13the expression is
v_{av}=\sqrt{\frac{2kT}{m}}
10 Answers
1ya i think yes vectorially...........
as they will be roaming in all directions...........its average velocity will be zero as all of them cancel away......VECTORIALLY.....
1can u tell if the Q is complete or not...coz, in case the Q says in a container then , we can say that it returns to its mean position....
so no displacement and so 0 velocity
1wel mue i dont think os its done like that............the number of particles going out of a plane and coming into it is taken equal....so the net velocity is ZERO..............
and abhirup i think tht exp is for avg speed.......
1but mue...........even if it was in a container wat gurantee can u give tht d molecule will come into its mean position?????
1we can guarantee, coz they r always in random motion.........
kab tak bach ke bhagta rahega us position se...........
when all the positions will be covered up, then atleast it will come to its mean position.......[3]
1hehe.....its not necesari tht if u have crossed a particular position....then u cant come back to it untill u crossed all spaces rite???
1i said even if it does not come b4 crossing.........
means, if it comes earlier, then itz ok, but if not, at last it will have to.....now, 1 thing'z there, u agreed that it will cm and so in a container itz 0, but the Q boy'z gone...he did not ans my doubt