3C Gay–Lussac’s Law—How the Pressure and Temperature of a Gas Are Related
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GAS LAWS THAT RELATE PRESSURE, VOLUME, AND TEMPERATURE
V1
T1
V1T2
T1
[3]
199
=
=
V2
Solve for V2 by multiplying both sides by T2.
T2
V2
Solve the problem.
• Substitute the three known quantities into the equation and solve for V2.
V2
=
V1T2
T1
=
(0.50 L)(77 K)
=
298 K
Kelvins cancel.
0.13 L
Answer
• Since the temperature has decreased, the volume of gas must decrease as well.
PROBLEM 7.10
A volume of 0.50 L of air at 37 °C is expelled from the lungs into cold surroundings at 0.0 °C.
What volume does the expelled air occupy at this temperature?
PROBLEM 7.11
(a) A volume (25.0 L) of gas at 45 K is heated to 450 K. What volume does the gas now
occupy? (b) A volume (50.0 mL) of gas at 400. °C is cooled to 50. °C. What volume does the
gas now occupy?
PROBLEM 7.12
Calculate the Kelvin temperature to which 10.0 L of a gas at 27 °C would have to be heated to
change the volume to 12.0 L.
Charles’s law can be used to explain how wind currents form at the beach (Figure 7.5). The air
above land heats up faster than the air above water. As the temperature of the air above the land
increases, the volume that it occupies increases; that is, the air expands, and as a result, its density
decreases. This warmer, less dense air then rises, and the cooler denser air above the water moves
toward the land as wind, filling the space left vacant by the warm, rising air.
▼
FIGURE 7.5
Focus on the Environment: How Charles’s Law Explains
Wind Currents
The same number of air molecules
occupy a smaller volume, giving
cooler air a higher density.
Warmer air rising
Heated air expands (Charles’s law), so
its density decreases. The hot air rises.
to take warm air’s place
Cooler air moving
Land
Sea
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GAS LAWS THAT RELATE PRESSURE, VOLUME, AND TEMPERATURE
[3]
201
Solve the problem.
• Substitute the three known quantities into the equation and solve for P2.
P2
P1T2
=
=
T1
(80. psi)(316 K)
291 K
=
Kelvins cancel.
87 psi
Answer
• Since the temperature has increased, the pressure of the gas must increase as well.
PROBLEM 7.13
A pressure cooker is used to cook food in a closed pot. By heating the contents of a pressure
cooker at constant volume, the pressure increases. If the steam inside the pressure cooker is
initially at 100. °C and 1.00 atm, what is the final temperature of the steam if the pressure is
increased to 1.05 atm?
PROBLEM 7.14
The temperature of a 0.50-L gas sample at 25 °C and 1.00 atm is changed to each of the
following temperatures. What is the final pressure of the system?
a. 310. K
PROBLEM 7.15
b. 150. K
c. 50. °C
d. 200. °C
Use Gay–Lussac’s law to answer the question posed at the beginning of the chapter: Why does
a lid pop off a container of food when it is heated in a microwave?
CONSUMER NOTE
7.3D THE COMBINED GAS LAW
All three gas laws—Boyle’s, Charles’s, and Gay–Lussac’s laws—can be combined in a single
equation, the combined gas law, that relates pressure, volume, and temperature.
P1V1
T1
initial conditions
Food cooks faster in a pressure
cooker because the reactions
involved in cooking occur at a faster
rate at a higher temperature.
SAMPLE PROBLEM 7.5
P2V2
=
T2
new conditions
The combined gas law contains six terms that relate the pressure, volume, and temperature of
an initial and final state of a gas. It can be used to calculate one quantity when the other five are
known, as long as the amount of gas is constant. The combined gas law is used for determining the
effect of changing two factors—such as pressure and temperature—on the third factor, volume.
We solve this type of problem by following the same three steps in the How To in Section 7.3A,
using the equation for the combined gas law in step [2]. Sample Problem 7.5 shows how this is
done. Table 7.2 summarizes the equations for the gas laws presented in Section 7.3.
A weather balloon contains 222 L of helium at 20 °C and 760 mm Hg. What is the volume of
the balloon when it ascends to an altitude where the temperature is –40 °C and 540 mm Hg?
ANALYSIS
Since this question deals with pressure, volume, and temperature, the combined gas law is
used to determine a final volume (V2) because five quantities are known—the initial pressure,
volume, and temperature (P1, V1, and T1), and the final pressure and temperature (P2 and T2).
SOLUTION
[1]
Identify the known quantities and the desired quantity.
P1 = 760 mm Hg
P2 = 540 mm Hg
T1 = 20 °C
T2 = –40 °C
V1 = 222 L
V2 = ?
known quantities
desired quantity
• Both temperatures must be converted to Kelvin temperatures.
• T1 = °C + 273 = 20 °C + 273 = 293 K
• T2 = °C + 273 = –40 °C + 273 = 233 K
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202
GASES, LIQUIDS, AND SOLIDS
[2]
Write the equation and rearrange it to isolate the desired quantity, V2, on one side.
Use the combined gas law.
P1V1
T1
P1V1T2
T1P2
[3]
P2V2
=
T2
=
Solve for V2 by multiplying both sides by
T2
P2
.
V2
Solve the problem.
• Substitute the five known quantities into the equation and solve for V2.
V2
P1V1T2
=
T1P2
(760 mm Hg)(222 L)(233 K)
=
(293 K)(540 mm Hg)
=
248.5 L rounded to 250 L
Answer
Kelvins and mm Hg cancel.
TABLE 7.2
Summary of the Gas Laws That Relate Pressure, Volume,
and Temperature
Law
Boyle’s law
Charles’s law
Gay–Lussac’s law
Combined gas law
PROBLEM 7.16
Equation
Relationship
P1V1 = P2V2
As P increases, V decreases for constant T and n.
V1
V2
=
T1
T2
As T increases, V increases for constant P and n.
P1
T1
P1V1
T1
=
P2
T2
As T increases, P increases for constant V and n.
=
P2V2
T2
The combined gas law shows the relationship of P,
V, and T when two quantities are changed.
The pressure inside a 1.0-L balloon at 25 °C was 750 mm Hg. What is the pressure inside the
balloon when it is cooled to –40 °C and expands to 2.0 L in volume?
7.4 AVOGADRO’S LAW—HOW VOLUME AND MOLES
ARE RELATED
Each equation in Section 7.3 was written for a constant amount of gas; that is, the number of moles
(n) did not change. Avogadro’s law describes the relationship between the number of moles of a
gas and its volume.
• Avogadro’s law: When the pressure and temperature are held constant, the volume of a
gas is proportional to the number of moles present.
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AVOGADRO’S LAW—HOW VOLUME AND MOLES ARE RELATED
203
As the number of moles of a gas increases, its volume increases as well. Thus, dividing the
volume by the number of moles is a constant (k). The value of k is the same regardless of the
identity of the gas.
When the number of moles increases...
∝
V
n
...the volume increases.
V
n
=
k
Thus, if the pressure and temperature of a system are held constant, increasing the number of
moles increases the volume of a gas.
Increase the
number of moles.
V=1 L
V=2 L
Volume increases.
n = 1 mol
n = 2 mol
Since dividing the volume of a gas by the number of moles is a constant, knowing the volume and
number of moles initially (V1 and n1) means we can calculate a new volume or number of moles
(V2 and n2) when one of these quantities is changed.
V1
V2
=
n1
initial conditions
n2
new conditions
To solve a problem of this sort, we follow the same three steps listed in the How To outlined in
Section 7.3A, using Avogadro’s law in step [2].
SAMPLE PROBLEM 7.6
The lungs of an average male hold 0.25 mol of air in a volume of 5.8 L. How many moles of air
do the lungs of an average female hold if the volume is 4.6 L?
ANALYSIS
This question deals with volume and number of moles, so Avogadro’s law is used to determine
a final number of moles when three quantities are known—the initial volume and number of
moles (V1 and n1), and the final volume (V2).
SOLUTION
[1]
Identify the known quantities and the desired quantity.
V1 = 5.8 L
V2 = 4.6 L
n1 = 0.25 mol
known quantities
[2]
n2 = ?
desired quantity
Write the equation and rearrange it to isolate the desired quantity, n2, on one side.
• Use Avogadro’s law. To solve for n2, we must invert the numerator and denominator on
both sides of the equation, and then multiply by V2.
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