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The rate of effusion of two gases a and ...

The rate of effusion of two gases `a` and `b` under identical conditions of temperature and pressure is in the ratio of 2 : 1. What is the ratio of rms velocity of their molecules if `T_(a)` and `T_(b)` are in the ratio of 2 : 1?

A

`2 : 1`

B

`sqrt(2) : 1`

C

`2sqrt(2) : 1`

D

`1 : sqrt(2)`

Text Solution

Verified by Experts

The correct Answer is:
C
`(r_a)/(r_b)=sqrt((M_b)/(M_a)),(2)/(1)= sqrt((M_b)/(M_a))" or "(M_b)/(M_a)= (4)/(1)`
`(C_a)/(C_b)=(sqrt((3RT_a)/(M_a)))/(sqrt((3RT_b)/(M_b)))implies (C_a)/(C_b)=sqrt((T_(a)M_(b))/(T_(b)M_(a)))=sqrt((2)/(1)xx(4)/(1))`
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