = 0

• _{Integration of this function can be}

accomplished using integral from the Table of Integral. Here, a = 2 and b = 0















    ax b  C

a b

ax a

dx b

ax 3

3 _sin

3 1 sin

1 cos



x dx  x  sin (2x)  C

6 1 )

2 sin( 2

1 )

2 (

(20)

Partial Fractions

• _{Consider an integral of the type}

constants are

and where

, ) (

)

(a x b x a b



_ _

• This type of integral can be trasformed into simpler integral by the following method. Let A = (a – x) and B = (b –x). Then

AB A B

AB A AB

B B

  

  1

(21)

• _Therefore, • _Therefore,          B A A B AB 1 1 ) ( 1 1              ( ) 1 ) ( 1 ) ( 1 ) )( ( 1 x b x a a b x b x a

• _{Which can be integrated to give,} • _{Which can be integrated to give,}



             ( ) ( ) ( ) 1 ) )(

( b x

dx x a dx a b x b x a dx







            

 a x C

x b a b C x b x a a b x b x a dx ) ( ) ( ln ) ( 1 ) ln( ) ln( ) ( 1 ) )( (

(22)

2.5 The Integral as a summation of

infinitesimally

2.5 The Integral as a summation of

infinitesimally

• _{In the previous sections we have}

considered integration as the purely mathematical operation of finding

antiderivatives.

• _{Let us now turn to the more physical}

aspects of integration in order to

understand the physical importance of the integral

• _{In the previous sections we have}

considered integration as the purely mathematical operation of finding

antiderivatives.

• _{Let us now turn to the more physical}

aspects of integration in order to

understand the physical importance of the integral

(23)

• _{Consider, as an example:}

1. The expansion of an ideal gas from a volume of V₁ to a volume V₂ against a

constant external pressure P_ext,

• _{Consider, as an example:}

1. The expansion of an ideal gas from a volume of V₁ to a volume V₂ against a

constant external pressure P_ext,

The plot seen in the indicator

diagram is not a graph of P_ext as a function of V.

There is no functional dependence between the

external pressure on the gas and the volume of the gas

The plot seen in the indicator

diagram is not a graph of P_ext as a function of V.

There is no functional dependence between the

external pressure on the gas and the volume of the gas

(24)

• _{The diagram shows what the external}

pressure is doing on one axis and

what the volum is doing on the other axis

• _{Both can vary independently}

• _{The work done by the gas does}

depend on both the external prssure and the volume change

• _{For expansion, the work is}_w_{= -}

P_ext(V₂ – V₁)

• _{W is just the negative of the area}

under the P_ext versus V diagram

• _{The diagram shows what the external}

pressure is doing on one axis and

what the volum is doing on the other axis

• _{Both can vary independently}

• _{The work done by the gas does}

depend on both the external prssure and the volume change

• _{For expansion, the work is}_w_{= -}

P_ext(V₂ – V₁)

• _{W is just the negative of the area}

(25)

2. The external prssure changes, in some fashion, as the volume changes

– _{The external pressure is not changing as a}

function of volume

– _{To emphasize this point, let us assume that}

the external prsessure actually goes up as the volumes changes (see indicator diagram )

– _{A gas could be expanding against atmospheric}

pressure which perhaps is increasing over the period of time that the expansion take place

2. The external prssure changes, in some fashion, as the volume changes

– _{The external pressure is not changing as a}

function of volume

– _{To emphasize this point, let us assume that}

the external prsessure actually goes up as the volumes changes (see indicator diagram )

– _{A gas could be expanding against atmospheric}

pressure which perhaps is increasing over the period of time that the expansion take place

(26)

Indicator diagram showing PV work done by a gas expanding against a variable external pressure

(27)

• _{The approximate area under the}

curve, then, is just the sum of the four rectangles

• _{The approximate area under the}

curve, then, is just the sum of the four rectangles





 

 

 

1 4

3 2

1 approx

i V

P V

P A

• If we extend this process-that is if we

divide the area under the curve into more and more rectangles of smaller and smaller

V-the sum approaches a fixed value as N approaches infinity

• If we extend this process-that is if we

divide the area under the curve into more and more rectangles of smaller and smaller

V-the sum approaches a fixed value as N approaches infinity

(28)

• _{Hence, we can write} • _{Hence, we can write}



     N i i N V P A 1

lim

• _{However, since as N approaches infinity,}__V

approaches zero, we also can write

• _{However, since as N approaches infinity,}__V

approaches zero, we also can write



     N i i V V P A 1 0

lim

• _{But, by definition} • _{But, by definition}





     2 1 1 0

lim

V V i N i i V dV P V P

(29)

• _{Where the symbol is read “the}

integral from V₁ to V₂”, and V₁ and V₂ are called the limits of integration. Hence,

• _{Where the symbol is read “the}

integral from V₁ to V₂”, and V₁ and V₂ are called the limits of integration.

Hence,



dV

P

A

ext





This integral is called a

definite integral

This integral is called a

definite integral

(30)

2.6 Line Integral

• _{Integral of the general type}





dx y

Are called line integral, because such integral represent the area under the specific curve (path) connecting x1 to x2

(31)

• _{Such integral can be evaluated}

analytically (i.e., by finding the

antiderivative) only if an equation, the path, y = f(x) is known, since under these circumstances the

integral

contains only one variable

• _{Such integral can be evaluated}

analytically (i.e., by finding the

antiderivative) only if an equation, the path, y = f(x) is known, since under these circumstances the

integral

contains only one variable



)

(

dx

x

(32)

• _{If y is not a function of x (as in the}

case of Pext versus V), or

• _{if y is a function of x, but the equation}

relating y and x is not known or can

not be integrated (as in the case ), or

• _{If y is not a function of x (as in the}

case of Pext versus V), or

• _{if y is a function of x, but the equation}

relating y and x is not known or can

not be integrated (as in the case ), or

e

y



(33)

•

if y is a function of more than

one variable that changes with x,

then the line integral

cannot

be

evaluated analytically, and one

must resort to

a graphical or

numeric method of

integration

in order

to

evaluate the integral

•

if y is a function of more than

one variable that changes with x,

then the line integral

cannot

be

evaluated analytically, and one

must resort to

a graphical or

numeric method of

integration

in order

to

evaluate the integral

(34)

1. The integral in below can be evaluated analytically if we assume that P_ext is a

constant, hence, it can be brought out of the integral

1. The integral in below can be evaluated analytically if we assume that Pext is a

constant, hence, it can be brought out of the integral

) ( ₂ ₁

1 2

V V

P dV

A _ext

V ext V

ext   



_

work = -A = - P_ext (V₂ - V₁)

(35)

2. A second way to evaluate the above integral is found in the concept of

reversibility. If the expansion of the gas

is reversible, then for all practical

purposes the external is equal to the gas pressure, Pext = Pgas. This gives

2. A second way to evaluate the above integral is found in the concept of

reversibility. If the expansion of the gas

is reversible, then for all practical

purposes the external is equal to the gas pressure, Pext = Pgas. This gives





1 2

) ,

(

gas V

ext dV P T V dV

P A

It can be integrated if P is a function of V only so T must be constant

(isothermal condition)

It can be integrated if P is a function of V only so T must be constant

(36)





1 2 1 2

ln

2 1 2 1 2 1 2 1

V

nRT

V

nRT

V

dV

nRT

dV

V

nRT

dV

P

dV

P

A

V V V V V V gas V V ext









We can write We can write

(37)

Example

• _{The change in enthalpy as a function}

of temperature is given by the equation

• _{The change in enthalpy as a function}

of temperature is given by the equation

dT C



 

Find the change in enthalpy for one mole of real gas when the temperature of the gas is increased from, say, 298.2 K to 500.0 K

(38)

Solution

• _{We first recognize that the above integral is}

a line integral. This integral cannot be evaluated unless CP as a function of T is known

• We could assume that CP is a constant

and evaluate the integral that way, but over such a large temperature range the approximation would be poor

• _{Another approach would be to determine} the integral numerically

• _{We first recognize that the above integral is}

a line integral. This integral cannot be evaluated unless CP as a function of T is known

• _{We could assume that}_C_P_{is a constant} and evaluate the integral that way, but over such a large temperature range the approximation would be poor

• _{Another approach would be to determine} the integral numerically

(39)

• One analytical approach is to expand CP

as a power series in temperature

CP = a + bT + cT2

The constant a, b, and c are known for many common gases. Substituting this into the H equation gives

• One analytical approach is to expand CP

as a power series in temperature

CP = a + bT + cT2

The constant a, b, and c are known for many common gases. Substituting this into the H equation gives



 

 

)

( 2

dT cT

bT a

) (

3 )

( 2 )

(T₂ T₁ b T₂2 T₁2 c T₂3 T₁3 a

H      

(40)

2.7 Double and Triple

Integrals

2.7 Double and Triple

Integrals

• _{Functions could be differentited more}

than once

• _{How about the determination of}

multiple integrals

Example,

• _{The volume of cylinder}_{is a}

function of both the radius and the

height of the cylinder. That is, V = f(r, h)

• _{Functions could be differentited more}

than once

• _{How about the determination of}

multiple integrals Example,

• _{The volume of cylinder}_{is a}

function of both the radius and the

height of the cylinder. That is, V = f(r, h)

(41)

• _{Let us suppose that we allow the}

height of the cylinder, h, to change while holding the radius, r, constant

• _{The integral from h = 0 to h = h,}

then, could be expressed as

• _{Let us suppose that we allow the}

height of the cylinder, h, to change while holding the radius, r, constant

• _{The integral from h = 0 to h = h,}

then, could be expressed as



dh

h

r

f

)

,

(42)

• _{But the value of this line integral}

depends on the value of the radius, r, and hence the integral could be

considered to be a function of r.

• _{But the value of this line integral}

depends on the value of the radius, r, and hence the integral could be

considered to be a function of r.



 h dh h r f r g 0 ) , ( ) (

• _{If we allow r to vary from r = 0 to r = r and}

integrate over the change, we can write

• _{If we allow r to vary from r = 0 to r = r and}

integrate over the change, we can write



 h r r dr dh h r f dr r g 0 0 0 ) , ( ) (

(43)

• _{To evaluate the above double integral, we}

integrate first while holding

r a constant, which gives us g(r).

• _{Then we integrate next while}

holding h constant. Such a process is known as successive partial

integration

• _{To evaluate the above double integral, we}

integrate first while holding

r a constant, which gives us g(r).

• _{Then we integrate next while}

holding h constant. Such a process is known as successive partial

integration



dh h

r f

) , (



dr r

g 0

) (

(44)

• _{For example, let us evaluate} • _{For example, let us evaluate}

_

h r dr dh r 0 0 2



  h h r dh r r g 0 2 2 ) (  



  r r h r dr rh dr r g 0 2 0 2 ) (   First, First,

Next, we integrate

That is the volume of a cylinder

(45)

• _{The above argument can be}

extended to the triple integral. For example, let us evaluate the triple

• _{The above argument can be}

extended to the triple integral. For example, let us evaluate the triple



x y z

dz dy

0 0 0

• _{First, evaluate}

• _{Substituting this back into the above}

equation gives

• _{First, evaluate}

• _{Substituting this back into the above}

equation gives

x dx





y z

dz dy

0 0

(46)

• _{Next, evaluate}

• _{Substituting this back into the above}

equation gives

• _{Integrating this gives}

• _{Which is the volume of a rectangular}

box x by y by z

• _{Next, evaluate}

• _{Substituting this back into the above}

equation gives

• _{Integrating this gives}

• _{Which is the volume of a rectangular}

box x by y by z

xy dy





dz xy

xyz dz

xy z





(47)

• _{Problem. The differential volume element in}

spherical polar coordinates is dV = r2sin d d dr.

Given that  goes from 0 to 2, and r goes from 0 to r, evaluate the triple integral

• _{Problem. The differential volume element in}

spherical polar coordinates is dV = r2sin d d dr.

Given that  goes from 0 to 2, and r goes from 0 to r, evaluate the triple integral



  







2 0 2 0 0

sin d d dr r V r Solutio n



      2 0 2

2 _sin _d ₂ _r _sin

_



     0 2 2 _sin ₄

2 r d r



  r V r dr r 0 3 2 3 4

(48)

PROBLEMS

1. Evaluate the following integral

(consider all uppercase letters to be constant):

1. Evaluate the following integral

(consider all uppercase letters to be constant):



x dx a) 4 2

(



P d

f ) (



x b) 1₂ (



dp p

RT g)

(



x dx c) sin 3

(



M d

h) (



x  x dx

d₎ ₍₃ ₅₎2

(



e dx e) 4 x

(



W t dt i) cos(2 )

(49)

2. Evaluate the following integrals

using the Table of Integral (consider all uppercase letters to be

constant):

2. Evaluate the following integrals

using the Table of Integral (consider all uppercase letters to be

constant):



_e _dx

a₎ 4x

(

x  x  x dx

b₎ ₍ 4 ₂ 2 ₄₎ 3

(



x  A dx

c) ( )

( 2 2

       _dx A x N d) sin2  (

x  A dx

e₎ ₍ 2 2₎1/2

(      

 _x_dx

A x N f ₎ _sin2 

(



e x dx g) x cos

(

 Wt dt h) sin (2 ) ( 2 _

   d

i) cos sin ( 3

  d

k₎ _cos4

(

 x dx l) sin (3 4) ( 6

₍₄_ ₎₍₃_ ₎

) ( x x dx m

(50)

3. Evaluate the following definite integrals using the Table of

indefinite and definite interals, as needed

3. Evaluate the following definite integrals using the Table of

indefinite and definite interals, as needed 2 1 2 . T T d

a a bT cT dT T �_ _ _ � � � � � � 2 1 . P P RT b dP P

�

2 0 . c d  

�

2 1 2 1 2 2 2 /2 2 0 . .

. sin cos

T T V V H d dT RT

nRT n a

e dV

V nb V

f d      � �  � _ � � �

�

2 2 0 . sin a n x

g x dx

a 

�

2 2 0 . ax

h x e dx

�



(51)

2 / 0

/ 2 3

( )

.

. (2

1)

r a

m kT

a J J

i e

r dr

j e

d

k

J

e

dJ



_

�



�



�

 



�

(52)

4. Consider the ideal gas law equation P = nRT/V, where in the case n, R, and T are assumed to be constant. Prepare a graph of P versus V,

choosing suitable coordinates, for n = 1 mole, R = 0,0821 liter atm/mol K and T = 298 K from a volume of V = 1.00 liters to a volume of V = 10.0 liters. Consider now the area under the P versus V curve from V = 2.00 liters to V = 6.00 liters. Determine the approximate area

graphically by breaking up the area into four rectangles of equal width V; compare your answer to that found by analytically

integrating the function between these limits of integration

4. Consider the ideal gas law equation P = nRT/V, where in the case n, R, and T are assumed to be constant. Prepare a graph of P versus V,

graphically by breaking up the area into four rectangles of equal width V; compare your answer to that found by analytically

integrating the function between these limits of integration

(53)

5. Evaluate the following multiple integrals using the Table of

Integrals, as needed

5. Evaluate the following multiple integrals using the Table of

Integrals, as needed

2 2 2 2 2 . . ( ) . ln

. ln x

a yx dx dy

b x y dx dy c y x dx dy

d x y e dx dy dz



�

2 2 2 2

2 2 2

/2 2 0 0 2

2 0 0 0

8 0 0 0

. cos . sin . y x z r n n n h

mkT a b c

x y z

e r dr d

f r dr d d

g e dn dn dn

        � � � � � � � _ _ _ � � � �

��

(54)

6. The equation of a straight line passing through the origin of a

cartesian coordinate system is y = mx, where m is the slope of the line. Show that the area of a triangle

made up of this line and the x axis between x = 0 and x = a is A = ½ ay.

6. The equation of a straight line passing through the origin of a

cartesian coordinate system is y = mx, where m is the slope of the line. Show that the area of a triangle

made up of this line and the x axis between x = 0 and x = a is A = ½ ay.

(55)

7. The Kirchoff equation for a chemical

reaction relating the variation of H of a reaction with absolute temperature is where CP as a power series in

T, ∆CP = a + bT + cT2. Derive an

equation for H as a function of

temperature. (Hint: Write the above derivative in differential form)

7. The Kirchoff equation for a chemical

reaction relating the variation of H of a reaction with absolute temperature is where CP as a power series in

T, ∆CP = a + bT + cT2. Derive an

equation for H as a function of

temperature. (Hint: Write the above derivative in differential form)

( )

P P

C T

 �

� �_{ } � _� � � �

(56)

8. The Gibbs-Helmholtz equation for a chemical reaction is

( / )

G T H

T T

 

�

� �_{ } � _� � � �

Where G is the Gibbs free energy

change attending the reaction, and T is absolute temperature. Expressing

H in a power series in T, ∆H = a + bT

+ cT2. where a, b, and c are experimentally

determined constants, derive an expression for G as a function of temperature

Where G is the Gibbs free energy

change attending the reaction, and T is absolute temperature. Expressing

H in a power series in T, ∆H = a + bT

+ cT2. where a, b, and c are experimentally

determined constants, derive an expression for G as a function of temperature

(57)

9. Find the probability of finding a

particle confined to e field-free one-dimensional box in the state n = 1 at x = L/2 in a range L/2 ± 0.05 L, where L is the width of box, given 9. Find the probability of finding a

particle confined to e field-free one-dimensional box in the state n = 1 at x = L/2 in a range L/2 ± 0.05 L, where L is the width of box, given

dx L

x L

L L









05 . 0 2

2 sin 2

(58)

10. Find the probability of finding an electron in the 1s-state of the

hydrogen atom at r = a₀ in a range a₀ ± 0.005a₀, where a₀ is the Bohr radius, given

10. Find the probability of finding an electron in the 1s-state of the

hydrogen atom at r = a₀ in a range a₀ ± 0.005a₀, where a₀ is the Bohr radius, given

















0 0

0 005

. 0

005 . 0

2 0

1

4 obability

Pr

a a

_r

_dr

e

a

(59)

11. Find the expectation value <x> for an electron in the 1s-state of the

hydrogen atom, given that

11. Find the expectation value <x> for an electron in the 1s-state of the

hydrogen atom, given that

2 / 3

0 0

4 r a

x e r dr

�



� �



��

� �

�

12.The differential volume element in cylindrical coordinates is dV = r d dr dz. Show that if r goes from 0 to r,  from 0 to 2, and z from 0 to h, the volume of a cylinder is V =  r2 h

(1)

6. The equation of a straight line passing through the origin of a

cartesian coordinate system is y = mx, where m is the slope of the line. Show that the area of a triangle

made up of this line and the x axis between x = 0 and x = a is A = ½ ay.

6. The equation of a straight line passing through the origin of a

cartesian coordinate system is y = mx, where m is the slope of the line. Show that the area of a triangle

made up of this line and the x axis between x = 0 and x = a is A = ½ ay.

(2)

7. The Kirchoff equation for a chemical

reaction relating the variation of H of a reaction with absolute temperature is where CP as a power series in

T, ∆CP = a + bT + cT2. Derive an

equation for H as a function of

temperature. (Hint: Write the above derivative in differential form)

7. The Kirchoff equation for a chemical

reaction relating the variation of H of a reaction with absolute temperature is where CP as a power series in

T, ∆CP = a + bT + cT2. Derive an

equation for H as a function of

temperature. (Hint: Write the above derivative in differential form)

( )

P P

C T



�

� �_{ } � _� �

(3)

8. The Gibbs-Helmholtz equation for a chemical reaction is

( / )

G T H

T T

 

�

� �_{ } � _� �

� �

Where G is the Gibbs free energy

change attending the reaction, and T is absolute temperature. Expressing

H in a power series in T, ∆H = a + bT + cT2. where a, b, and c are experimentally

determined constants, derive an expression for G as a function of temperature

Where G is the Gibbs free energy

change attending the reaction, and T is absolute temperature. Expressing

H in a power series in T, ∆H = a + bT + cT2. where a, b, and c are experimentally

determined constants, derive an expression for G as a function of temperature

(4)

9. Find the probability of finding a

particle confined to e field-free one-dimensional box in the state n = 1 at x = L/2 in a range L/2 ± 0.05 L, where L is the width of box, given 9. Find the probability of finding a

particle confined to e field-free one-dimensional box in the state n = 1 at x = L/2 in a range L/2 ± 0.05 L, where L is the width of box, given

dx L

x L

L L









05 . 0 2

sin 2

(5)

10. Find the probability of finding an electron in the 1s-state of the

hydrogen atom at r = a₀ in a range a₀ ± 0.005a₀, where a₀ is the Bohr radius, given

10. Find the probability of finding an electron in the 1s-state of the

hydrogen atom at r = a₀ in a range a₀ ± 0.005a₀, where a₀ is the Bohr radius, given



  











0 0 0 0 0 005 . 0 005 . 0 2 0

1

4 obability

Pr

a a a a a

_r

_dr

e

a

(6)

11. Find the expectation value <x> for an electron in the 1s-state of the

hydrogen atom, given that

11. Find the expectation value <x> for an electron in the 1s-state of the

hydrogen atom, given that

2 / 3 0 0

4 r a

x e r dr

�



� �



��

� �

�

12.The differential volume element in cylindrical coordinates is dV = r d dr dz. Show that if r goes from 0 to r,  from 0 to 2, and z from 0 to h, the volume of a cylinder is V =  r2 h