Question

The average velocity of molecules of a gas of molecular weight$M$at temperature$T$is.

& A)\sqrt{\dfrac{3RT}{M}} \\\ & B)\sqrt{\dfrac{8RT}{\pi M}} \\\ & C)\sqrt{\dfrac{2RT}{M}} \\\ & D)\text{Zero} \\\ \end{aligned}$$

Answer 1

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:
$\left\langle v \right\rangle =\int\limits_{0}^{\infty }{vP\left( v \right)}dv$

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 $\infty$.
Therefore, the mean velocity is given by,
$\left\langle v \right\rangle =\int\limits_{0}^{\infty }{vP\left( v \right)}dv$
Where, $P\left( v \right)$ is the Maxwell-Boltzmann distribution given as,
$P\left( v \right)={{\left( \dfrac{m}{2\pi kT} \right)}^{\dfrac{3}{2}}}4\pi {{v}^{2}}{{e}^{\left( -\dfrac{m{{v}^{2}}}{2kT} \right)}}$
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,
$\begin{aligned} & \left\langle v \right\rangle =\int\limits_{0}^{\infty }{vP\left( v \right)}dv \\\ & \left\langle v \right\rangle =4\pi {{\left( \dfrac{m}{2\pi kT} \right)}^{\dfrac{3}{2}}}\int\limits_{0}^{\infty }{{{v}^{3}}{{e}^{\left( -\dfrac{m{{v}^{2}}}{2kT} \right)}}}dv \\\ \end{aligned}$
The result of integral in this form is given as,
$\int\limits_{0}^{\infty }{{{v}^{3}}}\exp \left( -\alpha {{v}^{2}} \right)dv=\dfrac{1}{2{{\alpha }^{2}}}$
So, here we have $\alpha =\dfrac{m}{2kT}$,

& \Rightarrow \left\langle v \right\rangle =4\pi {{\left( \dfrac{m}{2\pi kT} \right)}^{\dfrac{3}{2}}}\dfrac{4{{k}^{2}}{{T}^{2}}}{2{{m}^{2}}} \\\ & \Rightarrow \left\langle v \right\rangle =\sqrt{\dfrac{8kT}{\pi m}} \\\ \end{aligned}$$ Here, $$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$$ is the gas constant, we can modify the equation as, $$\left\langle v \right\rangle =\sqrt{\dfrac{8RT}{\pi M}}$$ So, we have found the average velocity of a gas molecule as $$\sqrt{\dfrac{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{\dfrac{2RT}{M}}$$ and most probable velocity is given as $${{v}_{mp}}=\sqrt{\dfrac{3RT}{M}}$$. These three velocities are related as $${{v}_{rms}}:{{v}_{average}}:{{v}_{mp}}=\sqrt{2}:\sqrt{\dfrac{8}{\pi }}:\sqrt{3}$$.