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3C Gay–Lussac’s Law—How the Pressure and Temperature of a Gas Are Related

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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