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As history of thermodynamics say, it was a mystery that what is the required condition for a given energy conversion to take place? Like there are two possible events each conserving energy but only one is chosen. So, in order to resolve this Clausius introduced a quantity called entropy which was given by $\int dq/T$. But can I know the reason for which Clausius chose this integral or quantity, why not any other quantity (changes in whom, positive or negative, would decide the occurrence of a given event)? I hope there lies an explanation to this which does not use statistical mechanics.

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If you don't want to use statistichal mechanics, you can view it as a completely mathematical thing. When you write the differential form $\delta Q$, you are not speaking of an exact differential, i.e. it is not really the differential of any function of the thermodynamical state. Temperature is, in this case, called the integrating factor, which means that $\delta Q/T$ is an exact form, in particular, it's $dS$, the differential of Entropy. This is a way to let entropy come out.

On the other hand, much more physical explanations can be given. The first uses obviously stat mech, but without making calculations, i can just tell you that $S$ turns out to be very closely connected with the number of possible microscopical states a thermodynamical (thus macroscopical) state can admit.

Finally, a reason is that the quantity $\int \delta Q/T$ is never negative in normal thermodynamical transformations, that is, it allows a simple formulation of the second principle.

I hope this is what you were looking for.

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  • $\begingroup$ Thanks a lot for your reply. Well, actually I was trying to seek an explanation without stats mech because I guess Clasius did not think of stats mech while providing a way to resolve the problem. By the way could you provide me few more physical explanations ? $\endgroup$ Commented Jan 30, 2013 at 15:08
  • $\begingroup$ Actually I absolutely don't know what historically speaking Clausius was thinking about, probably it was the last reason I wrote. I don't have anything else in mind right now, but perhaps someone else more expert than me can help you. $\endgroup$
    – Martino
    Commented Jan 30, 2013 at 16:22

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