Question

The ratio between the root mean square velocity of ${H_2}$ at 50K and that of ${O_2}$ at 800K is:
(a) 0.25
(b) 1
(c) 2
(d) 4

Answer 1

Hint: The root mean square velocity is the measure of the speed of a particle in a gas. We can solve the above problem by using the formula $\sqrt {\dfrac{{3RT}}{M}} $

Complete step by step answer:
1: The root mean square velocity or ${v_{rms}}$ is directly proportional to the square root of temperature and is inversely proportional to the square root of molecular weight.
2: Now, we calculate ${v_{rms}}$ of ${H_2}$ at 50K using the formula $\sqrt {\dfrac{{3RT}}{M}} $
Where, R is the universal gas constant (8.314 J mol^-1 K^-1 )
     T is the temperature, here it is 50K
     M is the molecular mass which is 2 g/mol for H₂
∴ ${v_{rms}}$for ${H_2}$ is $ = \sqrt {\dfrac{{3R \times 50}}{2}} $
$ \Rightarrow {v_{{H_2}}} = \sqrt {75R} $
(Since R is common in both and we need to calculate the ratio, we will leave it as R for simplicity and it will get cancelled at the end.)
3: Similarly,${v_{rms}}$for ${O_2}$ is $ = \sqrt {\dfrac{{3R \times 800}}{{32}}} $ (Since molecular mass of ${O_2}$ IS 32 g/mol)
$ \Rightarrow {v_{{O_2}}} = \sqrt {75R} $
4: Now taking the ratio between ${v_{{H_2}}}$ and ${v_{{O_2}}}$ , we get:
$
\dfrac{{{v_{{H_2}}}}}{{{v_{{O_2}}}}} = \dfrac{{\sqrt {75R} }}{{\sqrt {75R} }} \\
   \Rightarrow\dfrac{{{v_{{H_2}}}}}{{{v_{{O_2}}}}} = 1 \\
 $
Thus, the correct option is (a).

Note:
We mostly use ${v_{rms}}$instead of${v_{avg}}$because for a typical gas sample, the net velocity is zero. This is because the particles are moving randomly constantly in all directions. This is a key formula as the velocity of the particles is what determines both the diffusion and effusion rates. The faster the root mean square velocity, the faster the diffusion. Effusion occurs by a difference of pressures while diffusion occurs due to difference in concentrations.