InterviewSolution
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InterviewSolution
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Two molecules of a gas have speeds of `9xx10^(6) ms^(-1)` and `1xx10^(6) ms^(-1)`, respectively. What is the root mean square speed of these molecules. |
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Answer» For n- molecules we know that `V_("rms") =sqrt((V_(1)^(2)+V_(2)^(2)+V_(3)^(2)+.........+V_(n)^(2))/(n))" "[V_("rms")underset("square velocity")(= "root mean"]]` Where`V_(1),V_(2),V_(3) ..................V_(n)` are individual velocities of n-molecules of the gas. For two molecules, `V_("rms") =sqrt((V_(1)^(2)+V_(2)^(2))/(2)) " "[V_(1),V_(2),V_(3).............V_(n)"are individual velocity"]` `" Given " " "V_(1) 9xx 10^(6)m//s` `"and "" "V_(2) 1xx 10^(6) m//s` `:. " " V_("rms") =sqrt(((9xx10^(6))^(2)xx(1 xx 10^(6))^(2))/(2))` `=sqrt((81 xx 10^(12) + 1xx10^(12))/(2))` `=sqrt(((81 +1)xx10^(12))/(2))` `=sqrt((82 xx 10^(12))/(2))` `=sqrt(41)xx10^(6) m//s` |
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