Question

The temperature at which the average speed of oxygen molecules is double that of the same molecule at $0^\circ C$ is _______
(A) 546 K
(B) 1092 K
(C) 277 K
(D) $1911^\circ C$

Answer 1

Hint
From the formula of the average speed of gasses, we can find the average speed of the oxygen molecule at a certain temperature. In the second case the average speed doubles, so by taking the ratio of the average speed in the 2 cases, we can find the temperature in the second case.
In this solution, we will be using the following formula,
$v = \sqrt {\dfrac{{8RT}}{{\pi M}}}$
Where $v$ is the average speed
$R$ is the universal gas constant
$T$ is the temperature of the gas
and $M$ is the molar mass.

Complete step by step answer
In the question, it is given that the oxygen molecule is at a temperature of $0^\circ C$. So on the Kelvin scale, this temperature is equal to $273K$
So the average speed of the gas molecules is given by the formula,
$v = \sqrt {\dfrac{{8RT}}{{\pi M}}}$
So in the first case by substituting the temperature the average speed is,
${v_1} = \sqrt {\dfrac{{8R \times 273}}{{\pi M}}}$
For the second case let the temperature be ${T_2}$ and the average speed is given by,
${v_2} = \sqrt {\dfrac{{8R \times {T_2}}}{{\pi M}}}$
From the question, it is given that the average speed in the second case is double of that in the first case. Therefore,
${v_2} = 2{v_1}$
So substituting the values of the average speed we get,
$\sqrt {\dfrac{{8R \times {T_2}}}{{\pi M}}} = 2\sqrt {\dfrac{{8R \times 273}}{{\pi M}}}$
The constants are cancelled out from both sides. So we get,
$\sqrt {{T_2}} = 2\sqrt {273}$
On squaring both the sides we get
${T_2} = {\left( {2\sqrt {273} } \right)^2}$.
On calculating this gives us,
${T_2} = 4 \times 273$
$\Rightarrow {T_2} = 1092K$
Therefore the temperature of the oxygen molecules for which the average speed is twice of that of the molecules at $0^\circ C$ is $1092K$
So the correct answer is option B.

Note
As the temperature of a molecule increases, its average speed also increases. This is because, as the temperature increases, the kinetic energy of the gas molecules increases. As a result, the random motion of the molecules also increases. So the average speed of the oxygen molecule at $1092K$ is more than at $273K$.