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 ×

$10^{3}$ K

Therefore, the temperature of the argon atom is 2.52 ×

$10^{3}$ K.

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