Calculate the work done when 5 moles of an ideal gas expand isothermally and reversibly from a volume of 2 L to 20 L at 27^ C . (Given R = 8.314 J K ⁻¹ mol ⁻¹ )

Step-by-step solution

For an isothermal reversible expansion of an ideal gas, the work done is given by: W = -2.303 nRT ( V₂ V₁ ) Given: n = 5 mol R = 8.314 J K ⁻¹ mol ⁻¹ T = 27^ C = 27 + 273 = 300 K V₁ = 2 L V₂ = 20 L Substituting the values into the formula: W = -2.303 5 8.314 300 ( 20 2 ) W = -2.303 12471 (10) W = -28720.7 J W -28.72 kJ Answer: -28.72 kJ