Question

State the law of equipartition of energy. Use it to obtain the value of gamma(=Cp/Cv) for a monoatomic gas​

Answer 1

A single atom is free to move in space along the X, Y and Z axis. However, each of these movements requires energy. This is derived from the energy held by the atom. The Law of Equipartition of Energy defines the allocation of energy to each motion of the atom (translational, rotational and vibrational). Before we understand this law, let’s understand a concept called ‘Degrees of Freedom’.

Answer 2

$\underline{\underline{\textbf{Law \:of \:Equipartition \:of\: Energy}}}$

According to it, for nay dynamical system in thermal equilibrium the total energy is distributed equally amongst all degrees of freedom and energy associated with each molecule per degree of freedom is $\dfrac{1}{2}$ $K_{B}$T.

• $K_{B}$ = Boltz means constant.

• T = Absolute temperature.

$\underline{\textbf{Gamma\:for\:Monoatomic\:Gas\::}}$

Consider 1 mole of monoatomic gas is heated to temperature dT.

Here degree of freedom = 3

[A = number of particle in system.

R = number of relation among particles.

N = number of degree of freedom.

N = 3A - R

» Here;

A = 1, R = 0

N = 3(1) - 0 = 3]

Therefore energy per particle = 3 × $\dfrac{1}{2}$ $K_{B}$T

Energy per particle of mole = $(\dfrac{3}{2})$ $K_{B}$T $N_{A}$

U = $\dfrac{3}{2}$ RT

Differentiate with respect to time (T)

$\dfrac{dU}{dT}$ = $\dfrac{3R}{2}$ $( \dfrac{DT}{DT} )$

$\dfrac{dU}{dT}$ = $\dfrac{3R}{2}$

dU = $\dfrac{3}{2}$ RdT ....(1)

At constant volume

dQ = $C_{v}$.1.dT

Also;

dW = PdV = 0

From first law of thermodynamics

dQ = dU + dW

$C_{v}$.1.dT = $\dfrac{3RdT}{2}$ + 0

$C_{v}$ = $\dfrac{3R}{2}$

Now..

$C_{P}$ - $C_{V}$ = R

$C_{P}$ = $C_{V}$ + R

$C_{P}$ = $\dfrac{3R}{2}$ + R

$C_{P}$ = $\dfrac{5}{2}$

$\gamma$ = $\dfrac{C_{P}}{C_{V}}$

= $\dfrac{ \frac{5R}{2} }{ \frac{3R}{2} }$

= $\dfrac{5}{3}$

= $\boxed{1.66}$