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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 -

Given

Temperature of the helium atom, THe = -200 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 speedof argon and

Let (Vrms)He be the rms speedof helium

The rms speed of argon is given by:

(Vrms)Ar = √3RTAr / MAr ………… (i)

Where,

R is the universal gas constant

TAr istemperature of argon gas

The rms speed of helium is given by:

(Vrms)He = √3RTHe / MHe ………… (ii)

Given that,

(Vrms)Ar = (Vrms)He

√3RTAr /MAr = √3RTHe / MHe

TAr /MAr = THe / MHe

TAr =THe / MHe x MAr

= (253 / 4) 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

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