Kinetic Theory

First of all how will any external agent do any work.You need some hand to move it.But how can some hand come here?And even then you are able to move then how did you calculate final temperatures?I do not know how to calculate final temperatures.But I consider that you have calculated right.But you have done one mistake in calculating the internal energy.The mistake have been done in calculating change in the temperature.
Let me calculate the change in temperature
Final temperature:Suppose you have a metallic stick.Half part of it is at the temperature T and the other half is at T' then will the temperature of the stick be the sum of these two temperature values.Certainly not.So calculating this temperature is out of our iitjee course.
Initial temperature:Consider the system initially to be a rod.Then half of this rod is at the temperature T and the other half is at the temperature T.So temperature of the rod has to be just T not 2T.
I think you have understood all my points.So in my way the problem is wrong and it is not so important for iitjee.
And also your gamma given is wrong.For mono it s 1.33 ,for di 1.4 and for ti 1.7.The value given by is in between 1.4 and 1.7.So which kind of molecule is this.And this value is not in our course.It may be in higher classes.

firstly this is a very good question(bhai tu question lata kaha se hai)

consider an external agent applying a force F in the left direction then P1 n P2 be pressure in the container since the process is carried out slowly clearly system shall remain at equilibrium which implies that P1a+F=P2a

F=∫_V1^V2(P2-P1)a

now consider the piston has travelled a distance x then volume V1=V0+ax
V2=V0-ax

also V0+ax=n(V0-ax)
x=(n-1)v0/(n+1)A

now

∫(nRT/v0-Ax-nRT/v0+Ax)adx...........................................3

now this work done expression has a unknown T

this work done shall be utilized in in ncresing the internal energy.

hence
∫(nRT/v0-Ax-nRT/v0+Ax)adx=∫2nC_vdt

solve to get T=T_initial[(n+1)²/4n]^(y-1)/2 where y is adiabatic exponent.

now calculate work done by substituting this T in 3

phew damn good question .

What is n here.And what about division in the ratio 3:1.My question was how did you get the value of gamma to be 1.5.Is it some two and half atomic gas.If it is so then for which kind of molecule is this?

n is the ratio mantioned and vinay i think you are the most complaining person i have seen what the hell does the molecule matter

@ JEE Topper -- you made a simple problem very complex.
Whats the answer you are getting by your method?
I got W = 2PV(√2/√3 + √2 - 2)

:D