Question

The energy of radiation which is at thermal equilibrium within an enclosure depends only on the volume of the enclosure and on the wall temperature, T. It is also known that the pressure of Th q T c q’ w 3 radiation is equal to one third of the energy per unit volume. Show that the energy per unit volume, u , and entropy per unit volume, s, of the radiation are given by

uH' and s=4ET3 and S=- 4-3 ξ

where ξ is a constant (note that u = ξT^4 is Stefan's law for black body radiation)

Answer 1

It is given that energy of radiation is in equilibrium within an enclosure, this implies that the enclosure is black body chamber. In addition, it is also given that the pressure of the radiation is one-third of volume, which is Maxwell’s theory of radiation. This can be stated as
P=1/3 u ----(1)
This implies rate of change of Pressure w.r.t temperature can be stated as
dP/dT = 1/3 dU/dT-----(2)
This means we can use law of thermodynamics to find the energy of radiation.
As per thermodynamic law, the energy of a radiation can be expressed as:
dU=TdS-PdV
Here, U is the energy of radiation, T is the temperature of enclosure, which is wall, dS is change in entropy, P is the pressure of radiation and v is volume
Now, energy per unit volume will be:
dU/dv = T(dS/dV) – P ------divided the entire equation with dV
This implies, dU/dv = T(dS/dV) – 1/3 u ----- using (1)
This implies, dS/dv= T dU/dV + 1/3 u
Since, we have to find energy per unit volume dU/dV and u are one and the same.
Therefore
dS/dv = T u + (1/3)u -----(3)
Now dS/dv is change in entropy per unit volume and it can be written dP/dT using Maxwell’s relation
This implies dP/dT = u(4/3 T)
This implies dU/dT= T/4u ----using equation (2)
This implies dU/4u= dT/T
Now integrating on both sides will give
u= ξT4   
Where ξ is constant of integration, which is known as boltzman constant
Now, substitung this value in (3)
Will give dS/dv = ξT⁴ (4/3 T)
Therefore, S= 4/3 ξT³