Equilibrium, Kp constant, Partial Pressures

A system in equilibrium contains 
$$I\_{2\_(g)}\ \text{and} \ I\_{(g)}$$
under pressures $$P\_1=0,21 \,\text{atm. and}\ P\_2=0,23 \,\text{atm.}$$
While its temperature remains constant, we reduce the system's volume to half its original value.


I have to determine $P\_1'$ when the system's equilibrium is reestablished.


I know I can calculate $K\_p$ and I know I somehow have to use $PV =NRT$, but from there on I am lost, I cannot figure out how to come up with enough equations to find all of my unknowns. Can anyone show me an explicit solution to the problem?


What I tried, but got really nowhere:


$$2I\_{(g)} \leftrightarrow I\_{2\_{(g)}}, K\_p=\frac{P\_1}{P\_2^2}=3,97.$$
$$P\_t=P\_1 + P\_2 = 0,44 \,\text{atm.}$$


When the new equilibrium is established :


$$Kp=\frac{P\_1'}{P\_2'^2}=3,97$$


$$\frac{2P\_1 }{P\_1'}=\frac{n\_1}{n\_1'}$$
and same with $P\_2, P\_2', n\_2, n\_2'$, but I don't know where to go from there.




The method referred to in the comment is not correct and is a common mistake. One may think,


$$p\_1'+p\_2'=p\_t\frac{V\_t}{V\_t'}=2p\_t=0.88 \rm atm,$$


but this is invalid since $n\_1+n\_2\neq n\_1'+n\_2'$, thus $p\_tV\_t\neq p\_t'V\_t'$


The correct way is to use the fact that the number of iodine atoms is constant, from which we get,
$$2n\_1+n\_2=2n\_1'+n\_2',$$
$$(2p\_1+p\_2)V\_t=(2p\_1'+p\_2')V\_t',$$
$$1.3 {\rm atm}=4p\_1+2p\_2=2p\_1'+p\_2',$$


Then you can go on by combining this with the equation you have already obtained from $K\_p$ (listed in your question) and solve for $p\_1'$ and $p\_2'$.


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