Really how does the entropy of the universe increase?
Why does the OP say that the change in entropy of the surrounding is negative of the change of entropy of the system. I know that Q(surrounding) will be equal to -nRTln(V2/V1) (where T is the temperature of the system and V2 and V1 are final and initial volumes respectively) but it has to be divided by the temperature of the surrounding ,not the temperature of the system !
The OP in the link is making erroneous statements, such as "Since the the surroundings remain at constant pressure". For an isothermal expansion it is the temperature of the surroundings that is constant. Now, getting to your questions.
Why does the OP say that the change in entropy of the surrounding is negative of the change of entropy of the system.
The OP is referring to a reversible isothermal expansion where the surroundings is a thermal reservoir. For a reversible process the total change in entropy (system + surroundings) is zero. Therefore the change in entropy of the surroundings has to be the negative of the change in entropy of the system.
...but it has to be divided by the temperature of the surrounding ,not the temperature of the system !
For a reversible isothermal expansion, the temperature of the surroundings is considered the same as the temperature of the system, since the difference is infinitesimal.
Hope this helps.
If the expansion is reversible, then the temperature of the system is the same as the temperature of the surroundings. Reversibility implies no entropy generation, but heat transfer is occurring into the system (to avoid the temperature decrease that would otherwise occur with expansion), and heat transfer down a temperature gradient generates entropy. Therefore, there must be no temperature difference.
Note that reversibility is an idealization; in reality, a temperature difference is required to drive heat transfer. However, we can come arbitrarily close to reversibility at the expense of slowing down the process.
