0.885 kg of O
2
.
Calculate the pressure of O
2
at 21
o
C.
SOLUTION:
PLAN:
We are given
V
,
T
and mass, which can be converted to
moles (
n
).
Use the ideal gas law to find
P
.
V
= 438 L
T
= 21°C = 294 K
n
= 0.885 kg O
2
(convert to mol)
P
is unknown
= 27.7 mol O
2
P
=
nRT
V
=
27.7 mol
294.15 K
atm·L
mol·K
0.0821
x
x
438 L
= 1.53 atm
0.885 kg O
2
x
10
3
g
1 kg
1 mol O
2
32.00 g O
2
x
525
Sample Problem 5.6
Using Gas Laws to Determine a Balanced
Equation
PROBLEM:
The pistoncylinders is depicted before and after a
gaseous reaction that is carried out at constant pressure.
The temperature is 150 K before the reaction and 300 K
after the reaction. (Assume the cylinder is insulated.)
Which of the following balanced equations describes the reaction?
(1)A
2
(
g
) + B
2
(
g
) → 2AB(
g
)
(2) 2AB(
g
) + B
2
(
g
) → 2AB
2
(
g
)
(3) A(
g
) + B
2
(
g
) → AB
2
(
g
)
(4) 2AB
2
(
g
)
A
2
(
g
) + 2B
2
(
g
)
526
SOLUTION:
PLAN:
We are told that
P
is constant for this system, and the depiction
shows that
V
does not change either. Since
T
changes, the
volume could not remain the same unless the amount of gas in
the system also changes.
n
1
T
1
=
n
2
T
2
Sample Problem 5.6
Since
T
doubles, the total number of moles of gas must halve –
i.e., the moles of product must be half the moles of reactant.
This relationship is shown by equation (3).
T
1
T
2
=
n
2
n
1
=
150 K
300 K
=
½
A(
g
) + B
2
(
g
) → AB
2
(
g
)
527
The Ideal Gas Law and Gas Density
The density of a gas is

directly proportional to its molar mass and

inversely proportional to its temperature.
M
x
P
RT
=
d
=
m
V
m
M
moles =
and
m
V
density =
m
M
RT
PV
=
528
Sample Problem 5.7
Calculating Gas Density
(a) At STP, or 273 K and 1.00 atm:
PLAN:
We can use the molar mass of CO
2
to find its density from
the ideal gas equation.
PROBLEM:
Find the density (in g/L) of CO
2
(
g
) and the number of
molecules per liter
(a)
at STP and
(b)
at room conditions (20.°C and 1.00 atm).
SOLUTION:
M
x
P
RT
d
=
=
44.01 g/mol
x 1.00
atm
0.0821
x 273 K
mol·K
atm·L
= 1.96 g/L
1.96 g CO
2
1 L
1 mol CO
2
44.01 g CO
2
x
x
6.022 x 10
23
molecules
1 mol
= 2.68 x 10
22
molecules CO
2
/L
529
Sample Problem 5.7
(b) At 20.°C and 1.00 atm:
SOLUTION:
M
x
P
RT
d
=
=
44.01 g/mol
x 1.00
atm
0.0821
x 293 K
mol·K
atm·L
= 1.83 g/L
1.83 g CO
2
1 L
1 mol CO
2
44.01 g CO
2
x
x
6.022 x 10
23
molecules
1 mol
= 2.50 x 10
22
molecules CO
2
/L
T
= 20.°C + 273.15 = 293 K
530
Molar Mass from the Ideal Gas Law
m
M
n
=
PV
RT
=
mRT
PV
M
=
531
Determining the molar
mass of an unknown
volatile liquid.
based on the method of
J.B.A. Dumas (18001884)
532
Sample Problem 5.8
Finding the Molar Mass of a Volatile Liquid
PROBLEM:
An organic chemist isolates a colorless liquid from a petroleum
sample. She uses the Dumas method and obtains the following
data:
PLAN:
SOLUTION:
Use unit conversions, mass of gas, and density
M
relationship.
Volume of flask = 213 mL
Mass of flask + gas = 78.416 g
T = 100.0
o
C
Mass of flask = 77.834 g
P = 754 torr
Calculate the molar mass of the liquid.
m = (78.416  77.834) g = 0.582 g
M
=
mRT
VP
=
0.582 g
atm*L
mol*K
0.0821
373 K
x
x
0.213 L x
0.992 atm
=
84.4 g/mol
533
•
Gases mix homogeneously in any proportions.
–
Each gas in a mixture behaves as if it were the only gas present.
•
The pressure exerted by each gas in a mixture is called
its
partial pressure
.