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

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Heat & Thermodynamics

The kinetic energy of translation of the molecules in 50 g of

CO_{2}

gas at

1 7^{\circ} C

is :

(A)3986.3 J
(B)4102.8 J✓ Correct
(C)4205.5 J
(D)3582.7 J

Answer: (B)

Solution

The mean translational kinetic energy of an ideal gas is

K E_{t r an s} = \frac{3}{2} n R T

Given mass of CO

_{2}

= 50 g, molar mass = 44 g/mol, so

n = \frac{50}{44} = \frac{25}{22} mol

Temperature

T = 1 7^{\circ} C = 290 K

, and using

R = 8.3 J/(mol\cdotK)

K E_{t r an s} = \frac{3}{2} \times \frac{25}{22} \times 8.3 \times 290

= \frac{3}{2} \times \frac{25 \times 8.3 \times 290}{22} = \frac{3}{2} \times \frac{60175}{22} = \frac{3}{2} \times 2735.2

\approx 4102.8 J

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Single correct option

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