1.00 mol of perfect gas at 25 C is expanded isothermally from an initial pressure of 5.00 atm to a final pressure of 1.00 atm in two ways:(a) reversibly (b) against a constant external pressure of 1.00 atm. Calculate the values for q, w, dU, dH, dG, dSsys, dSsur, dStot for each path. (I pretty much know how to calculate all of them, but for Gibbs free energy for part (a) I got a different answer. I got -3.98kJ, but the answer is +3.98kJ. The only way I can get that is if I plug in dSsur in the dG equation instead of dSys. Do you use dSsur to get dG for an isothermal reversible process? I thought dG= dH-TdSsys Please explain.) Thanks. This is what I did to solve for (a) w= -nRT ln Vf/Vi w= -3.98 kJ q= -w for isothermal process b/c dU is equal to 0 so q=+3.98kJ dU=0 dH= dU + d(nRT), for isothermal process T=0 Thus, dH=0 dSsys= q/T= 3.98kJ/298 K= +13.39 J/K dSsur= -13.39 J/K This is where my problem is dG= dH-TdS dG= 0-(298K)(13.39J/K) dG= -3.99 kJ <----- this is what I got, but my professor's answer to this problem is positive. I think he used dSsur, but I don't understand why? Isn't Gibbs free energy in regards to the "system" ?