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

Temperature of the helium atom, T_He = –20°C= 253 K 

Atomic mass of argon, M_Ar = 39.9 u 

Atomic mass of helium, M_He = 4.0 u 

Let, (v_rms)_Ar be the rms speed of argon. 

Let (v_rms)_He be the rms speed of helium. 

The rms speed of argon is given by:

\((v_{rms})_{Ar}=\sqrt\frac{3RT_{Ar}}{M_{Ar}}.........(i)\)

Where,

R is the universal gas constant

T_Ar is temperature of argon gas

The rms speed of helium is given by:

\((v_{rms})_{He}=\sqrt\frac{3RT_{He}}{M_{He}}.........(ii)\)

It is given that:

(v_rms)Ar = (v_rms)He

\(\sqrt\frac{3RT_{Ar}}{M_{Ar}}=\sqrt\frac{3RT_{He}}{M_{He}}\)

\(\frac{T_{Ar}}{M_{Ar}}=\frac{T_{He}}{M_{He}}\)

\(T_{Ar}=\frac{T_{He}}{M_{He}}×M_{Ar}\)

\(=\frac{253}{4}×39.9\)

= 2523.675 = 2.52 × 10³ K

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