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The molecule of a given mas of gas have r.m.s. speed `200 ms^(-1)` at `27^(@)C` and `10^(5)Nm^(-2)` pressure. When the absolute temperature is doubled and the pressure is halved, then find rms speed of the molecules of the same gas. |
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Answer» Correct Answer - `200sqrt(2)ms^(-1)` `V_(rms) = sqrt((3RT)/(m)), (V_(rms_(2)))/(V_(rms_(1))) = sqrt((T_(2))/(T_(1))) = sqrt((2xx300)/(300))` `:. V_(rms2) = V_(rms1) sqrt(2) = 200 sqrt(2)` Note: There is no effect of change in pressure on r.m.s. velocity. This is because density also varies in this case. |
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