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N molecules each of mass m of gas A and ...

`N` molecules each of mass `m` of gas `A` and `2 N` molecules each of mass `2m` of gas `B` are contained in the same vessel which is maintained at a temperature `T`. The mean square velocity of the molecules of `B` type is denoted by `v^(2)` and the mean square of the x-component of the velocity of `A` type is denoted by `omega^(2)`. What is the ratio of `omega^(2)//v^(2) = ?`

A

`3:2`

B

`1:3`

C

`2:3`

D

`1:1`

Text Solution

Verified by Experts

The correct Answer is:
C
The mean sequare velocity of gas molecules is giben by `v^(2)=(3kT)/(m).`
For gas A, `v_(A)^(2)=(3kT)/(m)" "...(i)`
For a gas molecule.
`v^(2)-v_(x)^(2)+v_(y)^(2)+v_(z)^(2)=3v_(x)^(2)" "(thereforev_(x)^(2)=v_(y)^(2)=v_(z)^(2))`
or `v_(x)^(2)=(v^(2))/(3)`
From eqn. (i), we get
`w^(2)=v_(x)^(2)=[(3kT)/(m/(3))]=(kT)/(m)" "...(ii)`
For gas B, `v_(B)^(2)=v^(2)=(3kT)/(2m)" "...(iii)`
Dividing eqn. (ii) by eqn. (iii), we get
`(w^(2))/(v^(2))=(KT)/((m/(3kT))/(2m))=2/3`
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N molecules each of mass m of gas A and 2 N molecules each of mass 2m of gas B are contained in the same vessel which is maintined at a temperature T. The mean square of the velocity of the molecules of B type is denoted by v^(2) and the mean square of the x-component of the velocity of a tye is denoted by omega^(2) . What is the ratio of omega^(2)//v^(2) = ?

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Knowledge Check

  • N molecules , each of mass m, of gas A and 2 N molecules , each of mass 2m, of gas B are contained in the same vessel which is maintained at a temperature T. The mean square velocity of molecules of B type is denoted by V_(2) and the mean square velocity of A type is denoted by V_(1) then (V_(1))/(V_(2)) is

    A
    `2`
    B
    `1`
    C
    `1//3`
    D
    `2//3`

  • Root mean square velocity of a gas molecule is proportional to

    A
    `m^(1//2)`
    B
    `m^(0)`
    C
    `m^(-1//2)`
    D
    m

  • Root mean square velocity of a gas molecule is proprotional to

    A
    `m^(1//2)`
    B
    `m^(0)`
    C
    `m^(-1//2)`
    D
    `m`

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