Question

At what temperature the RMS velocity of oxygen will be the same at that of methane at $27^\circ C$?
A. $54^\circ C$
B. $327K$
C. $600K$
D. $573K$

Answer 1

We know that root mean square velocity could be defined as the measure of the particle speed in gas and is defined as square root of the average velocity-squared of the molecules found in a gas. We can write the expression of RMS velocity as,
$RMS = \sqrt {\dfrac{{3RT}}{M}}$
Here, the universal gas constant is given as R.
Temperature (in Kelvin) is given as T.
Molar mass of the gas molecule is given as M.

Complete answer:
Consider the velocity of methane as ${V_m}$ and ${V_o}$ as the velocity of oxygen. We can write the RMS velocity of methane as,
${V_m} = \sqrt {\dfrac{{3R{T_1}}}{{{M_1}}}}$ ………. (1)
Here, ${T_1}$ represents the temperature of methane gas and ${M_1}$ is the molar mass of methane gas.
The RMS velocity of oxygen as,
${V_0} = \sqrt {\dfrac{{3R{T_2}}}{{{M_2}}}}$ ………. (2)
Here, ${T_2}$ represents the temperature of oxygen gas and ${M_2}$ is the molar mass of oxygen gas.
The equations (1) and (2) are equated.
$\sqrt {\dfrac{{3R{T_1}}}{{{M_1}}}} \,\,$=$\sqrt {\dfrac{{3R{T_2}}}{{{M_2}}}}$ ……… (3)
The values of temperature and molar mass of oxygen and methane are substituted.
$\sqrt {\dfrac{{3R\left( {300K} \right)}}{{16}}} \,\,$=$\sqrt {\dfrac{{3R{T_2}}}{{32}}} \,$
We have to square on both sides to get temperature ${T_2}$
$\dfrac{{3R\left( {300K} \right)}}{{16}} = \dfrac{{3R{T_2}}}{{32}}$
${T_2} = \dfrac{{32\left( {300K} \right)}}{{16}}$
On simplification we get,
${T_2} = 600K$
So, Temperature the RMS velocity of oxygen will be the same as that of methane at $27^\circ C$ is $600K$.

So, the correct answer is “Option C”.

Note:
We have to convert the temperature in Celsius to kelvin with the help of a conversion factor. The conversion factor is $0^\circ C = 273K$. We have to know about the average velocity. Average velocity of a gas is the arithmetic mean of velocities of several molecules of gas at mentioned temperature. We have to know the average velocity is,
$AV = \sqrt {\dfrac{{8RT}}{M}}$
Here, the universal gas constant is given as R.
Temperature (in Kelvin) is given as T.
Molar mass of the gas molecule is given as M.