Question

How much work (in j) is involved in a chemical reaction if the volume decreases from 5.00 to 1.26 l against a constant pressure of 0.857 atm?

Answer 1

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
$W = \int p dV = (86836 \, \frac{N}{m^{2}} )*(-3.74 \times 10^{-3} \, m^{3}) = -324.77 \, J$

Answer:
The work involved is 324.8 J