Calculate the work done in kJ when 3 moles of an ideal gas at 27^ C expand isothermally and reversibly from 10 atm . to 1 ~atm [ R =8.314 ~J ~K ⁻¹ ~mol ⁻¹ ]

Step-by-step solution

The work done in an isothermal reversible expansion of an ideal gas is given by W = -nRT ( V₂ V₁ ) . For an isothermal process, Boyle's law implies P₁V₁ = P₂V₂ , so V₂ V₁ = P₁ P₂ . Substituting gives W = -nRT ( P₁ P₂ ) . Using the values n = 3 mol, T = 300 K, R = 8.314 J K ⁻¹ mol ⁻¹ , P₁ = 10 atm, and P₂ = 1 atm: W = -(3)(8.314)(300) (10) Since (10) 2.303 , we compute: W = -3 8.314 300 2.303 = -17230.9 J Converting to kilojoules: W = -17230.9 1000 = -17.2309 kJ, which rounds to -17.23 kJ. The value -17.23 kJ corres