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) = 3x 2e− βx
(β is a constant)
(
)
f "(x) = 6x − 3β x 2 e− βx
€
€
b.) Y (x) = A cos( π x )
(A is a constant)
Y "(x) = − π A sin( π x )
€
€
1 − 2y 2
c.) g(y) =
(
g"(y) = − 2y 1− 2y 2
€
€
)
2
d.) H (T ) = a + bT + cT +
H "(T ) = b + 2cT −
€
€
e.) u(r) =
A
r
12
u"(r) = −
€
€
€
€
−1 / 2
f.)
−
12A
r
13
B
(a, b, c, and d are constants)
d
T2
(A and B are constants)
r6
+
d
T
6B
r7
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
b.)
V2
∫V
A3
+ C .
3
3V dV
1
V2
∫V
1
V2
3V dV = 3 ∫ V dV
V1
V
"V 2 % 2
= 3 $ ' .
#2& V
1
Here, to complete the solution, we must evaluate the result at the limits of integration. The final result is
therefore
V2
∫V
1
c.)
#V 2
V2 &
3V dV = 3 % 2 − 1 ( .
2 '
$2
1
∫ V + b dV
(b is a constant)
Here, we can let u = V + b . Then, du = dV (since b is a constant). Substituting, the integral becomes
1
∫ u du = ln (u) + C.
Expressing the result in terms of V gives
1
∫ V + b dV = ln (V + b) + C.
3
2.
continued
d.)
∫ eaRT dT
(a and R are constants)
Here, we can let u = aRT . Then, du = aR dT , or alternately, dT =
1
du . Substituting, the integral
aR
becomes
1
∫ eaRT dT = aR ∫ eu du
= 1 u
e + C .
aR
Expressing the result in terms of T gives
1
∫ eaRT dT = aR eaRT + C .
e.)
500 !
a
b$
∫ 300 #" T 2 + T 3 &% dT
(a =250, b=5.0×104)
The easiest way to tackle this integral is to break it into the sum of two integrals,
500 !
a
b$
500
1
500
1
500
T −3 dT
∫ 300 #" T 2 + T 3 &% dT = a ∫ 300 T 2 dT + b ∫ 300 T 3 dT = a ∫
500 !
a
500
300
b$
T −2 dT + b ∫
500
300
500
∫ 300 #" T 2 + T 3 &% dT = a (−T −1 ) 300 + b (− 12 T −2 ) 300 .
The next step is to evaluate the result at the limits and substitute the numerical values of a and b,
500 !
a
b$
500
500
∫ 300 #" T 2 + T 3 &% dT = a (−T −1 ) 300 + b (− 12 T −2 ) 300 !
$
! 1
1 $
b
1
1 &
= a # −
+ + & + ## −
2
2&
" 500 300 %
2 " ( 500 )
(300) %
!
$
! 1
1 $
1
1 &
4
= 250 # −
+ + & + 2.5 ×10 ## −
2
2
" 500 300 %
(300) &%
" ( 500 )
(
)
(
= 250 1.333×10 −3 + 2.5 ×10 4 7.111×10 −6
= 0.333 + 0.178
$
a
b
∫ 300 #" T 2 + T 3 &% dT = 0.511 .
500 !
)
4
3.
For each of the following functions of x and y, determine the partial derivatives
# ∂2 f &
#∂ f & #∂ f & #∂ 2 f & #∂ 2 f & # ∂ 2 f &
(.
((, and %%
%
( ,%
( , %% 2 (( , %% 2 (( , %%
$ ∂x 'y $ ∂y 'x $ ∂x ' $ ∂y ' $ ∂x ∂y '
∂y ∂x ('
$
y
x
a.) f (x, y) = x 2 y + 3y
#∂ f &
% ( = 2xy
$ ∂ x 'y
€
#∂ f &
2
%
( = x + 3
$ ∂ y 'x
€
#∂2 f &
%%
( = 2y
2(
$ ∂ x 'y
#∂2 f &
%%
( = 0
2(
$ ∂ y 'x
€
# ∂2 f &
%%
(( = 2x
$∂ x ∂ y '
# ∂2 f &
%%
(( = 2x
$∂ y ∂ x '
€
€
b.) f (x, y) = 5e x y + y
€
#∂ f &
x
% ( = 5e y
$ ∂ x 'y
€
#∂ f &
x
% ( = 5e + 1
$ ∂ y 'x
€
#∂2 f &
%%
( = 5e x y
2(
$ ∂ x 'y
#∂2 f &
%%
( = 0
2(
$ ∂ y 'x
€
# ∂2 f &
%%
(( = 5e x
∂
x
∂
y
$
'
# ∂2 f &
%%
(( = 5e x
∂
y
∂
x
$
'
c.) f (x, y) = y ℓn(x) + x ℓn(x)
€
#∂ f &
y
+ 1 + ℓn(x)
%
( =
∂
x
x
$
'y
€
€
#∂2 f &
y
1
%%
( = −
+
2(
2
x
x
$ ∂ x 'y
€
# ∂2 f &
1
%%
(( =
x
$∂ x ∂ y '
€
#∂ f &
%
( = ℓn(x)
$ ∂ y 'x
#∂2 f &
%%
( = 0
2(
$ ∂ y 'x
# ∂2 f &
1
%%
(( =
x
$∂ y ∂ x '
5
3.
continued
d.) f (x, y) = 6x 3
#∂ f &
2
% ( = 18x
$ ∂ x 'y
#∂ f &
% ( = 0
$ ∂ y 'x
€
#∂2 f &
%%
( = 36x
2(
$ ∂ x 'y
#∂2 f &
%%
( = 0
2(
$ ∂ y 'x
€
# ∂2 f &
%%
(( = 0
$∂ x ∂ y '
€
€e.) f (x, y) = ( xy )
1/ 2
#∂ f &
%
( =
$ ∂ x 'y
€
# ∂2 f &
%%
(( = 0
$∂ y ∂ x '
1
2
x −1 / 2 y1 / 2
€
#∂2 f &
%%
( = −
2(
$ ∂ x 'y
€
# ∂2 f &
%%
(( =
$∂ x ∂ y '
1
4
1
4
x −3 / 2 y1 / 2
x −1 / 2 y −1 / 2
#∂ f &
%
( =
$ ∂ y 'x
1
2
x1 / 2 y −1 / 2
#∂2 f &
%%
( = −
2(
$ ∂ y 'x
1
4
# ∂2 f &
%%
(( =
$∂ y ∂ x '
x −1 / 2 y −1 / 2
1
4
x1 / 2 y −3 / 2
f.) f (x, y) = 3x 2 cosy + xy 3
€
#∂ f &
3
%
( = 6x cos y + y
∂
x
$
'y
#∂ f &
2
2
%
( = − 3x sin y + 3xy
∂
y
$
'x
€
#∂2 f &
% 2 ( = 6 cos y
$ ∂ x 'y
#∂2 f &
% 2 ( = − 3x 2 cos y + 6xy
$ ∂ y 'x
€
# ∂2 f &
%
( = − 6x sin y + 3y 2
∂
x
∂
y
$
'
€
€
# ∂2 f &
%
( = − 6x sin y + 3y 2
∂
y
∂
x
$
'
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.]
3T
a.) H (T , P) = 23 R ℓnT − PℓnP +
2P
€
#∂ H &
3R
3
+
%
( =
2T
2P
$ ∂ T 'P
€
#∂ H &
3T
%
( = − 1 − ℓnP −
$ ∂ P 'T
2P 2
€
b.) s(v, t) = 12 vt 2 + ve−v
€
#∂ s &
% ( =
$ ∂ v 't
€
#∂ s &
% ( = vt
$ ∂ t 'v
€
1 t2
2
+ e−v − ve−v
(
)
c.) g(x, y) = e−3x 1− x 2 y 3ℓny
€
#∂ g &
- 3
−3x *
2
% ( = e +,−3 1− x − 2x /. y ℓny
$ ∂ x 'y
€
#∂ g &
−3x
1 − x 2 3y 2 ℓny + y 2
% ( = e
$ ∂ y 'x
€
(
(
d.) P(V ,T ) =
)
)(
RT
(1 + bV )
V
€
#∂ P &
RT
%
( = − 2
$ ∂ V 'T
V
€
#∂ P &
R
(1 + bV )
%
( =
∂
T
V
$
'V
€
e.) u(r,θ ) = 23 r 2 cosθ − re r sinθ
€
#∂ u &
r
r
% ( = 3rcosθ − sinθ e + re
$ ∂ r 'θ
€
$∂ u '
2
r
& ) = − 23 r sinθ − re cosθ
∂
θ
% (r
€
)
(
)
7
4.
continued
+ RT 2 Pe−3P
H (T ,P) =
3 RT
2
€
#∂ H &
%
( =
$ ∂ T 'P
3R
2
€
#∂ H &
2 −3P
− 3Pe−3P
%
( = RT e
∂
P
$
'T
f.)
€
+ 2RTPe−3P
(
)
g.) P(V ,T ) = RT + RTVℓnV
€
#∂ P &
%
( = RT ( ℓnV + 1)
$ ∂ V 'T
€
#∂ P &
%
( = R + RVℓnV
$ ∂ T 'V
€
5.
For each of the following functions of three variables, evaluate the requested partial derivatives.
a.) r =
x 2 + y 2 + z 2 ; evaluate #% ∂ r &( .
$ ∂x 'y,z
#∂ r &
= x x2 + y2 + z2
% (
$ ∂ x 'y,z
€
(
€
)
−1 / 2
€
$ '
b.) y = r sin θ cos φ ; evaluate & ∂ y ) .
% ∂φ (r,θ
$∂ y '
= − r sin θ sin φ
& )
% ∂ φ (r,θ
€
€
€
=
x
r
8
6.
Evaluate the following expressions using the ideal gas equation of state.
#∂ P &
a.) %
(
$ ∂ 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 =
€
nRT
RT
=
.
€
V
Vm
Then, the partial derivative may be evaluated,
€
#
&
b.) % ∂ P (
$ ∂ Vm 'T
€
#∂ P &
R
=
.
%
(
∂
T
V
$
'Vm
m
€
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 =
nRT
RT €
=
. €
V
Vm
Then, the partial derivative may be evaluated,
€
#∂ T &
c.) %
(
$ ∂ P 'Vm
€
# ∂P &
RT
%
( = − 2 .
Vm
$ ∂ Vm 'T
€
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 =
PV
PVm€
=
. €
nR
R
Then, the partial derivative may be evaluated,
€
€
#∂ T &
%
(
$ ∂ P 'V
=
m
Vm
.
R
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 =
PV
PVm €
=
. €
nR
R
Then, the partial derivative may be evaluated,
€
# ∂T &
P
.
%
( =
R
$ ∂ Vm 'P
€
7. The isothermal compressibility κ is defined by the relation
κ = −
1
V
%∂ V (
'
* ,
& ∂ P )T
and the expansion coefficient α is given by
€
1 $∂ V '
&
)
V % ∂ T (P
α =
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 =
nRT
.
P
Next, the partial derivative may be evaluated,
€
#∂ V &
nRT
%
( = − 2 .
P
$ ∂ P 'T
Finally, the partial derivative may be substituted into the expression for the isothermal compressibility and
simplified,
€
κ = −
1
V
%∂ V (
1
'
* = −
∂
P
V
&
)T
% nRT (
P
⋅ '− 2 * =
& P )
nRT
κ =
€
€
1
.
P
% nRT (
⋅' 2 *
& P )
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 =
nRT
.
P
Next, the partial derivative may be evaluated,
€
#∂ V &
nR
.
%
( =
∂
T
P
$
'P
Finally, the partial derivative may be substituted into the expression for the expansion coefficient and
simplified,
€
1 $∂ V '
1 $ nR '
P $ nR '
α =
⋅& ) =
⋅& )
&
) =
V % ∂ T (P
V %P (
nRT % P (
α =
€
1
T
€
8.
"
a %
The van der Waals equation for a real gas is defined as $ P +
'(Vm − b) = RT , where Vm is the
Vm2 &
#
molar volume, R is the gas constant, and a and b are van der Waals constants. For the van der Waals
equation, determine
€
#∂ P &
a.) %
(
$ ∂ T 'Vm
€
In order to evaluate the partial derivative, we must first solve the van der Waals equation for P,
€
"
a %
$ P + 2 '(Vm − b) = RT
Vm &
#
a
RT
P + 2 =
V
Vm
m − b
RT
a
P =
− 2 .
Vm − b
Vm
Using that expression, the partial derivative may be evaluated,
€
#∂ P &
R
=
.
%
(
∂
T
V
$
'Vm
m −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 =
RT
a
− 2 ,
Vm − b
Vm
the partial derivative may be evaluated,
€
€
# ∂P &
RT
2a
+ 3 .
%
( = −
2
Vm
$ ∂ Vm 'T
(Vm − b)