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, THe = –20°C= 253 K

Atomic mass of argon, MAr = 39.9 u

Atomic mass of helium, MHe = 4.0 u

Let, (vrms)Ar be the rms speed of argon.

Let (vrms)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

TAr 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:

(vrms)Ar = (vrms)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