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JEE MAIN 2025
28-01-2025 SHIFT-2
Question
The kinetic energy of translation of the molecules in 50 g of $\mathrm{CO}_{2}$ gas at $17^{\circ} \mathrm{C}$ is
Select the correct option:
A
3582.7 J
B
4102.8 J
C
3986.3 J
D
4205.5 J
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\quad(\mathrm{KE})_{\text {Trandational }}=\left[\frac{3}{2} \mathrm{KT}\right] \times$ no. of molecule
No. of molecule $=\left[\frac{50}{44} \times 6.023 \times 10^{23}\right] ;(\mathrm{KE})_{\text {Tnnsataional }}=4108.644 \mathrm{~J}$
Question Tags
JEE Main
Physics
Medium
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