Question

At what temperature the RMS velocity of oxygen will be the same as that of methane at ${27^ \circ }C$.
A)${54^ \circ }C$.
B)$227K$.
C)$600\,K$.
D)$573\,K$.

Answer 1

Hint: We know that the root-mean-square velocity is the measure of the speed of particles in a gas and is defined as the square root of the average velocity-squared of the molecules in a gas.
Mathematically expression for RMS velocity is,
$RMS = \sqrt {\dfrac{{3RT}}{M}} $
Where,
R is the universal gas constant.
T is the temperature in Kelvin.
M is the molar mass of the gas molecule.

Complete step by step answer:

Let's take the velocity of methane as ${V_m}$ and the velocity of oxygen as ${V_o}$.
The RMS velocity of methane ${V_m} = \sqrt {\dfrac{{3R{T_1}}}{{{M_1}}}} \xrightarrow{{}}1$
Where ${T_1}\,\& {M_1}$ are temperature and molar mass of methane gas.
The RMS velocity of oxygen ${V_o} = \sqrt {\dfrac{{3R{T_2}}}{{{M_2}}}} \xrightarrow{{}}2$
Where ${T_2}\,\& {M_2}$ are temperature and molar mass of Oxygen.
Now, equate the equation 1 and 2.
$\sqrt {\dfrac{{3R{T_1}}}{{{M_1}}}} = \sqrt {\dfrac{{3R{T_2}}}{{{M_2}}}} \xrightarrow{{}}3$
We know that the molar mass of methane and oxygen is 16g/mol & 32g/mol respectively.
Now, Substitute the value of temperature and molar mass in equation 3. We get,
$\sqrt {\dfrac{{3R\left( {300K} \right)}}{{16}}} = \sqrt {\dfrac{{3R{T_2}}}{{32}}} $
Square on both sides,
$\dfrac{{\not 3\not R\left( {300\,K} \right)}}{{16}} = \dfrac{{\not 3\not R{T_2}}}{{32}}$
${T_2} = \dfrac{{32\left( {300\,K} \right)}}{{16}}$
${T_2} = 600\,K$
Thus, the temperature the RMS velocity of oxygen will be the same as that of methane is $600\,K$.

$\therefore $Option C is the correct answer.


Note:
Always we have to remember to convert temperature in Celsius to Kelvin, using the conversion factor.
${0^ \circ }C = 273\,K$
Also we have to know about average velocity,
Average velocity:
Average velocity of a gas is the arithmetic mean of the velocities of different molecules of a gas at a given temperature.
Mathematic expression for Average velocity is,
$AV = \sqrt {\dfrac{{8RT}}{{\pi M}}} $
Where,
R is the universal gas constant.
T is the temperature in Kelvin.
M is the molar mass of the gas molecule.