[LucaZombini]

25,62g of N₂O₄are added to an isolated system, whose volume is 37,3 dm3 and temperature held at 298K. determinate the dissociation degree. determinate also at which volume the system must be taken to halve the dissociation degree (Kp = 1,5*10^4 Pa, which is about 0,148 atm) . the results should be:

a = 0.3601
V = 7,27 dm³

this is how i did it

n_N2O4 = 25,62/92 = 0,278
PV = nRT -> P = nRT/V

P = 0,0278*0,0821*298/37,3 = 0,18

N₂O₄  2NO₂

Kp = p2NO₂/pN₂O₄
0,148 = (2*0,18*α)² /0,18(1-α)
α = 0,3621

honestly, i don't know if i have calculated the α correctly since the 0,002 shouldn't be there. Also, i don't know how to proceed to the second part of the problem. do i have to i put the α as 0,3621/2 and calculate the volume?