1. Reversible Processes
A reversible thermodynamic process is one in which the universe (i.e. the system and its surroundings) can be returned to their initial conditions.
Because heat only flows spontaneously from hot to cold, for a process to be reversible, it must either be:
a) adiabatic – no heat flow (i.e. have perfect thermal insulation)
b) the heat must flow between two objects at the same temperature. In this case, the
temperatures won’t change: isothermal process
Furthermore, there must be no friction (i.e. mechanical energy loss) or turbulence – i.e. it must be infinitely slow.
All of these conditions are impossible, so there is no perfectly reversible process. However, processes can be considered practically reversible if they are sufficiently slow for the pressure and temperature to stay well defined (i.e. quasi-static) and approximately adiabatic or isothermal processes (with negligible friction).

2. 1824: Carnot suggested that “No heat engine operating between two energy reservoirs can be more efficient than a reversible engine operating between the same two reservoirs.” (Carnot’s theorem)
A reversible engine operating between two energy reservoirs is called a Carnot Engine. Since the only steps possible for a reversible engine are adiabatic or isothermal, here are the steps for a Carnot engine cycle:
Isothermal absorption of Qh from hot reservoir (Th).
Adiabatic work out (i.e. Eint <0) until at temperature of cold reservoir (Tc).
Isothermal exhaust of Qc at cold reservoir.
Adiabatic work done on system (i.e. Eint >0) until back to Th.

3. Proof of Carnot’s Theorem
We will prove that the Clausius statement of the 2nd Law  Carnot’s Theorem.
We will do this by assuming that Carnot’s Theorem is incorrect and show that this contradicts the Clausius statement.
i.e. If not B  A, then A  B.
(proof by contrapositive)
Assume there is an engine operating between Th and Tc that is more efficient than a Carnot engine: e > eC  |W| / |Qh| > |WC| / |QhC|
Match the work outputs of the two engines so that |WC| = |W|. (You can do this by running them different numbers of cycles until the works are equal.)
So: |QhC| > |Qh|
But |W| = |WC|  |Qh| - |Qc| = |QhC| - |QcC| (First Law)
|QcC| - |Qc|= |Qhc| - |Qh| > 0

4. Consider the Carnot engine running backwards as a heat pump (allowable, since it is reversible). Consider the combined system of the heat engine and Carnot heat pump. Since the W’s cancelled, no net work is being done (either on or by the combined system).
But |QcC| - |Qc|= |Qhc| - |Qh| > 0
i.e. there is net heat flowing from cold reservoir to hot reservoir!
This breaks the Clausius statement of 2nd Law.
Therefore our original assumption must be false  There is no engine more efficient than the Carnot engine. (QED)

5. Consider n moles of an ideal gas with CP/CV = 
Carnot’s Theorem
There is no engine operating between two heat baths that is more efficient than a Carnot (i.e. reversible) engine operating between those heat baths.
Corollary: All reversible engines (i.e. Carnot engines) operating between the same heat baths have the same efficiency.
Consider n moles of an ideal gas with CP/CV = 
Recall: adiabatic curves are steeper than isothermal

6. 0 (since isothermal)
Qh = Eint(B)- Eint(A) – WAB
|Qh| = |WAB| = nRTh ln(VB/VA)
Similarly |Qc| = nRTc ln(VC/VD)
eC = 1 - |Qc/Qh|
eC = 1 – (Tc/Th) ln(VC/VD)/ln(VB/VA)
But on adiabatic curves, PV = constant
nRTV/V = constant
T V-1 = constant
Therefore (VC/VB)-1 = Th/Tc = (VD/VA)-1
VC/VB = VD/VA
VC/VD = VB/VA
 eC = 1 – Tc/Th

7. eC = 1 – Tc/Th
True for any Carnot (i.e. reversible) engine operating between Tc and Th
(not just ideal gas engine).
[Tc and Th must be in Kelvin!]
For example, consider a cavity with no matter inside. The walls, at temperature T, will emit and absorb electromagnetic waves (“photons”) at different wavelengths. The radiation pressure P = Eint/3V = BT4, where B = constant = (4/3) /c = 2.5 x N/m2K4.
Also, in an adiabatic process, T  V-1/3 for photons.
Blackbody Spectrum

8. Consider the blackbody radiation in a cavity undergoing a Carnot cycle (isotherm-adiabat-isotherm-adiabat) [not very practical].
Blackbody radiation: 1) P = BT4  isothermal step: P = constant
2) adiabatic step: TV1/3 = constant
3) Eint = 3PV
Qh = Q(12) = Eint(12) –W(12)
Qh = Eint(2) – Eint(1) + P1(V2-V1)
Qh = 4P1(V2-V1) = 4BTh4(V2-V1)
|Qc| = 4BTc4(V3-V4)
Efficiency e = 1 – (Tc/Th)4 (V3-V4)/(V2-V1)
But from adiabatic steps,
Tc3V3=Th3V2 and Tc3V4=Th3V1
 V3/V4 = V2/V1 and V3/V2 = (Th/Tc)3
 e = 1 – (Tc/Th) Th Tc 1 2 3 4

9. For any engine operating between Tc and Th e  eC = 1 – Tc/Th
[= if reversible, < if not reversible]
Let’s check for some of the engines we’ve analyzed:
Ideal gas engine with  = 9/7  e = 0.22
The highest temperature in this cycle is Ta
and the lowest is Tc [why?] ,
and we found Ta/Tc = 5.94.
Therefore, eC = 1 – Tc/Ta = 0.83 > e 
adiabatic

10. For ideal gas with  = 7/5, we found
e = [(a-1) (b-1)] / [7/2 (a-1)b + 5/2(b-1)]
The highest temperature in this cycle is T2 and the lowest is T4 so
eC = 1 – T4/T2 = 1 – 1/ab
To compare e and eC, let’s plug in some values for a and b; e.g. a=3, b=5. V0
aV0 P0
bP0
eC = 1 – 1/15 = 0.933
e = 8 / [35+10] = 

11. Otto cycle: e = 1 – (V2/V1)-1 = 1 – TA/TB
(since AB is adiabatic, TV-1 = constant)
TA is the minimum temperature of the cycle, but TC is the maximum temperature, so
eC = 1 – TA/TC > e. 
[Finding a value for eC, however, requires specifying either the energy released in combustion (BC) or exhausted in DA.]

12. Reversible Processes
A reversible thermodynamic process is one in which the universe (i.e. the system and its surroundings) can be returned to their initial conditions.
Because heat only flows spontaneously from hot to cold, for a process to be reversible, it must either be:
a) adiabatic – no heat flow (i.e. have perfect thermal insulation)
b) the heat must flow between two objects at the same temperature. In this case, the
temperatures won’t change: isothermal process
Furthermore, there must be no friction (i.e. mechanical energy loss) or turbulence – i.e. it must be infinitely slow.
All of these conditions are impossible, so there is no perfectly reversible process. However, processes can be considered practically reversible if they are sufficiently slow for the pressure and temperature to stay well defined (i.e. quasi-static) and approximately adiabatic or isothermal processes (with negligible friction).

13. What about heat flow in constant –P or constant-V processes?
These can be achieved approximately reversibly by bringing the system (S) into contact with an “infinite” series of heat baths, each at slightly higher (or lower) temperatures, so there is a series of “quasi-isothermal” heat exchanges.
So to do a constant-P or constant-V process reversibly, you must use an “infinite” number of heat baths; cannot do reversibly with a single heat bath (as generally done experimentally).
For each step, dQ = C dT (for constant-P or constant-V or whatever process was used.)

14. Carnot’s Theorem: There is no engine operating between two heat baths that is more efficient than a Carnot (i.e. reversible) engine operating between those heat baths.
Since a cycle with constant V and/or constant P step(s) must either be irreversible or done with an infinite number of heat baths, such a cycle CANNOT be a Carnot cycle and must be less efficient than a Carnot cycle operating between its maximum and minimum temperatures.

15. S = S(f) – S(i)  dQrev/T
In 1855, Clausius proved the following (it is actually a corollary to “Clausius’ Theorem”):
If a system changes between two equilibrium states, i and f, then the integral
dQrev/T is the same for any reversible path between i and f.
Therefore there is a quantity, Clausius named “entropy” (S), that only depends on the endpoints, and not the path:
S = S(f) – S(i)  dQrev/T
dS = dQrev/T f i

16. S = dQ/T = (1/TA) dQ = Q/TA = (nRTA ln(Vhi/Vlo))/TA
Example
Suppose n moles of an ideal gas, with CP/CV= , undergoes the isothermal expansion from A to B.
One can go along the reversible isotherm (with a single heat bath at TA=TB):
S = dQ/T = (1/TA) dQ = Q/TA = (nRTA ln(Vhi/Vlo))/TA
S = nR ln(Vhi/Vlo)
Alternatively, one can go AC and then CB, using
infinite numbers of heat baths (to keep reversible)
with each dQ = C dT:
S =n CV dT/T + nCP dT/T TC TB TA TC TA
S = n (CP-CV)  dT / T = nR ln (TA/TC) [Here we switched limits and changed
sign of AC integral, used TB = TA for CB integral and
used CP – CV = R for any ideal gas.]
But from isobar, TA/Tc = Vhi/Vlo, so S = nR ln(Vhi/Vlo) and we obtained same change in entropy for both routes.  TC