Name CHEMISTRY 372
Stud. ID. # r EXAM 1
Wednesday
11 February 2016
Directions:
. , Write your name and Student Identification Number on this page.
. This is NOT a multiplechoice exam. You are therefore expected to show yourwork.
Credit will not be given f
Solutions to UPC Chapter #9
July 3, 2013
1a) We rst calculate the energy of a particle in a box with length LH = 5a0 2.65= 0.265 nm to
A
approximate the hydrogen atom in its ground state (n=1):
EH =
h2
= 8.62 1019 J
8me L2
H
For the hydrogen atom, the box
Solutions to UPC Chapter #10
July 3, 2013
1) The number of thermally accessible states is given by the partition function.
For N2 , we have ( = 2):
qr =
T
300 K
=
52
r
2(2.9 K)
qr =
T
300 K
=
107
r
(2.8 K)
For CO ( = 1):
Note that CO has more accessible
Solutions to UPC Chapter #8
July 3, 2013
1a) The energy is obtained from the formula
En =
hcRH
(6.63 1034 J s)(3.0 108 m/s)(1.0974 107 m1 )
=
n2
n2
En =
2.18 1018 J
n2
and
L2 =
l(l + 1)
The results are summarized in the following table:
En (J)
2.18 1018
Solutions to UPC Chapter #7
July 3, 2013
1)
nx
2
sin
b
b
n (x) =
|n|2(x)
a)
0
0.2
0.4
0.6
0.8
1
x/b
Figure 1: Particle in a box probability density functions (p.d.f.); the curve at the bottom is the ground state
p.d.f. for n = 1, the curve above the groun
Solutions to UPC Chapter #6
July 3, 2013
1a) To calculate the force constant we employ Eq. 6.5
=
1
=
c
2c
f
m
therefore
f = 4 2 mc2 2
= 4(3.142)2 (1.7 1027 kg)(3 1010 cm/s)2 (3000cm1 )2
kg
N
= 5.4 102 2 = 5.4 102
s
m
b) We can note that
h =
hc
thus, = c/
CHM 373
Solutions to UPC Chapter #5
July 3, 2013
1. a) P = 0.1 MPa
P
T
=
P
0.0033876
=
= 3.37 104 MPa K1
T
10.049
=
P
0.0040214
= 4.167 103 MPa K1
=
T
0.96484
=
P
0.0048880
=
= 9.038 103 MPa K1
T
0.54082
V
P = 1.1 MPa
P
T
V
P = 2.1 MPa
P
T
V
b)
U
V
T
U
V
CHM 373
Solutions to UPC Chapter #4
July 3, 2013
1.
P
T
V
V
T
=
V
P
2.
S
V
U
=
V P
P
=
V T
T
T
U
V
=
U
S
U
S
Note: This is true since dU = T dS P dV =
U
V
P
S
=
P
T
V
dS +
V
U
V
= P
S
3. a) Using the cross derivative relationship from dG = SdT + V dP ,
S
CHM 373
Solutions to UPC Chapter #3
July 3, 2013
1. For ideal diatomic gases with thermally active translational and rotational motions, D = 5.
a) Recall: U =
b) If U =
D
nRT . So
2
U
T
5
nR
2
=
V
U
V
D
nRT and T is constant,
2
=
5
nR.
2
= 0.
T
nRT
. So i
CHM 373
Solutions to UPC Chapter #2
July 3, 2013
1. The system does work by lifting a 1lb weight 1ft using an internal battery with a power output of 1W
(which produces both work and heat).
a)
W = F D = mgd = (0.454kg)(9.8m/s2 )(0.305m)
W = 1.36J
The sign
Understanding Physical Chemistry
Solutions to Homework Problems in Chapter 1
September 11, 2013
1.
K
V
b)
a)
E
c)
&me
&me
&me
The total energy, E is the sum of kinetic, K and potential, V energies. Unless the ball is thrown straight up,
Name , - \ _ . . . CHEMISTRY 372
Stud. iD. # ._ , EXAM 1
Tuesday
17 February 2015
Directions:
- Write your name and Student Identification Number on this page.
- This is NOT a multiple-choice exam. You are therefore expected to show yourwork.
Credit will
Name,"_q ., _ CHEMISTRY 372
Stud. Id. # m 7 . EXAM 2
Wednesday
25 March 2015
Directions:
PRINT your name and Student Identification Number on this page.
This is NOT a multiple-choice exam. You are therefore expected to show yourwork.
The TAs will not turn
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Name _ _. 7 - CHEMISTRY 372
Student ID # _ EXAM 3
Thursday
16 April 2015
Directions:
. Write your name and Student Identification Number on this page.
. This isnt a multiple-choice exam; you are expected to show your work. The TAs
won't look for answers o
Name 7 _ CHEMISTRY 372
Stud. ID # EXAM 2
Wednesday
9 March 2016
Directions:
Write your name, Student Identification Number on this page.
This is NOT a multiple-choice exam. You are therefore expected to show your work.
The TAs will not turn over a page to
Name CHEMISTRY 372
Student ID # EXAM 3
Wednesday
13 April 2016-
Directions:
- Write your name and Student Identification Number on this page.
. This isnt a multiple-choice exam; you are expected to show your work. The TAs
won't look for answers on the bac