If \( \mathrm{He}(\mathrm{g}) \) has an average kinetic energy of \( 6650 \mathrm{~J} / \mathrm{mol} \) under certain conditions, what is the root mean square speed of \( \mathrm{Cl}_{2}(\mathrm{~g}) \) molecules under the same conditions? root mean square speed:

To find the root mean square speed (\(v_{\text{rms}}\)) of \( \mathrm{Cl}_2(g) \), we can use the relationship between average kinetic energy and \(v_{\text{rms}}\). The average kinetic energy per mole is given by the formula: \[ \text{Average K.E.} = \frac{1}{2} M v_{\text{rms}}^2 \] where \(M\) is the molar mass in kg/mol. Given the average kinetic energy for \( \mathrm{He} \), we should also consider the molar mass of \( \mathrm{Cl}_2 \), which is about \( 70.9 \, \mathrm{g/mol} \) or \( 0.0709 \, \mathrm{kg/mol} \). Setting the average kinetic energy equal for both gases, we get: \[ 6650 \, \mathrm{J/mol} = \frac{1}{2} (0.0709 \, \mathrm{kg/mol}) v_{\text{rms}}^2 \] Solving for \(v_{\text{rms}}\): \[ v_{\text{rms}}^2 = \frac{2 \times 6650 \, \mathrm{J/mol}}{0.0709 \, \mathrm{kg/mol}} \approx 187,177.75 \] Taking the square root gives us: \[ v_{\text{rms}} \approx 433.55 \, \mathrm{m/s} \] So, the root mean square speed of \( \mathrm{Cl}_2(g) \) under the same conditions is approximately \( 433.6 \, \mathrm{m/s} \).