Question

Find the temperature at which molecules of oxygen gas have the same rms speed as that of Helium atoms at $240\text{K}$.
$\{\text{Molecular mass of}{O}_2=32\text{g/mol}\&\text{He}=4\text{g/mol}\}$

• A. $1920\text{K}$
• B. ${2400}^{\circ }\text{C}$
• C. $24\text{K}$
• D. $758.95\text{K}$

Answer 1

The correct option is A $1920\text{K}$

RMS speed of a molecule of mass $M$ is given by
$V_{rms}=\sqrt{\frac{3RT}{M}}$
RMS speed of oxygen is given by $\left(V_{O_2}\right)=\sqrt{\frac{3R{T}_{O_2}}{M_{O_2}}}......\left(1\right)$
RMS speed of helium gas $\left(V_{He}\right)=\sqrt{\frac{3R{T}_{He}}{M_{He}}}......\left(2\right)$
Given that, RMS speed of oxygen gas and helium gas are same.
So, equating $\left(1\right)$ and $\left(2\right)$, we get
$\sqrt{\frac{3R{T}_{O_2}}{M_{O_2}}}=\sqrt{\frac{3R{T}_{He}}{M_{He}}}$
$\Rightarrow \frac{T_{O_2}}{M_{O_2}}=\frac{T_{He}}{M_{He}}$
$\therefore T_{O_2}=T_{He}\times \frac{M_{O_2}}{M_{He}}=240\times \frac{32}{4}=1920\text{K}$
Thus, option (a) is the correct answer.