Step 1: Calculate individual \(C_V\) values.
\[
C_{V1} = \frac{R}{\gamma_1 - 1} = \frac{R}{0.6} = 1.67R, C_{V2} = \frac{R}{\gamma_2 - 1} = \frac{R}{0.4} = 2.5R
\]
The average molar \(C_V\) for the mixture is
\[
C_V = \frac{C_{V1} + C_{V2}}{2} = \frac{1.67R + 2.5R}{2} = 2.08R
\]
Step 2: Calculate \(C_P\).
\[
C_P = C_V + R = 2.08R + R = 2.9R
\]
Step 3: Conclusion.
Thus, \(C_V = 2.08R\) and \(C_P = 2.9R\).

At a particular temperature T, Planck's energy density of black body radiation in terms of frequency is \(\rho_T(\nu) = 8 \times 10^{-18} \text{ J/m}^3 \text{ Hz}^{-1}\) at \(\nu = 3 \times 10^{14}\) Hz. Then Planck's energy density \(\rho_T(\lambda)\) at the corresponding wavelength (\(\lambda\)) has the value \rule{1cm}{0.15mm} \(\times 10^2 \text{ J/m}^4\). (in integer)
[Speed of light \(c = 3 \times 10^8\) m/s]
(Note: The unit for \(\rho_T(\nu)\) in the original problem was given as J/m³, which is dimensionally incorrect for a spectral density. The correct unit J/(m³·Hz) or J·s/m³ is used here for the solution.)