2.6.3: The Gas Laws (Problems)
- Page ID
- 210665
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)PROBLEM \(\PageIndex{1}\)
Sometimes leaving a bicycle in the sun on a hot day will cause a blowout. Why?
- Answer
-
As temperature of a gas increases, pressure will also increase based on the ideal gas law. The volume of the tire can only expand so much before the rubber gives and releases the build up of pressure.
PROBLEM \(\PageIndex{2}\)
Explain how the volume of the bubbles exhausted by a scuba diver change as they rise to the surface, assuming that they remain intact.
- Answer
-
As the bubbles rise, the pressure decreases, so their volume increases as suggested by Boyle’s law.
PROBLEM \(\PageIndex{3}\)
One way to state Boyle’s law is “All other things being equal, the pressure of a gas is inversely proportional to its volume.”
(a) What is the meaning of the term “inversely proportional?”
(b) What are the “other things” that must be equal?
- Answer a
-
The pressure of the gas increases as the volume decreases
- Answer b
-
amount of gas, temperature
PROBLEM \(\PageIndex{4}\)
An alternate way to state Avogadro’s law is “All other things being equal, the number of molecules in a gas is directly proportional to the volume of the gas.”
- What is the meaning of the term “directly proportional?”
- What are the “other things” that must be equal?
- Answer a
-
The number of particles in the gas increases as the volume increases
- Answer b
-
temperature, pressure
PROBLEM \(\PageIndex{5}\)
A spray can is used until it is empty except for the propellant gas, which has a pressure of 1344 torr at 23 °C. If the can is thrown into a fire (T = 475 °C), what will be the pressure in the hot can?
- Answer
-
3.40 × 103 torr
PROBLEM \(\PageIndex{6}\)
What is the temperature of an 11.2-L sample of carbon monoxide, CO, at 744 torr if it occupies 13.3 L at 55 °C and 744 torr?
- Answer
-
276°K
3°C
PROBLEM \(\PageIndex{7}\)
A 2.50-L volume of hydrogen measured at –196 °C is warmed to 100 °C. Calculate the volume of the gas at the higher temperature, assuming no change in pressure.
- Answer
-
12.1 L
PROBLEM \(\PageIndex{8}\)
A balloon inflated with three breaths of air has a volume of 1.7 L. At the same temperature and pressure, what is the volume of the balloon if five more same-sized breaths are added to the balloon?
- Answer
-
2.83 L
PROBLEM \(\PageIndex{9}\)
A weather balloon contains 8.80 moles of helium at a pressure of 0.992 atm and a temperature of 25 °C at ground level. What is the volume of the balloon under these conditions?
- Answer
-
217 L
PROBLEM \(\PageIndex{10}\)
How many grams of gas are present in each of the following cases?
- 0.100 L of CO2 at 307 torr and 26 °C
- 8.75 L of C2H4, at 378.3 kPa and 483 K
- 221 mL of Ar at 0.23 torr and –54 °C
- Answer a
-
7.24 × 10–2 g
- Answer b
-
23.1 g
- Answer c
-
1.5 × 10–4 g
PROBLEM \(\PageIndex{11}\)
A high altitude balloon is filled with 1.41 × 104 L of hydrogen at a temperature of 21 °C and a pressure of 745 torr. What is the volume of the balloon at a height of 20 km, where the temperature is –48 °C and the pressure is 63.1 torr?
- Answer
-
1.2741 × 104 L
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PROBLEM \(\PageIndex{12}\)
While resting, the average 70-kg human male consumes 14 L of pure O2 per hour at 25 °C and 100 kPa. How many moles of O2 are consumed by a 70 kg man while resting for 1.0 h?
- Answer
-
0.0565 mol
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PROBLEM \(\PageIndex{13}\)
A balloon that is 100.21 L at 21 °C and 0.981 atm is released and just barely clears the top of Mount Crumpet in British Columbia. If the final volume of the balloon is 144.53 L at a temperature of 5.24 °C, what is the pressure experienced by the balloon as it clears Mount Crumpet?
- Answer
-
0.644 atm
PROBLEM \(\PageIndex{14}\)
If the temperature of a fixed amount of a gas is doubled at constant volume, what happens to the pressure?
- Answer
-
Temperature and Pressure are directly proportional. Pressure will also have to increase (doubling).
PROBLEM \(\PageIndex{15}\)
If the volume of a fixed amount of a gas is tripled at constant temperature, what happens to the pressure?
- Answer
-
Volume and pressure are inversely proportional. The pressure decreases by a factor of 3.
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Contributors and Attributions
Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/85abf193-2bd...a7ac8df6@9.110).
- Adelaide Clark, Oregon Institute of Technology
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