Ideal Gases: General Gas Law, Temperature and Volume Calculation (almost ready for review, July 9, 2025)
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 |
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| 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 HintR, the ideal gas constant does not change, so consider the impact of decreasing the temperature of the balloon: [latex]\mathrm{R}=\frac{\mathrm{PV}}{\mathrm{nT}}[/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.
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Solution
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Check out this solution.
Show/Hide Solution
Constant n and P, T is decreased [remove italics in formulas and text below please, SB July 9]
[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]
[latex]\begin{gathered} \frac{V_i}{T_i}=\frac{V_f}{T_f}(\text { constant } n, P) \end{gathered}[/latex]
[latex]V_f=\frac{V_i T_f}{T_i}[/latex]
Vi = 506 mL
Ti = 22.00oC + 273.15 = 295.15 K
Tf = 4.00oC + 273.15 = 277.15 K
[latex]\mathrm{V}_f=\frac{(506 \mathrm{~mL})(277.15 \mathrm{~K})}{(295.15 \mathrm{~K})}=475 \mathrm{~mL}[/latex]
Answer: Volume = 475 mL
Guided Solution
Do you want more help?
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
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| Your Instructor blows up a balloon so that it is filled with 506 mL of Helium at 22.00oC and 1.00 atm pressure. What volume will the balloon have at 4.00oC?
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 exerting force on the walls of the container which results in pressure. When the temperature is decreased the gas molecules will have less kinetic energy so are moving slower, and collide with the balloon walls with less force and less often, so we expect the volume to decrease.
Show/Hide Watch Out!If we were working with a fixed volume container, pressure would decrease when 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 [remove italics in formulas and text below please, SB July 9] [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{gathered} \frac{V_i}{T_i}=\frac{V_f}{T_f}(\text { constant } n, P) \end{gathered}[/latex]
[latex]V_f=\frac{V_i T_f}{T_i}[/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, as 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 [remove italics in formulas and text below please, SB July 9] |
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| Since this is a changing conditions ideal gas problem, we can start with the general form of the ideal gas law
[latex]\mathrm{R}=\frac{\mathrm{PV}}{\mathrm{nT}}[/latex] [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] 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{gathered} \frac{V_i}{T_i}=\frac{V_f}{T_f}(\text { constant } n, P) \end{gathered}[/latex] [latex]V_f=\frac{V_i T_f}{T_i}[/latex]
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| Since using Temperature in a calculation units must be Kelvin:
Vi = 506 mL, our calculate Vf will also be in mL [latex]\mathrm{V}_f=\frac{(506 \mathrm{~mL})(277.15 \mathrm{~K})}{(295.15 \mathrm{~K})}=475 \mathrm{~mL}[/latex] Answer: Volume = 475 mL [latex]\begin{gathered} \frac{V_i}{T_i}=\frac{V_f}{T_f}(\text { constant } T, P) \end{gathered}[/latex] In order for the relationship to be valid, Vf must be less than Vi
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| For an ideal gas volume is proportional to absolute temperature (in Kelvin) when amount and pressure are constant. |
| Answer: Volume is predicted to: decrease
Explain your reasoning in clear and complete statements: Temperature changes impact the average kinetic energy of gas molecules. Lowering the temperature of a gas means molecules have less kinetic energy and move more slowly, which means collisions of the ideal gas molecules with the walls of the container happen with less force and less often in a given period of time. Because the pressure of the balloon and the amount of gas remain constant, when the temperature is lowered, the volume of the balloon should decrease. If this same example was done with a fixed volume container, the pressure inside the container would decrease as the volume would be constant.
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We have enough information to check our prediction by solving for Vf in terms of Vi.
Show/Hide Watch Out!
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| To review your understanding, try this quiz:
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Check Your Work
Make sure you are using the correct form of the ideal gas law for changing conditions. Our question was a qualitative thinking question where we are asked to explain our answer by relating it to our theory. You can use the provided data to check your prediction by solving for Vf in terms of Vi, but you also need to explain your reasoning in sentences.
Show/Hide Don’t Forget!
If you wish to use the question data to check your prediction make sure you remember to convert temperatures to Kelvin. [remove italics in formulas and text below please, and make font size consistent in formulas, SB July 9]
[latex]\begin{gathered} \frac{V_i}{T_i}=\frac{V_f}{T_f}(\text { constant } n, P) \end{gathered}[/latex]
[latex]\mathrm{V}_f=\frac{\mathrm{V}_i \mathrm{T}_f}{\mathrm{T}_i}[/latex]
Ti = 22.0 + 273.15 = 295.15 K
Tf = -10.0 + 373.15 = 263.15 K
[latex]\begin{gathered} \mathrm{V}_f=\mathrm{V}_i\left(\frac{263.15 \mathrm{~K}}{295.15 \mathrm{~K}}\right) \\ \mathrm{V}_f=0.892 \mathrm{~V}_i \end{gathered}[/latex] therefore Vf < Vi
Does your answer make chemical sense?
Show/Hide Answer
The temperature of the ideal gas was decreased and we were told the pressure remained constant. When temperature decreases the gas moves more slowly since it has less kinetic energy, so there are fewer and less energetic collisions with the walls over the same time period. The volume of the balloon decreases so that the pressure is constant.
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.
