Question

At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20 °C? (atomic mass of Ar = 39.9 u, of He = 4.0 u).

Answer 1

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

Answer 2

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: