Chapter 6.4, Problem 29E

a.

To determine

Derive the distribution of $W=\frac{m{V}^2}{2}$, the kinetic energy of the molecule.

Answer to Problem 29E

The probability density function for $W=\frac{m{V}^2}{2}$ is $f_W\left(w\right)=\frac{1}{\Gamma \left(\frac{3}{2}\right){\left(kT\right)}^{\frac{3}{2}}}{w}^{\frac{1}{2}}{\text{e}}^{\frac{-w}{kT}},w\ge 0$.

Explanation of Solution

Calculation:

Gamma distribution:

If the random variable X follows Gamma distribution with $\left(\alpha ,\beta \right)$,

• The probability density function for X is $f\left(x\right)=\frac{1}{\Gamma \left(\alpha \right){\beta }^{\alpha }}{x}^{\alpha -1}{e}^{-\frac{x}{\beta }},x>0$.
• The mean if X is $E\left(X\right)=\alpha \beta$

The probability density function for V is, $f\left(v\right)=a{v}^2{\text{e}}^{-b{v}^2}\text{, }v>0$.

From the given information, the random variable W is defined as $W=\frac{m{V}^2}{2}$.

Consider,

$\begin{array}{c}W=\frac{m{V}^2}{2}\\ m{V}^2=2W\\ V^2=\frac{2W}{m}\\ V=\sqrt{\frac{2W}{m}}\end{array}$

 $\begin{array}{c}\frac{dV}{dw}=\frac{1}{2}{\left(\frac{2W}{m}\right)}^{\frac{-1}{2}}\times \frac{2}{m}\\ =\frac{1}{\sqrt{2mw}}\end{array}$

The probability density function for W is,

$\begin{array}{c}{f}_W\left(w\right)=f\left(v\right)\left|\frac{dv}{dw}\right|\\ =a{v}^2{\text{e}}^{-b{v}^2}\times \left|\frac{1}{\sqrt{2mw}}\right|\\ =a{\left(\sqrt{\frac{2w}{m}}\right)}^2{\text{e}}^{-\frac{m}{2kT}\times {\left(\sqrt{\frac{2w}{m}}\right)}^2}\times \frac{1}{\sqrt{2mw}}\\ =\frac{a\sqrt{2w}}{m^{\frac{3}{2}}}{\text{e}}^{\frac{-w}{kT}},w\ge 0\end{array}$

The density function is in the form of Gamma density function and the constant ‘a’ is obtained by equating the integral of density function to 1.

Therefore,

$\begin{array}{c}{\displaystyle \underset{0}{\overset{1}{\int }}{f}_W\left(w\right)dw}=1\\ {\displaystyle \underset{0}{\overset{1}{\int }}\frac{a\sqrt{2w}}{m^{\frac{3}{2}}}{\text{e}}^{\frac{-w}{kT}}dw}=1\\ \frac{a\sqrt{2}}{m^{\frac{3}{2}}}{\displaystyle \underset{0}{\overset{1}{\int }}{w}^{\frac{1}{2}}{\text{e}}^{\frac{-w}{kT}}dw}=1\\ \frac{a\sqrt{2}}{m^{\frac{3}{2}}}\times \Gamma \left(\frac{3}{2}\right){\left(kT\right)}^{\frac{3}{2}}=1\end{array}$

                           $\frac{a\sqrt{2}}{m^{\frac{3}{2}}}=\frac{1}{\Gamma \left(\frac{3}{2}\right){\left(kT\right)}^{\frac{3}{2}}}$

Therefore, the density function for W is,

 $\begin{array}{c}{f}_W\left(w\right)=\frac{a\sqrt{2w}}{m^{\frac{3}{2}}}{\text{e}}^{\frac{-w}{kT}}\\ =\frac{1}{\Gamma \left(\frac{3}{2}\right){\left(kT\right)}^{\frac{3}{2}}}{w}^{\frac{1}{2}}{\text{e}}^{\frac{-w}{kT}},w\ge 0\end{array}$

b.

To determine

Find the value of $E\left(W\right)$.

Answer to Problem 29E

The value of $E\left(W\right)$ is $\frac{3}{2}kT$.

Explanation of Solution

Calculation:

The probability density function for W is $f_W\left(w\right)=\frac{1}{\Gamma \left(\frac{3}{2}\right){\left(kT\right)}^{\frac{3}{2}}}{w}^{\frac{1}{2}}{\text{e}}^{\frac{-w}{kT}},w\ge 0$.

The probability density function $f_W\left(w\right)=\frac{1}{\Gamma \left(\frac{3}{2}\right){\left(kT\right)}^{\frac{3}{2}}}{w}^{\frac{1}{2}}{\text{e}}^{\frac{-w}{kT}},w\ge 0$ is similar to the standard Gamma probability density function, $f\left(x\right)=\frac{1}{\Gamma \left(\alpha \right){\beta }^{\alpha }}{x}^{\alpha -1}{e}^{-\frac{x}{\beta }},x>0$ with $\alpha =\frac{3}{2}$ and $\beta =kT$.

Therefore, the mean of the random variable W is,

$\begin{array}{c}E\left(W\right)=\alpha \beta \\ =\frac{3}{2}\times kT\\ =\frac{3}{2}kT\end{array}$