Ncert Exercises & Solutions

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).

The temperature of the helium atom,

T_{He}

= –20°C= 253 K

Atomic mass of argon,

M_{Ar}

= 39.9 u

The 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:

{(v_{rms})}_{Ar} = \sqrt{\frac{3 {RT}_{Ar}}{M_{Ar}}}

… (i)

The RMS speed of helium is:

{(v_{rms})}_{He} = \sqrt{\frac{3 {RT}_{He}}{M_{He}}}

… (ii)

According to the question:

{(v_{rms})}_{Ar} = {(v_{rms})}_{He}

\sqrt{\frac{3 {RT}_{Ar}}{M_{Ar}}} = \sqrt{\frac{3 {RT}_{He}}{M_{He}}}

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

T_{Ar} = \frac{T_{He}}{M_{He}} \times M_{Ar}

= \frac{253}{4} \times 39.9

= 2523.675 = 2.52 ×

10^{3}

Therefore, the temperature of the argon atom is 2.52 ×

10^{3}

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