Q 11-12-037JEE MainJEE Main 2025 (28 Jan, Shift 2)Easy
The kinetic energy of translation of the molecules in $50\ \text{g}$ of $\text{CO}_2$ gas at $17^\circ\text{C}$ is
Answer: (B) $4102.8\ \text{J}$
Moles: $n = \dfrac{50}{44}$; $T = 290\ \text{K}$.
$$KE = \frac{3}{2}nRT = \frac{3}{2}\times\frac{50}{44}\times8.3\times290 \approx 4103\ \text{J}$$
This matches $4102.8\ \text{J}$. (With $R = 8.314$ it comes to about $4110\ \text{J}$, still closest to this option.)
Solution by Sreeraj P, M.Sc Physics