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 .
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