The kinetic energy of translation of the molecules in 50 g of $ \text{CO}_2 $ gas at 17°C is :
Solution
<p>$$\begin{aligned}
& (\mathrm{KE})_{\text {Transsational }}=\left[\frac{3}{2} \mathrm{KT}\right] \times \text { no. of molecule } \\
& \text { No. of molecule }=\left[\frac{50}{44} \times 6.023 \times 10^{23}\right] \\
& (\mathrm{KE})_{\text {Transsational }}=4108.644 \mathrm{~J}
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
This question is part of PrepWiser's free JEE Main question bank. 271 more solved questions on Thermodynamics are available — start with the harder ones if your accuracy is >70%.