Question

At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the $rmsspeed$ of a helium gas atom at $-{20}^{\circ }C?$ (atomic mass of $Ar=39.9u,$ atomic mass of $He=4.0u).$

Answer 1

Here, atomic mass of argon $M_1=39.9u$

Atomic mass of helium $M_2=4.0u$

Let $C_1$ and $C_2$ be the rms velocities of argon and helium at temperature $T_{1}K$ and $T_{2}K$ respectively.

We know that rms speed is given by,

$C=\sqrt{\frac{3RT}{M}}$

$C_1=\sqrt{\frac{3R{T}_1}{M_1}}=\sqrt{\frac{3R{T}_1}{39.9}}$

$C_2=\sqrt{\frac{3R{T}_2}{M_2}}=\sqrt{\frac{3R\times 253}{4}}$

Since $C_1=C_2$

$\sqrt{\frac{3R{T}_1}{39.9}}=\sqrt{\frac{3R\times 253}{4}}$

$T_1=\frac{39.9\times 253}{4}=2523.7K$.