Question

The lowest possible temperature in outer space is $2.7K$. What is the rms speed of hydrogen molecules at this temperature? (The molar mass of hydrogen is $2 \times {10^{ - 3}}Kg/mol$).

Answer 1

Hint: According to the kinetic theory of gases, the square root mean velocity of the molecules of a gas is proportional to the square root of its absolute temperature.
Root mean square velocity of a gas is the square root of the mean of the squares of the velocities of individual molecules.

Complete step by step answer:
We know that the r.m.s. speed of a gas, ${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $
This is also known as kinetic interpretation of temperature.
Where $R$ is the universal gas constant.
$R = 8.311Jmo{l^{ - 1}}{K^{ - 1}}$
$T$ is the absolute temperature
$M$ is molar mass of hydrogen
It is given that $T = 2.7K$ and $M = 2 \times {10^{ - 3}}Kg/mol$
The r.m.s. speed of hydrogen molecules is ${v_{rms}} = \sqrt {\dfrac{{3 \times 8.311 \times 2.7}}{{2 \times {{10}^{ - 3}}}}} $
Or ${v_{rms}} = 183.45m{s^{ - 1}}$
Or ${v_{rms}} = 1.83 \times {10^2}m{s^{ - 1}}$
Hence, the molecular speed of hydrogen molecules at $2.7K$ is $1.83 \times {10^2}m{s^{ - 1}}$.

Note:
Alternative approach to solve the given problem:
We know that the kinetic energy per molecule, $\dfrac{1}{2}m{v^2}_{rms} = \dfrac{3}{2}kT$
Or ${v_{rms}} = \sqrt {\dfrac{{3kT}}{m}} $
The above formula is the expression for r.m.s. speed of a gas molecule in terms of Boltzmann constant.
Where, $k$ is the Boltzmann constant
$k = 1.38 \times {10^{ - 23}}J{K^{ - 1}}$
$m$ is the mass of one gas molecule.
$m = \dfrac{M}{{{N_A}}}$
$M$ is the molar of a gas in $gmo{l^{ - 1}}$
${N_A}$ is Avogadro's number. ${N_A} = 6.022 \times {10^{ - 23}}$
It is given that $M = 2 \times {10^{ - 3}}Kgmo{l^{ - 1}}$
The r.m.s. speed of hydrogen molecules is
${v_{rms}} = \sqrt {\dfrac{{3{N_A}kT}}{M}} $
Substitute the required values in the above formula. We got,
${v_{rms}} = \sqrt {\dfrac{{3 \times 2.022 \times {{10}^{23}} \times 1.38 \times {{10}^{ - 23}} \times 2.7}}{{2 \times {{10}^{ - 3}}}}} $
Calculate the above mathematical expression.
${v_{rms}} = 183.45m{s^{ - 1}}$
Or ${v_{rms}} = 1.83 \times {10^2}m{s^{ - 1}}$

> The r.m.s. speed in terms of pressure $\left( P \right)$ of gas, ${v_{rms}} = \sqrt {\dfrac{{3P}}{\rho }} $
Where $\rho $ is the density of the gas.
> The kinetic energy $\left( {\dfrac{3}{2}kT} \right)$ per molecule is independent of the mass of the molecule. It only depends upon the absolute temperature of the gas.