The kinetic energy of translation of the molecules in 50 g of CO ₂ gas at 17°C is: (a) 4205.5 J (b) 3582.7 J (c) 3986.3 J (d) 4102.8 J

Answer: (d). Translational KE = 3/2nRT. n = 50/44 = 1.136 mol, T = 17 + 273 = 290 K KE = 1.5 × 1.136 × 8.3 × 290 ≈ 4102.8 J So the answer is (d) 4102.8 J.

The kinetic energy of translation of the molecules in 50 g of CO ₂… | JEE Main 2025 (28 Jan, Shift 2)

Q12JEE Main 2025 · 28 JAN · SHIFT 2Medium

Pressure & Kinetic Energy of a Gas

The kinetic energy of translation of the molecules in 50 g of CO ${}_{2}$ gas at 17°C is:

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Answer: (d)

Translational KE $= \frac{3}{2}nRT$ .
$n = \frac{50}{44} = 1.136$ mol, $T = 17 + 273 = 290$ K
$KE = 1.5 \times 1.136 \times 8.3 \times 290 \approx 4102.8$ J
So the answer is (d) 4102.8 J.

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