## Past quiz 2015 question 9

$\Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}$

$\Delta G^{\circ}= -RT\ln K$

$\Delta G^{\circ}= \sum \Delta G_{f}^{\circ}(products) - \sum \Delta G_{f}^{\circ}(reactants)$

Natalie Yakobian
Posts: 45
Joined: Fri Sep 25, 2015 3:00 am

### Past quiz 2015 question 9

Hi,
For the past quiz numbr 9:
A sample of 1 mol of gas initially at 1 atm and 298 K is heated at constant pressure to 350 K then the gas is compressed isothermally to its initial volume and finally it is cooled to 298 K at constant volume.

I understand why Delta G and Delta S of system are 0 because they are state funcitons but [b]why is the Delta S of surrounding not also 0?
[/b]

Thanks!

Chem_Mod
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### Re: Past quiz 2015 question 9

If you make a plot of Temp versus Vol of the process happening, then the plot looks like a triangle whereby the ending point equals the starting point, and you know that the change in entropy for the system is equal to zero. Now, for an irreversible process, the system is not in equilibrium, so the deltaStotal will be positive if the process happened spontaneously or negative if it did not. Let's say that the expansion at constant pressure was irreversible. Then you know that deltaSsurr=deltaStotal-deltaSsys for that particular step, whereby deltaStotal was not zero. Then the total change in the surroundings would be different than the total change in the system.