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Question #42561

Difficulty: 8/10 | 🟢 Free

The Question

A perfectly reflecting closed cavity of volume $V$ contains a photon gas in thermal equilibrium at temperature $T$ . The total internal energy of the photon gas is $U_0$ . An identical cavity of volume $V$ contains an ideal monatomic gas, which also has a total internal energy $U_0$ . Let $P_{rad}$ be the radiation pressure of the photon gas and $P_{gas}$ be the pressure of the ideal monatomic gas. Which of the following represents the correct expression for $P_{rad}$ and the ratio $\frac{P_{rad}}{P_{gas}}$ ?

Options

A.
$P_{rad} = \frac{U_0}{3V}, \; \frac{P_{rad}}{P_{gas}} = \frac{1}{2}$
B.
$P_{rad} = \frac{2U_0}{3V}, \; \frac{P_{rad}}{P_{gas}} = 1$
C.
$P_{rad} = \frac{U_0}{3V}, \; \frac{P_{rad}}{P_{gas}} = 1$
D.
$P_{rad} = \frac{U_0}{V}, \; \frac{P_{rad}}{P_{gas}} = \frac{3}{2}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Equipartition Theorem|Radiation Pressure|Electromagnetic Waves

Why This Question Is Tricky

Common Mistakes

This question tests the distinction between the thermodynamics of an ultra-relativistic photon gas and a classical ideal gas.

Students often incorrectly apply the classical equipartition theorem ( $P = 2U/3V$ ) to photons, assuming they behave like a standard gas with 3 degrees of freedom.

This trap leads them to conclude that both gases exert the same pressure for a given internal energy density, missing the fundamental $P = U/3V$ relation derived from the $E = pc$ momentum-energy relation of massless photons.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Thermodynamics - Class 11 Part 2
Concept	Equipartition Theorem
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free