Q 11-12-121JEE MainJEE Main 2020 (6 Sep, Shift 1)Easy
Molecules of an ideal gas are known to have three translational degrees of freedom and two rotational degrees of freedom. The gas is maintained at a temperature $T$. The total internal energy $U$ of a mole of this gas, and the value of $\gamma = \dfrac{C_p}{C_v}$ are, respectively:
Answer: (C) $U = \tfrac52RT$ and $\gamma = \tfrac75$
Degrees of freedom $f = 3 + 2 = 5$.
Internal energy of one mole: $U = \dfrac f2RT = \dfrac52RT$.
$\gamma = 1 + \dfrac2f = 1 + \dfrac25 = \dfrac75$.
Solution by Sreeraj P, M.Sc Physics