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The average distance traveled by a molecule between successive collisions is called mean free path. For a given amount of a gas, how does the mean free path of a gas depend on (a) density, (b) temperature at constant volume, \((\mathrm{c})\) pressure at constant temperature, \((\mathrm{d})\) volume at constant temperature, and (e) size of the atoms?

Short Answer

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(a) decreases with increasing density; (b) decreases with increasing temperature at constant volume; (c) decreases with increasing pressure at constant temperature; (d) increases with increasing volume at constant temperature; (e) decreases with larger atom size.

Step by step solution

01

Understanding Mean Free Path

The mean free path \( \lambda \) is the average distance a molecule travels between collisions. It is given by the formula: \( \lambda = \frac{kT}{\sqrt{2} \pi d^2 P} \), where \( k \) is Boltzmann's constant, \( T \) is the temperature, \( d \) is the diameter of the molecules, and \( P \) is the pressure.
02

Effect of Density

Density \( \rho \) is related to pressure \( P \) and volume \( V \) by \( \rho = \frac{mP}{kT} \), where \( m \) is the mass of the gas molecules. Increasing density increases pressure due to more molecules per volume. Since \( \lambda \propto \frac{1}{P} \), the mean free path decreases with increasing density.
03

Effect of Temperature at Constant Volume

At constant volume, increasing temperature causes particles to move faster, resulting in higher pressure, \( P \propto T \). Since \( \lambda \propto \frac{1}{P} \), increases in temperature at constant volume decrease the mean free path.
04

Effect of Pressure at Constant Temperature

When pressure is increased while keeping temperature constant, it results in more molecules in a given volume (increased density). As such, the mean free path \( \lambda \), which is inversely related to pressure, decreases.
05

Effect of Volume at Constant Temperature

At constant temperature, increasing volume reduces the pressure and density \( (P \propto 1/V) \). Thus, because \( \lambda \propto \frac{1}{P} \), increasing the volume increases the mean free path.
06

Effect of Size of Atoms

The mean free path \( \lambda \) is inversely proportional to the square of the diameter \( d \) of the molecules (\( \lambda \propto \frac{1}{d^2} \)). Thus, increasing the size of the atoms reduces the mean free path.

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Key Concepts

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

Density Effect on Mean Free Path
Density plays a crucial role in determining the mean free path of gas molecules. The mean free path is the average distance a molecule can travel without colliding with another molecule.
  • Higher density means more molecules are packed into a given volume.
  • This results in a higher probability of collisions occurring.
  • Thus, as density increases, the mean free path shortens.
In mathematical terms, mean free path is inversely proportional to density, which means: \( \lambda \propto \frac{1}{\rho} \) This relationship indicates that the more tightly packed the gas molecules are, the less distance they can travel freely.
Temperature and Mean Free Path
Doing a little mental stretching, let's consider how temperature affects mean free path when volume is held constant. As temperature rises, molecules move faster.
  • Faster-moving molecules bump into each other more often.
  • This results in increased pressure in the room available since speed creates momentum.
  • At constant volume, increasing temperature leads to a decrease in the mean free path.
The mean free path is inversely proportional to this increased pressure:\( \lambda \propto \frac{1}{P}\)So, warmer environments, with energy-packed molecules, have molecules traveling shorter distances before nudging their neighbors.
Pressure and Mean Free Path
Next, let’s look at what happens to mean free path when pressure changes while temperature remains steady. Pressure directly influences the mean free path.
  • An increase in pressure means more gas molecules are squeezed into a confined space.
  • This increased concentration leads to more frequent collisions.
  • As pressure rises (while keeping temperature the same), the mean free path decreases.
Thus, we can relate mean free path inversely to pressure:\( \lambda \propto \frac{1}{P}\)In essence, when more molecules share the same space, their opportunities to journey freely diminish.
Volume and Mean Free Path
Imagine keeping a constant temperature but playing around with the volume of the container. Here’s how volume influences the mean free path:
  • Expanding the volume reduces density and the collisions that occur within the space.
  • Hence, the mean distance a molecule travels between collisions—the mean free path—grows.
  • Conversely, compressing the volume clumps molecules together, shrinking the mean free path.
Mathematically, if pressure is inversely related to volume \( (P \propto 1/V) \), the mean free path is proportional to volume:\( \lambda \propto V\)So in larger spaces, molecules enjoy a longer cruise before their next bump.
Molecular Size and Mean Free Path
Finally, consider how the size of molecules affects the mean free path. The mean free path is not only about speed and space; it's also about size.
  • Larger molecules have bigger diameters, increasing their chance of colliding with others.
  • This naturally shortens the mean free path.
  • In a gas mix with molecules of varied sizes, bigger molecules will generally have shorter mean free paths.
This is mathematically described by:\( \lambda \propto \frac{1}{d^2}\)So, tiny molecules have the advantage of more wiggle room, translating to a lengthier mean free path.

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Most popular questions from this chapter

A student tries to determine the volume of a bulb like the one shown in Figure \(10.12 .\) These are her results: mass of the bulb filled with dry air at \(23^{\circ} \mathrm{C}\) and \(744 \mathrm{mmHg}=91.6843 \mathrm{~g} ;\) mass of evacuated bulb \(=\) \(91.4715 \mathrm{~g}\). Assume the composition of air is 78 percent \(\mathrm{N}_{2}, 21\) percent \(\mathrm{O}_{2},\) and 1 percent argon by volume. What is the volume (in mL) of the bulb? (Hint: First calculate the average molar mass of air, as shown in Problem \(3.129 .)\)

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