Figure 19-28 shows a hypothetical speed distribution for particles of a certain gas: P(v)= Cv² for$\text{0}<\mathbf{v}\le {\mathbf{v}}_0$and P(v) = 0 for$\mathbf{v}\mathbf{>}{\mathbf{v}}_0$ . Find

(a) an expression for Cin terms of v₀,

(b) the average speed of the particles, and

(c) their rms speed.

The distributionof speed for particles of the gas is,

$$\begin{gathered} P\left(v\right)=C{v}^2\text{for}0<v\le v_0 \\ P\left(v\right)=0\text{\hspace{0.17em}\hspace{0.17em}for}v>v_0 \end{gathered}$$

If v is the speed of the particles of the gas, then the mean of the speeds of all the particles is represented by the average speed and the rms speed of the particles is given as the square root of the mean of squaresof the speed of the particles of the gas.

The speed distribution must satisfy the condition-

${\int }_0^{\infty }P\left(v\right)dv=1$

The average of the speed of the particles of the gas is given as-

$v_{avg}={\int }_0^{\infty }vP\left(v\right)dv$

The average of the squares of speed of the particles of the gas is given as-

${\left(v^2\right)}_{avg}={\int }_0^{\infty }{v}^{2}P\left(v\right)dv$

The rms speed of the particles is given as-

.$v_{rms}=\sqrt{{\left(v^2\right)}_{avg}}$

where, v is velocity and P represents the distribution of speed.

The distribution function must satisfy the relation,

$\begin{array}{l}{\int }_0^{\infty }P\left(v\right)dv=1\\ {\int }_0^{v_0}P\left(v\right)dv+{\int }_{v_0}^{\infty }P\left(v\right)dv=1\\ {\int }_0^{v_0}P\left(v\right)dv+{\int }_{v_0}^{\infty }\left(0\right)dv=1\\ {\int }_0^{v_0}P\left(v\right)dv=1\end{array}$

For $P\left(v\right)=C{v}^2\text{,for}0<v\le v_0$

${\int }_0^{v_0}C{v}^{2}dv=1$

$$\begin{gathered} C{\left(\frac{v^3}{3}\right)}_0^{v_0}=1 \\ C\left(\frac{{v_0}^3}{3}\right)=1 \end{gathered}$$

Hence, an expression for c in terms of $v_0$ is .$c=\frac{3}{v_0^3}$

The average of the speed of the particles is given as-

$$\begin{gathered} v_{avg}={\int }_0^{\infty }vP\left(v\right)dv \\ {v}_{avg}={\int }_0^{v_0}v\left(C{v}^2\right)dv \\ {v}_{avg}={\int }_0^{v_0}{v}^3\left(\frac{3}{{v_0}^3}\right)dv \\ {v}_{avg}=\left(\frac{3}{{v_0}^3}\right){\int }_0^{v_0}{v}^{3}dv \\ {v}_{avg}=\left(\frac{3}{{v_0}^3}\right){\int }_0^{v_0}{v}^{3}dv \\ {v}_{avg}=\left(\frac{3}{{v_0}^3}\right)\frac{{v_0}^4}{4} \\ {v}_{avg}=\frac{3v_0}{4} \end{gathered}$$

Hence, the value of average speed of the particles is .$\frac{3}{4}{v}_0$

The rms speed is defined as the square root of averageof the square of speed of the particles,

$$\begin{gathered} {\left(v^2\right)}_{avg}={\int }_0^{v_0}{v}^{2}p\left(v\right)dv \\ {\left(v^2\right)}_{avg}={\int }_0^{v_0}{v}^2\left(C{v}^2\right)dv \\ {\left(v^2\right)}_{avg}={\int }_0^{v_0}{v}^4\left(\frac{3}{{v_0}^3}\right)dv \\ {\left(v^2\right)}_{avg}=\left(\frac{3}{{v_0}^3}\right){\int }_0^{v_0}{v}^{4}dv \\ {\left(v^2\right)}_{avg}=\left(\frac{3}{{v_0}^3}\right){\left(\frac{v^5}{5}\right)}_0^{v_0} \\ {\left(v^2\right)}_{avg}=\left(\frac{3}{{v_0}^3}\right)\frac{{v_0}^5}{5} \\ {\left(v^2\right)}_{avg}=\frac{3{v_0}^2}{5} \end{gathered}$$

The rms speed is the square root of the average speed of the particles,

$$\begin{gathered} v_{rms}=\sqrt{{\left(v^2\right)}_{avg}} \\ {v}_{rms}=\sqrt{\frac{3{v_0}^2}{5}} \\ {v}_{rms}=\sqrt{\frac{3}{5}}{v}_0 \\ {v}_{rms}=0.775v_0 \end{gathered}$$

Hence, the rms speed is $v_{rms}=0.775v_0$.

Therefore, the expression of C is$c=\frac{3}{v_0^3}$ , average speed of the particles is $v_{avg}=\frac{3v_0}{4}$, and the rms speedis $v_{rms}=0.775v_0$.