Question

At what temperature of a gas will the number of molecules, whose velocities fall within given internal from $v$ to $v+dV$, be greatest? The mass of each molecules is equal to $m$.

• A. $T=\frac{1}{3}m{v}^{2}k$
• B. $T=\frac{1}{2}m{v}^{2}k$
• C. $T=\frac{m{v}^2}{3k}$
• D. $T=\frac{2}{3}m{v}^{2}k$

Answer 1

The correct option is C $T=\frac{m{v}^2}{3k}$

velocities intemal $\to v$ to $v+dv$

mass of each molecule is equal to $\underset{–}{m}$

$dN\left(v\right)=\frac{N4}{\sqrt{\pi }}\frac{v^{2}dv}{V_p^2}\cdot e^{-\frac{v^2}{V_p^2}}$

For a given $v$ to $v+dv$ liegiven $v$ and $dv$ this is maximum.

when

$\begin{array}{c}\frac{\delta }{\delta v_p}\frac{dN\left(v\right)}{N{v}^{2}dv}=0\\ (-3v{p}^{-4}+\frac{2V^2}{V_p^{\sigma }})e^{-\frac{v^2}{v{p}^2}}=0\\ V^2=\frac{3}{2}\cdot V_p^2=\frac{3kT}{m}\\ T=\frac{1}{3}\frac{m{v}^2}{k}][T=\frac{1}{3}\frac{m{v}^2}{k}]\end{array}$

Correct Answer -C