i think since pressure is low, then volume is high.. so volume correction will be neglected...
??
Q2) since pressure is low so pressure correction n2aV2 can be neglected.
So equation becomes, P(V - nb) = nRT Dividing by nRT througout we get,
PVnRT - PnbnRT = 1,
Z - bRTp = 1,
Z = 1 + bRTp ................which is option(b)
(not sure of the ans.)
i think since pressure is low, then volume is high.. so volume correction will be neglected...
??
Ans 3 A..see when NaoH is added buffer is formed at point A ..then at point B first equivalence..then at C again buffer is formed then second equivalence at D
nicely explained...
i was thinking all the time it was monobasic acid....
yes agree with Asish volume correction shud be neglected wen volume is very high
@Manmay we say 1 + α ≈ 1 wen α < < 1 ..so since volume is high so an2/V2 will be low and presuure is also given very less..so we cant neglect it..
then what should be the answer govind??
btw solved the first one?
ya. sorry i did the opposite thing.
So by neglecting Volume correction I guess ans will be (d)