Q 11-12-135JEE MainJEE Main 2019 (10 Jan, Shift 2)Easy
$2\ \text{kg}$ of a monoatomic gas is at a pressure of $4\times10^4\ \text{N m}^{-2}$. The density of the gas is $8\ \text{kg m}^{-3}$. What is the order of energy of the gas due to its thermal motion?
Answer: (C) $10^4\ \text{J}$
Volume $V = \dfrac{2}{8} = 0.25\ \text{m}^3$. For a monoatomic gas
$$E = \frac32nRT = \frac32PV = \frac32\times4\times10^4\times0.25 = 1.5\times10^4\ \text{J}$$
which is of the order of $10^4\ \text{J}$.
Solution by Sreeraj P, M.Sc Physics