Calculate the work energy, w, gained or lost by the system when a gas expands from 15 l to 35 l against a constant external pressure of 1.5 atm

In physical chemistry or in thermodynamics, the work done on the system or by the system (depending on the sign convention) can be determined in several ways. When assumptions like ideal gas behavior is applied, then the formula for work is

W = Δ (PV)

which is the change of the product of Pressure and Volume. But since it was specified that Pressure is constant, the work could be simplified into

W = PΔV = P (V₂ - V₁)

Since we already know the constant pressure and the volumes of the ideal gas before and after the change, we could now solve for work. But let's establish first the units of work which is in Joules. When simplified, Joules is equal to m³*Pa. So, we first change the unit of pressure from atm to Pascals (1 atm = 101,325 Pa) and the unit of volume from liters to m³ (1 m³ = 1000 L),

1.5 atm * 101325 Pa/1 atm = 151987.5 Pa

15 L * 1 m³/1000 L = 0.015 m³

35 L * 1 m³/1000 L = 0.035 m³

Then, they are now ready for substitution,

W = 151987.5 Pa (0.035 m³ - 0.015 m³)

W = 3,039.75 Joules