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The molecule of a given mas of gas have ...

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.

Text Solution

Verified by Experts

The correct Answer is:
`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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Knowledge Check

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