For dissociation of PCl5 if relative molar mass is M and total pressure at equilibrium is p at a density d and temperature d,find an expression for degree of dissociation.If vapor density has a value 62 when temperature is 230°C what is value of p/d!
1If you succeed to solve the one above then u would take this one with a little more care
At temp. T: 2AB2(g) =2AB(g)+B2(g)
with a degree of dissociation x.Find expression of x in terms of Kp and total pressure p!
1final blow:
30 cm long tube is partition by thin weakly conducting frictionless separator.Ideal gases filled in two parts.In beginning temperature of two parts A and B is 400K and 100K.Separator slides to a momentary rest when length of A is 20cm.Find final final equilibrium position!
493) the eqm (i think) will be 10cm for side A!
And ye kya hai bhai?
why 2 of us??
humne kya bigada??! :P
492) i am getting,
KP=4x3P(1+x)(1-2x)2
approximating x<<1
we get,
x=\left( {\frac{K_P}{4P}}\right)^{\frac{1}{3}}
tell me if i am wrong! [1]
1bigada kuch nai....koi attempt karta nahin hai mere doubts seedhe se toh thought of a twist[3][3]
bhai u r right for 1st one plz post full solution for third one[1]....i mean 3rd one[1]
493)
in initial case, pressure must be equated at both ends!
thus,
p(2V)=n1R(4T)
p(V)=n2R(T)
gives n1:n2=1:2
n1=k
n2=2k
at second case,
the p and T on both sides will be same!
using the data we get A=10 cm side!
