A Tale of Two Gases: The Ultimate Answer Key
Introduction
When chemistry teachers hand out worksheets titled “A Tale of Two Gases,” students often stare at the equations and feel a chill run down their spines. But the worksheet is designed to test understanding of gas behavior, ideal gas law, partial pressures, and the kinetic molecular theory. This answer key breaks down every problem step by step, explains the underlying concepts, and offers tips for remembering the key facts. Whether you’re a student looking for guidance or a teacher wanting a ready‑made resource, this complete walkthrough will illuminate every corner of the worksheet.
1. Problem 1 – Ideal Gas Law Basics
Problem Statement
A 2.00 L container holds 0.500 mol of an ideal gas at 298 K. What is the pressure inside the container?
Step‑by‑Step Solution
-
Identify the known variables.
- Volume, (V = 2.00,\text{L})
- Moles, (n = 0.500,\text{mol})
- Temperature, (T = 298,\text{K})
-
Recall the ideal gas law.
[ PV = nRT ] where (R = 0.08206,\text{L·atm·K}^{-1}\text{·mol}^{-1}). -
Rearrange for pressure.
[ P = \frac{nRT}{V} ] -
Plug in the numbers.
[ P = \frac{(0.500)(0.08206)(298)}{2.00} ] -
Calculate.
[ P \approx \frac{12.226}{2.00} = 6.11,\text{atm} ]
Answer: (6.11,\text{atm})
Key Takeaway
The ideal gas law is a quick way to link pressure, volume, temperature, and the amount of gas. Remember the units: (R) must be in the same volume and pressure units you’re using.
2. Problem 2 – Partial Pressures in a Mixture
Problem Statement
A mixture of 2 L of nitrogen and 3 L of oxygen at 1 atm total pressure and 300 K is mixed in a 5 L container. What are the partial pressures of N₂ and O₂ after mixing?
Step‑by‑Step Solution
-
Determine the mole fractions.
Since (PV = nRT), the number of moles of each gas is proportional to its volume at the same (P) and (T).
[ n_{\text{N₂}} \propto 2,\text{L}, \quad n_{\text{O₂}} \propto 3,\text{L} ] -
Calculate mole fraction of each gas.
[ X_{\text{N₂}} = \frac{2}{2+3} = 0.400, \quad X_{\text{O₂}} = \frac{3}{5} = 0.600 ] -
Use Dalton’s Law.
Partial pressure (P_i = X_i P_{\text{total}}).
[ P_{\text{N₂}} = 0.400 \times 1,\text{atm} = 0.400,\text{atm} ] [ P_{\text{O₂}} = 0.600 \times 1,\text{atm} = 0.600,\text{atm} ]
Answer: (P_{\text{N₂}} = 0.400,\text{atm}); (P_{\text{O₂}} = 0.600,\text{atm})
Key Takeaway
When gases mix, the total pressure remains the same if the volume and temperature are constant. Partial pressures are simply the product of the mole fraction and the total pressure.
3. Problem 3 – Temperature Change and Pressure
Problem Statement
A sealed 3.00 L container holds 1.00 mol of an ideal gas at 300 K and 1.00 atm. If the temperature is raised to 400 K, what is the new pressure?
Step‑by‑Step Solution
-
Use the combined gas law.
[ \frac{P_1}{T_1} = \frac{P_2}{T_2} ] -
Solve for (P_2).
[ P_2 = P_1 \times \frac{T_2}{T_1} = 1.00,\text{atm} \times \frac{400}{300} ] -
Calculate.
[ P_2 = 1.33,\text{atm} ]
Answer: (1.33,\text{atm})
Key Takeaway
Temperature and pressure are directly proportional when volume and moles are constant. A 33 % increase in temperature yields a 33 % increase in pressure And that's really what it comes down to..
4. Problem 4 – Density of a Gas
Problem Statement
What is the density of carbon dioxide (CO₂) at STP (0 °C, 1 atm)?
Step‑by‑Step Solution
-
Find the molar mass of CO₂.
C = 12.01 g/mol, O = 16.00 g/mol.
[ M = 12.01 + 2(16.00) = 44.01,\text{g/mol} ] -
Use the ideal gas law to find number of moles in 1 L.
At STP, (n = \frac{PV}{RT}).
[ n = \frac{(1,\text{atm})(1,\text{L})}{(0.08206)(273)} = 0.0446,\text{mol} ] -
Calculate density.
[ \rho = \frac{nM}{V} = \frac{0.0446 \times 44.01}{1} \approx 1.97,\text{g/L} ]
Answer: (1.97,\text{g/L})
Key Takeaway
Density of a gas at a given temperature and pressure can be derived from its molar mass and the ideal gas law. CO₂ is much denser than air, which explains why it can accumulate in low‑lying areas.
5. Problem 5 – Kinetic Molecular Theory
Problem Statement
Explain why the kinetic energy of gas molecules increases with temperature.
Answer
According to the kinetic molecular theory, gas molecules are in constant, random motion. Temperature is a measure of the average kinetic energy of these molecules. When heat is added to a gas, the energy is transferred to the molecules, causing them to vibrate and collide more vigorously. That said, the average kinetic energy, ( \frac{3}{2}kT ), is directly proportional to temperature. Which means, as temperature rises, so does the speed and energy of the gas molecules Most people skip this — try not to. Turns out it matters..
Key Takeaway
Temperature is a direct indicator of molecular motion; higher temperature = higher kinetic energy That alone is useful..
6. Problem 6 – Real Gas Deviations
Problem Statement
Why does nitrogen deviate from ideal behavior at high pressure?
Answer
Nitrogen molecules experience intermolecular attractions (Van der Waals forces) that are not accounted for in the ideal gas law. This causes the actual pressure to be lower than predicted by the ideal gas law. At high pressures, molecules are forced closer together, increasing the significance of these attractive forces and reducing the effective volume they occupy. Additionally, the finite size of nitrogen molecules becomes a factor, further deviating from ideal behavior Easy to understand, harder to ignore..
Key Takeaway
Real gases deviate from ideality when intermolecular forces or molecular volumes become significant, especially under high pressure or low temperature conditions Simple as that..
7. Problem 7 – Boyle’s Law Application
Problem Statement
A 1.50 L sample of helium at 1.00 atm is compressed to 0.750 L at the same temperature. What is the new pressure?
Step‑by‑Step Solution
-
Apply Boyle’s Law.
[ P_1V_1 = P_2V_2 ] -
Solve for (P_2).
[ P_2 = \frac{P_1V_1}{V_2} = \frac{(1.00)(1.50)}{0.750} = 2.00,\text{atm} ]
Answer: (2.00,\text{atm})
Key Takeaway
Halving the volume of a gas at constant temperature doubles its pressure. Boyle’s Law is a simple yet powerful tool for quick calculations Surprisingly effective..
8. Problem 8 – Charles’s Law
Problem Statement
A 2.00 L sample of argon is heated from 25 °C to 75 °C at constant pressure. What is its new volume?
Step‑by‑Step Solution
-
Convert temperatures to Kelvin.
(T_1 = 298,\text{K}), (T_2 = 348,\text{K}). -
Use Charles’s Law.
[ \frac{V_1}{T_1} = \frac{V_2}{T_2} ] -
Solve for (V_2).
[ V_2 = V_1 \times \frac{T_2}{T_1} = 2.00 \times \frac{348}{298} \approx 2.34,\text{L} ]
Answer: (2.34,\text{L})
Key Takeaway
Temperature changes at constant pressure cause proportional changes in volume. The ratio of volume to temperature stays constant.
9. Problem 9 – Avogadro’s Number and Moles
Problem Statement
How many molecules are in 3.00 mol of hydrogen gas (H₂)?
Step‑by‑Step Solution
-
Use Avogadro’s number.
(N_A = 6.022 \times 10^{23},\text{molecules/mol}). -
Multiply.
[ N = nN_A = 3.00 \times 6.022 \times 10^{23} = 1.81 \times 10^{24},\text{molecules} ]
Answer: (1.81 \times 10^{24}) molecules
Key Takeaway
Moles provide a bridge between the microscopic world of molecules and the macroscopic world of measurable amounts Not complicated — just consistent..
10. Problem 10 – Real‑World Application: Gas Transfer
Problem Statement
A 10 L tank contains 0.250 mol of methane at 298 K. If the tank is vented entirely to the atmosphere (1 atm), how many moles of methane escape?
Step‑by‑Step Solution
-
Determine initial pressure.
[ P_{\text{initial}} = \frac{nRT}{V} = \frac{(0.250)(0.08206)(298)}{10} \approx 0.612,\text{atm} ] -
Since the tank is vented, all methane escapes.
The final amount of methane in the tank is zero. -
Moles that escape = initial moles.
[ 0.250,\text{mol} ]
Answer: (0.250,\text{mol})
Key Takeaway
When a gas is released into the atmosphere, the quantity of gas that escapes equals the initial amount present, regardless of the pressure difference.
FAQ
| Question | Answer |
|---|---|
| What is the ideal gas law? | (PV = nRT) – a relationship between pressure, volume, temperature, and amount of gas. Even so, |
| **When does the ideal gas law fail? ** | Under high pressure or low temperature where intermolecular forces and finite molecular size matter. Worth adding: |
| **How do you remember R’s value? ** | Think of “R = 0.Think about it: 08206 L·atm·K⁻¹·mol⁻¹” – the first two digits are 0. Think about it: 08, a handy cue. |
| What is Dalton’s Law? | The total pressure of a gas mixture equals the sum of the partial pressures of its components. In real terms, |
| **Why do gases expand when heated? ** | Increased kinetic energy causes molecules to move faster, colliding more often and pushing against the container walls. |
Conclusion
“A Tale of Two Gases” is more than a worksheet; it’s a gateway to mastering the behavior of gases in all their forms. But by dissecting each problem, we’ve reinforced core principles—ideal gas law, partial pressures, temperature effects, and real‑gas deviations—while providing practical strategies for solving common questions. Keep this key handy, revisit the concepts regularly, and soon the mysteries of gas behavior will feel as natural as breathing itself.
