**1**
kunl
**·**2011-03-05 03:14:52
If you succeed to solve the one above then u would take this one with a* little more care*

At temp. T: 2AB_{2}(g) =2AB(g)+B_{2}(g)

with a degree of dissociation x.Find expression of x in terms of K_{p} and total pressure p!

**1**
kunl
**·**2011-03-05 03:30:04
final 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!

**49**
Subhomoy Bakshi
**·**2011-03-05 07:09:10
3) the eqm (i think) will be 10cm for side A!

And ye kya hai bhai?

why 2 of us??

humne kya bigada??! :P

**49**
Subhomoy Bakshi
**·**2011-03-05 07:19:43
2) i am getting,

K_{P}=4x^{3}P(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]

**1**
kunl
**·**2011-03-05 07:51:10
bigada 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]

**1**
kunl
**·**2011-03-05 07:53:11
second ka answer is little wrong[1]

**49**
Subhomoy Bakshi
**·**2011-03-05 07:57:34
3)

in initial case, pressure must be equated at both ends!

thus,

p(2V)=n_{1}R(4T)

p(V)=n_{2}R(T)

gives n_{1}:n_{2}=1:2

n_{1}=k

n_{2}=2k

at second case,

the p and T on both sides will be same!

using the data we get A=10 cm side!

**49**
Subhomoy Bakshi
**·**2011-03-05 08:02:11
ohh shit!! :P

i took 2x in place of ur x! :D :D :D

**1**
kunl
**·**2011-03-05 08:23:14
anyone for 1st question??

**1**
swordfish
**·**2011-03-06 01:15:38
Is the answer for first part of first question

PM/dRT - 1 ?