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.

Answer

`200sqrt(2)ms^(-1)`

Explanation

`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.