# 3.4: The thermodynamic existence of entropy

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There is a fact that’s hidden in the solution to our Carnot cycle in section 3.3, and we need a formal derivation in order to dig that fact out.

Let’s return to the definition of efficiency:

$$e=-\frac{w}{{q}_{ABS}}$$

We already have made statements about what work and heat should be for this engine in general; because we’re studying a closed cycle for which Δ*U* = 0, *q* = -*w* for the full cycle. Let’s rewrite the definition of efficiency specific to the full cycle:

$$e=\frac{{q}_{OUT}+{q}_{IN}}{{q}_{IN}}$$

The heat input in this cycle is the heat transfer from stage 12, which was positive. The heat output from this cycle was the heat transfer from stage 34, which was negative:

$${q}_{IN}=nR{T}_{1}\mathrm{ln}\left(\frac{{V}_{2}}{{V}_{1}}\right)$$

$${q}_{OUT}=nR{T}_{3}\mathrm{ln}\left(\frac{{V}_{4}}{{V}_{3}}\right)$$

If we substitute these expressions into the efficiency definitions, we’ll be able to cancel out *n* and *R* - but the natural logarithm expressions won’t cancel. But the adiabatic steps in the Carnot cycle connect temperature and volume, and there are only two temperatures - the high temperature *T*_{1}, and the low temperature *T*_{3}. (The temperature at state 1, after all, is isothermal to the temperature at state 2; likewise the temperatures at state 3 and state 4.) We can relate the two volume ratios using the adiabatic relations…

$${T}_{1}{V}_{2}^{\gamma -1}={T}_{3}{V}_{3}^{\gamma -1}\to \frac{{T}_{3}}{{T}_{1}}=\frac{{V}_{2}^{\gamma -1}}{{V}_{3}^{\gamma -1}}$$

$${T}_{1}{V}_{1}^{\gamma -1}={T}_{3}{V}_{4}^{\gamma -1}\to \frac{{T}_{3}}{{T}_{1}}=\frac{{V}_{1}^{\gamma -1}}{{V}_{4}^{\gamma -1}}$$

We can set both right sides equal to one another, eliminate the exponents, and rearrange for equality’s sake…

$$\frac{{V}_{2}^{\gamma -1}}{{V}_{3}^{\gamma -1}}=\frac{{V}_{1}^{\gamma -1}}{{V}_{4}^{\gamma -1}}\to \frac{{V}_{2}}{{V}_{1}}=\frac{{V}_{3}}{{V}_{4}}$$

This would have been *perfectly convenient* if *V*_{2} / *V*_{1} = *V*_{4} / *V*_{3}. As it stands, though, we’re still okay because we can rearrange *q _{IN}* with a negative log:

$${q}_{IN}=nR{T}_{1}\mathrm{ln}\left(\frac{{V}_{2}}{{V}_{1}}\right)=-nR{T}_{1}\mathrm{ln}\left(\frac{{V}_{4}}{{V}_{3}}\right)$$

Now we substitute *q _{IN}* and

*q*into the efficiency definition and simplify:

_{OUT}$$e=\frac{{q}_{OUT}+{q}_{IN}}{{q}_{IN}}=\frac{nR{T}_{3}\mathrm{ln}({V}_{4}/{V}_{3})-nR{T}_{1}\mathrm{ln}({V}_{4}/{V}_{3})}{-nR{T}_{1}\mathrm{ln}({V}_{4}/{V}_{3})}=\frac{{T}_{3}-{T}_{1}}{-{T}_{1}}$$

Here we have our demonstration that the Carnot engine’s efficiency only depends on temperature. Again, remember that there are only two temperatures - the “HOT” temperature *T*_{1}, and the “COLD” temperature *T*_{3}. Further simplification gets us a classic expression:

$$e=-\frac{{T}_{3}}{{T}_{1}}+\frac{{T}_{1}}{{T}_{1}}=1-\frac{{T}_{3}}{{T}_{1}}$$

$$e=1-\frac{{T}_{COLD}}{{T}_{HOT}}$$

That expression is the ultimate statement of the futility of chasing a perfect engine. *T*_{3}/*T*_{1}* *or *T _{COLD}*/

*T*must be zero if the efficiency of the engine is to be a perfect 1, and all the input heat is to be transformed into useful work. There are only two ways we can get that ratio to zero; either the hot-temperature reservoir must have an infinite temperature (lol), or the cold-temperature reservoir must be absolute zero (which is physically impossible).

_{HOT}*There are no perfect engines*. Every engine that you could ever create would contribute heat energy to the universe.

This, ultimately, is the thermodynamic significance of entropy; you can extend this thinking to the understanding that *every transfer of energy* adds heat energy, unusable thermal chaos, to the universe.