5.0 moles of an Ideal gas at 3.0 atm pressure and 27^ C is compressed isothermally to half its volume by application of an external pressure of 3.5 ~atm . What is the amount of work done (in joules) on the gas? Given: 1 ~L ~atm =101.3 ~J : R =0.082 ~L ~atm ~K ⁻¹ ~mol ⁻¹

Step-by-step solution

The initial state of the gas is given by n = 5.0 mol, P₁ = 3.0 atm, and T = 27^ C = 300 K. Using the ideal gas law PV = nRT , the initial volume V₁ is calculated as: V₁ = nRT P₁ = 5.0 0.082 300 3.0 = 41 L. The gas is compressed isothermally to half its volume, so the final volume V₂ is: V₂ = V₁ 2 = 41 2 = 20.5 L. The change in volume V = V₂ - V₁ = 20.5 - 41 = -20.5 L. The work done on the gas during an irreversible isothermal compression against a constant external pressure P_ ext is given by W = -P_ ext V . Given