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The process by which a gas passes throug...

The process by which a gas passes through a small hole into vacuum is called effusion. The rate of change of pressure(p) of a gas at constant temperature due to effusion of gas from a vessel of constant volume can be related to rate of change of number of molecules by the expression:
`(dp)/(dt)=(kT)/V((dN)/(dt))`
where rate of change of number of molecules
`Rightarrow -((dN)/(dt))=(pA_(0))/(2pimkT)^(1//2)`
where k=Boltzmann constant
`N_(A)` = Avogadro's number
T= Temperature (in K)
V= volume of vessel
N=Number of molecules
`A_(0)` = Area of aperture
m=Mass of single molecule
`gamma=V/A_(0)sqrt((2pim)/(kT)`
In 1 m long tube at one end He is introduced while from other end `SO_(2)` is introduced under identical conditions Gas will first meet from He and at a distance:

A

`1/2` m

B

`1/5` m

C

`3/5` m

D

`4/5` m

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The correct Answer is:
d
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The process by which a gas passes through a small hole into vacuum is called effusion. The rate of change of pressure(p) of a gas at constant temperature due to effusion of gas from a vessel of constant volume can be related to rate of change of number of molecules by the expression: (dp)/(dt)=(kT)/V((dN)/(dt)) where rate of change of number of molecules Rightarrow -((dN)/(dt))=(pA_(0))/(2pimkT)^(1//2) where k=Boltzmann constant N_(A) = Avogadro's number T= Temperature (in K) V= volume of vessel N=Number of molecules A_(0) = Area of aperture m=Mass of single molecule gamma=V/A_(0)sqrt((2pim)/(kT) If 2 g of SO_(2) effuses from given container in 10 sec then, mass of He effusing out in 30 seconds under identical conditions will be:

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