# Exercise 9 | Properties of Gases

Calculate the temperature at which a carbon dioxide molecule would have the same root-mean-square speed as an argon atom at 60.0°C.

vrms = $\sqrt{\frac{3\mathrm{RT}}{{\mathrm{M}}_{kg}}}$

vrms (Ar at 60.0°C) =  = 4.56 x 102 m.s-1

vrms (CO2) = $\sqrt{\frac{3\mathrm{RT}}{{\mathrm{M}}_{\mathrm{kg}}}}$

⇒ T = vrms (CO2)2

⇒ T = (4.56 x 102)2 x

⇒ T = 367 K = 93.8°C