Question

The value of Boltzmann constant is: (In erg $K^{-1}$ molecul$e^{-1}$)
A. $1.38\times {10}^{-16}$
B. $1.38\times {10}^{-23}$
C. $8.314\times {10}^7$
D. $6.023\times {10}^{-16}$

Answer 1

Hint: It is a proportionality factor that relates average kinetic energy of particles in gas with thermodynamic temperature of gas.

Complete step by step solution:
It is known that Boltzmann constant ($k_b$), is a physical constant relating the average kinetic energy of particles in a gas with the temperature of the gas.
It is sort of a conversion type.
For simple ideal gases whose molecules are of mass m and have only kinetic energy, the Boltzmann constant k relates the average kinetic energy per molecule to the absolute temperature. The relationship can be given by: ${\displaystyle \frac{m{v}^2}{2}}={\displaystyle \frac{3}{2}}kT$ where $v^2$ is the average of the squared velocity of gas molecules and $T$is the absolute temperature(in kelvin).
Also, it is the gas constant R divided by the Avogadro number NA : $K_b={\displaystyle \frac{R}{N_A}}$.
Now we can calculate the value of Kb by using the formula: $K_b={\displaystyle \frac{R}{N_A}}$

Calculation:
We know value of gas constant, $R=8.3144J/K/mol$
Also, value of Avogadro number, $N_A=6.02214\times {10}^{23}$
Therefore, Boltzmann constant, $K_b={\displaystyle \frac{R}{N_A}}={\displaystyle \frac{8.3144}{6.02214\times {10}^{23}}}=1.3806\times {10}^{-23}$ J/K/molecule
Now to convert the above calculated value of $K_b$ from J/K/molecule to erg $K^{-1}$ molecul$e^{-1}$, we have to multiply the above calculated value by 107:
$k_b=\left(1.3806\times {10}^{-23}\right)\left({10}^7\right)=1.3806\times {10}^{-16}$ erg $K^{-1}$ molecul$e^{-1}$.
Hence, from above points we can now easily conclude that option A is the correct option.

Note: It should be remembered that Boltzmann constant is measured by measuring atomic speed of gas or speed of sound of gas. Also, one should remember the dimensional formula for Boltzmann’s constant which is $M^2{L}^2{T}^{-2}{\theta }^{-1}$.