Question

En un proceso isobárico, la presión del gas es de 105 Pa. Halle el desplazamiento del pistón, cuya área es de 0,4 m2, cuando el gas desarrolla un trabajo de 8 x 104 J

Answer 1

The displacement of the piston is 1904.76 m

Step-by-step explanation:

Here we have an isobaric or constant pressure process, therefore;

Work done, W = P × (V₂ - V₁)

Where:

W = 8 × 10⁴ J

P = Pressure of the gas = 105 Pa

V₁ = Initial volume of the cylinder of the piston

V₂ = Final volume of the cylinder of the piston

We note that volume of the cylinder of the piston = Base area, A × h

Therefore;

V₁ = A × h₁

V₂ = A × h₂

Which gives;

V₂ - V₁ = A × h₂ - A × h₁ = A×(h₂ - h₁)

Where:

A = Base area of the piston = 0.4 m²

h₂ - h₁ = Displacement of the piston

∴ 8 × 10⁴ J = 105 Pa × (V₂ - V₁) = 105 Pa × 0.4 m² × (h₂ - h₁)

h₂ - h₁ = (8 × 10⁴ J)/(105 Pa × 0.4 m²)

h₂ - h₁ = 1904.76 m

Hence the displacement of the piston = 1904.76 m.