Swieter; p 1 of 5
X:\~Natsci\ADJUNCT\bhaskar\pdf\1041\ANS_EX#8.DOC
CHM1040
Answer to Exercise Set 8
QUESTIONS ABOUT GASES
1.
a.
atm
0.714
=
torr
760
atm
1
torr x
543
b.
torr
10
x
1.33
=
atm
1
torr
760
x
atm
1.75
3
c.
mL
10
x
4.32
=
L
1
mL
10
x
L
43.2
4
3
d.
g/L
10
x
1.004
=
L
1
mL
10
x
mL
g
1.004
3
3
e.
f.
K
293
=
273.15
+
C
20
°
C
173
_
=
273.15
_
K
100
°
2.
Since a torr is defined as a mm Hg at 0
°
C and sea level, the mercury would be more dense (it would
contract at the sub-zero temperatures at the north pole) and thus one would need to add mm Hg to
the reading to get the correct pressure. Conversely, at the equator (here the temperature is much
higher than zero and the mercury would expand) one would need to subtract mm Hg from the
reading to get the correct pressure.
3.
a.
L
1.12
=
torr
800
torr
760
x
L
1.18
=
P
P
V
=
V
or
V
P
=
V
P
2
1
1
2
2
2
1
1
•
•
•
b.
L
2.24
=
torr
760
atm
1
x
atm
0.526
torr
760
x
L
1.18
=
P
P
V
=
V
or
V
P
=
V
P
2
1
1
2
2
2
1
1
•
•
•
c.
2
P
=
V
2
V
P
=
V
V
P
=
P
or
V
P
=
V
P
1
1
1
1
2
1
1
2
2
2
1
1
•
•
•
•
4.
a.
mL
133
=
K
.
250
K
.
333
x
mL
100
=
T
T
V
=
V
or
T
V
=
T
V
15
15
1
2
1
2
2
2
1
1
•
b.
C
_
112
or
K
.
385
=
mL
100
K
.
250
x
mL
154
=
V
T
V
=
T
or
T
V
=
T
V
23
15
1
1
2
2
2
2
1
1
•
c.
V
2
=
T
T
2
x
V
=
T
T
V
=
V
or
T
V
=
T
V
1
1
1
1
1
2
1
2
2
2
1
1
•
•
5.
a.
mL
10
x
2.00
=
mol
0.250
mol
1.25
x
mL
400
=
n
n
V
=
V
or
n
V
=
n
V
3
1
2
1
2
2
2
1
1
•
b.
mol
0.200
=
mL
400
mL
320
x
mol
0.250
=
V
V
n
=
n
or
n
V
=
n
V
1
2
1
2
2
2
1
1
•
c.
V
2
=
n
n
2
x
1
Vsub
=
n
n
V
=
V
or
n
V
=
n
V
1
1
1
1
2
1
2
2
2
1
1
•
•
•
6.
3
P
2
=
T
n
)
V
x
(3
)
T
x
(4
)
n
x
(‰
V
P
=
T
n
V
T
n
V
P
=
P
or
T
n
V
P
=
T
n
V
P
1
1
1
1
1
1
1
1
1
1
2
2
2
1
1
2
2
2
2
2
1
1
1
1
•
•
•
•
•
•
•
•
•
•
•

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