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) = 3x2e−βx (β is a constant)

€

" f (x) = 6x − 3β x2( )e−βx

b.)

€

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

€

" Y (x) = − π A sin π 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 − B

r6 (A and B are constants)

€

" u (r) = − 12Ar13 + 6B

r7

f.)

€

s(t) = e−3t 1 − ℓn(t)[ ]

€

" s (t) = − 3e−3t 1 − ℓn(t)[ ] − e−3t

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 dV

V1

V2∫    =   3 V dVV1

V2∫

=  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

2  − V1

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 = aR dT , or alternately, dT = 1aR du . Substituting, the integral

becomes

eaRT∫  dT   =   1aR  eu∫  du

=   1aR eu  + 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=5.0×104)

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 1

T 2 dT

300

500∫    +   b 1

T 3 dT

300

500∫  

=  a T −2  dT300

500∫    +   b T −3  dT

300

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

!

"#

$

%&    +  2.5×104 −

1500( )2

+  1300( )2

!

"##

$

%&&

=  250 1.333×10−3( )  +  2.5×104 7.111×10−6( )=  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

€

∂ f∂x#

$ %

&

' ( y, ∂ f

∂y#

$ %

&

' ( x, ∂

2 f∂x2

#

$ % %

&

' ( ( y

, ∂2 f∂y2

#

$ % %

&

' ( ( x

, ∂2 f∂x∂y

#

$ % %

&

' ( ( , and

€

∂2 f∂y∂x

#

$ % %

&

' ( ( .

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

#

$ % %

&

' ( ( = 5ex ∂2 f

∂ y∂ x

#

$ % %

&

' ( ( = 5ex

c.)

€

f (x, y) = y ℓn(x) + x ℓn(x)

€

∂ f∂ x

#

$ %

&

' ( y

= yx

+ 1 + ℓn(x) ∂ f∂ y

#

$ %

&

' ( x

= ℓn(x)

€

∂2 f∂ x2

#

$ % %

&

' ( ( y

= − yx2 + 1

x ∂

2 f∂ y2

#

$ % %

&

' ( ( x

= 0

€

∂2 f∂ x∂ y

#

$ % %

&

' ( ( = 1

x ∂2 f

∂ y∂ x

#

$ % %

&

' ( ( = 1

x

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 x

−1 / 2 y1 / 2 ∂ f∂ y

#

$ %

&

' ( x

= 12 x

1 / 2 y−1 / 2

€

∂2 f∂ x2

#

$ % %

&

' ( ( y

= − 14 x

−3 / 2 y1 / 2 ∂2 f

∂ y2

#

$ % %

&

' ( ( x

= − 14 x

1 / 2 y−3 / 2

€

∂2 f∂ x∂ y

#

$ % %

&

' ( ( = 1

4 x−1 / 2y−1 / 2 ∂2 f

∂ y∂ x

#

$ % %

&

' ( ( = 1

4 x−1 / 2y−1 / 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 ℓnT − PℓnP + 3T

2P

€

∂H∂T

#

$ %

&

' ( P

= 3R2T

+ 32P

€

∂H∂ P

#

$ %

&

' ( T

= −1 − ℓnP − 3T2P2

b.)

€

s(v, t) = 12 vt

2 + ve−v

€

∂ s∂ v#

$ %

&

' ( t

= 12 t

2 + e−v − ve−v

€

∂ s∂ t#

$ %

&

' ( v

= vt

c.)

€

g(x, y) = e−3x 1− x2( ) y3ℓny

€

∂ g∂ x

#

$ %

&

' ( y

= e−3x −3 1− x2( ) − 2x* + ,

- . / y

3ℓny

€

∂ g∂ y

#

$ %

&

' ( x

= e−3x 1− x2( ) 3y2ℓny + y2( )

d.)

€

P(V ,T ) = RTV

1 + bV( )

€

∂ P∂V#

$ %

&

' ( T

= − RTV 2

€

∂ P∂T#

$ %

&

' ( V

= RV

1 + bV( )

e.)

€

u(r,θ ) = 32 r

2cosθ − rersinθ

€

∂ u∂ r#

$ %

&

' ( θ

= 3rcosθ − sinθ er + rer( )

€

∂ u∂θ

$

% &

'

( ) r

= − 32 r

2sinθ − rercosθ

7

4. continued

f.)

€

H (T ,P) = 32 RT + RT 2Pe−3P

€

∂H∂T

#

$ %

&

' ( P

= 32 R + 2RTPe−3P

€

∂H∂ P

#

$ %

&

' ( T

= RT 2 e−3P − 3Pe−3P( )

g.)

€

P(V ,T ) = RT + RTVℓnV

€

∂ P∂V#

$ %

&

' ( T

= RT ℓnV + 1( )

€

∂ P∂T#

$ %

&

' ( V

= R + RVℓnV

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

a.)

€

r = x2 + y2 + z2 ; evaluate

€

∂ r∂x#

$ %

&

' ( y,z

.

€

∂ r∂ x

#

$ %

&

' ( y,z

= x x2 + y2 + z2( )−1 / 2 = x

r

b.)

€

y = r sinθ cosφ ; evaluate

€

∂ y∂φ

$

% &

'

( ) r,θ

.

€

∂ y∂φ

$

% &

'

( ) r,θ

= − r sinθ sinφ

8

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

a.)

€

∂ P∂T#

$ %

&

' ( 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,

€

∂ P∂T#

$ %

&

' ( Vm

= RVm

.

b.)

€

∂ P∂Vm

#

$ %

&

' ( T

The partial derivative required involves P and also requires a derivative of

€

Vm to be evaluated. 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,

€

∂ P∂Vm

#

$ %

&

' ( T

= − RTVm

2 .

c.)

€

∂T∂ P#

$ %

&

' ( Vm

In this case, the partial derivative required involves T and also requires

€

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

€

Vm before the partial derivative is evaluated,

€

T = PVnR

= PVmR

.

Then, the partial derivative may be evaluated,

€

∂T∂ P#

$ %

&

' ( Vm

= VmR

.

9

6. continued

d.)

€

∂T∂Vm

#

$ %

&

' ( P

.

The partial derivative required involves T and also requires a derivative of

€

Vm to be evaluated. Therefore, the ideal gas equation of state should be solved for T and written in terms of

€

Vm before the partial derivative is evaluated,

€

T = PVnR

= PVmR

.

Then, the partial derivative may be evaluated,

€

∂T∂Vm

#

$ %

&

' ( P

= PR

.

7. The isothermal compressibility κ is defined by the relation

€

κ = − 1V

∂V∂ P%

& '

(

) * T

,

and the expansion coefficient α is given by

€

α = 1V

∂V∂T$

% &

'

( ) P

Evaluate these quantities for an ideal gas (assume that n is constant).

For the isothermal compressibility, the partial derivative required involves V. Therefore, the ideal gas equation of state should be solved for V before the partial derivative is evaluated,

€

V = nRTP

.

Next, the partial derivative may be evaluated,

€

∂V∂ P#

$ %

&

' ( T

= − nRTP2 .

Finally, the partial derivative may be substituted into the expression for the isothermal compressibility and simplified,

€

κ = − 1V

∂V∂ P%

& '

(

) * T

= − 1V⋅ −

nRTP2

%

& '

(

) * = P

nRT⋅nRTP2

%

& '

(

) *

€

κ = 1P

.

10

7. continued For the expansion coefficient, the partial derivative required also involves V. Therefore, the ideal gas equation of state should be solved for V before the partial derivative is evaluated,

€

V = nRTP

.

Next, the partial derivative may be evaluated,

€

∂V∂T#

$ %

&

' ( P

= nRP

.

Finally, the partial derivative may be substituted into the expression for the expansion coefficient and simplified,

€

α = 1V

∂V∂T$

% &

'

( ) P

= 1V⋅nRP

$

% &

'

( ) = P

nRT⋅nRP

$

% &

'

( )

€

α = 1T

8. The van der Waals equation for a real gas is defined as

€

P + aVm

2

"

# $

%

& ' Vm − b( ) = RT , where

€

Vm is the

molar volume, R is the gas constant, and a and b are van der Waals constants. For the van der Waals equation, determine

a.)

€

∂ P∂T#

$ %

&

' ( Vm

In order to evaluate the partial derivative, we must first solve the van der Waals equation for P,

€

P + aVm

2

"

# $

%

& ' Vm − b( ) = RT

P + aVm

2 = RTVm − b

P = RTVm − b

− aVm

2 .

Using that expression, the partial derivative may be evaluated,

€

∂ P∂T#

$ %

&

' ( Vm

= RVm − b

.

11

8. continued

b.)

€

∂ P∂Vm

#

$ %

&

' ( T

.

In order to evaluate this partial derivative, the van der Waals equation must again be solved for P. Using the same expression obtained in part (a),

€

P = RTVm − b

− aVm

2 ,

the partial derivative may be evaluated,

€

∂ P∂Vm

#

$ %

&

' ( T

= − RTVm − b( )2 + 2a

Vm3 .