JEE Main · 2015 · 04 Apr 2015 · Offline · Heat & Thermodynamics

WhatsApp Email

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 = \frac{U}{V} \propto T^{4}

and pressure

p = \frac{1}{3} (\frac{U}{V})

. If the shell now undergoes an adiabatic expansion the relation between T and R is:

(A) $T \propto e^{- 3 R}$
(B) $T \propto \frac{1}{R}$ ✓ Correct
(C) $T \propto \frac{1}{R ^{3}}$
(D) $T \propto e^{- R}$

Answer: (B)

Solution

For a reversible adiabatic change the photon gas does work only at the expense of its own internal energy, so with

d Q = 0

we have

d U = - p d V

. Feeding in the radiation pressure

p = \frac{1}{3} \frac{U}{V}

turns this into

\frac{d U}{U} = - \frac{1}{3} \frac{d V}{V}

, which integrates to

U V^{1/3} = constant

. Now bring in the fact peculiar to black-body radiation, that the energy density

U / V \propto T^{4}

and hence

U \propto V T^{4}

. Substituting,

V T^{4} \cdot V^{1/3} = V^{4/3} T^{4}

stays constant, and for a spherical cavity with

V \propto R^{3}

this reads

R^{4} T^{4} = constant

. Taking the fourth root leaves the clean scaling

T \propto 1/ R

JEE Main · 2015 · 04 Apr 2015 · Offline · Single correct option

Practise more: all JEE Main PYQs on Heat & Thermodynamics →

JEE Main · 2015 · 04 Apr 2015 · Offline · Heat & Thermodynamics

Solution

More JEE Main questions on Heat & Thermodynamics