Question

The average velocity of molecules of a gas of molecular weight$$M$$at temperature$$T$$is.
$$\begin{array}{rl}& A)\sqrt{{\displaystyle \frac{3RT}{M}}}\\ & B)\sqrt{{\displaystyle \frac{8RT}{\pi M}}}\\ & C)\sqrt{{\displaystyle \frac{2RT}{M}}}\\ & D)\text{Zero}\end{array}$$

Answer 1

Hint: We will need to derive the expression for average velocity of a gas molecule by considering that it follows Maxwell-Boltzmann distribution. If we have a probability density function $$P\left(x\right)$$, then the expectation value of a quantity $$f\left(x\right)$$ is given by the integral of product of $$f\left(x\right)$$ and $$P\left(x\right)$$ over $$dx$$. This integral is evaluated over limits of probability density function.

Formula used:
$$⟨v⟩=\underset{0}{\overset{\mathrm{\infty }}{\int }}vP\left(v\right)dv$$

Complete step by step answer:
In this case, the probability density function $$P\left(v\right)$$ is the Maxwell-Boltzmann distribution and the quantity we want to find the expectation value is simply the velocity $$v$$. We know velocity can only be positive. So we have the limit from zero to $$\mathrm{\infty }$$.
Therefore, the mean velocity is given by,
$$⟨v⟩=\underset{0}{\overset{\mathrm{\infty }}{\int }}vP\left(v\right)dv$$
Where, $$P\left(v\right)$$ is the Maxwell-Boltzmann distribution given as,
  $$P\left(v\right)={\left({\displaystyle \frac{m}{2\pi kT}}\right)}^{{\displaystyle \frac{3}{2}}}4\pi v^2{e}^{(-{\displaystyle \frac{m{v}^2}{2kT}})}$$
Where, $$m$$ is the mass of the gas molecule.
              $$k$$ is the Boltzmann constant
           $$T$$ is the given temperature.
           $$v$$ is the velocity of one molecule.
So, the average velocity is given as,
    $$\begin{array}{rl}& ⟨v⟩=\underset{0}{\overset{\mathrm{\infty }}{\int }}vP\left(v\right)dv\\ & ⟨v⟩=4\pi {\left({\displaystyle \frac{m}{2\pi kT}}\right)}^{{\displaystyle \frac{3}{2}}}\underset{0}{\overset{\mathrm{\infty }}{\int }}{v}^3{e}^{(-{\displaystyle \frac{m{v}^2}{2kT}})}dv\end{array}$$
The result of integral in this form is given as,
  $$\underset{0}{\overset{\mathrm{\infty }}{\int }}{v}^3\exp (-\alpha v^2)dv={\displaystyle \frac{1}{2{\alpha }^2}}$$
So, here we have $$\alpha ={\displaystyle \frac{m}{2kT}}$$,
   $$\begin{array}{rl}& \Rightarrow ⟨v⟩=4\pi {\left({\displaystyle \frac{m}{2\pi kT}}\right)}^{{\displaystyle \frac{3}{2}}}{\displaystyle \frac{4k^2{T}^2}{2m^2}}\\ & \Rightarrow ⟨v⟩=\sqrt{{\displaystyle \frac{8kT}{\pi m}}}\end{array}$$
Here, $$m$$ is the mass of one molecule. If we take the mass of the compound as $$M$$, they are related by $$M=N_{A}m$$. Where,$$N_A$$ is the Avogadro number.
Since, $$R=N_{A}k$$, where $$R$$ is the gas constant, we can modify the equation as,
       $$⟨v⟩=\sqrt{{\displaystyle \frac{8RT}{\pi M}}}$$
So, we have found the average velocity of a gas molecule as $$\sqrt{{\displaystyle \frac{8RT}{\pi M}}}$$.

So, the correct answer is “Option B”.

 Note:
We can also solve this question easily if we know the expression for rms velocity or most probable velocity of a gas molecule with same mass. The rms velocity of a gas with mass $$M$$is given as $$v_{rms}=\sqrt{{\displaystyle \frac{2RT}{M}}}$$ and most probable velocity is given as $$v_{mp}=\sqrt{{\displaystyle \frac{3RT}{M}}}$$. These three velocities are related as $$v_{rms}:v_{average}:v_{mp}=\sqrt{2}:\sqrt{{\displaystyle \frac{8}{\pi }}}:\sqrt{3}$$.