The value of Boltzmann constant is: (In erg $K^{-1}$ molecul$e^{-1}$)
A. $1.38 \times \mathop {10}\nolimits^{ - 16}$
B. $1.38 \times \mathop {10}\nolimits^{ - 23}$
C. $8.314 \times \mathop {10}\nolimits^7$
D. $6.023 \times \mathop {10}\nolimits^{ - 16}$
Answer
248.7k+ views
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: $\dfrac{{m{v^2}}}{2} = \dfrac{3}{2}kT$ 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 : ${K_b} = \dfrac{R}{{{N_A}}}$.
Now we can calculate the value of Kb by using the formula: ${K_b} = \dfrac{R}{{{N_A}}}$
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, ${K_b} = \dfrac{R}{{{N_A}}} = \dfrac{{8.3144}}{{6.02214 \times {{10}^{23}}}} = 1.3806 \times {10^{ - 23}}$ 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 ${M^2}{L^2}{T^{ - 2}}{\theta ^{ - 1}}$.
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: $\dfrac{{m{v^2}}}{2} = \dfrac{3}{2}kT$ 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 : ${K_b} = \dfrac{R}{{{N_A}}}$.
Now we can calculate the value of Kb by using the formula: ${K_b} = \dfrac{R}{{{N_A}}}$
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, ${K_b} = \dfrac{R}{{{N_A}}} = \dfrac{{8.3144}}{{6.02214 \times {{10}^{23}}}} = 1.3806 \times {10^{ - 23}}$ 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 ${M^2}{L^2}{T^{ - 2}}{\theta ^{ - 1}}$.
Recently Updated Pages
JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Isoelectronic Definition in Chemistry: Meaning, Examples & Trends

Ionisation Energy and Ionisation Potential Explained

Iodoform Reactions - Important Concepts and Tips for JEE

Introduction to Dimensions: Understanding the Basics

Instantaneous Velocity Explained: Formula, Examples & Graphs

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Electric Field of a Uniformly Charged Ring

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Derivation of Equation of Trajectory Explained for Students

Understanding the Angle of Deviation in a Prism

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

CBSE Notes Class 11 Chemistry Chapter 9 - Hydrocarbons - 2025-26

CBSE Notes Class 11 Chemistry Chapter 5 - Thermodynamics - 2025-26

CBSE Notes Class 11 Chemistry Chapter 8 - Organic Chemistry Some Basic Principles And Techniques - 2025-26

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

