Let point 1 be the small pipe and 2 be the larger pipe. Using the continuity equation,
ρ₁A₁v₁ = ρ₂A₂v₂
Since density is the same,
A₁v₁ = A₂v₂
A(1) = (2A)(v₂)
v₂ = 0.5 m/s
Then, we use the mechanical energy balance:
(P₂ - P₁)/ρ + (v₂² - v₁²)/2 = 0
(P₂ - 1000 Pa)/ 1000 kg/m³ + (0.5² - 1²)/2 = 0
P₂ = 1375 Pa
Thus, the pressure of the larger pipe is 1,375 Pa.