Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

First assertion

If a system is already in a high temperature, adding energy, will increment the entropy in a low amount (compared with a system in a lower temperature).

Question (if assertion is right)

What if the heat is enough that let molecules breaks (activation energy), this would lead to new multiplicity (more freedom) so more entropy. It is a higher entropy grow than if the temperature were lower! that seems to be in contradiction with first assertion.

I see there is something wrong here, but I don't know where.

share|improve this question
add comment

2 Answers

up vote 2 down vote accepted

The assertion is based on the assumption that you either have only ‘small’ increases in temperature (and hence small increases in entropy, think of all the $dS$ and $dT$ you encounter in standard thermodynamics) or that your system is sufficiently homogenous that the change in entropy is a continous function of the change in temperature. This obviously breaks down if your molecules start to break up.

share|improve this answer
    
+1 I think that's the key, system is out of equilibrium, those are not small increases, then, it's a mistake to maintain temperature constantly "low" while increasing energy, that won't happen –  HDE Nov 16 '12 at 18:52
add comment

By definition of temperature

$$\frac{1}{T} = \left( \frac{\partial S}{\partial U} \right)_{N_j}$$

If temperature is higher adding the same amount of energy $\delta U$ at constant composition $N_j$ results in a lower change $\delta S$ in the entropy. But if composition is changing due to chemical reaction $\mathrm{AB} \rightarrow \mathrm{A} + \mathrm{B}$ then there is an extra variation in the entropy due to change in composition

$$\frac{\mu_j}{T} = - \left( \frac{\partial S}{\partial N_j} \right)_U$$

The total change in the entropy is given by the variation of energy plus the variation on composition

$$\mathrm{d}S = \frac{1}{T}\mathrm{d}U - \sum_j \frac{\mu_j}{T}\mathrm{d}N_j $$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.