JEE Main 2025 Jan 28 Shift 2, Physics Q5: Kinetic Theory of an Ideal Gases

JEE Main2025Jan 28, Shift 2Physics

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
C
$4205.5$ J
D
$3582.7$ J

Solution

For an ideal gas, the average translational kinetic energy per molecule is $\frac{3}{2} k_{B} T$ , so total translational KE is $\frac{3}{2} n R T$ where $n$ is moles.

$n = \frac{50}{44}$ mol; $T = 290$ K; $R = 8.314$ J/mol/K.

$K E = \frac{3}{2} (\frac{50}{44}) (8.314) (290) \approx 4103$ J.

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JEE Main 2025 Jan 28 Shift 2, Physics Q5: Kinetic Theory of an Ideal Gases

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