Question

At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of helium gas at 20 degree Celsius

Answer 1

Answer:

Root mean square speed of Argan atom

( Vrms1 ) = √{3RT1/M1} -----(1)

Root mean square spped of Helium atom ( Vrms2) = √ { 3RT2/M2}-------(2)

Divide equations (1) and (2)

Vrms1/Vrms2 = √{3RT1/M1}/√{3RT2/M2}

= √{ T1 ×M2/T2×M2}

But ,

Vrms1 = Vrms2 [ A/C to question , ]

1 = √{ T1 × M2/M1 × T2}

M1/M2 = T1/T2

So,

T1 = T2 × { M1/M2}

Here,

T2 = -20°C = -20+273 = 253 K

M1 = 40g/mol

M2 = 4 g/mol

T1 = ?

T1 = 253 × { 40 /4 }

= 2530 K

Explanation: