Root mean square speed

Chemistry Level 2

Using kinetic theory, compare the root mean square speeds of a helium atom, an oxygen molecule and a xenon atom at 298 298 K.

μ r m s ( He ) > μ r m s ( O 2 ) > μ r m s ( Xe ) \mu_{rms}(\text{He})>\mu_{rms}(\text{O}_2)>\mu_{rms}(\text{Xe}) μ r m s ( Xe ) > μ r m s ( O 2 ) > μ r m s ( He ) \mu_{rms}(\text{Xe})>\mu_{rms}(\text{O}_2)>\mu_{rms}(\text{He}) μ r m s ( O 2 ) > μ r m s ( He ) > μ r m s ( Xe ) \mu_{rms}(\text{O}_2)>\mu_{rms}(\text{He})>\mu_{rms}(\text{Xe}) μ r m s ( Xe ) > μ r m s ( He ) > μ r m s ( O 2 ) \mu_{rms}(\text{Xe})>\mu_{rms}(\text{He})>\mu_{rms}(\text{O}_2)

Krishna Singh by Krishna Singh , 22, India

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1 solution

Bleach Byakuya
Jun 3, 2014

Xenon atom is the heaviest , followed by oxygen atom and finally helium atom ...

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Wont the fact that Oxygen is diatomic will change the formula here. So, v r m s = 5 R T 32 v_{rms} = \sqrt{ \frac{5RT}{32} } instead of 3 R T 16 \sqrt{ \frac{3RT}{16} } as 5 is degrees of freedom for diatomic molecule .
Please correct me if I am wrong.

Avi Aryan - 7 years ago

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