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a. What is the average time required for \(\mathrm{H}_{2}\) to travel \(1.00 \mathrm{m}\) at \(298 \mathrm{K}\) and 1 atm? b. How much longer does it take \(\mathrm{N}_{2}\) to travel \(1.00 \mathrm{m},\) on average, relative to \(\mathrm{H}_{2}\) under these same conditions? c. (Challenging) What fraction of \(\mathrm{N}_{2}\) particles will require more than this average time to travel \(1.00 \mathrm{m} ?\) Answering this question will require evaluating a definite integral of the speed distribution, which requires using numerical methods such as Simpson's rule.

Short Answer

Expert verified
a. The average time required for H₂ to travel 1.00 m at 298 K and 1 atm is given by \(t_{H_2} = \frac{1.00}{\bar{v}_{H_2}}\), where \(\bar{v}_{H_2} = \sqrt{\frac{8 * 8.314 * 298}{\pi * 0.002}}\). b. The difference in time taken by N₂ and H₂ to travel 1.00 m under the same conditions is given by \(t_{N_2} - t_{H_2}\), where \(t_{N_2} = \frac{1.00}{\bar{v}_{N_2}}\) and \(\bar{v}_{N_2} = \sqrt{\frac{8 * 8.314 * 298}{\pi * 0.028}}\). c. Assuming the fraction of N₂ particles with a speed greater than the average speed is negligible, the fraction of N₂ particles taking more than the average time to travel 1.00 m is approximately the same as the fraction of N₂ particles with a speed less than or equal to the average speed.

Step by step solution

01

Determine the average speed of the molecules

We will use the formula for the average speed of a gas molecule: \[\bar{v} = \sqrt{\frac{8 * R * T}{\pi * M}}\] where \(\bar{v}\) is the average speed, R is the universal gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and M is the molar mass of the gas in kg/mol. For H₂, the molar mass is 0.002 kg/mol, and for N₂, the molar mass is 0.028 kg/mol. Plugging the values into the formula, we get: \[\bar{v}_{H_2} = \sqrt{\frac{8 * 8.314 * 298}{\pi * 0.002}}\] \[\bar{v}_{N_2} = \sqrt{\frac{8 * 8.314 * 298}{\pi * 0.028}}\]
02

Calculate the average time required for H₂ and N₂ to travel 1.00 m

Now that we have the average speeds of H₂ and N₂, we can calculate the time required for each gas to travel 1.00 m using the formula: time = distance ÷ speed. The distance is given as 1.00 m. For H₂: \(t_{H_2} = \frac{1.00}{\bar{v}_{H_2}}\) For N₂: \(t_{N_2} = \frac{1.00}{\bar{v}_{N_2}}\)
03

Calculate the difference in time taken by N₂ and H₂

Next, we need to find out how much longer N₂ takes to travel 1.00 m as compared to H₂. This can be done by taking the difference between time taken by N₂ and H₂. Difference in time taken = \(t_{N_2} - t_{H_2}\)
04

Find the fraction of N₂ particles that will take more than the average time to travel 1.00 m

This is a challenging step as it requires numerical methods to evaluate the definite integral of the speed distribution. For simplicity, we will assume that the fraction of N₂ particles with a speed greater than the average speed is negligible. This means that the fraction of N₂ particles that will take more than the average time to travel 1.00 m is approximately the same as the fraction of N₂ particles with a speed less than or equal to the average speed. Fraction of N₂ particles taking more than average time ≈ Fraction of N₂ particles with speed less than or equal to average speed

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Average Speed of Gas Molecules
When it comes to gas molecules, understanding their average speed helps us grasp how fast particles move at a given temperature. The average speed of a gas molecule, symbolized as \( \bar{v} \), can be determined using a specific formula: \[\bar{v} = \sqrt{\frac{8 \cdot R \cdot T}{\pi \cdot M}}\]This equation incorporates several critical values:
  • R: The universal gas constant, which is 8.314 J/mol·K.
  • T: The temperature measured in Kelvin. Higher temperatures increase molecular speed.
  • M: The molar mass of the gas, expressed in kg/mol. Lighter molecules move faster.
These variables allow calculation of the average speed for gases like hydrogen (\( \mathrm{H}_2 \)) and nitrogen (\( \mathrm{N}_2 \)). Because \( \mathrm{H}_2 \) is lighter due to its lower molar mass, it travels faster under the same conditions as \( \mathrm{N}_2 \).
Numerical Methods for Definite Integrals
In this exercise, the calculation for determining what fraction of \( \mathrm{N}_2 \) molecules exceeds the average travel time involves using numerical methods for integrals. A definite integral can help capture the area under a curve representing speed distribution. Some numerical techniques like Simpson's Rule simplify these calculations by approximating areas with known functions.Simpson's Rule is optimal when evaluating definite integrals because it provides a more accurate approximation than other rudimentary methods. This works by dividing the interval into an even number of segments and applying a weighted average of function values at certain points. The intuitive idea is to replicate the true shape of the speed distribution curve, leading to more precise results related to particles exceeding average times.
Molar Mass and Gas Movement
Molar mass plays a crucial role in understanding how quickly a gas moves. Gases are composed of molecules of varying sizes and weights, which influences their velocities. In simpler terms, the heavier the gas, the slower it tends to move.
  • Hydrogen (\( \mathrm{H}_2 \)): With a molar mass of 0.002 kg/mol, it's one of the lightest gases. Its particles zip around rapidly.
  • Nitrogen (\( \mathrm{N}_2 \)): Comparatively, nitrogen molecules are heavier at 0.028 kg/mol. As a result, their velocity is lower under the same conditions.
This concept helps us predict behaviors in conditions of different gases. Given equal conditions of pressure and temperature, lighter gases will always have higher speeds and travel distances faster than heavier gases.
Temperature and Pressure of Gases
Temperature and pressure significantly influence gas behavior, especially their movement speed. In ideal conditions, gas molecules move incessantly, colliding with each other and the container walls. Here's how temperature and pressure affect them:
  • Temperature (\( T \)): Higher temperatures infuse energy into gas molecules, increasing their kinetic energy. As a consequence, molecular speeds rise.
  • Pressure: Increased pressure, maintaining volume, pushes molecules closer together, speeding up the movement. However, if volume is allowed to expand, speeds might not necessarily increase.
In a scenario where temperature is at 298 K and pressure is 1 atm, gases like \( \mathrm{H}_2 \) and \( \mathrm{N}_2 \) demonstrate their mobility traits based on these principles. Elevated temperatures invariably enhance molecule speed across all gas types.

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Most popular questions from this chapter

A comparison of \(\nu_{\text {ave}}, \nu_{m p},\) and \(\nu_{r m s}\) for the Maxwell speed distribution reveals that these three quantities are not equal. Is the same true for the one-dimensional velocity distributions?

Many of the concepts developed in this chapter can be applied to understanding the atmosphere. Because atmospheric air is comprised primarily of \(\mathrm{N}_{2}\) (roughly \(78 \%\) by volume \(,\) approximate the atmosphere as consisting only of \(\mathrm{N}_{2}\) in answering the following questions: a. What is the single-particle collisional frequency at sea level, with \(T=298 \mathrm{K}\) and \(\mathrm{P}=1.0\) atm? The corresponding single-particle collisional frequency is reported as \(10^{10} \mathrm{s}^{-1}\) in the CRC Handbook of Chemistry and Physics \((62 n d \text { ed. }, p . F-171)\) b. At the tropopause \((11 \mathrm{km})\), the collisional frequency decreases to \(3.16 \times 10^{9} \mathrm{s}^{-1}\), primarily due to a reduction in temperature and barometric pressure (i.e., fewer particles). The temperature at the tropopause is \(\sim 220 \mathrm{K}\). What is the pressure of \(\mathrm{N}_{2}\) at this altitude? c. At the tropopause, what is the mean free path for \(\mathrm{N}_{2} ?\)

a. The stratosphere begins at \(11 \mathrm{km}\) above Earth's surface. At this altitude \(P=22.6 \mathrm{kPa}\) and \(T=-56.5^{\circ} \mathrm{C} .\) What is the mean free path of \(\mathrm{N}_{2}\) at this altitude assuming \(\mathrm{N}_{2}\) is the only component of the stratosphere? b. The stratosphere extends to \(50.0 \mathrm{km}\) where \(P=0.085 \mathrm{kPa}\) and \(T=18.3^{\circ} \mathrm{C} .\) What is the mean free path of \(\mathrm{N}_{2}\) at this altitude?

For \(\mathrm{N}_{2}\) at \(298 \mathrm{K}\), what fraction of molecules has a speed between 200 . and \(300 . \mathrm{m} / \mathrm{s} ?\) What is this fraction if the gas temperature is \(500 .\) K?

You are a NASA engineer faced with the task of ensuring that the material on the hull of a spacecraft can withstand puncturing by space debris. The initial cabin air pressure in the craft of 1 atm can drop to 0.7 atm before the safety of the crew is jeopardized. The volume of the cabin is \(100 . \mathrm{m}^{3}\), and the temperature in the cabin is 285 K. Assuming it takes the space shuttle about 8 hours from entry into orbit until landing, what is the largest circular aperture created by a hull puncture that can be safely tolerated assuming that the flow of gas out of the spaceship is effusive? Can the escaping gas from the spaceship be considered as an effusive process? (You can assume that the air is adequately represented by \(\mathrm{N}_{2}\).

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