The change in volume is
V₂ - V₁ = 1.26 - 5 L = - 3.74 L = - 3.74 x 10⁻³ m³
The constant pressure is
p = 0.857 atm = 0.857*101325 Pa = 86836 Pa
By definition, the work done is
[tex]W = \int p dV = (86836 \, \frac{N}{m^{2}} )*(-3.74 \times 10^{-3} \, m^{3}) = -324.77 \, J[/tex]
Answer:
The work involved is 324.8 J
