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Q.

If the root mean square velocity of the molecules of gas enclosed in a vessel is doubled, the pressure becomes

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a

4 times

b

double

c

half

d

same

answer is A.

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Detailed Solution

$P_{1} = p; V_{{rms}_{2}} = V_{{rms}_{1}}$
To find $P_{2}$ : We Know that
$V_{rms} = \sqrt{\frac{3 P}{d}} V_{rms} \propto \sqrt{P} V_{{rms}_{2}} = 2 V_{{rms}_{1}} Therefore, \frac{V_{{rms}_{2}}}{V_{{rms}_{2}}} = 2 \sqrt{\frac{P_{2}}{P_{1}}} = 2 \Rightarrow P_{2} = 4 P_{1}$

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