University of Washington
Department of Chemistry
Chemistry 452/456
Summer Quarter 2011
Homework Assignment 1
Due at 5 p.m. on 6/27/11.
This homework is worth a total of 10 points. This homework set is intended as a quick
and relatively painless way for you to brush up on your math skills, and for me to
understand your math background. Questions of these types will appear in lectures and
exams. In fact some of the calculus problems cover Lecture 1 material.
QUESTION SET A: Algebra
Properties of exponential functions and logarithms are widely exploited in this course.
Product and summation symbols are also widely used in chapters on quantum and
statistical mechanics. Simultaneous equations are solved in the course of obtaining
molecular orbital wave functions.
1A)
(a2bc-1)(ab-3) = a3b-2c-1
2A)
(a2/3)4/3 = a8/9
3A)
log10100 = log10102=2log1010 = 2
4A)
loge(e5/3)=ln(e5/3) = 5/3ln(e)=5/3
5A)
simplify e2ln3= eln9=9
6A)
express as one term: ln 9 – ln 4 = ln(9/4)=2ln(3/2)
7A)
use the quadratic equation to solve 2x2 – x – 3 = 0
x=
−b ± b 2 − 4ac 1 ± 1 − 4 ( 2 )( −3) 1 ± 1 + 24 1 ± 25 1 ± 5 6 4 3
= ,− = ,−1
=
=
=
=
4 4 2
2a
2 ( 2)
4
4
4
1
1
1
1 1 1 1
⎛ 1 ⎞
⎛ 1 ⎞
×⎜
×
×
= × × =
⎟ ×⎜
⎟ =
3 − 1 4 − 1 5 − 1 2 3 4 24
x =3
x =3 ⎝ x − 1 ⎠ x = 4 ⎝ x − 1 ⎠ x =5
5
1
1 1 1 12 + 8 + 6 26 13
⎛ 1 ⎞
⎛ 1 ⎞
⎛ 1 ⎞
9A) ∑
=⎜
=
=
⎟ +⎜
⎟ +⎜
⎟ = + + =
24
24 12
⎝ x − 1 ⎠ x =3 ⎝ x − 1 ⎠ x = 4 ⎝ x − 1 ⎠ x =5 2 3 4
x =3 x − 1
5
8A)
⎛ 1 ⎞ ⎛ 1 ⎞
∏ ⎜⎝ x − 1 ⎟⎠ = ⎜⎝ x − 1 ⎟⎠
QUESTION SET B: Derivatives and their Applications
The derivative is the calculus operation used most frequently in this course. Derivatives
are used generally throughout mechanics. Derivatives are used in Taylor expansions to
approximate values of functions and are used to find the minima and maxima of
functions. Derivatives are fundamental components of differential equations. Differential
equations are used to solve mechanical problems.
1B)
df
d
=2
( cos θ sin θ )
dθ
dθ
df
d
d
= 2sin θ
( cos θ ) + 2 cos θ ( sin θ ) = −2sin 2 θ + 2 cos 2 θ
dθ
dθ
dθ
2
2
= 2 ( cos θ − sin θ ) = 2 cos 2θ
f (θ ) = 2 cos θ sin θ
or …
df
d
d
d sin 2θ d 2θ
=2
= 2 cos 2θ
( cos θ sin θ ) = ( sin 2θ ) =
dθ
dθ
dθ
d 2θ dθ
2B)
f ( x ) = (1 – x 4 ) / x
df
d
= ⎡⎣ x −1 (1 – x 4 ) ⎤⎦
dx dx
df
d
d
∴ = ⎡⎣ x −1 (1 – x 4 ) ⎤⎦ = ( x −1 + x 3 ) = − x −2 + 3 x 2
dx dx
dx
3B)
1− e x
d2 f
=?
dx 2
2x
1− e x
−1
−1
f (x ) =
= (2x ) − (2x ) e−x
2x
df
d
−1
−1
−2
−2
−1
∴
=
(2 x ) − (2 x ) e−x = −(2 x ) + (2 x ) e−x + (2 x ) e−x
dx dx
−2
−2
−1
= −(2x ) + e−x (2x ) + (2x )
f (x ) =
[
2
[
]
]
[
[
]]
[
] [
d
d f
−2
−2
−1
−3
−2
−1
−3
−2
=
−(2x ) + e−x (2x ) + (2x ) = 2(2x ) − e−x (2x ) + (2x ) + e−x −2(2x ) − (2x )
2
dx
dx
[
= x −3 − e−x (2x ) + x −2 + x −3
4B)
−1
]
∂ ⎛ 4 6 3 ⎞
2 x y + 2 ⎟ = 8 x 3 y 6 − 6 x −3
x ⎠
∂ x ⎜⎝
∂ ⎛ 4 6 3 ⎞
2 x y + 2 ⎟ = 12 x 4 y 5
x ⎠
∂ y ⎜⎝
5B) Using a Taylor expansion show that close to 1, ln(x) ≈ x -1. Use a Taylor
expansion to approximate ln(1.05) = ? to first order.
]
( x − 1) ⎛ d 2 ln x ⎞ + …
x − 1 ⎛ d ln x ⎞
ln ( x ) = ln (1) +
+
⎜
⎟
⎜
⎟
1! ⎝ dx ⎠ x =1
2! ⎝ dx 2 ⎠ x =1
x − 1 ⎛ d ln x ⎞
⎛1⎞
≈ ln (1) +
⎜
⎟ = 0 + ( x − 1) ⎜ ⎟ = x − 1
1! ⎝ dx ⎠ x =1
⎝ x ⎠ x =1
2
∴ ln (1.05 ) ≈ 1.05 − 1 = 0.05
Find the local maximum and minimum of the equation f(x) = x3 – 27x
(ignore any max or min at ± ∞)
2
Minimum occurs at df/dx = 0 and d f/dx2 > 0
2
Maximum occurs at df/dx = 0 and d f/dx2 < 0
df
d
= ( x 3 − 27 x ) = 3 x 2 − 27 = 0
dx dx
∴ x = ±3
6B)
d2 f
d
= ( 3x 2 − 27 ) = 6 x
2
dx
dx
∴ x=3 is a minimum, x=-3 is a maximum
QUESTION SET C Differentials and Path Integrals
1C) Determine which of the following differentials is exact and which is inexact. Prove
your answer using Euler’s test for exactness.
a) 6xy3 dx + 9x2y2 dy
M ( x, y ) = 6xy3 and N ( x, y ) =9x 2 y 2
∂M
∂N
= 18 xy 2 and
=18xy 2
∂y
∂x
∂M ∂N
∴
=
→ exact
∂y
∂x
b) 4xy dx + 3x2y dy
M ( x, y ) = 4xy and N ( x, y ) = 3x 2 y
∂M
∂N
= 4 x and
= 6xy
∂y
∂x
∂M ∂N
∴
≠
→ inexact
∂y
∂x
c) 16xy3z dx + 24 x2y2z dy + 8x2y3 dz
M ( x, y, z ) =16xy3 z; N ( x, y, z ) = 24 x 2 y 2 z; P ( x, y, z ) = 8x 2 y3
∂M
∂M
∂N
∂N
∂P
∂P
= 48 xy 2 z;
= 16xy3 ;
= 48 xy 2 z;
= 24x 2 y 2 ;
= 16 xy 3 ;
= 24x 2 y 2
∂y
∂z
∂x
∂z
∂x
∂y
∂M ∂N
∂M ∂P ∂N ∂P
∴
=
=
=
→ exact
;
;
∂y
∂x
∂z
∂x
∂z ∂y
2C) For 6xy3 dx + 9x2y2 dy determine the path integrals along x=y and x=y2 from (0,0) to
(1,1).
1
1
1
1
1
1
6 x5
9 y5
+
⎯⎯→
+
=
+
=3
6xy
dx
9x
y
dy
6x
dx
9y
dy
∫0
∫0
∫0
∫0
5 0
5 0
3
2
1
x= y
2
1
4
4
1
1
1
1
12 x 7/2
9 y7
6xy
dx
9x
y
dy
6x
dx
9y
dy
+
⎯⎯⎯
→
+
=
+
=3
∫0
∫0
∫0
∫0
7 0
7 0
3
2
x= y2
2
5/ 2
6
3C) For 4xy dx + 3x2y dy determine the path integrals along x=y and x=y2 from (0,0) to
(1,1).
1
1
1
1
1
1
4 x3
3y4
25
+
⎯⎯→
+
=
+
=
4xy
dx
3x
y
dy
4x
dx
3y
dy
∫0
∫0
∫0
∫0
3 0
4 0 12
x= y
2
1
1
2
1
3
1
1
1
8 x 5/ 2
3 y6
63
+
⎯⎯⎯
→
+
=
+
=
4xy
dx
3x
y
dy
4x
dx
3y
dy
∫0
∫0
∫0
∫0
5 0
6 0 30
x= y2
2
3/2
5
QUESTION SET D: More Integrals and their Applications
Integrals and derivatives go hand-in-hand in mechanics courses. In this course integrals
are commonly used to calculate averages of various types. Although we expect you to be
able to perform simple integrations from scratch, it is far more often the case that you are
required to evaluate an integral that has standard form. Standard form means the integral
has a form with a generally recognized answer, which you can find in books or on the
Web. In this course very often the challenge is to correctly frame a physical problem in
terms of an integral with standard form.
3
3
1D)
2 x3
2
52
2
x
dx
=
= ( 27 − 1) =
∫x=1
3 1 3
3
2
Vf
2D)
∫
Vi
1
1
1 ⎛V ⎞
dv = ( ln V f − ln Vi ) = ln ⎜ f ⎟
3v
3
3 ⎝ Vi ⎠
2 2a / 3 2 ⎛ πx ⎞
3D) Evaluate the integral;
∫ sin ⎜⎝ L ⎟⎠dx where L is a constant, using the standard
L a /3
x 1
form ∫ sin 2 axdx = − sin 2ax + C , where a and C are constants.
2 4a
Solution:
2 a /3
2⎛x L
⎛πx ⎞
⎛ 2π x ⎞ ⎞
∫a /3 sin ⎜⎝ L ⎟⎠dx = L ⎜⎝ 2 − 4π sin ⎜⎝ L ⎟⎠ ⎟⎠
a /3
⎡
⎛
⎛
⎛
⎛
⎞
⎞
⎞
2 2a
4 πa
2πa ⎞⎤
L
a
L
= ⎢⎜ ⎟ −
sin⎜
sin⎜
⎟ −⎜ ⎟ +
⎟⎥
L ⎣⎝ 6 ⎠ 4π ⎝ 3L ⎠ ⎝ 6 ⎠ 4 π ⎝ 3L ⎠⎦
⎛ a ⎞ 2 ⎛ ⎛ 4 πa ⎞
⎛ 2πa ⎞⎞
=⎜ ⎟−
⎟ − sin⎜
⎟⎟
⎜sin⎜
⎝ 3L ⎠ 4 π ⎝ ⎝ 3L ⎠
⎝ 3L ⎠⎠
2
L
2 a /3
2
+∞
4D)
∫e
Evaluate the integrals
−κ x 2 /2 kT
+∞
dx and
−∞
∫e
− p 2 /2 mkT
dp using the standard form
−∞
1/2
∞
2
⎛π ⎞
integral: ∫ e − cx dx = ⎜ ⎟ . Note κ, k, T, and m are constants. In the standard form c
⎝ 4c ⎠
0
indicates a constant. Note the function being integrated is an even function: f(x)=f(-x).
Solution:
+∞
−κ x
∫e
2
+∞
/2 kT
−∞
+∞
∫e
dx = 2 ∫ e −κ x
− p 2 /2 mkT
−∞
2
/2 kT
dx =
0
+∞
dp = 2 ∫ e − p
2
/2 mkT
2π kT
κ
dp = 2π mkT
0
5D) The average kinetic energy of a very small spherical particle moving in one
+∞
1
p 2 − p 2 /2 mkT
dp , where m, k, T are constants. Calculate
dimension is: E =
∫ 2me
2π mkT −∞
∞
E using the standard form:
∫x
2 n − ax 2
dx =
1⋅ 3 ⋅ 5
( 2n − 1) π
, where n is a positive
2 an
a
integer and a is a constant greater than zero. Note the function being integrated is even.
0
Solution:
e
n +1
1
E =
2π mkT
=
+∞
1
m 2π mkT
∞
∫
+∞
p 2 − p 2 / 2 mkT
2
dp = E =
∫−∞ 2me
2π mkT
p 2e− p
/2 mkT
∫
0
p 2 − p 2 /2 mkT
e
dp
2m
dp
0
But ∫ x 2 n e − ax dx =
2
0
∴ E =
2
+∞
1
1⋅ 3 ⋅ 5 ( 2n − 1) π
2n +1 a n
a
2mkT
m 2π mkT 4
2π mkT =
kT
2
Gas Laws
6) Calculate the pressure exerted by benzene for a molar volume of 1.42 L at 790 K
using the Redlich-Kwong equation of state:
P=
RT
a
1
nRT
n2 a
1
−
=
−
Vm − b
T Vm (Vm + b) V − nb
T V (V + nb)
The Redlich-Kwong parameters a and b for benzene are 452.0 bar dm6 mol–2 K1/2 and
0.08271 dm3 mol–1, respectively. Is the attractive or repulsive portion of the potential
dominant under these conditions?
7) About how many oxygen molecules arrive each second at the mitochondrion of
an active person? The following data are available:
•
•
•
Oxygen consumption is about 40.mL of O2 per minute per kilogram of body
mass measured at T=300. K and P=1.0 atm.
An adult with a body mass of 64. kilograms has about 1.0x1012 cells.
Each cell has about 800. mitochondria.
Solution:
First, calculate the number of moles of oxygen consumed per minute per kg of body
weight;
-1
−3
-1
-1
3 -1
PV ′ (1atm)(101325Pa ⋅ atm )(40 ×10 L ⋅ min kg )(0.01m L )
′
=
= 1.63 ×10−3 mol ⋅ min -1 ⋅ kg -1
n =
-1
-1
RT
(8.31J ⋅ K ⋅ mol )(300K)
⎛ 1min ⎞
1min
−3
-1
= (1.63 ×10−3 mol⋅ min -1 ⋅ kg -1 )(64kg)⎜
⎟ = 1.73 ×10 mol⋅ s
⎝ 60s ⎠
60s
moleculesO2 = (1.73 ×10−3 mol⋅ s-1 )(6.023 ×10 23 mol-1 )= 1.04 ×10 21 s−1
n = n ′ × 64kg ×
(1.04 ×1021s-1)
moleculesO2
=
= 1.30 ×10 6 s-1mitochondrion-1
mitochondrion (1.0 ×1012 cells)(800mitochondrion ⋅ cell-1 )
Calculation of Work, Heat and ∆U
8) 1-mol sample of an ideal gas for which CV,m=3/2R undergoes the following twostep process: (1) From an initial state of the gas described by T = 28.0°C and P =
2.00 × 104 Pa, the gas undergoes an isothermal expansion against a constant
external pressure of 1.00 × 104 Pa until the volume has doubled. (2) Subsequently,
the gas is cooled at constant volume. The temperature falls to –40.5°C. Calculate
q, w, ∆U for each step and for the overall process.
Solution:
a) Isothermal expansion of an ideal gas means ∆T=0 therefore
∆U1 = nCV ∆T = q1 + w1 = 0 so q1 = − w1 = Pext ∆V . The initial volume is
V1 =
−1
−1
RT ( 8.31JK mol ) ( 301K )
=
= 0.125m 3
4
−2
P1
2.00 × 10 Ntm
The final volume is double so V2 = 2V1 = 0.250m 3 . Therefore
q1 = − w1 = Pext ∆V = (1.00 × 104 Pa )( 0.250m 3 − 0.125m 3 ) = 1250 J
Also ∆H1 = nCP ∆T = 0
b) At constant volume the work is zero w2=0. Then
3R
3 × 8.31JK −1
∆U 2 = nCv ∆T = q2 =
× (−68.5K ) = −854J
(232.5K − 301K ) =
2
2
5R
5 × 8.31JK −1
∆H 2 = nC p ∆T = n(Cv + R)∆T =
× (−68.5K ) = −1423J
(232.5K − 301K ) =
2
2
c) Total changes:
∆U = ∆U1 + ∆U 2 = 0 − 854 J = −854 J
∆H = ∆H1 + ∆H 2 = 0 − 1423 J = −1423J
q = q1 + q2 = 1250 J − 854 J = 396 J
w = w1 + w2 = −1250 J + 0 = −1250 J
9) Count Rumford observed that using cannon boring machinery a single horse
could heat 11.6 kg of ice water (T=273K) to T=355K. in 2.5 hours. Assuming the
same rate of work, how high could a horse raise a 150 kg weight in one minute?
Assume the heat capacity of water is 4180 J K-1 kg-1.
Solution: The heat produced by a single horse driving cannon boring equipment is
qP = MCwater ∆T = 11.6kg × 4180JK −1kg −1 × ( 355K − 273K ) = 3980kJ
3980kJ
≈ 1600kJ hr -1
2.5hr
In one minute the work performed by the horse is
1hr
w = 1.0 min×
× 1600kJhr -1 = 27kJ
60 min
This amount of work is equated to the expression
27000J = MGn ∆h = 150kg × 9.80655 × ∆h
The rate of heat production is r =
∴∆h =
27000J
= 18m
1500kgms-2
10) A major league pitcher throws a baseball with a speed of 150 kilometers per
hour. If the baseball weighs 220 grams and its heat capacity is 2.0 J g-1 K-1,
calculate the temperature rise of the ball when it is stopped by the catcher’s mitt.
Assume no heat is transferred to the catcher’s mitt. Assume also that the catcher’s
arm does not recoil when he catches the ball.
Solution:
When the ball stops kinetic energy is converted to heat so…
1
mV 2 = mC ∆T
2
2
⎡
⎛ 1hr ⎞ ⎤
150 × 103 m hr -1 ) ⎜
(
⎟
2
⎢
V
⎝ 3600s ⎠ ⎥⎦
∴∆T =
=⎣
= 0.44K
2C
( 2 ) ( 2000.0Jkg-1K -1 )