Sample Problem 5.4Applying the Volume-Amount RelationshipPROBLEM:A scale model of a blimp rises when it is filled with helium to a volume of 55.0 dm3. When 1.10 mol of He is added to the blimp, the volume is 26.2 dm3. How many more grams of He must be added to make it rise? Assume constant Tand
P
.
1
M

5-30
Sample Problem 5.4
SOLUTION:
n
1
= 1.10 mol
V
1
= 26.2 dm
3
n
2
= unknown
V
2
= 55.0 dm
3
T
and
P
are constant
P
1
V
1
n
1
T
1
P
2
V
2
n
2
T
2
=
= 1.10 mol x
55.0 dm
3
26.2 dm
3
= 2.31 mol He
n
2
=
V
2
V
1
n
1
x
V
1
n
1
V
2
n
2
=
Additional amount of He needed = 2.31 mol – 1.10 mol = 1.21 mol He
1.21 mol He x
4.003 g He
1 mol He
=
4.84 g He

Sample Problem 5.5
Solving for an Unknown Gas Variable at
Fixed Conditions
PROBLEM:A steel tank has a volume of 438 L and is filled with 0.885 kg of O2. Calculate the pressure of O2at 21oC.PLAN:We are given V, Tand mass, which can be converted to moles (n). Use the ideal gas law to find PV= 438 LT= 21°C = 294 Kn= 0.885 kg O2(convert to mol) Pis unknown
.
=
g

Sample Problem 5.6Using Gas Laws to Determine a Balanced EquationPROBLEM:The piston-cylinders 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?

5-33
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
)

P
=