Chapter 1, Problem 1.1PR

(a)

Interpretation Introduction

Interpretation:

The mean speed from Maxwell Boltzmann distribution for a gas of mass $\text{M}$ and temperature $\text{T}$ is $\sqrt{\left(\frac{\text{8RT}}{\text{πM}}\right)}$ that has to be proved.

Concept Introduction:

Maxwell-Boltzmann distribution:

Maxwell-Boltzmann distribution is the particular probability distribution.  It was first defined and used for describing the particle speeds in idealized gases where the particle movement is free without interaction with one another except very brief collision in which they exchange energy and momentum with each other.

Maxwell-Boltzmann distribution is the result of kinetic theory of gases.  Mathematically, this is the chi distribution with the three degrees of freedom i.e. the component of velocity vector in three dimensions with a scale parameter measuring speeds in units proportional to the square root of the ratio of the temperature and particle mass.

According to Maxwell-Boltzmann’s distribution the probability distribution in all three dimensions are given as,

${\text{p(v}}_{\text{x}}\text{)}{\text{dv}}_{\text{x}}\text{=}{\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{\text{1}}{\text{2}}\right)}{\text{e}}^{\left(-\frac{{\text{mv}}_{\text{x}}^{\text{2}}}{\text{2kT}}\right)}{\text{dv}}_{\text{x}}\mathrm{...}\text{(}\text{in}\text{1}\text{dimension}\text{velocity}\text{is}{\text{v}}_{\text{x}}\text{)}$

$\begin{array}{l}{\text{p(v}}_{\text{2d}}\text{)}{\text{dv}}_{\text{2d}}\text{=}{\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{\text{1}}{\text{2}}\right)}{\text{e}}^{\left(-\frac{{\text{mv}}_{\text{2d}}^{\text{2}}}{\text{2kT}}\right)}{\text{dv}}_{\text{2d}}\mathrm{...}\text{(}\text{in}\text{2}\text{dimension}\text{velocity}\text{is}{\text{v}}_{\text{2d}}\text{)}\\ \\ \text{(where}{\text{v}}_{\text{2d}}^{\text{2}}\text{=}{\text{v}}_{\text{x}}^{\text{2}}\text{+}{\text{v}}_{\text{y}}^{\text{2}}\text{)}\end{array}$

$\begin{array}{l}\text{p(v)}\text{dv=}{\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{\text{3}}{\text{2}}\right)}{\text{×4π×v}}^{\text{2}}{\text{×e}}^{\left(-\frac{{\text{mv}}^2}{\text{2kT}}\right)}\text{dv}\mathrm{...}\text{(}\text{in}\text{3}\text{dimension}\text{velocity}\text{is}\text{v}\text{)}\\ \\ \text{(where}{\text{v}}^{\text{2}}\text{=}{\text{v}}_{\text{x}}^{\text{2}}\text{+}{\text{v}}_{\text{y}}^{\text{2}}\text{+}{\text{v}}_{\text{z}}^{\text{2}}\text{)}\end{array}$

$\begin{array}{l}\text{T}\text{=}\text{temperature}\text{of}\text{the}\text{gas}\\ \\ \text{k}\text{=}\text{Boltzmann}\text{constant}\\ \\ \text{m}\text{=}\text{mass}\text{of}\text{the}\text{gas}\end{array}$

Mean velocity:

Mean velocity or average velocity is the arithmetic mean of the velocities possessed by the gas molecules at a given temperature.

Say no. of molecules in gas is $\text{n}$ and the velocities possessed by the molecules ${\text{n}}_1$ , ${\text{n}}_2$, ${\text{n}}_3$ etc. are respectively ${\text{c}}_{\text{1}}$, ${\text{c}}_2$, ${\text{c}}_3$ etc.

Hence the mean velocity is

$\langle \text{c}\rangle \text{=}\frac{{\text{n}}_{\text{1}}{\text{c}}_{\text{1}}\text{+}{\text{n}}_{\text{2}}{\text{c}}_{\text{2}}{\text{+n}}_{\text{3}}{\text{c}}_{\text{3}}\text{+}\mathrm{...}}{{\text{n}}_{\text{1}}{\text{+n}}_{\text{2}}{\text{+n}}_{\text{3}}\text{+}\mathrm{...}}$

Explanation of Solution

Mean speed of molecules with a Maxwell-Boltzmann distribution is ${\displaystyle \underset{\text{0}}{\overset{\infty }{\int }}\text{vρ(v)dv}}$.

It has to be determined with the help of calculus that the mean speed is $\sqrt{\left(\frac{\text{8RT}}{\text{πM}}\right)}$.

Now applying calculus,

  $\langle \text{c}\rangle \text{=}{\displaystyle \underset{\text{0}}{\overset{\infty }{\int }}\text{vρ(v)dv}}$

From the value of the Maxwell-Boltzmann distribution in three dimension it can be concluded that,

  $\begin{array}{l}\text{p(v)}\text{dv=}{\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{\text{3}}{\text{2}}\right)}{\text{×4π×v}}^{\text{2}}{\text{×e}}^{\left(-\frac{{\text{mv}}^2}{\text{2kT}}\right)}\text{dv}\mathrm{...}\text{(}\text{in}\text{3}\text{dimension}\text{velocity}\text{is}\text{v}\text{)}\\ \\ \text{(where}{\text{v}}^{\text{2}}\text{=}{\text{v}}_{\text{x}}^{\text{2}}\text{+}{\text{v}}_{\text{y}}^{\text{2}}\text{+}{\text{v}}_{\text{z}}^{\text{2}}\text{)}\end{array}$

  $\begin{array}{l}\langle \text{c}\rangle \text{=}{\displaystyle \underset{\text{0}}{\overset{\infty }{\int }}\text{v}\times {\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{\text{3}}{\text{2}}\right)}{\text{×4π×v}}^{\text{2}}{\text{×e}}^{\left(-\frac{{\text{mv}}^2}{\text{2kT}}\right)}\text{dv}}\\ \end{array}$

Now applying calculus,

  $\begin{array}{l}\langle \text{c}\rangle \text{=}{\displaystyle \underset{\text{0}}{\overset{\infty }{\int }}{\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{\text{3}}{\text{2}}\right)}{\text{×4π×v}}^3{\text{×e}}^{\left(-\frac{{\text{mv}}^2}{\text{2kT}}\right)}\text{dv}}\\ \\ =\text{4π×}{\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{\text{3}}{\text{2}}\right)}{\displaystyle \underset{0}{\overset{\infty }{\int }}{\text{v}}^3{\text{×e}}^{\left(-\frac{{\text{mv}}^2}{\text{2kT}}\right)}\text{dv}}\\ \\ =\text{4π×}{\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{\text{3}}{\text{2}}\right)}{\displaystyle \underset{0}{\overset{\infty }{\int }}{\text{v}}^{2\times 1+1}{\text{×e}}^{\left(-\frac{{\text{mv}}^2}{\text{2kT}}\right)}\text{dv}}\end{array}$

According to the question the integral form that has to be used is,

  ${\displaystyle \underset{\text{0}}{\overset{\infty }{\int }}{\text{x}}^{\text{2n}\text{+}\text{1}}}{\text{e}}^{{\text{-ax}}^{\text{2}}}\text{dx=}\frac{\text{n!}}{{\text{2a}}^{\text{n+1}}}$

Now comparing the above two forms it can be concluded that,

  $\begin{array}{l}\text{n=1}\\ \\ \text{a}\text{=}\frac{\text{m}}{\text{2kT}}\end{array}$

Hence by putting the values in the form it is found that,

$\begin{array}{c}{\displaystyle \underset{0}{\overset{\infty }{\int }}{\text{v}}^{2\times 1+1}{\text{×e}}^{\left(-\frac{{\text{mv}}^2}{\text{2kT}}\right)}\text{dv}}\text{=}\frac{\text{1!}}{\text{2}{\left(\frac{\text{m}}{\text{2kT}}\right)}^{\text{1+1}}}\\ \\ \text{=}\frac{\text{1}}{\text{2}}\text{×}{\left(\frac{\text{2kT}}{\text{m}}\right)}^2\end{array}$

  $\begin{array}{l}\langle \text{c}\rangle =\text{4π×}{\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{\text{3}}{\text{2}}\right)}\times \frac{\text{1}}{\text{2}}\text{×}{\left(\frac{\text{2kT}}{\text{m}}\right)}^2\\ \\ =\text{4π}\times \left(\frac{\text{m}}{\text{2πkT}}\right)\times {\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{1}{\text{2}}\right)}\times \frac{\text{1}}{\text{2}}\text{×}{\left(\frac{\text{2kT}}{\text{m}}\right)}^2\\ \\ \text{=}{\left(\frac{\text{8kT}}{\text{πm}}\right)}^{\left(\frac{\text{1}}{\text{2}}\right)}\\ \\ \text{=}{\left(\frac{\text{8RT}}{\text{πM}}\right)}^{\left(\frac{\text{1}}{\text{2}}\right)}\mathrm{...}\left(\because \text{k=}\frac{\text{Rm}}{\text{M}}\right)\end{array}$

Hence the mean speed obtained from Maxwell-Boltzmann distribution is $\langle \text{c}\rangle ={\left(\frac{\text{8RT}}{\text{πM}}\right)}^{\left(\frac{\text{1}}{\text{2}}\right)}$.

(b)

Interpretation Introduction

Interpretation:

The root mean square velocity from Maxwell Boltzmann distribution for a gas of mass $\text{M}$ and temperature $\text{T}$ is $\sqrt{\left(\frac{\text{3RT}}{\text{M}}\right)}$ that has to be proved.

Concept Introduction:

Maxwell-Boltzmann distribution:

Maxwell-Boltzmann distribution is the particular probability distribution.  It was first defined and used for describing the particle speeds in idealized gases where the particle movement is free without interaction with one another except very brief collision in which they exchange energy and momentum with each other.

Maxwell-Boltzmann distribution is the result of kinetic theory of gases.  Mathematically, this is the chi distribution with the three degrees of freedom i.e. the component of velocity vector in three dimensions with a scale parameter measuring speeds in units proportional to the square root of the ratio of the temperature and particle mass.

According to Maxwell-Boltzmann’s distribution the probability distribution in all three dimensions are given as,

${\text{p(v}}_{\text{x}}\text{)}{\text{dv}}_{\text{x}}\text{=}{\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{\text{1}}{\text{2}}\right)}{\text{e}}^{\left(-\frac{{\text{mv}}_{\text{x}}^{\text{2}}}{\text{2kT}}\right)}{\text{dv}}_{\text{x}}\mathrm{...}\text{(}\text{in}\text{1}\text{dimension}\text{velocity}\text{is}{\text{v}}_{\text{x}}\text{)}$

$\begin{array}{l}{\text{p(v}}_{\text{2d}}\text{)}{\text{dv}}_{\text{2d}}\text{=}{\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{\text{1}}{\text{2}}\right)}{\text{e}}^{\left(-\frac{{\text{mv}}_{\text{2d}}^{\text{2}}}{\text{2kT}}\right)}{\text{dv}}_{\text{2d}}\mathrm{...}\text{(}\text{in}\text{2}\text{dimension}\text{velocity}\text{is}{\text{v}}_{\text{2d}}\text{)}\\ \\ \text{(where}{\text{v}}_{\text{2d}}^{\text{2}}\text{=}{\text{v}}_{\text{x}}^{\text{2}}\text{+}{\text{v}}_{\text{y}}^{\text{2}}\text{)}\end{array}$

  $\begin{array}{l}\text{p(v)}\text{dv=}{\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{\text{3}}{\text{2}}\right)}{\text{×4π×v}}^{\text{2}}{\text{×e}}^{\left(-\frac{{\text{mv}}^2}{\text{2kT}}\right)}\text{dv}\mathrm{...}\text{(}\text{in}\text{3}\text{dimension}\text{velocity}\text{is}\text{v}\text{)}\\ \\ \text{(where}{\text{v}}^{\text{2}}\text{=}{\text{v}}_{\text{x}}^{\text{2}}\text{+}{\text{v}}_{\text{y}}^{\text{2}}\text{+}{\text{v}}_{\text{z}}^{\text{2}}\text{)}\end{array}$

$\begin{array}{l}\text{T}\text{=}\text{temperature}\text{of}\text{the}\text{gas}\\ \\ \text{k}\text{=}\text{Boltzmann}\text{constant}\\ \\ \text{m}\text{=}\text{mass}\text{of}\text{the}\text{gas}\end{array}$

Root mean square velocity:

Root mean square velocity can be defined as the square root of the mean of the squares of the different velocities possessed by the molecules of the gas at a given temperature.

Say no. of molecules in gas is $\text{n}$ and the velocities possessed by the molecules ${\text{n}}_1$ , ${\text{n}}_2$, ${\text{n}}_3$ etc. are respectively ${\text{c}}_{\text{1}}$, ${\text{c}}_2$, ${\text{c}}_3$ etc.

Hence the mean velocity is

$\langle \text{c}\rangle \text{=}\frac{{\text{n}}_{\text{1}}{\text{c}}_{\text{1}}\text{+}{\text{n}}_{\text{2}}{\text{c}}_{\text{2}}{\text{+n}}_{\text{3}}{\text{c}}_{\text{3}}\text{+}\mathrm{...}}{{\text{n}}_{\text{1}}{\text{+n}}_{\text{2}}{\text{+n}}_{\text{3}}\text{+}\mathrm{...}}$

Hence the root mean square velocity is given by,

$\begin{array}{l}{\langle {\text{c}}^2\rangle }^{\frac{1}{2}}\text{=}{\left(\frac{{\text{n}}_{\text{1}}{\text{c}}_1^2\text{+}{\text{n}}_{\text{2}}{\text{c}}_2^2{\text{+n}}_{\text{3}}{\text{c}}_3^2\text{+}\mathrm{...}}{{\text{n}}_{\text{1}}{\text{+n}}_{\text{2}}{\text{+n}}_{\text{3}}\text{+}\mathrm{...}}\right)}^{\frac{1}{2}}\\ \end{array}$

Explanation of Solution

The mean speed of molecules with a Maxwell-Boltzmann distribution is ${\displaystyle \underset{\text{0}}{\overset{\infty }{\int }}{\text{v}}^2\text{ρ(v)dv}}$.

It has to be determined with the help of calculus that the root mean square velocity.is $\sqrt{\left(\frac{\text{3RT}}{\text{M}}\right)}$.

Now applying calculus,

  $\langle {\text{c}}^2\rangle \text{=}{\displaystyle \underset{\text{0}}{\overset{\infty }{\int }}{\text{v}}^2\text{ρ(v)dv}}$

From the value of the Maxwell-Boltzmann distribution in three dimensions it can be concluded that,

  $\begin{array}{l}\text{p(v)}\text{dv=}{\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{\text{3}}{\text{2}}\right)}{\text{×4π×v}}^{\text{2}}{\text{×e}}^{\left(-\frac{{\text{mv}}^2}{\text{2kT}}\right)}\text{dv}\mathrm{...}\text{(}\text{in}\text{3}\text{dimension}\text{velocity}\text{is}\text{v}\text{)}\\ \\ \text{(where}{\text{v}}^{\text{2}}\text{=}{\text{v}}_{\text{x}}^{\text{2}}\text{+}{\text{v}}_{\text{y}}^{\text{2}}\text{+}{\text{v}}_{\text{z}}^{\text{2}}\text{)}\end{array}$

$\langle {\text{c}}^2\rangle \text{=}{\displaystyle \underset{\text{0}}{\overset{\infty }{\int }}{\text{v}}^2\times {\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{\text{3}}{\text{2}}\right)}{\text{×4π×v}}^{\text{2}}{\text{×e}}^{\left(-\frac{{\text{mv}}^2}{\text{2kT}}\right)}\text{dv}}$

Now applying calculus,

$\begin{array}{l}\langle {\text{c}}^2\rangle \text{=}{\displaystyle \underset{\text{0}}{\overset{\infty }{\int }}{\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{\text{3}}{\text{2}}\right)}{\text{×4π×v}}^4{\text{×e}}^{\left(-\frac{{\text{mv}}^2}{\text{2kT}}\right)}\text{dv}}\\ \\ =\text{4π×}{\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{\text{3}}{\text{2}}\right)}{\displaystyle \underset{0}{\overset{\infty }{\int }}{\text{v}}^4{\text{×e}}^{\left(-\frac{{\text{mv}}^2}{\text{2kT}}\right)}\text{dv}}\\ \\ =\text{4π×}{\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{\text{3}}{\text{2}}\right)}{\displaystyle \underset{0}{\overset{\infty }{\int }}{\text{v}}^{2\times 2}{\text{×e}}^{\left(-\frac{{\text{mv}}^2}{\text{2kT}}\right)}\text{dv}}\end{array}$

According to the question the integral form that has to be used is,

  ${\displaystyle \underset{\text{0}}{\overset{\infty }{\int }}{\text{x}}^{\text{2n}}{\text{e}}^{{\text{-ax}}^{\text{2}}}\text{dx}}\text{=}\left(\frac{\text{(1×3×}\mathrm{..}\text{×(2n-1))}}{{\text{2}}^{\text{n+1}}{\text{×a}}^{\text{2}}}\right)\text{×}{\left(\frac{\text{π}}{\text{a}}\right)}^{\frac{\text{1}}{\text{2}}}$

Now comparing the above two forms it can be concluded that,

  $\begin{array}{l}\text{n=2}\\ \\ \text{a}\text{=}\frac{\text{m}}{\text{2kT}}\end{array}$

Hence by putting the values in the form it is found that,

  $\begin{array}{l}{\displaystyle \underset{\text{0}}{\overset{\infty }{\int }}{\text{v}}^{\text{2×2}}{\text{×e}}^{\left(\text{-}\frac{{\text{mv}}^2}{\text{2kT}}\right)}\text{dv}}\text{=}\left(\frac{\text{(1×3×}\mathrm{..}\text{×(2×2-1))}}{{\text{2}}^{\text{2+1}}\text{×}{\left(\frac{\text{m}}{\text{2kT}}\right)}^{\text{2}}}\right)\text{×}{\left(\frac{\text{π}}{\left(\frac{\text{m}}{\text{2kT}}\right)}\right)}^{\frac{\text{1}}{\text{2}}}\\ \\ \text{=}\left(\frac{\left(\frac{\text{(2×2)!}}{{\text{2}}^{\text{2}}\text{×2!}}\right)}{{\text{2}}^{\text{3}}\text{×}{\left(\frac{\text{m}}{\text{2kT}}\right)}^{\text{2}}}\right)\text{×}{\left(\frac{\text{π}}{\left(\frac{\text{m}}{\text{2kT}}\right)}\right)}^{\frac{\text{1}}{\text{2}}}\\ \\ \text{=}\left(\frac{\left(\frac{\text{4!}}{\text{4×2!}}\right)}{\text{8×}{\left(\frac{\text{m}}{\text{2kT}}\right)}^{\text{2}}}\right)\text{×}{\left(\frac{\text{π}}{\left(\frac{\text{m}}{\text{2kT}}\right)}\right)}^{\frac{\text{1}}{\text{2}}}\\ \\ \text{=}\frac{3\sqrt{\text{π}}}{8\times {\left(\frac{\text{m}}{\text{2kT}}\right)}^{\frac{\text{5}}{\text{2}}}}\end{array}$

  $\begin{array}{l}\langle {\text{c}}^{\text{2}}\rangle \text{=}\text{4π×}{\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{\text{3}}{\text{2}}\right)}\text{×}\frac{\text{3}\sqrt{\text{π}}}{\text{8×}{\left(\frac{\text{m}}{\text{2kT}}\right)}^{\frac{\text{5}}{\text{2}}}}\\ \\ \text{=}\frac{{\text{3×π}}^{\frac{\text{3}}{\text{2}}}}{{\text{2×π}}^{\frac{\text{3}}{\text{2}}}}\text{×}\frac{\text{1}}{\left(\frac{\text{m}}{\text{2kT}}\right)}\\ \\ \text{=}\frac{\text{3kT}}{\text{m}}\end{array}$

Now from this the root mean square velocity can be calculated as,

  $\begin{array}{l}{\langle {\text{c}}^{\text{2}}\rangle }^{\frac{1}{2}}\text{=}{\left(\frac{\text{3kT}}{\text{m}}\right)}^{\frac{1}{2}}\\ \\ ={\left(\frac{\text{3RT}}{\text{M}}\right)}^{\frac{1}{2}}\end{array}$

As $\text{k=}\frac{\text{Rm}}{\text{M}}$

Hence the root mean square velocity calculated from Maxwell-Boltzmann distribution is ${\langle {\text{c}}^{\text{2}}\rangle }^{\frac{1}{2}}\text{=}{\left(\frac{\text{3RT}}{\text{M}}\right)}^{\frac{1}{2}}$.

(c)

Interpretation Introduction

Interpretation:

An expression for the most probable speed has to be derived from Maxwell-Boltzmann distribution with the help of differentiation.

Concept Introduction:

Maxwell-Boltzmann distribution:

Maxwell-Boltzmann distribution is the particular probability distribution.  It was first defined and used for describing the particle speeds in idealized gases where the particle movement is free without interaction with one another except very brief collision in which they exchange energy and momentum with each other.

Maxwell-Boltzmann distribution is the result of kinetic theory of gases.  Mathematically, this is the chi distribution with the three degrees of freedom i.e. the component of velocity vector in three dimensions with a scale parameter measuring speeds in units proportional to the square root of the ratio of the temperature and particle mass.

According to Maxwell-Boltzmann’s distribution the probability distribution in all three dimensions are given as,

${\text{p(v}}_{\text{x}}\text{)}{\text{dv}}_{\text{x}}\text{=}{\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{\text{1}}{\text{2}}\right)}{\text{e}}^{\left(-\frac{{\text{mv}}_{\text{x}}^{\text{2}}}{\text{2kT}}\right)}{\text{dv}}_{\text{x}}\mathrm{...}\text{(}\text{in}\text{1}\text{dimension}\text{velocity}\text{is}{\text{v}}_{\text{x}}\text{)}$

$\begin{array}{l}{\text{p(v}}_{\text{2d}}\text{)}{\text{dv}}_{\text{2d}}\text{=}{\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{\text{1}}{\text{2}}\right)}{\text{e}}^{\left(-\frac{{\text{mv}}_{\text{2d}}^{\text{2}}}{\text{2kT}}\right)}{\text{dv}}_{\text{2d}}\mathrm{...}\text{(}\text{in}\text{2}\text{dimension}\text{velocity}\text{is}{\text{v}}_{\text{2d}}\text{)}\\ \\ \text{(where}{\text{v}}_{\text{2d}}^{\text{2}}\text{=}{\text{v}}_{\text{x}}^{\text{2}}\text{+}{\text{v}}_{\text{y}}^{\text{2}}\text{)}\end{array}$

$\begin{array}{l}\text{p(v)}\text{dv=}{\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{\text{3}}{\text{2}}\right)}{\text{×4π×v}}^{\text{2}}{\text{×e}}^{\left(-\frac{{\text{mv}}^2}{\text{2kT}}\right)}\text{dv}\mathrm{...}\text{(}\text{in}\text{3}\text{dimension}\text{velocity}\text{is}\text{v}\text{)}\\ \\ \text{(where}{\text{v}}^{\text{2}}\text{=}{\text{v}}_{\text{x}}^{\text{2}}\text{+}{\text{v}}_{\text{y}}^{\text{2}}\text{+}{\text{v}}_{\text{z}}^{\text{2}}\text{)}\end{array}$

$\begin{array}{l}\text{T}\text{=}\text{temperature}\text{of}\text{the}\text{gas}\\ \\ \text{k}\text{=}\text{Boltzmann}\text{constant}\\ \\ \text{m}\text{=}\text{mass}\text{of}\text{the}\text{gas}\end{array}$

Most probable velocity:

Most probable velocity can be defined as the velocity possessed by maximum number of molecules of gas at a given temperature.

Explanation of Solution

The maximum of Maxwell-Boltzmann distribution occurs when,

  $\frac{\text{dρ(v)}}{\text{dv}}\text{=0}$

Now according to the definition of most probable velocity, it is the velocity which is obtained by maximum no. of gas molecules.

Hence most probable velocity can be calculated as,

  $\frac{\text{dρ(v)}}{\text{dv}}\text{=0}$

From the value of the Maxwell-Boltzmann distribution in three dimension it can be concluded that,

  $\begin{array}{l}\text{p(v)}\text{dv=}{\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{\text{3}}{\text{2}}\right)}{\text{×4π×v}}^{\text{2}}{\text{×e}}^{\left(-\frac{{\text{mv}}^2}{\text{2kT}}\right)}\text{dv}\mathrm{...}\text{(}\text{in}\text{3}\text{dimension}\text{velocity}\text{is}\text{v}\text{)}\\ \\ \text{(where}{\text{v}}^{\text{2}}\text{=}{\text{v}}_{\text{x}}^{\text{2}}\text{+}{\text{v}}_{\text{y}}^{\text{2}}\text{+}{\text{v}}_{\text{z}}^{\text{2}}\text{)}\end{array}$

Thus from the above equations,

  $\frac{\text{d}{\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{\text{3}}{\text{2}}\right)}{\text{×4π×v}}^{\text{2}}{\text{×e}}^{\left(-\frac{{\text{mv}}^2}{\text{2kT}}\right)}}{\text{dv}}\text{=0}$

Applying rules of differentiation,

  $\begin{array}{l}{\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{\text{3}}{\text{2}}\right)}{\text{×4π×e}}^{\left(-\frac{{\text{mv}}^2}{\text{2kT}}\right)}\times \left(\text{2v}\text{-}\frac{\text{m}}{\text{kT}}\times {\text{v}}^3\right)=0\\ \\ \left(\text{2v}\text{-}\frac{\text{m}}{\text{kT}}\times {\text{v}}^3\right)=0\\ \\ \left(\text{2}\text{-}\frac{\text{m}}{\text{kT}}\times {\text{v}}^2\right)=0\\ \\ \frac{{\text{mv}}^2}{\text{kT}}=2\\ \\ \text{v}={\left(\frac{2\text{kT}}{\text{m}}\right)}^{\frac{1}{2}}\end{array}$

As $\text{k=}\frac{\text{Rm}}{\text{M}}$

Hence, $\text{v=}{\left(\frac{\text{2RT}}{\text{M}}\right)}^{\frac{\text{1}}{\text{2}}}$

Hence the most probable velocity of gas from the Maxwell-Boltzmann distribution is $\text{v=}{\left(\frac{\text{2RT}}{\text{M}}\right)}^{\frac{\text{1}}{\text{2}}}$.

(d)

Interpretation Introduction

Interpretation:

The fraction of ${\text{N}}_{\text{2}}$ molecules has to be estimated at the range of velocity of $\text{290}{\text{ms}}^{\text{-1}}$ to $\text{300}{\text{ms}}^{\text{-1}}$ at the temperature of $\text{500K}$.

Concept Introduction:

Maxwell-Boltzmann distribution:

Maxwell-Boltzmann distribution is the particular probability distribution.  It was first defined and used for describing the particle speeds in idealized gases where the particle movement is free without interaction with one another except very brief collision in which they exchange energy and momentum with each other.

Maxwell-Boltzmann distribution is the result of kinetic theory of gases.  Mathematically, this is the chi distribution with the three degrees of freedom i.e. the component of velocity vector in three dimensions with a scale parameter measuring speeds in units proportional to the square root of the ratio of the temperature and particle mass.

According to Maxwell-Boltzmann’s distribution the probability distribution in all three dimensions are given as,

${\text{p(v}}_{\text{x}}\text{)}{\text{dv}}_{\text{x}}\text{=}{\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{\text{1}}{\text{2}}\right)}{\text{e}}^{\left(-\frac{{\text{mv}}_{\text{x}}^{\text{2}}}{\text{2kT}}\right)}{\text{dv}}_{\text{x}}\mathrm{...}\text{(}\text{in}\text{1}\text{dimension}\text{velocity}\text{is}{\text{v}}_{\text{x}}\text{)}$

$\begin{array}{l}{\text{p(v}}_{\text{2d}}\text{)}{\text{dv}}_{\text{2d}}\text{=}{\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{\text{1}}{\text{2}}\right)}{\text{e}}^{\left(-\frac{{\text{mv}}_{\text{2d}}^{\text{2}}}{\text{2kT}}\right)}{\text{dv}}_{\text{2d}}\mathrm{...}\text{(}\text{in}\text{2}\text{dimension}\text{velocity}\text{is}{\text{v}}_{\text{2d}}\text{)}\\ \\ \text{(where}{\text{v}}_{\text{2d}}^{\text{2}}\text{=}{\text{v}}_{\text{x}}^{\text{2}}\text{+}{\text{v}}_{\text{y}}^{\text{2}}\text{)}\end{array}$

$\begin{array}{l}\text{p(v)}\text{dv=}{\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{\text{3}}{\text{2}}\right)}{\text{×4π×v}}^{\text{2}}{\text{×e}}^{\left(-\frac{{\text{mv}}^2}{\text{2kT}}\right)}\text{dv}\mathrm{...}\text{(}\text{in}\text{3}\text{dimension}\text{velocity}\text{is}\text{v}\text{)}\\ \\ \text{(where}{\text{v}}^{\text{2}}\text{=}{\text{v}}_{\text{x}}^{\text{2}}\text{+}{\text{v}}_{\text{y}}^{\text{2}}\text{+}{\text{v}}_{\text{z}}^{\text{2}}\text{)}\end{array}$

$\begin{array}{l}\text{T}\text{=}\text{temperature}\text{of}\text{the}\text{gas}\\ \\ \text{k}\text{=}\text{Boltzmann}\text{constant}\\ \\ \text{m}\text{=}\text{mass}\text{of}\text{the}\text{gas}\end{array}$

This probability density is actually the measure of the fraction of molecules present in a given velocity range at a given temperature.

  $\frac{\text{dN}}{\text{N}}\text{=p(v)dv}$

Where $\text{N}$ is the total no. of molecules present.

Answer to Problem 1.1PR

The fraction of ${\text{N}}_{\text{2}}$ molecules at the range of velocity of $\text{290}{\text{ms}}^{\text{-1}}$ to $\text{300}{\text{ms}}^{\text{-1}}$ at the temperature of $\text{500K}$ is $4.79\times {10}^{23}$.

Explanation of Solution

The fraction of ${\text{N}}_{\text{2}}$ molecules has to be estimated at the range of velocity of $\text{290}{\text{ms}}^{\text{-1}}$ to $\text{300}{\text{ms}}^{\text{-1}}$ at the temperature of $\text{500K}$.

From the value of the Maxwell-Boltzmann distribution in three dimensions it can be concluded that,

$\begin{array}{l}\text{p(v)}\text{dv=}{\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{\text{3}}{\text{2}}\right)}{\text{×4π×v}}^{\text{2}}{\text{×e}}^{\left(-\frac{{\text{mv}}^2}{\text{2kT}}\right)}\text{dv}\mathrm{...}\text{(}\text{in}\text{3}\text{dimension}\text{velocity}\text{is}\text{v}\text{)}\\ \\ \text{(where}{\text{v}}^{\text{2}}\text{=}{\text{v}}_{\text{x}}^{\text{2}}\text{+}{\text{v}}_{\text{y}}^{\text{2}}\text{+}{\text{v}}_{\text{z}}^{\text{2}}\text{)}\end{array}$

Hence,

  $\frac{\text{dN}}{\text{N}}\text{=p(v)}\text{dv=}{\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{\text{3}}{\text{2}}\right)}{\text{×4π×v}}^{\text{2}}{\text{×e}}^{\left(-\frac{{\text{mv}}^2}{\text{2kT}}\right)}\text{dv}$

Now putting the values of velocities and temperature in the above equation,

$\begin{array}{l}\text{T}\text{=}\text{500K}\\ \\ \text{k}\text{=}\text{Boltzmann}\text{constant=}\left(\text{1}{\text{.38×10}}^{\text{-23}}{\text{JK}}^{\text{-1}}\right)\\ \\ \text{m}\text{=}\text{28g}{\text{mol}}^{-1}\text{=}\left(28\times {10}^{-3}\text{kg}{\text{mol}}^{-1}\right)\end{array}$

$\begin{array}{l}\text{dv=}\text{(300-290)m}{\text{s}}^{-1}\\ =10\text{m}{\text{s}}^{-1}\end{array}$

$\begin{array}{l}\frac{\text{dN}}{\text{N}}\text{=}{\left(\frac{\text{m}}{\text{2πkT}}\right)}^{\left(\frac{\text{3}}{\text{2}}\right)}{\text{×4π×v}}^{\text{2}}{\text{×e}}^{\left(\text{-}\frac{{\text{mv}}^{\text{2}}}{\text{2kT}}\right)}\text{dv}\\ \\ \text{=}\left[\begin{array}{l}\frac{{\text{28×10}}^{\text{-3}}\text{kg}{\text{mol}}^{\text{-1}}}{\text{2π×(1}{\text{.38×10}}^{\text{-23}}{\text{JK}}^{\text{-1}}\text{)×500K}}\text{×4π×}\left({\text{300}}^{\text{2}}{\text{-290}}^{\text{2}}\right){\text{m}}^{\text{2}}{\text{s}}^{\text{-2}}\\ {\text{×e}}^{\left(\text{-}\frac{{\text{28×10}}^{\text{-3}}\text{kg}{\text{mol}}^{\text{-1}}\text{×}\left({\text{300}}^{\text{2}}{\text{-290}}^{\text{2}}\right){\text{m}}^{\text{2}}{\text{s}}^{\text{-2}}}{\text{2×(1}{\text{.38×10}}^{\text{-23}}{\text{JK}}^{\text{-1}}\text{)×500K}}\right)}{\text{×10ms}}^{\text{-1}}\end{array}\right]\\ \\ =4.79\times {10}^{23}\end{array}$

Hence the fraction of ${\text{N}}_{\text{2}}$ molecules at the range of velocity of $\text{290}{\text{ms}}^{\text{-1}}$ to $\text{300}{\text{ms}}^{\text{-1}}$ at the temperature of $\text{500K}$ is $4.79\times {10}^{23}$.