Physics - Heat & Thermo Question with Solution | TestHub

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