The helium and argon are put in the flask at the same room temperature (300 K). The ratio of average kinetic energies (per molecule) of helium and argon is:

(Give: Molar mass of helium = 4 g/mol, Molar mass of argon = 40 g/mol)

<p>Given the same room temperature of 300 K for helium and argon, we are tasked with finding the ratio of their average kinetic energies per molecule.</p> <p><strong>Formula for Kinetic Energy per Molecule:</strong></p> <p>$ \mathrm{K.E.} = \frac{\mathrm{f}}{2} \mathrm{kT} $</p> <p>In this equation:</p> <p><p>$\mathrm{f}$ is the degrees of freedom, which is 3 for both helium (He) and argon (Ar) since they are monatomic gases.</p></p> <p><p>$k$ is the Boltzmann constant.</p></p> <p><p>$T$ is the temperature in Kelvin.</p></p> <p><strong>Kinetic Energy Comparison:</strong></p> <p>For both helium and argon, since $\mathrm{f} = 3$, the expression simplifies:</p> <p>$ \frac{\mathrm{K} \cdot \mathrm{E}_{\mathrm{He}}}{\mathrm{K} \cdot \mathrm{E}_{\mathrm{Ar}}} = \frac{1}{1} $</p> <p>Thus, the average kinetic energy per molecule for both gases is the same at the same temperature, leading to a ratio of:</p> <p>$ 1:1 $</p> <p>This result implies that despite their differences in molar masses (helium being 4 g/mol and argon 40 g/mol), the average kinetic energy per molecule remains equal at the same temperature.</p>