Question

The ratio of total K.E. of $1.87{\text{g}}$ of ${{\text{H}}_2}$ and $5.53{\text{g}}$ of ${{\text{O}}_{\text{2}}}$at 300K is
A. $5:4:1$
B. $1:5.4$
C. $2:7:1$
D. $1:2.7$

Answer 1

Hint: To answer this question, you must recall the formula for Kinetic energy of a gas. The kinetic energy of a gas is the average of the kinetic energies of the molecules of the gas.
Formula used:
${\text{K}}{\text{.E}}{\text{.}} = \dfrac{3}{2}{\text{nRT}}$
Where, ${\text{K}}{\text{.E}}{\text{.}}$ is the total kinetic energy of the gas
${\text{n}}$ is the number of moles of gas present in the container
${\text{R}}$ is the gas constant
And, ${\text{T}}$ is the temperature of the gas

Complete step by step answer:
We are given the amount of gases taken at the given constant temperature.
So, to calculate the kinetic energies of the gases, we need to find the number of moles of the gases.
We know that the number of moles of a substance is given by the ratio of the given mass of the substance to its molar mass.
The given mass of hydrogen gas is $1.87{\text{g}}$ and we know the molar mass of bimolecular hydrogen gas is $2{\text{g}}/{\text{mol}}$.
So, the number of moles of hydrogen is given as,
${{\text{n}}_{\text{H}}} = \dfrac{{1.87}}{2} = 0.935{\text{mol}}$
The given mass of oxygen is $5.53{\text{g}}$ and we know that the molar mass of bimolecular oxygen gas is $32{\text{g}}/{\text{mol}}$.
We know that kinetic energy of gas is given by,
${\text{K}}{\text{.E}}{\text{.}} = \dfrac{3}{2}{\text{nRT}}$
Taking the ratio for kinetic energy of both the gases, we get,
$\dfrac{{{\text{K}}{{\text{E}}_{\text{H}}}}}{{{\text{K}}{{\text{E}}_{\text{O}}}}} = \dfrac{{\dfrac{5}{2}{{\text{n}}_{\text{H}}}{\text{RT}}}}{{\dfrac{5}{2}{{\text{n}}_{\text{O}}}{\text{RT}}}}$
Now, we substitute the values and solve to get,
$\dfrac{{{\text{K}}{{\text{E}}_{\text{H}}}}}{{{\text{K}}{{\text{E}}_{\text{O}}}}} = \dfrac{{\left( {0.935} \right) \times {\text{RT}}}}{{\left( {0.1728} \right) \times {\text{RT}}}}$
$ \Rightarrow \dfrac{{{\text{K}}{{\text{E}}_{\text{H}}}}}{{{\text{K}}{{\text{E}}_{\text{O}}}}} = \dfrac{{5.4}}{1}$
Hence, the ratio between the total kinetic energies of the given amount of hydrogen and oxygen is $5.4:1$.

The correct answer is A.

Note:
We know that at constant temperature, the kinetic energy of a gas is constant.
Thus, the ratio of kinetic energies of hydrogen and oxygen gas will be equal to the ratio of their number of moles.
$\dfrac{{{\text{K}}{{\text{E}}_{\text{H}}}}}{{{\text{K}}{{\text{E}}_{\text{O}}}}} = \dfrac{{{{\text{n}}_{\text{H}}}}}{{{{\text{n}}_{\text{O}}}}} = \dfrac{{5.4}}{1}$