Why can't a heat engine have 100% efficiency? Using the idea of isotherms and adiabats, I came up with the following argument to convince myself. First imagine the working gas within the system being heated along an isochore (hence all heat becomes internal energy). Then:
Adiabatic expansion will never do enough work to return the gas to it’s pre-heating internal energy state $\iff$ Fully efficient heat engines do not exist.
How good of an argument is this?
Thanks.
 A: What you are saying is correct and in fact it leads to one way among the many ways, Caratheodory's way, to phrase the 2nd law. Underlying it is the observation that if you plot the states that are accessible via a reversible adiabatic change then those states form a (hyper)surface in the space of the configuration coordinates and empirical temperature. The configuration coordinates, $X_k; k=1,2,..$ are the various mechanical, chemical, electrical, etc. parameters that describe the equilibrium of the system at some empirical temperature scale (this does not have to be the "absolute" temperature scale), say $\theta$.
A surface in those parameters are those values for which $f(\theta, X_1, X_2,...)=C$ for some function $f$ and arbitrary values of $C$. So the claim is that all adiabatic and reversible changes correspond to some function of $X_k$ and $\theta$ with a specific $C$. Now the really interesting part here is that these surfaces can be linearly ordered by their corresponding $C$ values. That is to any state $\mathcal A : X_1(\mathcal A), X_2(\mathcal A),...\theta(\mathcal A)$ there is a corresponding $C=C(\mathcal A)$

*

*such that all states that are accessible reversibly and
adiabatically from $\mathcal A$ have the same $C$, that is they are on
the same surface.

*any state that is not accessible adiabatically fall in to two categories: those
that are accessible by absorbing thermal energy fall on surfaces that have
larger $C$ (same $f$) and those that are only accessible by rejecting thermal
energy have smaller $C$ (same $f$).

We call the $C$ the empirical entropy.
A: I don't understand your argument. It is impossible to take a system following an isochoric (constant volume) heat addition and return it to its original state (original internal energy) by means of an adiabatic expansion alone. The adiabatic expansion can return the system to its original pressure, but not its original temperature and volume. Another process (e.g., a constant pressure compression) is needed, which will reject heat to restore the system to its original state. The efficiency is then the net work done divided by the heat added during the isochoric heat addition. That efficiency will be less than 100%.
That said, as far as a heat engine cycle efficiency is concerned, it is impossible to have a 100% efficient heat engine operating in a cycle as that would entail exchanging heat with a single thermal reservoir. That would violate the Kelvin-Planck statement of the second law which states:
No heat engine can can operate in a cycle while transferring heat with a single heat reservoir.
Hope this helps.
A: A simple way to demonstrate the engine efficiency is on the $TS$ graph.

The Carnot cycle on this graph is a rectangle ABCD. The area $a_{BCEF} = T_H(S_E-S_F)$ is the heat absorbed at $T_H$; the area $a_{ADEF} = -T_L(S_F-S_E)$ is the amount of heat rejeted at $T_L$; their difference is the area $a_{ABCD}$ of the cycle and is equal to the work produced. The efficiency is the the area inside the Carnot as a fraction of the area BCEF. This is always less than 100% unless $T_L=0$. Since $T_L$ can only be approached as a limit,  the efficiency is always less than 100%.
A: The reason why heat engine can't have a unit efficiency is that engine operate in cycle. Area under any process can gives work done but when engine return to its initial state via mechanical work done upon it, it never operate below the surrounding temperature which is always higher than absolute zero. So if adiabatic expansion after isochoric (ICE) process goes beyond surrounding temperature then it can't remove unused heat to surrounding. Why it is necessary, because when work is done on gas, its entropy increases and system need to be work done upon it. Also the amount of work done on expansion quickly decreases exponentially, so more mechanical work may be needed.
