Ideal Gases: General Gas Law, Temperature and Volume Calculation
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Question
Your instructor blows up a balloon so that it is filled with 506 mL of helium at 22.00°C and 1.00 atm pressure. What volume will the balloon have at 4.00°C?
Show/Hide Answer
Volume = 475 mL
Refer to Section 2.3 Relating Pressure, Volume, Amount, and Temperature – The Ideal Gas Law (1).
Strategy Map
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Check out the strategy map.
Show/Hide Strategy Map
| Strategy Map Steps |
|---|
| 1. Identify the information given in the question and what conditions are changing. |
| 2. Identify what conditions are staying the same. |
3. Recall how the changing variables are related.
Show/Hide HintThe ideal gas constant (R) does not change, so consider the impact of decreasing the temperature of the balloon: [latex]\begin{gathered} \mathrm{R}=\frac{\mathrm{PV}}{\mathrm{nT}} \end{gathered}[/latex] |
4. Simplify the general gas law to focus on your changing conditions.
Show/Hide Hint[latex]\begin{gathered} \frac{\mathrm{P}_\mathrm{i} \mathrm{V}_\mathrm{i}} {\mathrm{n}_\mathrm{i} \mathrm{T}_\mathrm{i}}=\frac{\mathrm{P}_\mathrm{f} \mathrm{V}_\mathrm{f}}{\mathrm{n}_\mathrm{f} \mathrm{T}_\mathrm{f}} \end{gathered}[/latex] Cancel terms that remain constant and use the provided data to calculate the new volume. |
Solution
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Check out this solution.
Show/Hide Solution
Constant n and P; T is decreased
[latex]\begin{aligned} \frac{\mathrm{P}_\mathrm{i} \mathrm{V}_\mathrm{i}} {\mathrm{n}_\mathrm{i} \mathrm{T}_\mathrm{i}}&=\frac{\mathrm{P}_\mathrm{f} \mathrm{V}_\mathrm{f}}{\mathrm{n}_\mathrm{f} \mathrm{T}_\mathrm{f}}\\ \frac{\mathrm{V}_\mathrm{i}}{\mathrm{T}_\mathrm{i}}&=\frac{\mathrm{V}_\mathrm{f}}{\mathrm{T}_\mathrm{f}}(\text{constant }\mathrm{n},\mathrm{P})\\ \mathrm{V}_\mathrm{f}&=\frac{\mathrm{V}_\mathrm{i} \mathrm{T}_\mathrm{f}}{\mathrm{T}_\mathrm{i}} \end{aligned}[/latex]
[latex]\begin{aligned} \mathrm{V}_\mathrm{i}&=506\mathrm{~mL}\\ \mathrm{T}_\mathrm{i}&=22.00^{\circ}\mathrm{C}+273.15=295.15\mathrm{~K}\\ \mathrm{T}_\mathrm{f}&=4.00^{\circ}\mathrm{C}+273.15=277.15\mathrm{~K} \end{aligned}[/latex]
[latex]\begin{gathered} \mathrm{V}_\mathrm{f}=\frac{(506 \mathrm{~mL})(277.15 \mathrm{~K})}{(295.15 \mathrm{~K})}=475 \mathrm{~mL} \end{gathered}[/latex]
Answer: Volume = 475 mL
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.
Show/Hide Guided Solution
| Guided Solution Ideas |
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This question is a qualitative problem where we use the ideal gas law to predict how the balloon volume will be impacted by decreasing the temperature at constant pressure.
Show/Hide Resource
|
| Your instructor blows up a balloon so that it is filled with 506 mL of helium at 22.00°C and 1.00 atm pressure. What volume will the balloon have at 4.00°C?
The question asks you to calculate the new balloon volume when temperature is decreased. Show/Hide Think About This!The question tells us that the pressure and amount of helium are constant. |
Gas molecules collide with the container walls, which exerts force on the walls of the container and results in pressure. When the temperature is decreased, the gas molecules will have less kinetic energy, so they move slower and collide with the balloon walls with less force and less often. Because of this we expect the volume to decrease.
Show/Hide Watch Out!If we were working with a fixed volume container, the pressure would decrease when the temperature decreases. |
Since this is a changing conditions problem, we use the general form of the ideal gas law.
Show/Hide Don’t Forget![latex]\begin{gathered} \frac{\mathrm{P}_\mathrm{i} \mathrm{V}_\mathrm{i}}{\mathrm{n}_\mathrm{i} \mathrm{T}_\mathrm{i}}=\frac{\mathrm{P}_\mathrm{f} \mathrm{V}_\mathrm{f}}{\mathrm{n}_\mathrm{f} \mathrm{T}_\mathrm{f}} \end{gathered}[/latex] Show/Hide Think About This!What conditions are constant? What conditions are changing? Once you recognize these, you can cancel values that stay constant during this problem. In this case, the pressure and temperature remain constant; this means we can remove them since they are the same on both sides. Constant: n and P T is decreased [latex]\begin{gathered} \frac{\mathrm{P}_\mathrm{i} \mathrm{V}_\mathrm{i}} {\mathrm{n}_\mathrm{i} \mathrm{T}_\mathrm{i}}=\frac{\mathrm{P}_\mathrm{f} \mathrm{V}_\mathrm{f}}{\mathrm{n}_\mathrm{f} \mathrm{T}_\mathrm{f}} \end{gathered}[/latex] constant n and P, Pi = Pf, and ni = nf [latex]\begin{aligned} \frac{\mathrm{V}_\mathrm{i}}{\mathrm{T}_\mathrm{i}}&=\frac{\mathrm{V}_\mathrm{f}}{\mathrm{T}_\mathrm{f}}(\text{constant } \mathrm{n}, \mathrm{P})\\ \mathrm{V}_\mathrm{f}&=\frac{\mathrm{V}_\mathrm{i} \mathrm{T}_\mathrm{f}}{\mathrm{T}_\mathrm{i}} \end{aligned}[/latex] |
Think about your units. What conversions are required?
Show/Hide Don’t Forget!Temperatures must always be absolute (convert to Kelvin scale) when dealing with ideal gas law calculations. This scale connects an absolute zero of temperature (0 K) to a situation where molecules have no kinetic energy and cannot collide with container walls because they are not moving (leading to zero volume). |
| Complete Solution |
|---|
| Since this is a changing conditions ideal gas problem, we can start with the general form of the ideal gas law:
[latex]\begin{aligned} \mathrm{R}&=\frac{\mathrm{PV}}{\mathrm{nT}}\\ \frac{\mathrm{P}_\mathrm{i} \mathrm{V}_\mathrm{i}} {\mathrm{n}_\mathrm{i} \mathrm{T}_\mathrm{i}}&=\frac{\mathrm{P}_\mathrm{f} \mathrm{V}_\mathrm{f}}{\mathrm{n}_\mathrm{f} \mathrm{T}_\mathrm{f}} \end{aligned}[/latex] What is given? Vi = 506 mL n is constant |
| Simplify and rearrange to get the formula you need to solve for your unknown:
Pi = Pf, ni = nf [latex]\begin{aligned} \frac{\mathrm{V}_\mathrm{i}}{\mathrm{T}_\mathrm{i}}&=\frac{\mathrm{V}_\mathrm{f}}{\mathrm{T}_\mathrm{f}}(\text{constant } \mathrm{n}, \mathrm{P})\\ \mathrm{V}_\mathrm{f}&=\frac{\mathrm{V}_\mathrm{i} \mathrm{T}_\mathrm{f}}{\mathrm{T}_\mathrm{i}} \end{aligned}[/latex] |
| Since we are using temperature in a calculation, the units must be in Kelvin:
[latex]\begin{aligned} \mathrm{V}_\mathrm{i}&=506\mathrm{~mL}, \text{our calculated }\mathrm{V}_\mathrm{f}\text{ will also be in mL}\\ \mathrm{T}_\mathrm{i}&=22.00^{\circ}\mathrm{C}+273.15=295.15\mathrm{~K}\\ \mathrm{T}_\mathrm{f}&=4.00^{\circ}\mathrm{C}+273.15=277.15\mathrm{~K} \end{aligned}[/latex] |
| To review your understanding, try this quiz:
If you are using a printed copy, you can scan the QR code with your digital device to go directly to the h5p:
|
| Plug in your values with units to your formula and solve for Vf.
[latex]\begin{aligned} \mathrm{V}_\mathrm{f}&=\frac{\mathrm{V}_\mathrm{i} \mathrm{T}_\mathrm{f}}{\mathrm{T}_\mathrm{i}}\\ \mathrm{V}_\mathrm{f}&=\frac{(506 \mathrm{~mL})(277.15 \mathrm{~K})}{(295.15 \mathrm{~K})}=475 \mathrm{~mL} \end{aligned}[/latex] Answer: Volume = 475 mL |
Check Your Work
Make sure you are using the correct form of the ideal gas law for changing conditions. Our question was a quantitative question where we are asked to calculate the new balloon volume when the temperature is decreased.
Show/Hide Check Your Work!
Make sure you remember to convert the temperatures to Kelvin.
Does your answer make chemical sense?
Show/Hide Answer
The temperature of the ideal gas in the balloon was decreased while the amount of gas pressure remained constant. When temperature decreases, the gas moves slower since it has less kinetic energy, resulting in fewer and less energetic collisions with the walls over the same time period. Because of this, the volume of the balloon decreases.
Attribution
- Question 2.E.27 is from LibreTexts TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520) (2) is used under a CC BY-NC-SA 4.0 license.
References
1. OpenStax. 2.3: Relating Pressure, Volume, Amount, and Temperature – The Ideal Gas Law. In TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520); LibreTexts, 2022. https://chem.libretexts.org/Courses/Thompson_Rivers_University/TRU%3A_Fundamentals_and_Principles_of_Chemistry_(CHEM_1510_and_CHEM_1520)/02%3A_Gases/2.03%3A_Relating_Pressure_Volume_Amount_and_Temperature_-_The_Ideal_Gas_Law.
2. OpenStax. 2.E: Gases (Exercises). In TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520); LibreTexts, 2025. https://chem.libretexts.org/Courses/Thompson_Rivers_University/TRU%3A_Fundamentals_and_Principles_of_Chemistry_(CHEM_1510_and_CHEM_1520)/02%3A_Gases/2.E%3A_Gases_(Exercises).
