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

Hey mate ^_^

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Answer:
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Root mean square speed of Argan atom
( Vrms1 )

= √{3RT1/M1} ----- (1)

Root mean square speed 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

Here,
T2 = 253 K
M1 = 40g/mol
M2 = 4 g/mol
T1 = ?

So,
T1 = T2 × { M1/M2}
T1 = 253 × { 40 /4 }
= 2530 K

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Final answer: 2530 K
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#Be Brainly❤️

Answer 2

Temperature of the helium atom, THe = –20°C= 253 K
= 2523.675 = 2.52 × 103 K

Therefore, the temperature of the argon atom is 2.52 × 103 K.