Question

Temperature at which rms speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at $-{20}^{\circ }\text{C}$ is
(Take molar mass of argon as $40\text{g}$ and for helium as $4\text{g}$)

• A. $2530\text{K}$
• B. $3000\text{K}$
• C. $2000\text{K}$
• D. $4000\text{K}$

Answer 1

The correct option is A $2530\text{K}$

Given,
Temperature of helium gas atom, $T_1=-{20}^{\circ }\text{C}=253\text{K}$
Let temperature of argon gas be $T_2$

We know that,
$V_{rms}=\sqrt{\frac{3RT}{M}}$
For helium gas, $(V_{rms}{)}_{He}=\sqrt{\frac{3\times 8.3\times 253}{4\times {10}^{-3}}}[\text{molar mass of helium}=4\text{g}]$

For Argon gas, $(V_{rms}{)}_{Ar}=\sqrt{\frac{3\times 8.3\times T_2}{40\times {10}^{-3}}}[\text{molar mass of argon}=40\text{g}]$

On equating both speeds,
$\sqrt{\frac{3\times 8.3\times 253}{4\times {10}^{-3}}}=\sqrt{\frac{3\times 8.3\times T_2}{40\times {10}^{-3}}}\Rightarrow \sqrt{\frac{253}{1}}=\sqrt{\frac{T_2}{10}}\Rightarrow T_2=2530\text{K}$