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Air at normal temperature and pressure contained in a closed tank adheres to the continuum hypothesis. Yet when sufficient air has been drawn from the tank, the hypothesis no longer applies to the remaining air. Why?

Short Answer

Expert verified
As air is removed, the mean free path increases. When it becomes comparable to the tank size, the continuum hypothesis no longer applies.

Step by step solution

01

- Understand the Continuum Hypothesis

The continuum hypothesis assumes that a fluid's properties (like density, temperature, and velocity) are continuously distributed and can be described by smooth functions, without considering the molecular nature of matter.
02

- Determine Conditions for Continuum Hypothesis Validity

The hypothesis is valid when the mean free path of the molecules (average distance a molecule travels before colliding with another molecule) is much smaller than the characteristic length of the system being studied (e.g., the size of the tank).
03

- Analyze the Effect of Drawing Air from the Tank

As air is drawn from the tank, the number of air molecules decreases, which increases the mean free path since there are fewer molecules to collide with.
04

- Relate Mean Free Path to Continuum Hypothesis Breakdown

When the mean free path becomes comparable to or larger than the characteristic length of the tank, the air can no longer be considered a continuous medium. This is because individual molecular movements and collisions become significant, and the continuum hypothesis breaks down.

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Key Concepts

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

Mean Free Path
The mean free path is the average distance a molecule travels before colliding with another molecule. This concept is crucial in understanding the behavior of gases.

The mean free path is influenced by factors like pressure and temperature.
  • High pressure means more molecules in a given space, leading to a shorter mean free path.
  • Lower pressure means fewer molecules, resulting in a longer mean free path.

As air is drawn from a tank, the pressure inside decreases. This increases the mean free path since there are fewer molecules to interact with each other. When the mean free path becomes significant compared to the size of the tank, the behavior of the air no longer follows the continuum hypothesis.
Fluid Properties
Fluid properties refer to characteristics like density, temperature, and velocity that define a fluid's state and behavior. These properties are crucial in fluid dynamics and thermodynamics.
  • Density: the mass per unit volume of a fluid. In a continuum, density is considered to be a continuous function.
  • Temperature: a measure of the average kinetic energy of molecules in the fluid. Temperature influences the pressure and, consequently, the mean free path.
  • Velocity: the speed and direction of fluid flow. Fluids can have velocity fields that vary over space and time.

The continuum hypothesis simplifies analysis by treating these properties as if they change smoothly, without accounting for individual molecular interactions. This simplification holds until the mean free path becomes significant compared to the characteristic length of the system.
Molecular Nature of Matter
The molecular nature of matter acknowledges that fluids are composed of many individual molecules. Each molecule moves and interacts based on physical laws.

In a gas, these molecules are far apart and move randomly, colliding with each other and the walls of the container.

This kinetic activity defines properties like pressure and temperature.
  • Pressure results from molecules colliding with the walls of the container.
  • Temperature is related to the average kinetic energy of the molecules.

The continuum hypothesis overlooks these individual molecular actions by assuming a smooth distribution of properties. However, as the mean free path grows due to reduced air, individual molecular behaviors become significant, leading to the breakdown of the continuum approximation.
Characteristic Length
The characteristic length is a scale parameter that defines the size of a system or a relevant feature within the system. This could be the diameter of a tank, the length of a pipe, or any dimension that is significant to the physical behavior of the fluid.

In the context of the continuum hypothesis, the characteristic length serves as a benchmark to compare with the mean free path. When the mean free path is much smaller than the characteristic length, the continuum assumption holds.

However, if the mean free path becomes comparable to or exceeds the characteristic length, the continuum hypothesis fails because molecular-level details become crucial.

In summary, maintaining the continuum hypothesis requires that the characteristic length remains significantly larger than the mean free path, ensuring a smooth distribution of fluid properties.

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Most popular questions from this chapter

A closed system consisting of \(5 \mathrm{~kg}\) of a gas undergoes a process during which the relationship between pressure and specific volume is \(p v^{1.3}=\) constant. The process begins with \(p_{1}=1\) bar, \(v_{1}=0.2 \mathrm{~m}^{3} / \mathrm{kg}\) and ends with \(p_{2}=0.25\) bar. Determine the final volume, in \(\mathrm{m}^{3}\), and plot the process on a graph of pressure versus specific volume.

A gas initially at \(p_{1}=1\) bar and occupying a volume of 1 liter is compressed within a piston-cylinder assembly to a final pressure \(p_{2}=4\) bar. (a) If the relationship between pressure and volume during the compression is \(p V=\) constant, determine the volume, in liters, at a pressure of 3 bar. Also plot the overall process on a graph of pressure versus volume. (b) Repeat for a linear pressure-volume relationship between the same end states.

One thousand \(\mathrm{kg}\) of natural gas at 100 bar and \(255 \mathrm{~K}\) is stored in a tank. If the pressure, \(p\), specific volume, \(v\), and temperature, \(T\), of the gas are related by the following expression $$ p=\left[\left(5.18 \times 10^{-3}\right) T /(v-0.002668)\right]-\left(8.91 \times 10^{-3}\right) / v^{2} $$ where \(v\) is in \(\mathrm{m}^{3} / \mathrm{kg}, T\) is in \(\mathrm{K}\), and \(p\) is in bar, determine the volume of the tank in \(\mathrm{m}^{3}\). Also, plot pressure versus specific volume for the isotherms \(T=250 \mathrm{~K}, 500 \mathrm{~K}\), and \(1000 \mathrm{~K}\).

A closed system consists of \(0.5 \mathrm{kmol}\) of liquid water and occupies a volume of \(4 \times 10^{-3} \mathrm{~m}^{3}\). Determine the weight of the system, in \(\mathrm{N}\), and the average density, in \(\mathrm{kg} / \mathrm{m}^{3}\), at a location where the acceleration of gravity is \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\).

The variation of pressure within the biosphere affects not only living things but also systems such as aircraft and undersea exploration vehicles. (a) Plot the variation of atmospheric pressure, in atm, versus elevation \(z\) above sea level, in \(\mathrm{km}\), ranging from 0 to \(10 \mathrm{~km}\). Assume that the specific volume of the atmosphere, in \(\mathrm{m}^{3} / \mathrm{kg}\), varies with the local pressure \(p\), in \(\mathrm{kPa}\), according to \(v=72.435 / p\). (b) Plot the variation of pressure, in atm, versus depth \(z\) below sea level, in \(\mathrm{km}\), ranging from 0 to \(2 \mathrm{~km}\). Assume that the specific volume of seawater is constant, \(v=0.956 \times\) \(10^{-3} \mathrm{~m}^{3} / \mathrm{kg}\) In each case, \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\) and the pressure at sea level is \(1 \mathrm{~atm}\)

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