Question

Given that the rms speed of a helium atom at a certain temperature is 1350 m/s1350 \mathrm{~m} / \mathrm{s}, find by proportion the rms speed of an oxygen (O2)\left(\mathrm{O}_2\right) molecule at this temperature. The molar mass of O2\mathrm{O}_2 is 32.0 g/mol32.0 \mathrm{~g} / \mathrm{mol}, and the molar mass of He\mathrm{He} is 4.00 g/mol4.00 \mathrm{~g} / \mathrm{mol}

Solution

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Answered 2 years ago
Answered 2 years ago
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To solve this problem we have to note that at the same temperature both species will have the same kinetic energy so it has to hold

mOvO2=mHevHe2m_Ov_O^2=m_{He}v_{He}^2

We can solve this for the oxygen speed

vO2=mHemOvHe2=432×13502v_O^2=\frac{m_{He}}{m_O}v_{He}^2=\frac{4}{32}\times 1350^2

Finally, we have that

vO=477m/sv_O=477\textrm{m/s}

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