0
$\begingroup$

My textbook says that $\delta S_{rev}= \frac{q_{rev}}{T}$. It also says that $\delta S_{surr}= \frac{-q_{rev}}{T}$. Logically, this should mean that the change in entropy in both the surroundings and the system are equal in magnitude but opposite in sign.

This is NOT true, though, as it would go against the Third Law of Thermodynamics. Also, my book proves the Law by comparing the equation for $\delta S_{surr}$ to the Entropy of the system found when subtracting the products' entropies from the reactants'.

Where am I going wrong? Why can't we compare the two $\delta S$ equations and disprove the Third Law?

Note: The first equation is $q_{rev}$ due to something to do with a reversible, isothermal reaction that I don't really understand, which might be the source of my confusion...

$\endgroup$
  • $\begingroup$ In fact it is "$\geqslant$" in both equations, not "$=$". $\endgroup$ – Ivan Neretin Jul 11 '18 at 20:55
  • $\begingroup$ No real world process is really thermodynamically reversible. $\endgroup$ – Mithoron Jul 11 '18 at 22:20
  • $\begingroup$ @Mithoron-I know...textbooks build on theoreticals. $\endgroup$ – Jo.P Jul 11 '18 at 22:53
  • $\begingroup$ Did you really mean the 3rd law of thermodynamics, or did you mean 2nd law of thermodynamics? $\endgroup$ – Chet Miller Jul 31 '18 at 12:01
  • $\begingroup$ What are the exact initial and final states of the system in this development, including temperatures, pressures, and chemical composition. $\endgroup$ – Chet Miller Jul 31 '18 at 18:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.