Question

The average velocity of molecules of a gas of molecular weight\[M\]at temperature\[T\]is.
\[\begin{align}
  & A)\sqrt{\dfrac{3RT}{M}} \\
 & B)\sqrt{\dfrac{8RT}{\pi M}} \\
 & C)\sqrt{\dfrac{2RT}{M}} \\
 & D)\text{Zero} \\
\end{align}\]

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

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,
    \[\begin{align}
  & \left\langle v \right\rangle =\int\limits_{0}^{\infty }{vP\left( v \right)}dv \\
 & \left\langle v \right\rangle =4\pi {{\left( \dfrac{m}{2\pi kT} \right)}^{\dfrac{3}{2}}}\int\limits_{0}^{\infty }{{{v}^{3}}{{e}^{\left( -\dfrac{m{{v}^{2}}}{2kT} \right)}}}dv \\
\end{align}\]
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 \[\alpha =\dfrac{m}{2kT}\],
   \[\begin{align}
  & \Rightarrow \left\langle v \right\rangle =4\pi {{\left( \dfrac{m}{2\pi kT} \right)}^{\dfrac{3}{2}}}\dfrac{4{{k}^{2}}{{T}^{2}}}{2{{m}^{2}}} \\
 & \Rightarrow \left\langle v \right\rangle =\sqrt{\dfrac{8kT}{\pi m}} \\
\end{align}\]
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 \[{{v}_{rms}}:{{v}_{average}}:{{v}_{mp}}=\sqrt{2}:\sqrt{\dfrac{8}{\pi }}:\sqrt{3}\].