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.(adsbygoogle = window.adsbygoogle || []).push({});

(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" ?

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# Using entropy to solve for Free Gibbs energy

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