
The ratio of root mean square velocity to average velocity of a gas molecule at a particular temperature is:
A. $1.086:1$
B. $1:1.086$
C. $2:1.086$
D. $1.086:2$
Answer
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Hint: The root-mean-square speed is a measure of the speed of particles in a gas used in kinetic theory of gases, and is defined as the square root of the average velocity-squared of the molecules in a gas. It is represented by the equation:
${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $ where ${v_{rms}}$ is the root-mean-square of the velocity, $M$ is the molar mass of the gas in kilograms per mole, $R$ is the molar gas constant, and $T$ is the temperature in Kelvin.
And average velocity is defined as the arithmetic mean of the velocities of different gas molecules at a specific given temperature and is represented by the equation:
${v_{avg}} = \sqrt {\dfrac{{8RT}}{{\pi M}}} $where ${v_{avg}}$ is the average velocity of gas molecules, $M$ is the molar mass of the gas and the units is kilograms per mole, $R$ is referred as molar gas constant, and $T$ is the temperature and the units are in Kelvin.
Complete step by step solution:
As we know, that Root mean square speed is represented by ${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $and the average speed is represented by, ${v_{avg}} = \sqrt {\dfrac{{8RT}}{{\pi M}}} $taking the ratio of these two terms we get,
$\dfrac{{{v_{rms}}}}{{{v_{avg}}}} = \sqrt {\dfrac{{\dfrac{{3RT}}{M}}}{{\dfrac{{8RT}}{{\pi M}}}}} $
$\dfrac{{{v_{rms}}}}{{{v_{avg}}}} = \sqrt {\dfrac{{3\pi }}{8}} $value of $\pi = 3.14$
$\dfrac{{{v_{rms}}}}{{{v_{avg}}}} = 1.086:1$
So, the correct option is A.
Additional Information: The Maxwell-Boltzmann Distribution describes the average molecular speeds for a collection of gas molecules at a specific given temperature, which includes our root mean square values , average speed values and the most probable speed values as well.
Note: The Maxwell-Boltzmann Distribution which describes the average molecular speeds is limited to ideal gases. As in the case of real gases, there are many effects that come into play, for example the van der Waals interactions, the relativistic speed limits and the quantum exchange interactions which can make their speed distribution different from the fundamental Maxwell–Boltzmann Distribution.
${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $ where ${v_{rms}}$ is the root-mean-square of the velocity, $M$ is the molar mass of the gas in kilograms per mole, $R$ is the molar gas constant, and $T$ is the temperature in Kelvin.
And average velocity is defined as the arithmetic mean of the velocities of different gas molecules at a specific given temperature and is represented by the equation:
${v_{avg}} = \sqrt {\dfrac{{8RT}}{{\pi M}}} $where ${v_{avg}}$ is the average velocity of gas molecules, $M$ is the molar mass of the gas and the units is kilograms per mole, $R$ is referred as molar gas constant, and $T$ is the temperature and the units are in Kelvin.
Complete step by step solution:
As we know, that Root mean square speed is represented by ${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $and the average speed is represented by, ${v_{avg}} = \sqrt {\dfrac{{8RT}}{{\pi M}}} $taking the ratio of these two terms we get,
$\dfrac{{{v_{rms}}}}{{{v_{avg}}}} = \sqrt {\dfrac{{\dfrac{{3RT}}{M}}}{{\dfrac{{8RT}}{{\pi M}}}}} $
$\dfrac{{{v_{rms}}}}{{{v_{avg}}}} = \sqrt {\dfrac{{3\pi }}{8}} $value of $\pi = 3.14$
$\dfrac{{{v_{rms}}}}{{{v_{avg}}}} = 1.086:1$
So, the correct option is A.
Additional Information: The Maxwell-Boltzmann Distribution describes the average molecular speeds for a collection of gas molecules at a specific given temperature, which includes our root mean square values , average speed values and the most probable speed values as well.
Note: The Maxwell-Boltzmann Distribution which describes the average molecular speeds is limited to ideal gases. As in the case of real gases, there are many effects that come into play, for example the van der Waals interactions, the relativistic speed limits and the quantum exchange interactions which can make their speed distribution different from the fundamental Maxwell–Boltzmann Distribution.
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