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

Question

Aicharya · Accepted Answer

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
$\begin{aligned}
& =\frac{3}{2}\left(\frac{25}{22}\right)(8.3)(290) \mathrm{J} \\ & =4102.8 \mathrm{~J}
\end{aligned}$

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

Solution