Chemistry Reference
In-Depth Information
P
1
V
1
T
2
P
2
T
1
(1.57 atm) (829 mL) (289 K)
(1.29
2
atm) (303 K)
The partial pressures are directly proportional to the num-
bers of moles present.
(e)
V
2
n
He
P
total
n
total
961 mL
(2.00 mol) (6.69 atm)
6.00 mol
P
He
2.23 atm
P
1
V
1
T
2
P
2
T
1
(717 torr) (602 mL) (320 K)
(2280 torr) (320 K)
(f )
V
2
189 mL
The pressures of the other gases can be calculated in the
same way:
P
2
V
2
T
1
P
1
V
1
(665 torr) (4.48 L) (382 K)
(101
1
torr) (6.12 L)
(g)
T
2
184 K
P
Ne
P
Ar
2.23 atm
As a check, calculate the total pressure:
P
1
V
1
T
2
V
2
T
1
(797 torr) (2.13 L) (351 K)
(9.11 L) (304 K)
(h)
P
2
215 torr
2.23 atm
2.23 atm
2.23 atm
6.69 atm
12.63
The numbers of moles of all the components are the same, so
the partial pressures are also the same. The volumes and
temperatures of all gases in any gaseous mixture are the
same, and in this mixture, the pressures are also the same.
Because and all the factors on the right side of
the equation are the same for all the gases, the value of
P
must also be the same for all the gases.
12.47
The pressure in torr is converted to atmospheres, and Kelvin
temperatures are used for each part:
1 atm
760 torr
a
b
(a)
696 torr
0.915
8
atm
P
nRT
V
,
P
1
V
1
T
2
T
1
P
2
(1.00 atm) (7.10 L) (306 K)
(273 K) (0.915
8
atm)
V
2
8.69 L
12.65
The partial pressure of water vapor in this system is
1 atm
760 torr
(b)
792 torr
a
b
1.04
2
atm
782 torr
752 torr
30 torr.
(0.977 mol) (0.0821 L
#
atm/mol
#
K) (306 K)
1.04
2
atm
The temperature is determined from Table 12.3 to be
29°C.
nRT
P
V
12.66
The vapor pressure of water is determined from Table 12.3.
23.6 L
P
O
2
P
total
P
H
2
O
12.48
The number of moles of gas before the change may be cal-
culated as follows:
781 torr
42.2 torr
739 torr
12.67
The equation for the reaction is
(1.00 atm) (7.10 L)
(0.0821 L
#
atm/mol
#
K) (273 K)
PV
RT
n
0.316
8
mol
Heat
2 KClO
3
(s)
¡
2 KCl(s)
3 O
2
(g)
a
1 mol KClO
3
122 g KClO
3
3 mol O
2
2 mol KClO
3
The same number of moles is present after the change be-
cause it is the same sample of gas. The new pressure in
atmospheres is
b
a
b
0.200 g KClO
3
2.45
9
10
3
mol O
2
1 atm
760 torr
The oxygen pressure is the total pressure minus the water va-
por pressure (from Table 12.3):
a
b
696 torr
0.915
8
atm
The new volume is
1 atm
760 torr
866.
4
torr
23.7
6
torr
842.
6
torr
a
b
(0.316
8
mol) (0.0821 L
#
atm/mol
#
K) (306 K)
0.915
8
atm
nRT
P
V
1.10
9
atm
nRT
P
8.69 L
V
PV
nR
(1.25 atm) (14.0 L)
(1.50 mol) (0.0821 L
#
atm/mol
#
K)
142 K
12.49
(a)
T
(2.45
9
10
3
mol O
2
) (0.0821 L
#
atm/mol
#
K) (298 K)
1.10
9
atm
(b)
The gas is more likely to be
H
2
because
H
2
O
would be a
0.0542 L
solid at that temperature.
(0.515 mol) (0.0821 L
#
atm/mol
#
K) (307 K)
8.05 L
nRT
V
12.70
The partial pressure of the oxygen is
12.51
P
P
O
2
P
total
P
H
2
O
769 torr
12.8 torr
756.
2
torr
1.61 atm
0.995 atm
PV
RT
(0.502
6
atm) (0.152 L)
(0.0821 L
#
atm/mol
#
K) (419 K)
0.00222 mol
P
O
2
V
RT
12.54
n
(0.995 atm) (1.00 L)
(0.0821 L
#
atm/mol
#
K) (288 K)
n
O
2
(0.300 mol) (0.0821 L
#
atm/mol
#
K) (303 K)
14.7 L
nRT
V
12.58
P
0.0420
8
mol O
2
Heat
0.508 atm
2 KClO
3
(s)
¡
2 KCl(s)
3 O
2
(g)
12.60
The total number of moles is 6.00 mol. The ideal gas law
works for both the gas mixture and the individual compo-
nents, so
a
2 mol KClO
3
3 mol O
2
a
122 g KClO
3
1 mol KClO
3
0.0420
8
mol O
2
b
b
3.42 g KClO
3
P
total
and
Dividing the second of these equations by the first yields
n
total
RT
V
P
He
n
He
RT
V
a
1 mol N
2
28.0 g N
2
12.73
(a)
0.901 g N
2
b
0.0321
8
mol N
2
P
He
P
total
n
He
RT
V
nRT
V
P
n
total
RT
V
(0.0321
8
mol) (0.0821 L
#
atm/mol
#
K) (298 K)
12.4 L
R
,
T
, and
V
all cancel out, leaving
P
He
P
total
n
He
n
total
0.0635 atm
