Question

From the relation \[{v_{rms}} = \sqrt {\left( {\dfrac{{3{K_B}T}}{m}} \right)} \], it follows that the constant ${K_B}$ should be expressed in units of……...

Answer 1

Hint: The unit of a physical quantity is an internationally accepted standard of measurement. Units are classified into two categories, namely, fundamental units and derived units. The unit of a particular physical quantity can be deduced from the known units of the other physical quantities. The units of all the digits given in the equation have to be known. Thereafter, using these units in the equation the required unit comes out.

Complete answer:
In the question, it is required to find the unit of ${K_B}$. The unit of ${K_B}$ can be deduced from the units of \[{v_{rms}}\], \[T\], and \[m\].
The equation \[{v_{rms}} = \sqrt {\left( {\dfrac{{3{K_B}T}}{m}} \right)} \] is an equation of kinetic molecular theory, where \[{v_{rms}}\] is the RMS velocity, \[T\] is the temperature, and \[m\]is the mass of the molecule.
The S.I. unit of \[{v_{rms}}\] is \[\dfrac{{\text{m}}}{{\text{s}}}\], \[T\] is \[{\text{s}}\]and \[m\]is \[{\text{kg}}\].
Square both sides of the equation\[{v_{rms}} = \sqrt {\left( {\dfrac{{3{K_B}T}}{m}} \right)} \]. Simplify and solve for ${K_B}$.
\[{\left( {{v_{rms}}} \right)^2} = {\left[ {\sqrt {\left( {\dfrac{{3{K_B}T}}{m}} \right)} } \right]^2}\]
\[ \Rightarrow v_{rms}^2 = \dfrac{{3{K_B}T}}{m}\]
\[ \Rightarrow 3{K_B}T = mv_{rms}^2\]
\[ \Rightarrow {K_B} = \dfrac{{mv_{rms}^2}}{{3T}}\]
To calculate the unit of ${K_B}$, substitute the units of \[{v_{rms}}\], \[T\], and \[m\]in the above equation.
Substitute \[\dfrac{{\text{m}}}{{\text{s}}}\] for \[{v_{rms}}\], \[{\text{s}}\] for \[T\], and \[{\text{kg}}\] for \[m\] and calculate the unit of ${K_B}$.
\[{\text{Unit of }}{{\text{K}}_{\text{B}}}{\text{ = }}\dfrac{{{\text{kg}}{{\left( {\dfrac{{\text{m}}}{{\text{s}}}} \right)}^{\text{2}}}}}{{\text{K}}}\]
\[ \Rightarrow {\text{Unit of }}{{\text{K}}_{\text{B}}}{\text{ = }}\dfrac{{{\text{kg}}{{\text{m}}^{\text{2}}}}}{{{{\text{s}}^2}{\text{K}}}}\]
\[ \Rightarrow {\text{Unit of }}{{\text{K}}_{\text{B}}}{\text{ = kg}}{{\text{m}}^{\text{2}}}{{\text{s}}^{ - 2}}{{\text{K}}^{ - 1}}\]
But\[{\text{kg}}{{\text{m}}^{\text{2}}}{{\text{s}}^{ - 2}}\] corresponds to the unit of energy, which is Joule $\left( {\text{J}} \right)$.
Substitute ${\text{J}}$ for \[{\text{kg}}{{\text{m}}^{\text{2}}}{{\text{s}}^{ - 2}}\] in the unit of ${K_B}$.
\[{\text{Unit of }}{{\text{K}}_{\text{B}}}{\text{ = J}}{{\text{K}}^{ - 1}}\].
Therefore, the unit of ${K_B}$ is \[{\text{J}}{{\text{K}}^{ - 1}}\].


Note:
The units of the fundamental quantities like distance, mass, time, etc are known as fundamental units. They cannot be deduced from other units and also cannot be resolved into any further simpler form.
The units that are deduced from the fundamental units are known as derived units. It is a grouping of the S.I. units.