Please Help! I do not know how to do this!! Please explain! A reaction vessel at 27 ∘C contains a mixture of SO2(P= 2.90 atm ) and O2?

Let y atm be decrease in partial pressure of O₂.

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
At equilibrium :
Partial pressure of SO₂, P(SO₂) = (2.90 - 2y) atm
Partial pressure of O₂, P(O₂) = (1.00 - y) atm
Partial pressure of SO₃ = 2y atm

Total pressure at equilibrium (atm) :
(2.90 - 2y) + (1.00 - y) + 2y = 3.65
2.90 - 2y + 1.00 - y + 2y = 3.65
3.90 - y = 3.65
y = 0.25

Hence, at equilibrium :
P(SO₂) = (2.90 - 2 × 0.25) atm = 2.40 atm
P(O₂) = (1.00 - 0.25) atm = 0.75 atm
P(SO₃) = 2 × 0.25 atm = 0.50 atm

PV = nRT
Concentration, n/V = P/RT

Kc = [SO₃]² / {[SO₂]² [O₂]}
Kc = [P(SO₃)/RT]² / {[P(SO₂)/RT]² × [P(O₂)/RT]}
Kc = {(P(SO₃))² / [(P(SO₂))² × P(O₂)]} × RT
Kc = {0.5² / [2.4² × 0.75]} × 0.0821 × (273 + 27)
Kc = 1.43