Chapter 19, Problem 19.22E

Interpretation Introduction

Interpretation:

The validation of the statement that the value of constant $K$, defined by $\frac{1}{v_x}\cdot \frac{\frac{\partial g_x\left(v_x\right)}{\partial v_x}}{g_x\left(v_x\right)}$ and $\frac{1}{v}\cdot \frac{\frac{\partial \Gamma \left(v\right)}{\partial v}}{\Gamma \left(v\right)}$ is the same for $\frac{1}{v_y}\cdot \frac{\frac{\partial g_y\left(v_y\right)}{\partial v_y}}{g_y\left(v_y\right)}$ and $\frac{1}{v_z}\cdot \frac{\frac{\partial g_z\left(v_z\right)}{\partial v_z}}{g_z\left(v_z\right)}$ is to be stated.

Concept introduction:

The probability distribution function of the velocities of the gas molecules in each dimension is given by $g_x\left(v_x\right)$, $g_y\left(v_y\right)$ and $g_z\left(v_z\right)$. This is a vector quantity. This probability function in the range of $-\infty$ to $+\infty$ is represented by,

${\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{g}_x\left(v_x\right)d{v}_x}=1$

Answer to Problem 19.22E

The given statement that the value of constant $K$, defined by $\frac{1}{v_x}\cdot \frac{\frac{\partial g_x\left(v_x\right)}{\partial v_x}}{g_x\left(v_x\right)}$ and $\frac{1}{v}\cdot \frac{\frac{\partial \Gamma \left(v\right)}{\partial v}}{\Gamma \left(v\right)}$ is the same for $\frac{1}{v_y}\cdot \frac{\frac{\partial g_y\left(v_y\right)}{\partial v_y}}{g_y\left(v_y\right)}$ and $\frac{1}{v_z}\cdot \frac{\frac{\partial g_z\left(v_z\right)}{\partial v_z}}{g_z\left(v_z\right)}$ has been validated.

Explanation of Solution

The distribution function $g_x$ is given by,

$g_x={\left(\frac{m}{2\pi kT}\right)}^{1/2}{e}^{-m{v}_x^2/2kT}$

It is given that the value of constant $K$ is defined by,

$\frac{1}{v_x}\cdot \frac{\frac{\partial g_x\left(v_x\right)}{\partial v_x}}{g_x\left(v_x\right)}$ …(1)

Substitute the value of $g_x$ in equation (1).

$K=\frac{1}{v_x}\cdot \frac{\frac{\partial {\left(\frac{m}{2\pi kT}\right)}^{1/2}{e}^{-m{v}_x^2/2kT}}{\partial v_x}}{{\left(\frac{m}{2\pi kT}\right)}^{1/2}{e}^{-m{v}_x^2/2kT}}$

Keep all the constant terms out of the differentiation.

$K=\frac{1}{v_x}\cdot \frac{{\left(\frac{m}{2\pi kT}\right)}^{1/2}\frac{\partial e^{-m{v}^2/2kT}}{\partial v_x}}{{\left(\frac{m}{2\pi kT}\right)}^{1/2}{e}^{-m{v}^2/2kT}}$

$K=\frac{1}{v_x}\cdot \frac{\frac{\partial e^{-m{v}^2/2kT}}{\partial v_x}}{e^{-m{v}^2/2kT}}$ …(2)

The differentiation of the exponential term with respect to $v_x$ is shown below.

$\begin{array}{l}\frac{\partial e^{-m{v}_x^2/2kT}}{\partial v_x}\\ =e^{-m{v}_x^2/2kT}.\left(\frac{-2m{v}_x}{2kT}\right)\end{array}$

$=e^{-m{v}_x^2/2kT}.\left(\frac{-m{v}_x}{kT}\right)$ …(3)

Substitute equation (3) into equation (2).

$K=\frac{1}{v_x}\cdot \frac{e^{-m{v}_x^2/2kT}.\left(\frac{-m{v}_x}{kT}\right)}{e^{-m{v}^2/2kT}}$

Cancel the common terms and thus, the value of $K$ is,

$K=\left(\frac{-m}{kT}\right)$

The value of $K$ obtained is a constant, since the terms that contain velocity $v_x$ cancels out.

Similarly for $g_y\left(v_y\right)$ and $g_z\left(v_z\right)$ the terms that contain velocity $v_y$ and $v_z$ respectively cancels out and the value of $K$ obtained is constant.

It is given that the value of constant $K$ is defined by,

$\frac{1}{v}\cdot \frac{\frac{\partial \Gamma \left(v\right)}{\partial v}}{\Gamma \left(v\right)}$ …(4)

The term $\Gamma \left(v\right)$ is given by,

$\Gamma \left(v\right)=g_x\left(v_x\right)\cdot g_y\left(v_y\right)\cdot g_z\left(v_z\right)$ …(5)

Since $\Gamma \left(v\right)$ is the product of three uni-dimensional distribution function the value of $\frac{1}{v}\cdot \frac{\frac{\partial \Gamma \left(v\right)}{\partial v}}{\Gamma \left(v\right)}$ is same for each uni-dimensional distribution function $g$. Thus, the value of $K$ is same for each uni-dimensional distribution function $g$.

Conclusion

The given statement that the value of constant $K$, defined by $\frac{1}{v_x}\cdot \frac{\frac{\partial g_x\left(v_x\right)}{\partial v_x}}{g_x\left(v_x\right)}$ and $\frac{1}{v}\cdot \frac{\frac{\partial \Gamma \left(v\right)}{\partial v}}{\Gamma \left(v\right)}$ is the same for $\frac{1}{v_y}\cdot \frac{\frac{\partial g_y\left(v_y\right)}{\partial v_y}}{g_y\left(v_y\right)}$ and $\frac{1}{v_z}\cdot \frac{\frac{\partial g_z\left(v_z\right)}{\partial v_z}}{g_z\left(v_z\right)}$ has been validated.