JEE Mains · Physics · STD 11 - Kinetic theory of gases
- JEE Mains 2025
- Easy
- MCQ
The kinetic energy of translation of the molecules in 50 g of \(\mathrm{CO}_2\) gas at \(17^{\circ} \mathrm{C}\) is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) 4102.8 J
Step-by-step Solution
Detailed explanation
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
\(\begin{aligned}
& =\frac{3}{2}\left(\frac{25}{22}\right)(8.3)(290) \mathrm{J} \\
& =4102.8 \mathrm{~J}
\end{aligned}\)
\(\begin{aligned}
& n=\frac{50 \mathrm{~g}}{44 \mathrm{~g}}=\frac{25}{22} \mathrm{~mol} \\
& T=17^{\circ} \mathrm{C}=290 \mathrm{~K}
\end{aligned}\)
\(\Rightarrow\) 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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