The number of bimolecular collisions per `cm^(3)` per second is `Z_(11)`. At constant temperature, by how much will `Z_(11)` change if the pressure is tripled in the vessel?

Options

A. Increase `3` times
B. Decrease `3` times
C. Increase `9` times
D. Decrease `9` times

Explanation

(`i`) `Z_(11)=(1)/(2)(Z_(1)N^(*))=(1)/(sqrt(2))pi sigme^(2)overline(c ) N^(*2)`
(` overline(c )=` average velocity)
(`N^(*)=` Number of molecules per `cm^(3)`)
If collision involves two unlike molecules, then `Z_(12) per cm^(3)` per second is
`Z_(12)=pi sigma_(12)^(2)overline(c )n_(1)n_(2)`, `overline(c )=sqrt((8KT)/(pi mu))`
`n_(1)` and `n_(2)` are the numbrs of molecules `per cm^(3)` of the two types of molecules, `sigma_(12)` is the average diameter of the two molecules.
i.e., `sigma_(12)=((sigma_(1)+sigma_(2))/(2))`
If `mu` is the reduced mass,
`(1)/(mu)=(1)/(m_(1))+(1)/(m_(2))`
(`ii`) `Z_(11) prop P^(2)` (if `T` is constant)
`Z_(11) prop (1)/(T^(3//2))` (if `P` is constant)
`Z_(11) prop sqrt(T)` (if `V` is constant)
`Z_(11) prop sqrt(P)` (if `V` is constant)