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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?A. Increase `3` timesB. Decrease `3` timesC. Increase `9` timesD. Decrease `9` times |
Answer» (`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) |
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