Exercise 9 | Properties of Gases

General Chemistry 2 - Exercise 9

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 = 3RTMkg

vrms (Ar at 60.0°C) = 3 × 8.314×33339.95 × 10-3 = 4.56 x 102 m.s-1


vrms (CO2) = 3RTMkg 

⇒ T = vrms (CO2)2 Mkg CO23R

⇒ T = (4.56 x 102)2 x 44.0 × 10-33 × 8.314 

⇒ T = 367 K = 93.8°C