Question

The temperature at which oxygen molecules have the same root mean square speed as helium atoms have at $300K$ is: Atomic masses: He = 4 u, O = 16 u)

• A. 300 K
• B. 600 K
• C. 1200 K
• D. 2400 K

Answer 1

The correct option is D 2400 K

$u_{rms}=\sqrt{\frac{3RT}{M}}$

$u_{rms}\left(O_2\right)=u_{rms}\left(He\right)$

$\sqrt{\frac{3R{T}_{O_2}}{M_{O_2}}}=\sqrt{\frac{3R{T}_{He}}{M_{He}}}$

$\frac{T_{O_2}}{M_{O_2}}=\frac{T_{He}}{M_{He}}$

$\therefore T_{O_2}=\frac{300\times 32}{4}=2400K$