# If He(g) has an average kinetic energy of 6670 J/mol under certain conditions, what is the root mean square speed of Cl2(g) molecules under the same conditions?

Nov 19, 2016

we know

${V}_{\text{rms}} = \sqrt{\frac{R T}{M}}$

where

${V}_{\text{rms"->"RMS velocity of the gas}}$

$T \to \text{Absolute temperature of the gas}$

$M \to \text{Molar mass of the gas}$

$R \to \text{Universal gas constant}$

So average molar kinetic energy of the gas

$E = \frac{1}{2} M {V}_{\text{rms}}^{2} = \frac{1}{2} R T$

This equation reveals that molar KE is independent of the nature of the gas . It only depends on temperature as ideal behavior is concerned. So both He(g) and $C {l}_{2} \left(g\right)$ will have same average $K E = 6670 \text{J/mol}$ .under the same condition of temperature.

So for $C {l}_{2} \left(g\right)$

$\frac{1}{2} {M}_{C {l}_{2} \left(g\right)} {V}_{r m s C {l}_{2}}^{2} = 6670$

$\textcolor{red}{\text{Taking atomic mass of Cl"=35.5"g/mol"=35.5xx10^-3"kg/mol}}$

$\implies {V}_{r m s C {l}_{2} \left(g\right)}^{2} = \frac{6670 \times 2}{2 \times 35.5 \times {10}^{-} 3}$

$\implies {V}_{r m s C {l}_{2} \left(g\right)} = \sqrt{\frac{6670000}{35.5}} \approx 433.5 m {s}^{-} 2$