If the mass of molecules of a gas in a closed vessel is halved and the speed doubled, then the ratio of initial to final pressure will be

4 : 1

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1 : 4

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1 : 2

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2 :1

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Solution

The correct option is B 1 : 2
V $_{r m s}$ ( root mean square velocity ) = $\sqrt{\frac{3 P V}{M}}$
Given that,mass of molecules is halved and speed is doubled
$\frac{v_{1}}{v_{2}}$ = $\sqrt{\frac{\frac{3 P_{1} V_{1}}{M_{1}}}{\frac{3 P_{2} V_{2}}{M_{2}}}}$
$v_{1} = v_{2}$
$M_{2} = \frac{M_{1}}{2}$ {new consideration}
$V_{1 r m s} = 2 V_{2 r m s}$
$\frac{1}{2} = \sqrt{\frac{\frac{P_{1}}{M_{1}}}{\frac{P_{2}}{M_{2}}}}$
$\frac{1}{2} = \sqrt{\frac{P_{1} M_{2}}{P_{2} M_{1}}}$
$M_{1} = 2 M_{2}$
$\frac{P_{1}}{P_{2}}$ = $\frac{1}{2}$

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If the mass of molecules of a gas in a closed vessel is halved and the speed doubled, then the ratio of initial to final pressure will be

The correct option is B 1 : 2Vrms( root mean square velocity ) = √3PVMGiven that,mass of molecules is halved and speed is doubledv1v2=    ⎷3P1V1M13P2V2M2v1=v2M2=M12 {new consideration}V1rms=2V2rms12=    ⎷P1M1P2M212=√P1M2P2M1M1=2M2P1P2=12