To calculate the maximum work obtained from a reversible isothermal expansion, we can use the formula:W = -nRT * ln Vf / Vi where W is the work done, n is the number of moles of the gas, R is the ideal gas constant, T is the temperature in Kelvin, and Vf and Vi are the final and initial volumes, respectively.First, we need to find the number of moles n of the gas. We can use the ideal gas law:PV = nRTwhere P is the pressure, V is the volume, and T is the temperature in Kelvin. Rearranging the equation to solve for n:n = PV / RTWe are given the initial pressure Pi = 4.00 atm , initial volume Vi = 3.00 L , and temperature 25C . First, we need to convert the temperature to Kelvin:T = 25C + 273.15 = 298.15 KNow, we can find the number of moles:n = 4.00 atm * 3.00 L / 0.0821 L atm/mol K * 298.15 K n 0.493 molesNext, we need to find the final volume Vf . Since the process is isothermal, we can use Boyle's law:Pi * Vi = Pf * Vfwhere Pf is the final pressure. We are given the final pressure Pf = 1.00 atm . Rearranging the equation to solve for Vf:Vf = Pi * Vi / PfVf = 4.00 atm * 3.00 L / 1.00 atmVf = 12.00 LNow, we can calculate the maximum work obtained from the reversible isothermal expansion:W = -nRT * ln Vf / Vi W = - 0.493 moles * 8.314 J/mol K * 298.15 K * ln 12.00 L / 3.00 L W -5709.4 JThe maximum amount of work that can be obtained from the reversible isothermal expansion is approximately -5709.4 J. The negative sign indicates that the work is done by the system the gas on the surroundings.