JEE MAIN - Physics (2015 (Offline) - No. 20)
Consider a spherical shell of radius $$R$$ at temperature $$T$$. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume $$u = {U \over V}\, \propto \,{T^4}$$ and pressure $$p = {1 \over 3}\left( {{U \over V}} \right)$$ . If the shell now undergoes an adiabatic expansion the relation between $$T$$ and $$R$$ is:
$$T\, \propto {1 \over R}$$
$$T\, \propto {1 \over {{R^3}}}$$
$$T\, \propto \,{e^{ - R}}$$
$$T\, \propto \,{e^{ - 3R}}$$
Explanation
As, $$P = {1 \over 3}\left( {{U \over V}} \right)$$
But $$\,\,\,\,$$ $${U \over V} = KT{}^4$$
So, $$\,\,\,\,\,P = {1 \over 3}K{T^4}$$
or $$\,\,\,\,{{uRT} \over V} = {1 \over 3}K{T^4}\,\,\,\,$$
$$\left[ \, \right.$$ As $$PV = uRT$$ $$\left. \, \right]$$
$${4 \over 3}\pi {R^3}{T^3} = $$ $$constant$$
$$\therefore$$ $$\,\,\,\,$$ $$T \propto {1 \over R}$$
But $$\,\,\,\,$$ $${U \over V} = KT{}^4$$
So, $$\,\,\,\,\,P = {1 \over 3}K{T^4}$$
or $$\,\,\,\,{{uRT} \over V} = {1 \over 3}K{T^4}\,\,\,\,$$
$$\left[ \, \right.$$ As $$PV = uRT$$ $$\left. \, \right]$$
$${4 \over 3}\pi {R^3}{T^3} = $$ $$constant$$
$$\therefore$$ $$\,\,\,\,$$ $$T \propto {1 \over R}$$
Comments (0)
