Ideal Gases: Use the Kinetic Molecular Theory to Explain Impact on a Gas as V and T Increased
Question
A 1 L sample of carbon monoxide (CO), initially at 0 °C and 1 atm, is heated to 546 K and its volume is increased to 2 L.
- What effect do these changes have on the number of collisions of the molecules of the gas per unit area of the container wall?
- What is the effect on the average kinetic energy of the molecules?
- What is the effect on the root mean square speed of the molecules?
Show/Hide Answer
- The number of collision per unit area of the container wall is constant.
- The average kinetic energy doubles.
- The root mean square speed increases √2 times its initial value; Urms is proportional to KEavg.
Refer to Section 2.5: The Kinetic-Molecular Theory (1).
Strategy Map
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| Strategy Map Steps |
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1. Recall the relationships between temperature, pressure, and volume.
Show/Hide Hint
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2. Recall the relationship between temperature and kinetic energy.
Show/Hide HintAs temperature increases, the kinetic energy of gas molecules increases. Show/Hide ResourceRefer to Section 2.5: The Kinetic-Molecular Theory (1). |
3. Recall the relationship to determine average kinetic energy and what change would cause it to increase or decrease.
Show/Hide Hint[latex]\begin{gathered} \mathrm{KE}_{\text {avg}}=\frac{3}{2}\mathrm{RT} \end{gathered}[/latex] |
4. Recall what root mean square speed is and what causes it to increase or decrease.
Show/Hide Hint[latex]\begin{gathered} \mathrm{U}_{\mathrm{rms}}=\sqrt{\frac{3\mathrm{RT}}{\mathrm{M}(\mathrm{x})}} \end{gathered}[/latex] |
5. Look at the question information and determine which parameters change and how they are related.
Show/Hide Hint[latex]\begin{gathered} \mathrm{MV}_1=1 \mathrm{~L};\mathrm{V}_2=2\mathrm{~L}\\ \mathrm{~T}_1=0^{\circ}\mathrm{C}=273\mathrm{~K};\mathrm{T}_2=546\mathrm{~K} \end{gathered}[/latex] n is constant. What happens to P? |
Solution
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a. What effect do these changes have on the number of collisions of the molecules of the gas per unit area of the container wall?
[latex]\begin{gathered}\frac{\mathrm{P}_1\mathrm{V}_1}{\mathrm{T}_1}=\frac{\mathrm{P}_2\mathrm{V}_2}{\mathrm{T}_2}\\\mathrm{T}_2=2\mathrm{T}_1\\\mathrm{V}_2=2\mathrm{V}_2\end{gathered}[/latex]
Substitute in to the equation for changing conditions:
[latex]\dfrac{\mathrm{P}_1\mathrm{V}_1}{\mathrm{T}_1}=\dfrac{\mathrm{P}_2 2\mathrm{V}_1}{2\mathrm{T}_1}[/latex]
Cancel terms that are the same:
[latex]\dfrac{\mathrm{P}_1\mathrm{V}_1}{\mathrm{T}_1}=\dfrac{\mathrm{P}_2 2\mathrm{V}_1}{2\mathrm{T}_1}[/latex]
Left with P1 = P2, no change in pressure.
Answer: Since pressure does not change, the number of collisions per unit area of the container wall is constant.
b. What is the effect on the average kinetic energy of the molecules?
As the temperature increases, so does the average kinetic energy. Looking at the impact of increased temperature on kinetic energy, as T doubles kinetic energy doubles.
[latex]\mathrm{KE}_{\text {avg}}=\frac{3}{2}\mathrm{RT}[/latex]
Answer: Since T2 is 2 times T1, the average kinetic energy doubles.
c. What is the effect on the root mean square speed of the molecules?
Looking at the impact of increased temperature on root mean square speed, as T doubles root mean square speed increase by √2 times.
[latex]\begin{gathered}\mathrm{U}_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M(x)}}\\\mathrm{U}_{\mathrm{rms}}=\sqrt{\frac{3 R T_1}{M(x)}}\text {or}\mathrm{U}_{\mathrm{rms}}=\sqrt{\frac{3 R\left(2 T_1\right)}{M(x)}}\end{gathered}[/latex]
Answer: The root mean square speed increases since the temperature has increased; it increases √2 times its initial value. Urms is proportional to KEavg.
Guided Solution
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The guided solution below will give you the reasoning for each step to get your answer, with reminders and hints.
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| Guided Solution Ideas |
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This question is a theory type problem that tests your knowledge on the kinetic molecular theory. In this problem, the temperature and volume of a gas is increased, so you must use your knowledge on the relationships of this chapter to predict the kinetics of the gas.
Show/Hide Resource
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A 1 L sample of carbon monoxide, CO, initially at 0°C and 1 atm, is heated to 546 K and its volume is increased to 2 L.
Show/Hide Think About This!As the temperature of the gas increases from 273 K to 546 K, the volume of the gas increases from 1 L to 2 L. |
Recall your relationships.
Show/Hide Don’t Forget!
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Recall the equations that demonstrate these relationships.
Show/Hide Don’t Forget![latex]\begin{aligned} &\mathrm{E}_{\mathrm{k}}=\frac{3}{2}\mathrm{RT}\\ & \mathrm{PV}=\mathrm{nRT}\\ &\mathrm{U}_{\mathrm{rms}}=\sqrt{\frac{3\mathrm{RT}}{\mathrm{M}(\mathrm{x})}} \end{aligned}[/latex] |
What is “root mean square speed” and what equation does it belong to?
Show/Hide Don’t Forget!The root mean square speed of a particle (Urms) is defined as the square root of the average of the squares of the velocities with n = the number of particles. [latex]\begin{gathered} \mathrm{U}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}(\mathrm{x})}} \end{gathered}[/latex] |
| Complete Solution |
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| a. What effect do these changes have on the number of collisions of the molecules of the gas per unit area of the container wall?.
[latex]\begin{gathered} \frac{\mathrm{P}_1\mathrm{V}_1}{\mathrm{T}_1}=\frac{\mathrm{P}_2\mathrm{V}_2}{\mathrm{T}_2}\\ \mathrm{T}_2=2\mathrm{T}_1 \\ \mathrm{V}_2=2\mathrm{V}_2 \end{gathered}[/latex] Substitute in to the equation for changing conditions: [latex]\begin{gathered} \frac{\mathrm{P}_1\mathrm{V}_1}{\mathrm{T}_1}=\frac{\mathrm{P}_2 2\mathrm{V}_1}{2\mathrm{T}_1} \end{gathered}[/latex] Cancel terms that are the same: [latex]\begin{gathered} \frac{\mathrm{P}_1\mathrm{V}_1}{\mathrm{T}_1}=\frac{\mathrm{P}_2 2\mathrm{V}_1}{2\mathrm{T}_1} \end{gathered}[/latex] Left with P1 = P2, no change in pressure. The pressure is the number of collisions per unit of area; if the pressure is constant, so are the number of collisions. Answer: Since pressure does not change, the number of collisions per unit area of the container wall is constant. |
| b. What is the effect on the average kinetic energy of the molecules?
[latex]\begin{gathered} \mathrm{KE}_{\mathrm{avg}}=\frac{3}{2} \mathrm{RT} \end{gathered}[/latex] As the temperature increases, so does the average kinetic energy. Answer: Since T2 is 2 times T1, the average kinetic energy doubles. |
| c. What is the effect on the root mean square speed of the molecules?
[latex]\begin{gathered} \mathrm{U}_{\mathrm{rms}}=\sqrt{\frac{3\mathrm{RT}}{\mathrm{M}(\mathrm{x})}}\\ \mathrm{U}_{\mathrm{rms}}=\sqrt{\frac{3\mathrm{RT}_1}{\mathrm{M}(\mathrm{x})}}\text{ or }\mathrm{U}_{\mathrm{rms}}=\sqrt{\frac{3\mathrm{R}\left(2\mathrm{T}_1\right)}{\mathrm{M}(\mathrm{x})}} \end{gathered}[/latex] If the temperature increases and the number of particles stays constant, the speed will increase. Answer: The root mean square speed increases since the temperature has increased; it increases √2 times its initial value. Urms is proportional to KEavg. |
Check Your Work
Summary of what we would expect based on the related chemistry theory.
Show/Hide Watch Out!
Make sure you look at the impact of increasing temperature and volume on the parameters you are asked about in the question.
To evaluate the impact on collisions, you must first think about impact on pressure using the appropriate form of the general gas law for changing conditions.
Since T increases, we expect KEavg and Urms to increase.
Does your answer make chemical sense?
Show/Hide Answer
If one factor causes an increase in pressure while the other causes a decrease in the pressure, they will counteract one another. In this case, each increased by a factor of 2, which cancels out. So, the pressure will stay constant. If factors cause the speed to increase and nothing causes the speed to decrease, the particles will move faster. This can be seen by the equations below.
This equation shows that kinetic energy and temperature are proportional:
[latex]\begin{gathered} \mathrm{E}_{\mathrm{k}}=\frac{3}{2} \mathrm{RT} \end{gathered}[/latex]
This equation shows that the root mean square speed is dependent on temperature:
[latex]\begin{gathered} \mathrm{U}_{\mathrm{rms}}=\sqrt{\frac{3\mathrm{RT}}{M(x)}} \end{gathered}[/latex]
PASS Attribution
- LibreTexts PASS Chemistry Book CHEM 1510/1520 (2).
- Question 2.E.45 from LibreTexts PASS Chemistry Book CHEM 1510/1520 (3) is used under a CC BY-NC-SA 4.0 license.
- Question 2.E.45 was adapted from question 9.5.6 from LibreTexts Chemistry 1e (OpenSTAX) (4), which is under a CC BY 4.0 license.
- Question 9.5.6 is question 95 from OpenStax Chemistry (5), which is under a CC BY 4.0 license. Access for free at https://openstax.org/books/chemistry/pages/1-introduction.
Reference List
1. OpenStax. 2.5: The Kinetic-Molecular Theory. In TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520); LibreTexts, 2023. https://chem.libretexts.org/Courses/Thompson_Rivers_University/TRU%3A_Fundamentals_and_Principles_of_Chemistry_(CHEM_1510_and_CHEM_1520)/02%3A_Gases/2.05%3A_The_Kinetic-Molecular_Theory.
2. Blackstock, L.; Brewer, S.; Jensen, A. PASS Chemistry Book CHEM 1510/1520; LibreTexts, 2023. https://chem.libretexts.org/Courses/Thompson_Rivers_University/PASS_Chemistry_Book_CHEM_1510_1520.
3. Blackstock, L.; Brewer, S.; Jensen, A. 2.2: 2.4: PASS Ideal Gases- Use the Kinetic Molecular Theory to Explain Impact on a Gas as V and T Increased (2.E.45) In PASS Chemistry Book CHEM 1510/1520; LibreTexts, 2024. https://chem.libretexts.org/Courses/Thompson_Rivers_University/PASS_Chemistry_Book_CHEM_1510_1520/02%3A_Gases/2.04%3A_2.4_PASS_Ideal_Gases-_use_the_Kinetic_Molecular_Theory_to_explain_impact_on_a_gas_as_V_and_T_increased_(2.E.45).
4. OpenStax. 9.E: Gases (Exercises). In Chemistry 1e (OpenSTAX); LibreTexts, 2023. https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/09%3A_Gases/9.E%3A_Gases_(Exercises).
5. Flowers, P.; Robinson, W. R.; Langley, R.; Theopold, K. Ch. 9 Exercises. In Chemistry 2e; OpenStax, 2019. https://openstax.org/books/chemistry-2e/pages/9-exercises.
