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### Transcript of Problem Set 1 Solutions - Illinois State University 360/Homework...Chemistry 360 Dr. Jean M....

• Chemistry 360 Dr. Jean M. Standard

Problem Set 1 Solutions 1. Determine the first derivatives of each of the following functions of one variable.

a.)

f (x) = 3x2ex ( is a constant)

" f (x) = 6x 3 x2( )ex b.)

Y (x) = A cos x( ) (A is a constant)

" Y (x) = A sin x( )

c.)

g(y) = 1 2y2

" g (y) = 2y 1 2y2( )1 / 2

d.)

H (T ) = a + bT + cT 2 + dT

(a, b, c, and d are constants)

" H (T ) = b + 2cT dT 2

e.)

u(r) = Ar12

Br6

(A and B are constants)

" u (r) = 12Ar13

+ 6Br7

f.)

s(t) = e3t 1 n(t)[ ]

" s (t) = 3e3t 1 n(t)[ ] e3t

t

• 2

2. Determine the indefinite or definite integrals of each of the following functions of one variable.

Note: For the solutions below involving indefinite integrals, the constant C is used to correspond to an overall arbitrary constant of integration. a.) kA2 dA (k is a constant)

kA2 dA =k A2 dA

=k A3

3+C .

b.) 3V dV

V1

V2

3V dVV1

V2 =3 V dVV1V2

=3 V2

2

"

#\$

%

&'

V1

V2

.

Here, to complete the solution, we must evaluate the result at the limits of integration. The final result is therefore

3V dVV1

V2 =3 V22

2V1

2

2

#

\$%

&

'(.

c.) 1V + b dV (b is a constant)

Here, we can let u =V + b . Then, du = dV (since b is a constant). Substituting, the integral becomes 1u du=ln u( )+C.

Expressing the result in terms of V gives

1V + b dV =ln V + b( )+C.

• 3

2. continued d.) eaRT dT (a and R are constants)

Here, we can let u = aRT . Then, du = aRdT , or alternately, dT = 1aRdu . Substituting, the integral

becomes

eaRT dT = 1aR eu du

= 1aReu +C .

Expressing the result in terms of T gives

eaRT dT = 1aR eaRT +C .

e.) aT 2+ b

T 3!

"#

\$

%&dT300

500 (a =250, b=5.0104)

The easiest way to tackle this integral is to break it into the sum of two integrals,

aT 2+ b

T 3!

"#

\$

%&dT300

500 =a 1T 2 dT300

500 +b 1T 3 dT300

500

=a T 2 dT300

500 +b T 3 dT300

500

aT 2+ b

T 3!

"#

\$

%&dT300

500 =a T 1( )

300

500+b 12 T

2( )300

500.

The next step is to evaluate the result at the limits and substitute the numerical values of a and b,

aT 2+ b

T 3!

"#

\$

%&dT300

500 =a T 1( )

300

500+b 12 T

2( )300

500

=a 1500

+ 1300

!

"#

\$

%& +

b2 1

500( )2+ 1300( )2

!

"##

\$

%&&

=250 1500

+ 1300

!

"#

\$

%&+2.5104

1500( )2

+ 1300( )2

!

"##

\$

%&&

=250 1.333103( )+2.5104 7.111106( )=0.333+0.178

aT 2+ b

T 3!

"#

\$

%&dT300

500 =0.511.

• 4

3. For each of the following functions of x and y, determine the partial derivatives

fx#

\$ %

&

' ( y, f

y#

\$ %

&

' ( x,

2 fx2#

\$ % %

&

' ( ( y

, 2 fy2#

\$ % %

&

' ( ( x

, 2 f

xy

#

\$ % %

&

' ( ( , and

2 fyx

#

\$ % %

&

' ( ( .

a.)

f (x, y) = x2y + 3y

f x

#

\$ %

&

' ( y

= 2xy f y

#

\$ %

&

' ( x

= x2 + 3

2 f x2#

\$ % %

&

' ( ( y

= 2y 2 f

y2#

\$ % %

&

' ( ( x

= 0

2 f x y

#

\$ % %

&

' ( ( = 2x

2 f y x

#

\$ % %

&

' ( ( = 2x

b.)

f (x, y) = 5ex y + y

f x

#

\$ %

&

' ( y

= 5ex y f y

#

\$ %

&

' ( x

= 5ex + 1

2 f x2#

\$ % %

&

' ( ( y

= 5ex y 2 f

y2#

\$ % %

&

' ( ( x

= 0

2 f x y

#

\$ % %

&

' ( ( = 5e

x 2 f

y x

#

\$ % %

&

' ( ( = 5e

x

c.)

f (x, y) = y n(x) + x n(x)

f x

#

\$ %

&

' ( y

= yx

+ 1 + n(x) f y

#

\$ %

&

' ( x

= n(x)

2 f x2#

\$ % %

&

' ( ( y

= yx2

+ 1x

2 f

y2#

\$ % %

&

' ( ( x

= 0

2 f x y

#

\$ % %

&

' ( ( =

1x

2 f

y x

#

\$ % %

&

' ( ( =

1x

• 5

3. continued

d.)

f (x, y) = 6x3

f x

#

\$ %

&

' ( y

= 18x2 f y

#

\$ %

&

' ( x

= 0

2 f x2#

\$ % %

&

' ( ( y

= 36x 2 f

y2#

\$ % %

&

' ( ( x

= 0

2 f x y

#

\$ % %

&

' ( ( = 0

2 f y x

#

\$ % %

&

' ( ( = 0

e.)

f (x, y) = xy( )1/ 2

f x

#

\$ %

&

' ( y

= 12 x1 / 2 y1 / 2 f

y

#

\$ %

&

' ( x

= 12 x1 / 2 y1 / 2

2 f x2#

\$ % %

&

' ( ( y

= 14 x3 / 2 y1 / 2

2 f y2#

\$ % %

&

' ( ( x

= 14 x1 / 2 y3 / 2

2 f x y

#

\$ % %

&

' ( ( =

14 x

1 / 2y1 / 2 2 f

y x

#

\$ % %

&

' ( ( =

14 x

1 / 2y1 / 2

f.)

f (x, y) = 3x2 cosy + xy3

f x

#

\$ %

&

' ( y

= 6x cos y + y 3 f y

#

\$ %

&

' ( x

= 3x 2 sin y + 3xy2

2 f x 2#

\$ %

&

' ( y

= 6 cos y 2 f

y 2#

\$ %

&

' ( x

= 3x 2 cos y + 6xy

2 f x y

#

\$ %

&

' ( = 6x sin y + 3y 2

2 f y x

#

\$ %

&

' ( = 6x sin y + 3y 2

• 6

4. For each of the following functions of two variables, evaluate the two first partial derivatives. [Where it appears in the expressions below, R corresponds to the gas constant.]

a.)

H (T ,P) = 32 R nT PnP + 3T2P

HT

#

\$ %

&

' ( P

= 3R2T

+ 32P

H P

#

\$ %

&

' ( T

= 1 nP 3T2P2

b.)

s(v, t) = 12 vt2 + vev

s v#

\$ %

&

' ( t

= 12 t2 + ev vev

s t#

\$ %

&

' ( v

= vt

c.)

g(x, y) = e3x 1 x2( ) y3ny

g x

#

\$ %

&

' ( y

= e3x 3 1 x2( ) 2x* + , - . / y3ny

g y

#

\$ %

&

' ( x

= e3x 1 x2( ) 3y2ny + y2( )

d.)

P(V ,T ) = RTV

1 + bV( )

PV#

\$ %

&

' ( T

= RTV 2

PT#

\$ %

&

' ( V

= RV

1 + bV( )

e.)

u(r, ) = 32 r2cos rersin

u r#

\$ %

&

' (

= 3rcos sin er + rer( )

u

\$

% &

'

( ) r

= 32 r2sin rercos

• 7

4. continued

f.)

H (T ,P) = 32 RT + RT2Pe3P

HT

#

\$ %

&

' ( P

= 32 R + 2RTPe3P

H P

#

\$ %

&

' ( T

= RT 2 e3P 3Pe3P( )

g.)

P(V ,T ) = RT + RTVnV

PV#

\$ %

&

' ( T

= RT nV + 1( )

PT#

\$ %

&

' ( V

= R + RVnV

5. For each of the following functions of three variables, evaluate the requested partial derivatives.

a.)

r = x2 + y2 + z2 ; evaluate

rx#

\$ %

&

' ( y,z

.

r x

#

\$ %

&

' ( y,z

= x x2 + y2 + z2( )1 / 2 = xr

b.)

y = r sin cos ; evaluate

y

\$

% &

'

( ) r,

.

y

\$

% &

'

( ) r,

= r sin sin

• 8

6. Evaluate the following expressions using the ideal gas equation of state.

a.)

PT#

\$ %

&

' ( Vm

The partial derivative required involves P and also requires

Vm to be held constant. Therefore, the ideal gas equation of state should be solved for P and written in terms of

Vm before the partial derivative is evaluated,

P = nRTV

= RTVm

.

Then, the partial derivative may be evaluated,

PT#

\$ %

&

' ( Vm

= RVm

.

b.)

PVm

#

\$ %

&

' (