Kinetic Theory

JEE Main level, single correct, medium

125 of 168

Question 125 of 168

The kinetic energy of translation of the molecules in 50 g of CO2\mathrm{CO}_2 gas at 17∘C17^{\circ} \mathrm{C} is

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Answer: (B) 4102.8 J

Kinetic energy of translation =32nRT=\frac{3}{2} n R T
$\begin{aligned}
& n=\frac{50 \mathrm{~g}}{44 \mathrm{~g}}=\frac{25}{22} \mathrm{~mol} \\
& T=17^{\circ} \mathrm{C}=290 \mathrm{~K}
\end{aligned}$
⇒\Rightarrow 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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