Version 052 – Exam Three: Shear – Shear – (52375)
1
This
printout
should
have
27
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
DrRuth says: In all cases, select the answer
that BEST answers the question asked. The
total number of points on this exam is 350.
Read the whole thing first and plan which to
answer first because you feel confident about
them. Notice the harder ones and save them
up till last so you don’t use up all your time
working on them.
Unless otherwise stated,
assume the gases in all questions are ideal.
Shear
52375
g
= 9
.
8 m
/
s
2
1 atm = 101325 Pa = 760 torr
R = 0.08206 L atm K
−
1
mol
−
1
R = 0.08314 L bar K
−
1
mol
−
1
R = 8.314 J K
−
1
mol
−
1
Effusion rate of gas 1
Effusion rate of gas 2
=
radicalbigg
M
2
M
1
Effusion rate at T
1
Effusion rate at T
2
=
radicalbigg
T
1
T
2
KE = 2
/
3
RT
u
rms
=
radicalbigg
3
RT
M
001
10.0 points
If it takes 15 s for a certain sample of neon
to effuse through a porous barrier, how much
time will it take for the same amount of ni
trogen gas to effuse through the barrier under
the same conditions?
1.
15 s
2.
less than 15 s
3.
greater than 15 s
correct
Explanation:
Rate of effusion
∝
1
√
MW
MWs for each gas:
Ne : 20
.
179 g
/
mol
N
2
: 28
.
0134 g
/
mol
The heaviest gas will have particles moving
at the lowest average speed so it will effuse
more slowly.
002
10.0 points
A 6.5 L sample of nitrogen at 25
◦
C and
1.5 atm is allowed to expand to 13.0 L. The
temperature remains constant.
What is the
final pressure?
1.
0.12 atm
2.
0.38 atm
3.
3.0 atm
4.
0.75 atm
correct
5.
0.063 atm
Explanation:
V
1
= 6.5 L
V
2
= 13.0 L
P
1
= 1.5 atm
Boyle’s law relates the volume and pressure
of a sample of gas:
P
1
V
1
=
P
2
V
2
P
2
=
P
1
V
1
V
2
=
(1
.
5 atm) (6
.
5 L)
13
.
0 L
= 0
.
75 atm
003
20.0 points
At STP a gas occupies 121 mL. How many
milliliters will this gas occupy at

52
◦
C and
1.18 atm?
1.

19.5 mL
2.
115 mL
3.
83.0 mL
correct
4.
127 mL
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Version 052 – Exam Three: Shear – Shear – (52375)
2
Explanation:
P
1
= 1 atm
P
2
= 1
.
18 atm
T
1
= 273
.
15 K
V
1
= 121 mL
T
2
=

52
◦
C + 273
.
15 = 221
.
15 K
P
1
V
1
T
1
=
P
2
V
2
T
2
V
2
=
P
1
V
1
T
2
T
1
P
2
=
(1 atm) (121 mL) (221
.
15 K)
(273
.
15 K) (1
.
18 atm)
= 83
.
0212 mL
004
15.0 points
Rank the molecules
CH
3
F
,
C
2
H
6
,
H
2
O
,
H
2
,
He
in terms of increasing viscosity.
1.
He
,
H
2
,
C
2
H
6
,
CH
3
F
,
H
2
O
correct
2.
C
2
H
6
,
H
2
O
,
CH
3
F
,
He
,
H
2
3.
H
2
,
He
,
CH
3
F
,
H
2
O
,
C
2
H
6
4.
None of the answers is correct.
5.
H
2
O
,
CH
3
F
,
C
2
H
6
,
H
2
,
He
Explanation:
The
intermolecular
attractions,
ranked
from weakest to strongest, are London forces
(also called van der Waals forces or induced
dipoles),
dipoledipole interactions,
hydro
gen bonding interactions, and ionion interac
tions. In general, the larger the molecule, the
greater the total intermolecular forces. Strong
intermolecular interactions cause molecules to
“stick” to one another, and are thus more
viscous than molecules that experience weak
intermolecular attractions.
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 Spring '09
 RUTH
 Hydrogen Bond, Intermolecular force, Van der Waals force

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