Q64 / 75 · JEE Main 2025 (28 Jan Shift 2) · Physics

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

  • A
    4205.5 J
  • Correct answer
    B
    4102.8 J
  • C
    3582.7 J
  • D
    3986.3 J

Solution

Kinetic energy of translation $=\frac{3}{2} n R T$ $\begin{aligned} & n=\frac{50 \mathrm{~g}}{44 \mathrm{~g}}=\frac{25}{22} \mathrm{~mol} \\ & T=17^{\circ} \mathrm{C}=290 \mathrm{~K} \end{aligned}$ $\Rightarrow$ Kinetic energy of translation $\begin{aligned} & =\frac{3}{2}\left(\frac{25}{22}\right)(8.3)(290) \mathrm{J} \\ & =4102.8 \mathrm{~J} \end{aligned}$

Hence, the correct answer is option: 'B'.

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