# 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.

$e equals minus StartFraction w Over q Subscript upper A upper B upper S Baseline EndFraction$

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 ΔU = 0, q = -w for the full cycle. Let’s rewrite the definition of efficiency specific to the full cycle:

$e equals StartFraction q Subscript upper O upper U upper T Baseline plus q Subscript upper I upper N Baseline Over q Subscript upper I upper N Baseline EndFraction$

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:

$q Subscript upper I upper N Baseline equals n upper R upper T 1 ln left-parenthesis StartFraction upper V 2 Over upper V 1 EndFraction right-parenthesis$

$q Subscript upper O upper U upper T Baseline equals n upper R upper T 3 ln left-parenthesis StartFraction upper V 4 Over upper V 3 EndFraction right-parenthesis$

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 T1, and the low temperature T3. (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…

$upper T 1 upper V 2 Superscript gamma minus 1 Baseline equals upper T 3 upper V 3 Superscript gamma minus 1 Baseline right-arrow StartFraction upper T 3 Over upper T 1 EndFraction equals StartFraction upper V 2 Superscript gamma minus 1 Baseline Over upper V 3 Superscript gamma minus 1 Baseline EndFraction$

$upper T 1 upper V 1 Superscript gamma minus 1 Baseline equals upper T 3 upper V 4 Superscript gamma minus 1 Baseline right-arrow StartFraction upper T 3 Over upper T 1 EndFraction equals StartFraction upper V 1 Superscript gamma minus 1 Baseline Over upper V 4 Superscript gamma minus 1 Baseline EndFraction$

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

$StartFraction upper V 2 Superscript gamma minus 1 Baseline Over upper V 3 Superscript gamma minus 1 Baseline EndFraction equals StartFraction upper V 1 Superscript gamma minus 1 Baseline Over upper V 4 Superscript gamma minus 1 Baseline EndFraction right-arrow StartFraction upper V 2 Over upper V 1 EndFraction equals StartFraction upper V 3 Over upper V 4 EndFraction$

This would have been perfectly convenient if V2 / V1 = V4 / V3. As it stands, though, we’re still okay because we can rearrange qIN with a negative log:

$q Subscript upper I upper N Baseline equals n upper R upper T 1 ln left-parenthesis StartFraction upper V 2 Over upper V 1 EndFraction right-parenthesis equals minus n upper R upper T 1 ln left-parenthesis StartFraction upper V 4 Over upper V 3 EndFraction right-parenthesis$

Now we substitute qIN and qOUT into the efficiency definition and simplify:

$e equals StartFraction q Subscript upper O upper U upper T Baseline plus q Subscript upper I upper N Baseline Over q Subscript upper I upper N Baseline EndFraction equals StartFraction n upper R upper T 3 ln left-parenthesis upper V 4 slash upper V 3 right-parenthesis minus n upper R upper T 1 ln left-parenthesis upper V 4 slash upper V 3 right-parenthesis Over minus n upper R upper T 1 ln left-parenthesis upper V 4 slash upper V 3 right-parenthesis EndFraction equals StartFraction upper T 3 minus upper T 1 Over minus upper T 1 EndFraction$

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 T1, and the “COLD” temperature T3. Further simplification gets us a classic expression:

$e equals minus StartFraction upper T 3 Over upper T 1 EndFraction plus StartFraction upper T 1 Over upper T 1 EndFraction equals 1 minus StartFraction upper T 3 Over upper T 1 EndFraction$

$e equals 1 minus StartFraction upper T Subscript upper C upper O upper L upper D Baseline Over upper T Subscript upper H upper O upper T Baseline EndFraction$

That expression is the ultimate statement of the futility of chasing a perfect engine. T3/T1 or TCOLD/THOT 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).

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.

3.4: The thermodynamic existence of entropy is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.