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).

Asked by Pragya Singh | 1 year ago |  160

1 Answer

Solution :-

Given

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 and

Let (Vrms)He be the rms speed of helium

The rms speed of argon is given by:

(Vrms)Ar = \(\sqrt{\dfrac{3RT_{Ar}}{M_{Ar}}}\) ………… (i)

Where,

R is the universal gas constant

TAr is temperature of argon gas

The rms speed of helium is given by:

(Vrms)He = \( \sqrt{\dfrac{3RT_{He}}{M_{He}}}\) ………… (ii)

Given that,

(Vrms)Ar = (Vrms)He

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

\( \dfrac{T_{Ar}}{M_{Ar} }\)\( \dfrac{T_{He}}{M_{He} }\)

TAr = \(\)\( \dfrac{T_{He}}{M_{He} \times M_{Ar}}\)

= (\( \dfrac{T_{Ar}}{M_{Ar} }\)) x 39.9

We get,

= 2523.675

= 2.52 x 103 K

Hence, the temperature of the argon atom is 2.52 x 103 K

Answered by Abhisek | 1 year ago

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