__Question 13.9__:
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).

__Solution__:*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}= (3RT

_{Ar}/ M

_{Ar})

^{1/2}… (

*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}= (3RT

_{He}/ M

_{He})

^{1/2}.....(ii)

It is given that:

(

*v*

_{rms})

_{Ar }= (

*v*

_{rms})

_{He}

_{}(3RT

_{Ar}/ M

_{Ar})

^{1/2}

_{}= (3RT

_{He}/M

_{He})

^{1/2}

T

_{Ar}/ M

_{Ar}= T

_{He}/ M

_{He}

T

_{Ar}= ( T

_{He}/ M

_{He}) X M

_{Ar}

= (253 / 4) X 39.9

= 2523.675 = 2.52 × 10

^{3}K

Therefore, the temperature of the argon atom is 2.52 × 10

^{3}K.

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