Ideal Gases: Calculate the Density of an Ideal Gas
Question
What is the density of laughing gas, dinitrogen monoxide (N2O), at a temperature of 325K and a pressure of 113.0 kPa?
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1.84 g/L
Refer to Section 2.4: Stoichiometry of Gaseous Substances, Mixtures, and Reactions (1).
Strategy Map
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| Strategy Map Steps |
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| 1. Identify what information the question gives you and do any necessary conversions. |
| 2. There are 2 approaches you could use:
a) To do this in 1 step, you can derive an equation that contains all known variables and the unknown density by manipulating the ideal gas equation to obtain an expression for density. Show/Hide Hint
b) To do this in 2 steps, first, use the ideal gas equation to solve for volume. Then, use the density equation to solve for density using the calculated volume. Show/Hide Hint[latex]\begin{aligned} \mathrm{V} & =\frac{\mathrm{n R T}}{\mathrm{P}} \\\mathrm{d} & =\frac{\mathrm{m}}{\mathrm{V}} \end{aligned}[/latex] |
3. Plug data into the equation and solve for your density.
Show/Hide HintMake sure to write your units in for all values. |
Solution
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[latex]\begin{gathered} 113.0\mathrm{kPa}=\frac{1\mathrm{~atm}}{101.325\mathrm{ kPa}}=1.115\mathrm{~atm} \end{gathered}[/latex]
Approach a) using a derived equation:
[latex]\begin{gathered} \mathrm{d}=\frac{\text { mass }}{\text { volume }}=\frac{\mathrm{m}}{\mathrm{v}}\\ \mathrm{ PV }=\mathrm{nRT}\\ \mathrm{n}=\frac{\text{ mass }}{\text{ molar mass }}=\frac{\mathrm{m}}{\mathrm{M}}\\ \mathrm{PV}=\frac{\mathrm{m}}{\mathrm{M}}\mathrm{RT}\\ \mathrm{d}=\frac{\mathrm{m}}{\mathrm{V}}=\frac{\mathrm{PM}}{\mathrm{RT}}\\ \mathrm{d}=\frac{(1.115\mathrm{~atm})(44.013\mathrm{~g}/\mathrm{mol})}{(0.08206\mathrm{~atm}\cdot\mathrm{L}/\mathrm{mol}\cdot\mathrm{k})(325\mathrm{k})}\\\mathrm{d}=\mathbf{1.84\mathrm{ g }/\mathrm{ L }} \end{gathered}[/latex]
Approach b) using 2 steps:
Step 1:
[latex]\begin{gathered} \mathrm{PV}=\mathrm{nRT}\\ \mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{P}}\\ \mathrm{V}=\frac{(1.000\mathrm{~mol})(0.08206\mathrm{~atm}\cdot\mathrm{L}/\mathrm{mol}\cdot\mathrm{k})(325\mathrm{k})}{1.115\mathrm{~atm}}\\ \mathrm{~V}=23.9\mathrm{~L}\\ \end{gathered}[/latex]
Step 2:
[latex]\begin{gathered} \mathrm{d}=\frac{\mathrm{m}}{\mathrm{V}}\\\mathrm{d}=\frac{44.013\mathrm{~g}}{23.9\mathrm{~L}}\\\mathbf{d}=\mathbf{1.84}\mathbf{g/L} \end{gathered}[/latex]
Answer: 1.84 g/L
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 calculation type problem where you are required to calculate the density of a given compound using 1 of the 2 possible methods.
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Looking at whats given in the question: “What is the density of laughing gas, dinitrogen monoxide (N2O), at a temperature of 325K and a pressure of 113.0 kPa?”
Show/Hide Think About This!We are given the temperature in Kelvin, the pressure in kilopascals and the chemical formula of dinitrogen monoxide. From this information, we are trying to calculate the density of the dinitrogen monoxide. |
Are these the correct units?
Show/Hide Think About This!You will need to convert from kilopascals to atmospheres. The conversion factor is: 101.325kPa = 1 atm. |
Recall how to determine density from mass and volume.
Show/Hide Don’t Forget![latex]\begin{gathered} \mathrm{d}=\frac{\mathrm{m}}{\mathrm{V}} \end{gathered}[/latex] |
How to manipulate the equation to derive a new one:
Show/Hide Think About This!We know the equation for density is: [latex]\begin{gathered} \mathrm{d}=\frac{\text{ mass }}{\text{ volume }}=\frac{\mathrm{m}}{\mathrm{v}} \end{gathered}[/latex] We know the ideal gas equation is: [latex]\mathrm{PV}=\mathrm{nRT}[/latex] We know that: [latex]\begin{gathered} \mathrm{n}=\frac{\text { mass }}{\text { molar mass }}=\frac{\mathrm{m}}{\mathrm{M}} \end{gathered}[/latex] Therefore, we can substitute [latex]\frac{\mathrm{m}}{\mathrm{M}}[/latex] in place of n which gives… [latex]\begin{gathered} \mathrm{PV}=\frac{\mathrm{m}}{\mathrm{M}}\mathrm{RT} \end{gathered}[/latex] Now, we can manipulate the equation to be equal to [latex]\frac{\mathrm{m}}{\mathrm{V}}[/latex], which is equal to density (d) and gives: [latex]\begin{gathered} \mathbf{d}=\frac{\mathbf{m}}{\mathbf{V}}=\frac{\mathbf{P M}}{\mathbf{R T}} \end{gathered}[/latex] |
| OR use the equations you already know:
Using the ideal gas equation, plug in the information you are given to solve for volume. THEN, using the density equation, plug in your calculated volume. What about the mass? Show/Hide Don’t Forget!Recall that you can find the mass by finding the molar mass of the compound. Look at the chemical formula. |
| Complete Solution |
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[latex]\begin{gathered} 113.0\mathrm{kPa}=\frac{1\mathrm{~atm}}{101.325\mathrm{kPa}}=1.115\mathrm{~atm} \end{gathered}[/latex] Approach a) using a derived equation: Deriving the equation: We know the equation for density is: [latex]\begin{gathered} \mathrm{d}=\frac{\text { mass }}{\text { volume }}=\frac{\mathrm{m}}{\mathrm{v}} \end{gathered}[/latex] We know the ideal gas equation is: [latex]\mathrm{PV}=\mathrm{nRT}[/latex] We know that: [latex]\begin{gathered} \mathrm{n}=\frac{\text{ mass }}{\text{ molar mass }}=\frac{\mathrm{m}}{\mathrm{M}} \end{gathered}[/latex] Therefore, we can substitute [latex]\frac{\mathrm{m}}{\mathrm{M}}[/latex] in place of n, which gives [latex]\mathrm{PV}=\frac{\mathrm{m}}{\mathrm{M}}{\mathrm{RT}}[/latex] Now, we can manipulate the equation to be equal to [latex]\frac{\mathrm{m}}{\mathrm{V}}[/latex], which is equal to density (d) and gives [latex]\mathbf{d}=\frac{\mathbf{m}}{\mathbf{V}}=\frac{\mathbf{P M}}{\mathbf{R T}}[/latex] Plugging into the equation: [latex]\begin{gathered} \mathrm{d}=\frac{\mathrm{PM}}{\mathrm{RT}}\\ \mathrm{M}=2(14.007)+15.999=44.013\mathrm{~g}/\mathrm{mol}\\ \mathrm{d}=\frac{(1.115\mathrm{~atm})(44.013\mathrm{~g}/\mathrm{mol})}{(0.08206\mathrm{~atm}\cdot\mathrm{L}/\mathrm{mol}\cdot\mathrm{k})(325\mathrm{k})}\\ \mathbf{d}=\mathbf{1.84}\mathbf{g/L} \end{gathered}[/latex] Answer: 1.84 g/L Show/Hide Don’t Forget!How to find M (molar mass) of N2O: Look up element masses on the periodic table. Then, add them up, making sure you use molecule stoichiometry. N = 14.007g/mol × 2 Molar mass = 2(14.007) + 15.999 = 44.013 g/mol |
| Approach b) using 2 steps:
1. Calculate the volume. Plug known information into the ideal gas equation: [latex]\mathrm{PV}=\mathrm{nRT}[/latex] [latex]\begin{gathered} \mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{P}}\\ \mathrm{V}=\frac{(1.000\mathrm{~mol})(0.08206\mathrm{~atm}\cdot\mathrm{L}/\mathrm{mol}\cdot \mathrm{k})(325\mathrm{k})}{1.115\mathrm{~atm}}\\ \mathrm{~V}=23.9\mathrm{~L} \end{gathered}[/latex] 2. Calculate the density. Recall the density equation: [latex]\begin{gathered} \mathrm{d}=\frac{\mathrm{m}}{\mathrm{V}} \end{gathered}[/latex] Plug the volume you calculated in the previous step and the molar mass of N2O into the density equation: [latex]\begin{aligned} &\mathrm{d}=\frac{44.013\mathrm{g}}{23.9\mathrm{L}}\\ &\mathrm{d}=1.84\mathrm{g/L} \end{aligned}[/latex] Answer: 1.84 g/L |
Check Your Work
Summary of what we would expect based on the related chemistry theory.
Show/Hide Watch Out!
Make sure your units cancel out when you plug them into your equation and that your final result is in the appropriate units for density.
Using a dimensional analysis approach and writing units beside values in equations let’s you see if the units cancel out to give you a result in appropriate units. Our gas density should have units of g/L.
Does your answer make chemical sense?
Show/Hide Answer
Density is the amount of matter that is packed into a given volume (mass divided by volume). It is a physical property that varies between materials. Gas density is the amount of matter in each volume at a specific temperature and pressure; therefore, it makes sense that we can calculate the density of dinitrogen monoxide gas in these specific conditions. It does not matter which stage you include the conditions in your calculation if they are used to achieve your final solution.
PASS Attribution
- LibreTexts PASS Chemistry Book CHEM 1510/1520 (2).
- Question 2.E.28 from LibreTexts PASS Chemistry Book CHEM 1510/1520 (3) is used under a CC BY-NC-SA 4.0 license.
References
1. OpenStax. 2.4: Stoichiometry of Gaseous Substances, Mixtures, and Reactions. 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/2.04%3A_Stoichiometry_of_Gaseous_Substances_Mixtures_and_Reactions.
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: PASS Ideal Gases – Calculate the Density of an Ideal Gas (2.E.28). 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/2.02%3A_2.2_PASS_Ideal_Gases_-_Calculate_the_density_of_an_ideal_gas_(2.E.28).
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.
