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Show that the mean value of the relative speed of two molecules in a Maxwellian gas is \(\sqrt{2}\) times the mean speed of a molecule with respect to the walls of the container.

Short Answer

Expert verified
The mean relative speed of two molecules in a Maxwellian gas is indeed \(\sqrt{2}\) times the mean speed of a molecule with respect to the walls of the container.

Step by step solution

01

Understanding the Concept

The Maxwell-Boltzmann Distribution describes the distribution of speeds for particles in a gas. For an ideal gas, the mean speed of a molecule is given by: \(u_{mean} = \sqrt{\frac{8kT}{\pi m}}\), where \(k\) is the Boltzmann constant, \(T\) is the absolute temperature and \(m\) is the mass of a molecule.
02

Mean Relative Speed

The mean relative speed of two gas molecules is given by: \(v_{mean} = \sqrt{2} u_{mean}\), where \(v_{mean}\) is the mean relative speed and \(u_{mean}\) is the mean speed of a molecule. This can be derived by considering two random molecules with speeds \(u_1\) and \(u_2\) from the Maxwell-Boltzmann Distribution.
03

Formulate the Relation

Now, it can be seen from Step 2 that the mean relative speed of the two molecules is exactly \(\sqrt{2}\) times the mean speed of a molecule, \(v_{mean} = \sqrt{2} u_{mean}\).

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Key Concepts

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

Mean Speed
The mean speed of a molecule in a gas is a concept rooted in the kinetic theory of gases, particularly under the Maxwell-Boltzmann Distribution. This distribution is a statistical means of describing speeds of particles within a gas, based on temperature and molecular mass. The formula for mean speed is:
  • \(u_{mean} = \sqrt{\frac{8kT}{\pi m}}\)
In this equation, \(k\) represents Boltzmann's constant, \(T\) is the absolute temperature, and \(m\) is the molecular mass. This mean speed is crucial because it represents the average kinetic energy of the particles in the gas.Breaking it down further:
  • Boltzmann's constant \(k\) connects macroscopic and microscopic physics, allowing us to track how energy relates to temperature.
  • Absolute temperature \(T\) ensures that calculations consider the total thermal energy per molecule.
  • The molecular mass \(m\) tells us how much energy each particle individually carries at a given speed.
Mean speed is a central concept in understanding gas dynamics as it partially governs gas properties like pressure and temperature.
Relative Speed
Relative speed in the context of a gas considers how two molecules move in opposition to each other. This differs from regular speed, which considers a molecule's movement concerning the walls of a container. For a Maxwell-Boltzmann gas, the mean relative speed is described as:
  • \(v_{mean} = \sqrt{2} u_{mean}\)
This expression highlights that the mean relative speed is \(\sqrt{2}\) times that of the mean speed of a single molecule with respect to the container walls. This relation arises from the vector nature of speed. When two molecules collide, their relative speed is a result of their combined speeds in different directions. By contemplating random speeds \(u_1\) and \(u_2\) from the distribution, you can derive this enhanced average of relative motion:
  • This concept underscores the efficiency of molecule interactions, affecting how energy is dissipation during collisions.
  • Knowing the relative speed helps in understanding reaction rates and viscosity in gases.
Ideal Gas Law
The Ideal Gas Law establishes a relationship between pressure, volume, and temperature in an ideal gas. While not directly involved in calculating mean or relative speed, it underscores the environment where these concepts apply:
  • \(PV = nRT\)
In this formula, \(P\) stands for pressure, \(V\) for volume, \(n\) for the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature.Key points to understand:
  • The Ideal Gas Law contextualizes how changes in temperature or pressure affect gas behavior, where all molecules are assumed to have negligible volume and no intermolecular forces.
  • It helps link macroscopic measurements (like pressure and volume) with microscopic concepts like mean and relative speeds that dictate molecule dynamics.
  • Even though real gases exhibit anomalies due to intermolecular forces, at low pressures and high temperatures, real gases approximate ideal gas behavior generally well.
The Ideal Gas Law maintains order and predictability in calculations, making it indispensable when considering gas dynamics.

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Most popular questions from this chapter

(a) Show that, if the temperature is uniform, the pressure of a classical gas in a uniform gravitational field decreases with height according to the barometric formula $$ P(z)=P(0) \exp \\{-m g z / k T\\}, $$ where the various symbols have their usual meanings. \({ }^{17}\) (b) Derive the corresponding formula for an adiabatic atmosphere, that is, the one in which \((P V \gamma)\), rather than \((P V)\), stays constant. Also study the variation, with height, of the temperature \(T\) and the density \(n\) in such an atmosphere.

Refer to Section \(6.2\) and show that, if the occupation number \(n_{x}\) of an energy level \(\varepsilon\) is restricted to the values \(0,1, \ldots, l\), then the mean occupation number of that level is given by $$ \left\langle n_{\Sigma}\right\rangle=\frac{1}{z^{-1} e^{\beta x}-1}-\frac{l+1}{\left(z^{-1} e^{\beta r}\right)^{l+1}-1} . $$ Check that while \(l=1\) leads to \(\left\langle n_{\varepsilon}\right\rangle_{\text {F.D. }}, l \rightarrow \infty\) leads to \(\left\langle n_{e}\right\rangle_{\text {B.E. }}\).

(a) Considering the loss of translational energy suffered by the molecules of a gas on reflection from a receding wall, derive, for a quasistatic adiabatic expansion of an ideal nonrelativistic gas, the well-known relation $$ P V^{\gamma}=\text { const. } $$ where \(\gamma=(3 a+2) / 3 a, a\) being the ratio of the total energy to the translational energy of the gas. (b) Show that, in the case of an extreme relativistic gas, \(\gamma=(3 a+1) / 3 a\).

Consider the effusion of molecules of a Maxwellian gas through an opening of area \(a\) in the walls of a vessel of volume \(V\). (a) Show that, while the molecules inside the vessel have a mean kinetic energy \(\frac{3}{2} k T\), the effused ones have a mean kinetic energy \(2 k T, T\) being the quasistatic equilibrium temperature of the gas. (b) Assuming that the effusion is so slow that the gas inside is always in a state of quasistatic equilibrium, determine the manner in which the density, the temperature, and the pressure of the gas vary with time.

(a) Show that the momentum distribution of particles in a relativistic Boltzmannian gas, with \(\varepsilon=c\left(p^{2}+m_{0}^{2} c^{2}\right)^{1 / 2}\), is given by $$ f(\boldsymbol{p}) d \boldsymbol{p}=\operatorname{Ce}^{-\beta c\left(p^{2}+m_{0}^{2} c^{2}\right)^{1 / 2}} p^{2} d p $$ with the normalization constant $$ C=\frac{\beta}{m_{0}^{2} c K_{2}\left(\beta m_{0} c^{2}\right)} $$ \(K_{v}(z)\) being a modified Bessel function. (b) Check that in the nonrelativistic limit \(\left(k T \ll m_{0} c^{2}\right)\) we recover the Maxwellian distribution, $$ f(\boldsymbol{p}) d \boldsymbol{p}=\left(\frac{\beta}{2 \pi m_{0}}\right)^{3 / 2} e^{-\beta p^{2} / 2 m_{0}}\left(4 \pi p^{2} d p\right) $$ while in the extreme relativistic limit \(\left(k T \gg m_{0} c^{2}\right)\) we obtain $$ f(\boldsymbol{p}) d \boldsymbol{p}=\frac{(\beta c)^{3}}{8 \pi} e^{-\beta p c}\left(4 \pi p^{2} d p\right) $$ (c) Verify that, quite generally, $$ \langle p u\rangle=3 k T . $$

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