Question

At what temperature is the rms speed of ${H_2}$ molecules the same as that of oxygen molecules at $1327^\circ C$?
A.$173K$
B.$100K$
C.$400K$
D.$523K$

Answer 1

Hint: The speed and temperature have a unique relation in them. It varies as our condition varies. For gaseous molecules also there is a relation between the temperature and their speed. The speed of molecules in a gas is proportional to temperature and inversely proportional to molar mass.

Complete step by step answer:
As we all know that the speed of molecules in a gas is proportional to temperature and inversely proportional to molar mass.
And we also know that
${V_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $ where,
${V_{rms}}$ is root mean square velocity.
$R$ is the gas constant.
$T$ is the temperature.
$M$ is the molecular mass of a given gaseous species.
So using this formula we will find our answer.
We have given that ${V_{rms}}$ of both the species, hydrogen and oxygen is equal.
Therefore,
${({V_{rms}})_O} = {({V_{rms}})_H}$
$\sqrt {\dfrac{{3R{T_O}}}{{{M_O}}}} = \sqrt {\dfrac{{3R{T_H}}}{{{M_H}}}} $ -------------------(1)
Where,
${M_O} = 32 = $ Molecular weight of oxygen
${M_H} = 2 = $ Molecular weight of hydrogen
${T_H} = ? = $ Temperature of hydrogen gas
${T_O} = 1327 + 273 = 1600K = $ Temperature of oxygen ($\therefore $ Temperature should be in$K$. )
Put all these values in equation (1)
$\sqrt {\dfrac{{3R \times 1600}}{{32}}} = \sqrt {\dfrac{{3R{T_H}}}{2}} $
$ \Rightarrow {T_H} = 100K$
Hence option (B) is correct, $100K$.

Note:
Approximately in every scientific or analytical calculation temperature is taken in Kelvin $(K)$, but to confuse you the examiner can give temperature in degree Celsius ($^\circ C$ ) or degree Fahrenheit ($^\circ F$) , so you have to convert that in Kelvin and then solve the numerical.