An HCl molecule has rotational, translational and vibrational motions. If the rms velocity of HCl molecules in its gaseous phase is $\bar{v}$, m is its mass and $\mathrm{k}_{\mathrm{B}}$ is Boltzmann constant, then its temperature will be:

(A) $\frac{m\,\bar{v}^{2}}{3 k_{B}}$

(B) $\frac{m\,\bar{v}^{2}}{7 k_{B}}$

(C) $\frac{m\,\bar{v}^{2}}{5 k_{B}}$

(D) $\frac{m\,\bar{v}^{2}}{6 k_{B}}$

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Heat & Thermodynamics

An HCl molecule has rotational, translational and vibrational motions. If the rms velocity of HCl molecules in its gaseous phase is

\overset{v}{ˉ}

, m is its mass and

k_{B}

is Boltzmann constant, then its temperature will be:

Answer: (A)

By the equipartition theorem the translational kinetic energy of one molecule is

\frac{1}{2} m \overset{v}{ˉ}^{2} = \frac{3}{2} k_{B} T

, where

\overset{v}{ˉ}

is the rms speed. Solving for the temperature gives

T = \frac{m v ˉ ^{2}}{3 k _{B}}

. The printed

v^{- 2}

is only a misprint of

\overset{v}{ˉ}^{2}

, so the intended choice is the one with

3 k_{B}

in the denominator.

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Single correct option