A Tale Of Two Gases Answer Key

Introduction: The Story Behind “A Tale of Two Gases”

The phrase “A Tale of Two Gases” instantly brings to mind a classic classroom problem that compares the behavior of two different gases under identical conditions. Think about it: this article serves as a complete answer key for that problem, guiding learners through every step of the solution while reinforcing the underlying concepts of the Ideal Gas Law, Dalton’s Law of Partial Pressures, and kinetic molecular theory. That said, students are asked to predict, calculate, and explain how variables such as pressure, volume, temperature, and molar mass influence each gas’s performance. By the end, you’ll not only have the final numbers but also a deeper intuition about why the two gases behave the way they do.

Problem Statement (Typical Version)

Two sealed containers, A and B, each hold 1.00 L of gas at 298 K.

• Container A contains oxygen (O₂), while container B contains helium (He).
  Still, > - Both gases are at the same initial pressure of 1. Day to day, 00 atm. > - The containers are then heated to 398 K while the volume remains constant.
  On the flip side, >
  Tasks

1. Calculate the final pressure in each container.
2. Because of that, determine the ratio of the average kinetic energies of the gas molecules after heating. > 3. Because of that, explain why the pressure change is the same for both gases even though their molar masses differ. Consider this: > 4. If a mixture of the two gases were placed in a single 2.Because of that, 00 L container at 298 K and 1. 00 atm total pressure, find the partial pressures of each gas.

1. Final Pressure After Heating (Using the Ideal Gas Law)

The Ideal Gas Law, (PV = nRT), can be rearranged for a constant‑volume process:

[ \frac{P_2}{P_1} = \frac{T_2}{T_1} ]

Because the number of moles ((n)) and the volume ((V)) stay the same, pressure is directly proportional to temperature (in Kelvin).

[ P_2 = P_1 \times \frac{T_2}{T_1}=1.00;\text{atm}\times\frac{398;\text{K}}{298;\text{K}}=1.34;\text{atm} ]

Answer: Both containers reach 1.34 atm after heating. The gas identity does not matter for this calculation because the Ideal Gas Law treats all ideal gases identically.

2. Ratio of Average Kinetic Energies

The average translational kinetic energy of a molecule in an ideal gas is given by:

[ \overline{KE} = \frac{3}{2}k_{\mathrm{B}}T ]

where (k_{\mathrm{B}}) is Boltzmann’s constant. Notice that temperature alone determines the average kinetic energy, independent of the gas’s mass.

Thus, after heating to 398 K, the average kinetic energies of O₂ and He are equal:

[ \frac{\overline{KE}{\text{O}2}}{\overline{KE}{\text{He}}}= \frac{3/2,k{\mathrm{B}}T}{3/2,k_{\mathrm{B}}T}=1 ]

Answer: The ratio is 1 : 1; both gases have the same average kinetic energy at the same temperature. That's the whole idea.

3. Why the Pressure Change Is Identical

Even though oxygen molecules are about 16 u each (32 u for O₂) and helium atoms are only 4 u, pressure depends on how often molecules strike the container walls, not on their mass. In a constant‑volume, constant‑mole scenario:

• Number of collisions per unit time is proportional to the average speed of the molecules.
• Average speed ((v_{\text{rms}})) varies with (\sqrt{T/m}). When temperature rises, both gases speed up, but the heavier O₂ molecules increase their speed less than He atoms.
• On the flip side, pressure also incorporates the momentum change per collision, which is proportional to mass × velocity. For ideal gases, the product of mass and the square of velocity yields a term proportional to temperature, canceling out the mass dependence.

Mathematically, the pressure derived from kinetic theory is:

[ P = \frac{1}{3}\frac{N}{V}m\overline{v^{2}} ]

Substituting (\overline{v^{2}} = \frac{3k_{\mathrm{B}}T}{m}) gives:

[ P = \frac{N}{V}k_{\mathrm{B}}T ]

All mass terms disappear, leaving pressure solely a function of temperature, number density, and universal constants. Hence, both gases experience the same pressure increase when heated at constant volume.

4. Partial Pressures in a Mixed‑Gas Container

When two ideal gases share a container, Dalton’s Law of Partial Pressures applies:

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[ P_{\text{total}} = P_{\text{O}2} + P{\text{He}} ]

First, determine the number of moles of each gas initially present in the 1.00 L containers:

[ n = \frac{PV}{RT} ]

Using (P = 1.00;\text{atm}), (V = 1.00;\text{L}), (T = 298;\text{K}), and (R = 0.

[ n = \frac{(1.00)(1.00)}{0.08206 \times 298}=0.0409;\text{mol} ]

Thus, each gas contributes 0.0409 mol. When combined into a 2.

[ n_{\text{total}} = 0.0409;\text{mol (O₂)} + 0.0409;\text{mol (He)} = 0.

Now apply the Ideal Gas Law again to find the total pressure in the 2.00 L vessel (still at 298 K):

[ P_{\text{total}} = \frac{n_{\text{total}}RT}{V_{\text{new}}} = \frac{0.08206 \times 298}{2.0818 \times 0.00} \approx 1.

Since the gases do not react, each retains its original mole fraction:

[ X_{\text{O}2}=X{\text{He}} = \frac{0.0409}{0.0818}=0.50 ]

Partial pressures are simply the mole fraction multiplied by the total pressure:

[ P_{\text{O}2}=X{\text{O}2},P{\text{total}}=0.50\times1.00;\text{atm}=0.50;\text{atm} ] [ P_{\text{He}}=X_{\text{He}},P_{\text{total}}=0.50\times1.00;\text{atm}=0.50;\text{atm} ]

Answer: In the 2.00 L mixture, oxygen exerts 0.50 atm and helium exerts 0.50 atm, summing to the original 1.00 atm total pressure.

5. Extending the Tale: Real‑World Connections

5.1. Why Engineers Care About Partial Pressures

In aerospace and diving industries, the partial pressure of oxygen dictates whether a breathing mixture is safe. The calculations above mirror the design of nitrox blends used by recreational divers, where a higher O₂ fraction reduces nitrogen absorption but must stay below toxic thresholds.

5.2. Atmospheric Science Insight

Our atmosphere is a natural “tale of two gases” (and many more). At sea level, nitrogen (N₂) and oxygen dominate, yet trace gases like helium, despite their minuscule concentrations, influence thermal conductivity and diffusion rates—properties that hinge on the kinetic arguments discussed earlier.

5.3. Laboratory Applications

When calibrating gas‑sensing equipment, technicians often use standard mixtures of known partial pressures. Understanding that pressure changes are temperature‑driven, not mass‑driven, ensures accurate sensor calibration across temperature ranges.

6. Frequently Asked Questions (FAQ)

Q1: Does the ideal‑gas assumption hold for real gases like O₂ at 1 atm?
Answer: At moderate pressures (≤ 1 atm) and room temperature, O₂ behaves nearly ideally, with deviations < 2 %. For high‑precision work, one would apply the Van der Waals equation, but the qualitative conclusions of this problem remain unchanged.

Q2: If the volume were allowed to change, would the pressure still be the same for both gases?
Answer: No. In a constant‑pressure heating scenario, the volume change depends on the molar mass through the gas’s compressibility factor. Heavier gases expand slightly less for a given temperature increase, leading to different final volumes.

Q3: How would adding a third gas, like nitrogen, affect the partial‑pressure calculation?
Answer: The total pressure would still be the sum of the three partial pressures. Each gas’s partial pressure equals its mole fraction times the total pressure. The presence of a third component simply reduces each individual mole fraction proportionally.

Q4: Can we use the same approach for liquids?
Answer: No. Liquids are incompressible on the scale of typical temperature changes, and their pressure–temperature relationship follows different thermodynamic equations (e.g., the Clausius‑Clapeyron relation for phase changes).

7. Key Takeaways

• Pressure in a sealed, constant‑volume container depends only on temperature and the number of moles, not on the identity or mass of the gas.
• Average kinetic energy is a function of temperature alone, giving different gases the same energy per molecule at equal temperatures.
• Partial pressures follow Dalton’s law; each component contributes proportionally to its mole fraction.
• The Ideal Gas Law provides a reliable framework for solving introductory‑level problems, while kinetic‑molecular theory explains why the math works.

By mastering these concepts, students can confidently tackle a wide range of thermodynamic questions—from classroom exercises to real‑world engineering challenges. The “tale of two gases” thus becomes more than a textbook problem; it turns into a foundational story that illustrates the elegance and universality of gas behavior.