Gases Chemistry Worksheet - Chapter 13, An Introduction To Chemistry Page 20
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Chapter 13
Gases
700.0 kPa. The reaction vessel has a constant volume, and we can assume that it has no
leaks. Therefore, the volume and moles of gas remain constant.
P
V
P
V
P
P
1
1
2
2
1
2
or
because V
= V
and n
= n
=
=
1
2
1
2
n
T
n
T
T
T
1
1
2
2
1
2
1000 kPa
= 1569 K or 1295 °C
T
= T
= 1098 K
2
1
700.0 kPa
Although Rebecca designs her pressure release valve to allow gas to escape when the
pressure reaches 1000 kPa, she hopes it will never need to be used. To avoid the release
of gases, she informs the rest of her team that they should be careful to design the
process to keep the temperature well below 1295 °C.
e
13.2 - Using the Combined Gas Law Equation
xerCise
A helium weather balloon is filled in Monterey, California, on a day when the
O
18
bjeCtive
atmospheric pressure is 102 kPa and the temperature is 18 °C. Its volume under
4
these conditions is 1.6 × 10
L. Upon being released, it rises to an altitude where the
temperature is -8.6 °C, and its volume increases to 4.7 × 10
4
L. Assuming that the
internal pressure of the balloon equals the atmospheric pressure, what is the pressure
at this altitude?
Equation Stoichiometry and Ideal Gases
13.3
Chemists commonly want to convert from the amount of one substance in a chemical
reaction to an amount of another substance in that reaction. For example, they might
want to calculate the maximum amount of a product that can be formed from a certain
amount of a reactant, or they might want to calculate the minimum amount of one
reactant that would be necessary to use up a known amount of another reactant.
Chapter 10 showed how to do conversions like these. In Example 10.1, we calculated
the minimum mass of water in kilograms necessary to react with 2.50 × 10
4
kg of
tetraphosphorus decoxide, P
O
, in the reaction
4
10
(s) + 6H
O(l ) → 4H
P
O
PO
(aq)
4
10
2
3
4
This kind of problem, an equation stoichiometry problem, is generally solved in the
following steps:
Measurable property of 1 → moles 1 → moles 2 → measurable property of 2
For substances that are pure solids and liquids, such as P
O
and H
O, mass is an
4
10
2
easily measured property, so the specific steps used in solving Example 10.1 were
→ moles P
→ moles H
O → mass H
Mass P
O
O
O
4
10
4
10
2
2
Amounts of gas, however, are more commonly described in terms of volume at a
particular temperature and pressure. (It is far easier to measure a gas’s volume than
to measure its mass.) Therefore, to solve equation stoichiometry problems involving
gases, we must be able to convert between the volume of a gaseous substance and
the corresponding number of moles. For example, if both substances involved in the
equation stoichiometry problem are gases, the general steps would be
Volume of gas 1 → moles 1 → moles 2 → volume of gas 2
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