
At what temperature will the average speed of $C{H_4}$ molecules have the same value as ${O_2}$ has at $300K$ ?
A. $1200K$
B. $150K$
C. $600K$
D. $300K$
Answer
504.9k+ views
Hint: We have to know that, the Kinetic Molecular Theory of Gases, a gas contains countless particles in fast movements. Every molecule has an alternate speed, and every crash between particles changes the rates of the particles. A comprehension of the properties of the gas requires a comprehension of the conveyance of molecule speeds.
Complete answer:
If we somehow managed to plot the quantity of particles whose speeds fall inside a progression of limited reaches, we would get a marginally hilter kilter bend known as a speed appropriation. The pinnacle of this bend would compare to the most likely speed. This speed circulation bend is known as the Maxwell-Boltzmann conveyance, yet is often alluded to simply by Boltzmann's name.
By using the following expression,
${u_{average}} = \sqrt {\dfrac{{8RT}}{{\pi M}}} $
Where,
$T$ is a temperature, $R$ is a universal gas constant, $M$ is the molar mass.
For $C{H_4}$ ,
The molar mass of methane = $16$
Substituting molar mass value in the average speed expression,
${u_{average}} = \sqrt {\dfrac{{8R{T_1}}}{{\pi 16}}} $
For ${O_2}$ ,
The molar mass of oxygen = $32$
Temperature = $300K$
Substituting both molar mass and temperature values in the average speed expression,
\[{u_{average}} = \sqrt {\dfrac{{8R\left( {300} \right)}}{{\pi \left( {32} \right)}}} \]
Both methane and oxygen expressions are equating,
\[\sqrt {\dfrac{{8R{T_1}}}{{\pi 16}}} = \sqrt {\dfrac{{8R\left( {300} \right)}}{{\pi \left( {32} \right)}}} \]
Then,
\[\dfrac{{{T_1}}}{{16}} = \dfrac{{300}}{{32}}\]
Therefore,
${T_1} = 150K$
So, the correct answer is “Option B”.
Note:
Average speed is the normal speed is the amount of the rates of the entirety of the particles separated by the quantity of particles. Most probable speed is the most likely speed is the speed related with the most elevated point in the Maxwell dispersion. Just a few particles may have this speed, however it is more probable than some other speed.
Complete answer:
If we somehow managed to plot the quantity of particles whose speeds fall inside a progression of limited reaches, we would get a marginally hilter kilter bend known as a speed appropriation. The pinnacle of this bend would compare to the most likely speed. This speed circulation bend is known as the Maxwell-Boltzmann conveyance, yet is often alluded to simply by Boltzmann's name.
By using the following expression,
${u_{average}} = \sqrt {\dfrac{{8RT}}{{\pi M}}} $
Where,
$T$ is a temperature, $R$ is a universal gas constant, $M$ is the molar mass.
For $C{H_4}$ ,
The molar mass of methane = $16$
Substituting molar mass value in the average speed expression,
${u_{average}} = \sqrt {\dfrac{{8R{T_1}}}{{\pi 16}}} $
For ${O_2}$ ,
The molar mass of oxygen = $32$
Temperature = $300K$
Substituting both molar mass and temperature values in the average speed expression,
\[{u_{average}} = \sqrt {\dfrac{{8R\left( {300} \right)}}{{\pi \left( {32} \right)}}} \]
Both methane and oxygen expressions are equating,
\[\sqrt {\dfrac{{8R{T_1}}}{{\pi 16}}} = \sqrt {\dfrac{{8R\left( {300} \right)}}{{\pi \left( {32} \right)}}} \]
Then,
\[\dfrac{{{T_1}}}{{16}} = \dfrac{{300}}{{32}}\]
Therefore,
${T_1} = 150K$
So, the correct answer is “Option B”.
Note:
Average speed is the normal speed is the amount of the rates of the entirety of the particles separated by the quantity of particles. Most probable speed is the most likely speed is the speed related with the most elevated point in the Maxwell dispersion. Just a few particles may have this speed, however it is more probable than some other speed.
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