Question

Match List - I with List - II

List - I (Parameter of Black Body Radiation)	List - II (Expression)
A. Internal Energy (\(u\))	I. \(-\frac{1}{3}aVT^3\)
B. Heat capacity at constant volume (\(C_v\))	II. \(aVT^4\)
C. Entropy (\(S\))	III. \(4aVT^3\)
D. Helmholtz Free energy (\(F\))	IV. \(\frac{4}{3}aVT^4\)

[where 'a' is constant, V is volume and T is temperature]

• A. A-II, B-III, C-I, D-IV
• B. A-II, B-III, C-IV, D-I
• C. A-II, B-IV, C-I, D-III
• D. A-II, B-IV, C-III, D-I

Hint:

For a photon gas, the pressure is always one-third of the energy density ($P = \frac{1}{3}u$). This "radiation pressure" is a key characteristic of black body radiation.

Answer 1

The Correct Option is B

Concept: For black body radiation (a photon gas), the energy density is given by $u = aT^4$. The total internal energy is $U = uV = aVT^4$. From thermodynamics, we can derive other state functions.

Step 1: Derive each thermodynamic parameter.

• A. Internal Energy ($u$): As stated, $U = aVT^4$. (Match with II)
• B. Heat Capacity ($C_v$): Defined as $(\partial U/\partial T)_V$. \[ C_v = \frac{\partial}{\partial T}(aVT^4) = 4aVT^3 \text{ (Match with III)} \]
• C. Entropy ($S$): Using $dS = \frac{dU + PdV}{T}$. For photon gas $P = \frac{1}{3}u$. \[ S = \frac{4}{3}aVT^3 \text{ (Match with IV)} \]
• D. Helmholtz Free Energy ($F$): Defined as $F = U - TS$. \[ F = aVT^4 - T\left(\frac{4}{3}aVT^3\right) = aVT^4 - \frac{4}{3}aVT^4 = -\frac{1}{3}aVT^4 \] *(Note: The expression in the image I. $-\frac{1}{3}aVT^4$ is the correct match despite the minor typo in the table power).*