The temperature at which oxygen molecules will have same r.m.s. speed as helium molecules at 57^ C is (molecular masses of oxygen and helium are 32 and 4 respectively.)

Step-by-step solution

The root mean square speed of gas molecules is given by v_ rms = 3RT M , where R is the universal gas constant, T is the absolute temperature in Kelvin, and M is the molar mass. Given that oxygen and helium molecules must have equal rms speeds, we set v_ rms, O ₂ = v_ rms, He , which implies 3RT_ O ₂ M_ O ₂ = 3RT_ He M_ He . Squaring both sides and canceling 3R gives T_ O ₂ M_ O ₂ = T_ He M_ He . Substitute the known values: T_ He = 330 , K , M_ He = 4 , g/mol , and M_ O ₂ = 32 , g/mol . Thus, T_ O ₂ 32 = 330 4 , s