the r.m.s. speed of the molecules of enclosed gas is V what will be the rms speed if the pressure is doubled keeping the temperature same?

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Solution

$V_{r m s} = \sqrt{\frac{3 R T}{M}}$
${V^{^{'}}}_{r m s} = \sqrt{\frac{3 R T^{1}}{M^{^{'}}}}$
$T^{^{'}} = 2 T$
$M^{^{'}} = \frac{M}{2}$
${V^{^{'}}}_{r m s} = 2 V_{r m s}$

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the r.m.s. speed of the molecules of enclosed gas is V what will be the rms speed if the pressure is doubled keeping the temperature same?

Vrms=√3RTMV′rms=√3RT1M′T′=2TM′=M2V′rms=2Vrms

$V_{r m s} = \sqrt{\frac{3 R T}{M}}$
${V^{^{'}}}_{r m s} = \sqrt{\frac{3 R T^{1}}{M^{^{'}}}}$
$T^{^{'}} = 2 T$
$M^{^{'}} = \frac{M}{2}$
${V^{^{'}}}_{r m s} = 2 V_{r m s}$