JEE Main 2025 — Kinetic Theory of Gases Question with Solution
JEE Main 2025 (28 Jan Shift 2)
Question
The kinetic energy of translation of the molecules in 50 g of gas at is
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Show full solutionCorrect option: B
Correct answer
B4102.8 J
Step-by-step explanation
Kinetic energy of translation
$\begin{aligned}
& n=\frac{50 \mathrm{~g}}{44 \mathrm{~g}}=\frac{25}{22} \mathrm{~mol} \\
& T=17^{\circ} \mathrm{C}=290 \mathrm{~K}
\end{aligned}$
Kinetic energy of translation
$\begin{aligned}
& =\frac{3}{2}\left(\frac{25}{22}\right)(8.3)(290) \mathrm{J} \\
& =4102.8 \mathrm{~J}
\end{aligned}$
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