
Relation between kinetic energy(E) and RMS speed is given by:
A. \[\mu = {(\dfrac{{2E}}{m})^{1/2}}\]
B. \[\mu = {(\dfrac{{3E}}{{2m}})^{1/2}}\]
C. \[\mu = {(\dfrac{E}{{2m}})^{1/2}}\]
D. \[\mu = {(\dfrac{E}{{3m}})^{1/2}}\]
Answer
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Hint: According to the kinetic molecular theory of gases, the average kinetic energy of molecules is directly proportional to the absolute temperature of the gas (\[K.E.\alpha T\] ). This can be represented by the following equation:
\[K.E. = \dfrac{3}{2}kT\]
K.E. represents Kinetic energy(average)
K represents Boltzmann constant
T represents Absolute temperature
Boltzmann constant is gas constant R divided by the Avogadro’s constant.
Kinetic energy is related to root mean square velocity by the following equation:
\[K.E. = \dfrac{1}{2}m{\mu ^2}\]
M represents mass
\[\mu \] represents Root mean square speed
K.E. represents Kinetic energy(average)
Complete step by step answer:
From the hint above we can see that: \[K.E. = \dfrac{3}{2}kT\] (equation 1)
We also know that root mean square velocity is given as: \[\mu = \sqrt {\dfrac{{3kT}}{m}} \] (equation 2)
Root mean square velocity can be defined as the square root of the average of the square of the velocity. When we measure the velocity of a particle at a given time it leads to a large distribution of values. Even the particles are moving in different directions so the velocity could be zero as some particles are moving slowly, and some are moving quickly. To properly assess the average velocity, the average of squares of velocities is taken and then the square root of that value is taken. This is defined as the root mean square velocity.
\[ \Rightarrow {\mu ^2} = \dfrac{{3kT}}{m}\] (Squaring both the sides in the above equation)
From equation 1
\[3kT = 2K.E.\]
Substituting this value in equation 2
\[\mu = \sqrt {\dfrac{{2K.E.}}{m}} \]
So, the correct answer is Option A.
Note: Gaseous particles move at random speeds in random directions. The Maxwell-Boltzmann distribution describes the average speeds of a collection of gaseous particles at a time. The average velocity of gases are generally expressed in the form of root mean square average value.
\[K.E. = \dfrac{3}{2}kT\]
K.E. represents Kinetic energy(average)
K represents Boltzmann constant
T represents Absolute temperature
Boltzmann constant is gas constant R divided by the Avogadro’s constant.
Kinetic energy is related to root mean square velocity by the following equation:
\[K.E. = \dfrac{1}{2}m{\mu ^2}\]
M represents mass
\[\mu \] represents Root mean square speed
K.E. represents Kinetic energy(average)
Complete step by step answer:
From the hint above we can see that: \[K.E. = \dfrac{3}{2}kT\] (equation 1)
We also know that root mean square velocity is given as: \[\mu = \sqrt {\dfrac{{3kT}}{m}} \] (equation 2)
Root mean square velocity can be defined as the square root of the average of the square of the velocity. When we measure the velocity of a particle at a given time it leads to a large distribution of values. Even the particles are moving in different directions so the velocity could be zero as some particles are moving slowly, and some are moving quickly. To properly assess the average velocity, the average of squares of velocities is taken and then the square root of that value is taken. This is defined as the root mean square velocity.
\[ \Rightarrow {\mu ^2} = \dfrac{{3kT}}{m}\] (Squaring both the sides in the above equation)
From equation 1
\[3kT = 2K.E.\]
Substituting this value in equation 2
\[\mu = \sqrt {\dfrac{{2K.E.}}{m}} \]
So, the correct answer is Option A.
Note: Gaseous particles move at random speeds in random directions. The Maxwell-Boltzmann distribution describes the average speeds of a collection of gaseous particles at a time. The average velocity of gases are generally expressed in the form of root mean square average value.
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