# Using entropy to solve for Free Gibbs energy

1. Sep 28, 2011

### jtruth914

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" ?

Last edited: Sep 28, 2011
2. Sep 29, 2011

### Puchinita5

Haha maybe you are in my class because I have an exam tomorrow and I was just working on this problem and also cannot figure out how he got a positive answer for G in part a either. I am hoping he made a mistake because I really don't understand why it would be positive.