Question

Show that rms velocity of an oxygen molecule is 2 times that of a sulfur dioxide molecule at S.T.P.

Answer 1

Given : \(\frac{m_o(SO_2)}{m_o(O_2)}\) = 64/32 =2 

The rms speed  v_rms = \(\sqrt{\frac{3RT}{M_o}}\) 

v_rms ∝ \(\frac{1}{\sqrt{M_o}}\) at constant T

\(\frac{v_{rms}(O_2)}{v_{rms}(SO_2)}\) = \(\sqrt{\frac{M_o(SO_2)}{M_o(O_2)}}=\sqrt{2}\)

Thus v_rms(O₂) = \(\sqrt{2}\) v_rms(SO₂