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At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the r.m.s. speed of a helium gas atom at `-20^(@) C` ? (Atomic mass of Ar = 39.9 u, of He = 4.0 u).A. `2.52xx10^(3)K`B. `2.52xx10^(2)K`C. `4.03xx10^(3)K`D. `4.03xx10^(2)K` |
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Answer» Correct Answer - A Let 1 and 2 represent for Argon atom and Helium atom. rms speed of Argon, `v_(rms_(1))=sqrt((3RT_(1))/(M_(1)))` rms speed of Helium `v_(rms_(2))=sqrt((3RT_(2))/(M_(2)))` According to question, `v_(rms_(1))=v_(rms_(2))` `thereforesqrt((3RT_(1))/(M_(1)))=sqrt((3RT_(2))/(M_(2))),(T_(1))/(M_(1))=(T_(2))/(M_(2))` `or T_(1)(T_(2))/(M_(2))xxM_(1)=(253)/(4)xx39.9=2.52xx10^(3)K` |
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