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JEE Main 202528 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
28 January
Shift
Evening Shift
Subject
Physics
Chapter
Kinetic Theory of Gases

JEE Main 2025 · 28 January · Evening Shift · Physics · Kinetic Theory of Gases

The kinetic energy of translation of the molecules in 50 g of CO_2...

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

Options

  1. A4205.5 J
  2. B4102.8 J
  3. C3582.7 J
  4. D3986.3 J

Correct answer

B. 4102.8 J

Step-by-step solution

Kinetic energy of translation =32nRT=\frac{3}{2} n R T n=50 g44 g=2522 molT=17∘C=290 K\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 =32(2522)(8.3)(290)J=4102.8 J\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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