Boltzmann Constant

The Boltzmann constant is vital for comprehending phenomena in thermodynamics, statistical mechanics, and quantum mechanics. It provides insights into concepts such as temperature, entropy, and the energy distribution among particles within a system. By applying the Boltzmann constant, physicists can illustrate how macroscopic properties arise from the interactions of individual particles, establishing it as a fundamental element in the study of physical systems.

1.0Definition of Boltzman Constant

• The Boltzmann constant serves as the proportionality factor that relates the thermodynamic temperature of a gas to the average K.E. of its particles.
• It is represented by k_b.

2.0Ideal Gas Equation In Terms of Boltzman Constant

PV=nRT …..(1)

If v is the volume of 1 gram mass of the gas and M₀ is the molecular mass ,then number of moles is given as,

$n=\frac{\text{ Mass of the gas }(\mathrm{gm})}{\text{ Molecular Mass }}=\frac{1}{M_0}$

PV=nRT

$PV=\frac{1}{M_0}RT\left(r=\frac{R}{M_0}\right)$

PV=rT, perfect gas equation for 1gm of the gas

kb= Boltzman Constant, it is the gas constant per molecule.

$k_b=\frac{R}{N_A}\Rightarrow R=k_b{N}_A$

$n=\frac{N}{N_A}=\frac{\text{ no.of molecules }}{\text{ Avogadro’s Number }}$

PV=nRT

$PV=\frac{N}{N_A}\cdot k_b{N}_A\cdot T=k_{b}NT$

PV=kbNT

3.0Formula of Boltzman Constant

$K_b=\frac{R}{N_A}=\frac{JMol{e}^{-1}K-1}{mol{e}^{-1}}=J{K}^{-1}$

Dimensional Formula: $\left[M{L}^2{T}^{-2}{K}^{-1}\right]$

4.0Value of Boltzman Constant

• One mole of gas at Standard Temperature Pressure
• Calculating Value of Universal Gas Constant,

$R=\frac{PV}{T}$

P=0.76 m of Hg column = 0.76 $\times 13.6\times 10^3\times 9.8{\mathrm{Nm}}^{-2}$

Standard Temperature(T) = 273.15 K

Volume of one mole of gas = $\text{ 22. }4\text{ litre }=22.4\times 10^{-3}{\mathrm{m}}^3$

$R=\frac{0.76\times 13.6\times 10^3\times 9.8\times 22.4\times 10^{-3}}{273.15}=8.31{\mathrm{Jmole}}^{-1}{\mathrm{K}}^{-1}$

• Value of R in C.G.S $R=\frac{8.31}{4.2}\mathrm{cal}{\mathrm{mole}}^{-1}{}^{\circ }{\mathrm{C}}^{-1}$
• $k_b=\frac{R}{N_A}=\frac{8.31{\mathrm{Jmole}}^{-1}{K}^{-1}}{6.023\times 10^{23}}=1.38\times 10^{-23}{\mathrm{JK}}^{-1}$

5.0Role in Thermodynamics

$E=\frac{3}{2}{k}_{b}T$

This equation implies that mean kinetic energy per molecule is commensurate to the absolute temperature of the gas. Faster the motion of the molecules of a gas, higher will be their kinetic energy and hence higher will be the temperature of the gas

6.0Application of Boltzman Constant

In Law of Equipartition of Energy

• This law realms that in any dynamical system in thermal equilibrium, the energy is equally disseminated amongst its various degree of freedom and the energy linked with each degree of freedom per molecule is $\frac{1}{2}{k}_{b}T$, where k_b = Boltzmann constant and T is the absolute temperature.
• The Law of Equipartition holds good for all degrees of freedom whether translational, rotational, vibrational.
• In the microscopic formulation of the ideal gas law, the Boltzmann constant connects macroscopic properties like pressure, volume, and temperature to the behavior of individual gas molecules.

7.0Sample Questions On Boltzman Constant

Q-1. In a definite region of space there are only 5 molecules per cm³ on an average. The temperature there is 3 K. What is the pressure of this gas?

Sol.

$PV=k_{b}NT$

$P=\frac{k_{b}NT}{V}=\left(\frac{N}{V}\right)k_{b}T$ 

$\left(\frac{N}{V}=5{\mathrm{cm}}^{-3}=5\times 10^6{\mathrm{m}}^{-3}\right)$

P=5 $\times 10^6\times 1.38\times 10^{-23}\times 3=20.7\times 10^{-17}{\mathrm{Nm}}^{-2}$

Q-2. There are N molecules of a gas in a vessel. If the number of molecules is intensified to 2N, what will be the total energy of the gas?

Sol.

Average K.E per molecule, $\frac{1}{2}m{v}^2=\frac{3}{2}{k}_{b}T$

Total energy of N molecules=, $\frac{1}{2}mN{v}^2=\frac{3}{2}{k}_{b}NT$

When the number of molecules is increased from N to 2N,total energy of the gas is doubled, though the average K.E. per molecule remains the same.

Q-3. At what temperature does all molecular motion cease. Explain?

Sol.

$E=\frac{3}{2}{k}_{b}T$

All molecular motion ceases at absolute zero or at zero Kelvin,

Absolute Temperature ∝ Average K.E of the molecules

T=0, Average K.E=0

Q-4. A prototype of an ideal gas occupies a volume V at a pressure P and absolute temperature T. The mass of each molecule is m, $k_b$ is Boltzman Constant, write the expression for the density of the gas.

Sol.

$P=\frac{1}{3}\rho V^2=\frac{2}{3}\frac{\rho }{m}\cdot \frac{1}{2}m{v}^2=\frac{2}{3}\cdot \frac{\rho }{m}\cdot E=\frac{3}{2}{k}_{b}T$

Density $\rho =\frac{Pm}{k_{b}T}$

Q5. What are the limitations of Boltzman Constant?

Solution:

1. It is applicable to ideal gases.

2.The Boltzmann constant is primarily applicable at standard and moderate temperatures. However, at extremely high temperatures (like those found in plasmas) or very low temperatures (close to absolute zero), different physical phenomena can arise that necessitate alternative descriptions.