Question

At what temperature is the rms velocity of a hydrogen molecule equal to that of an oxygen molecule at $ {47^ \circ }C $ ?

Answer 1

Hint: Gaseous particles, according to Kinetic Molecular Theory, are in a state of constant random motion; individual particles move at varying speeds, colliding and changing directions frequently. The movement of gas particles is described using velocity, which takes into consideration both speed and direction.
\[{v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} \]
 $ R = $ Universal Gas Constant $ = \begin{array}{*{20}{l}}
  {8.3144}&{J \cdot {K^ - }^1 \cdot mo{l^ - }^1}
\end{array} $
 $ T = $ Absolute temperature in Kelvin
 $ M = $ Molar mass of Gas molecules.

Complete answer:
The distribution of velocities does not vary despite the fact that the velocity of gaseous particles is continually changing. Because we can't determine the velocity of each individual particle, we frequently reason in terms of the average behaviour of the particles. The velocities of particles travelling in opposite directions have opposite signs. Because a gas's particles move at random, it's likely that there will be roughly as many moving in one direction as in the opposite, implying that the average velocity for a collection of gas particles equals zero; because this value is unhelpful, the average of velocities can be determined using a different method.
According to problem:
\[{v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} \]
Molar mass of Hydrogen $ = 2 $
Molar mass of oxygen $ = 32 $
Let $ T $ be the temperature at which the velocity of hydrogen equals that of oxygen at \[{47^ \circ }C.\]
\[ \Rightarrow \sqrt {\dfrac{{3RT}}{2}} = \sqrt {\dfrac{{3R \times (273 + 47)}}{{32}}} \]
 $ \Rightarrow T = 20\,K $ .

Note:
Let us know some little more about root mean square velocity. The root-mean-square speed considers both molecular weight and temperature, two elements that have a direct impact on a material's kinetic energy.