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 $^{\circ} C$ ? (atomic mass of Ar = 39.9 u, of He = 4.0 u)

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Solution

Temperature of the helium atom, $T_{H e} = - 20^{o} C = 253 K$
Atomic mass of argon, $M_{A r} = 39.9 u$
Atomic mass of helium, $M_{H e} = 4.0 u$
Let, $(V_{r m s})_{A r}$ be the rms speed of argon.
Let $(V_{r m s})_{H e}$ be the rms speed of helium.

The rms speed of argon is given by:
$(ν_{r m s})_{A r} = \sqrt{\frac{3 R T_{A r}}{M_{A r}}}$ ... (i)
Where,
R is the universal gas constant
$T_{A R}$ is temperature of argon gas

The rms speed of helium is given by:
$(ν_{r m s})_{H e} = \sqrt{\frac{3 R T_{H e}}{M_{H e}}}$ ... (ii)

It is given that:
$(ν_{r m s})_{A r} = (ν_{r m s})_{H e}$

$\sqrt{\frac{3 R T_{A r}}{M_{A r}}} = \sqrt{\frac{3 R T_{H e}}{M_{H e}}}$

$\frac{T_{A r}}{M_{A r}} = \frac{T_{H e}}{M_{H e}}$

$T_{A r} = \frac{T_{H e}}{M_{H e}} \times M_{A r}$

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

$= 2523.675 = 2.52 \times 10^{3} K$

Therefore, the temperature of the argon atom is $2.52 \times 10^{3} K$ .

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