InterviewSolution
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InterviewSolution
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The molecules of a given mass of a gas have rms velocity of `200 m//s at 27^(@)C and 1.0 xx 10^(5) N//m_(2)` pressure. When the temperature and pressure of the gas are respectively `127^(@)C and 0.05 xx 10^(5) Nm^(-2)`, the rms velocity of its molecules in `ms^(-1)` isA. `(400)/(sqrt(3))`B. `(100sqrt(2))/3`C. `100/3`D. `100 sqrt(2)` |
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Answer» Correct Answer - A Here, `upsilon_(1) = 200 m//s, T_(1) = 27^(@)C = 27+273` `300 K` `P_(1) = 1.0 xx 10^(5) N//m^(2), T_(2) = 127^(@)C = 127+273` `=400K` `P_(2) = 0.05 xx 10^(5) N//m^(2), upsilon_(2)=?` As `upsilon_(rms) = sqrt((3RT)/(m))` , therefore, `upsilon_(rms) prop sqrt(T)` There is no effect of change in pressure `:. (upsilon_2)/(upsilon_1) = sqrt((T_2)/(T_1)) = sqrt(400/300) = 2/(sqrt(3))` `upsilon_(2) = (2)/(sqrt(3)) upsilon_(1) = (2 xx 200)/(sqrt(3)) = (400)/(sqrt(3))m//s`. |
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