# Thermodynamics/The Second Law of Thermodynamics
## Introduction

* The 1st Law of Thermodynamics tells us that an increase in one form of energy, E, must be accompanied by a decrease in another form of energy, E.
* Likewise the 2nd Law of Thermodynamics tells us which processes in nature may or may not occur. For instance, with two objects in thermal contact, heat will spontaneously flow from a warmer to cooler object, but this effect does not spontaneously occur in the opposite direction.
* Irreversible process = processes that occur naturally in only one direction.
E

## Heat engines

Device that converts thermal E. (Q$_h$) into mechanical or electrical E.
During the cycle,
Heat (Q$_h$) is absorbed from a source at a high T.
Work is done by the engine.
Heat (Q$_c$) is expelled to a source at a lower T.
Since it is a cyclic process, ∆U = 0. Thus, W = Q$_{total}$ ---> W = Q$_h$ – Q$_c$
η = efficiency = ratio of work done to heat absorbed
η = W / Q$_h$ = (Q$_h$ – Q$_c$) / Q$_h$ = 1 – Q$_c$ / Q$_h$
η = 100% only if Q$_c$ = 0. i.e. a perfect heat engine would convert all of the absorbed heat Q$_h$ into mechanical work. From 2nd law of thermo, this is impossible: only a fraction of heat is converted to mechanical E.
E.g. η of automobile engines = 20%.
Kelvin–Planck form of the 2nd law of thermodynamics:
“It is impossible by a cycle to take heat from a hot reservoir and convert it into work without, at the same time, transferring an amount of heat from hot to cold reservoir.”

## Heat pumps

Device that converts work into thermal E. (Q$_h$).
Used to heat/cool homes.
During the cycle,
Heat (Q$_c$) is absorbed from a source at low T (e.g. outside air or food) by a circulating fluid.
Work is done on the engine by a compressor.
Heat (Q$_h$) is expelled to a source at higher T (e.g. room).
Since it is a cyclic process, ∆U = 0. Thus, W = Q$_{total}$ ---> W = Q$_h$ – Q$_c$
K = coefficient of performance = ratio of heat transferred to work required to transfer the heat.
K = Q$_h$ / W = Q$_h$ / (Q$_h$ – Q$_c$)
K can be much larger than 100% (denominator < 1).

## Refrigerators

Device that converts work into thermal E. (Qc).
During the cycle,
Heat (Qc) is absorbed from a source at low T (e.g. outside air or food) by a circulating fluid.
Work is done on the engine by a compressor.
Heat (Qh) is expelled to a source at higher T (e.g. room).
Since it is a cyclic process, ∆U = 0. Thus, W = Qtotal  W = Qh – Qc
K = coefficient of performance = ratio of heat transferred to work required to transfer the heat.
K = Qc / W = Qc / (Qh – Qc)
K can be much larger than 100% (denominator < 1).
A perfect refrigerator would transfer heat from a colder body to a hotter body without doing any work. From 2nd law of thermo, this is impossible.
E.g. K of a refrigerator = 5 or 6
Clausius form of the 2nd law of thermo:
“It is impossible to use a cyclic process to transfer heat from a colder to a hotter body without doing work on the system.”
i.e. heat will not flow spontaneously from a cold to a hot object.

## The Carnot Engine

* Carnot made the most efficient heat engine. Net work taken from the Carnot cycle is the largest possible for a given amount of heat supplied.
* Carnot’s theorem: No real (irreversible) engine operating between 2 heat reservoirs can be more efficient than a Carnot (reversible) engine operating between the same 2 reservoirs.
* Carnot used an ideal gas contained in a cylinder with a movable piston at one end. The cylinder walls and the piston are thermally non–conducting.
* The Carnot cycle consists of 2 adiabatic and 2 isothermal processes, all reversible:
A ---> B
Isothermal expansion of a gas placed in thermal contact with a heat reservoir at temp. Th.
During the process, the gas absorbs heat Qh from the base of the cylinder and does a work WAB in raising the piston.
B ---> C
Adiabatic expansion by replacing the base of the cylinder by a thermally non–conducting wall.
During the process, T falls from Th to Tc and the gas does work WBC in raising the piston.
C ---> D
Isothermal compression by placing the gas in thermal contact with a heat reservoir at temp. Tc.
During the process, the gas expels heat Qc to the reservoir and the work WCD is done on the gas by external agent.
D ---> A
Adiabatic compression by replacing the base of the cylinder by a non–conducting wall.
During the process, T increases from Tc to Th and the work WDA is done on the gas by external agent.
* Net work done in this reversible cyclic process = area enclosed by the path ABCDA = net heat transferred into the system, since ∆U = 0.
* Since the internal E. of an ideal gas depends only on absolute T, then in AB and CD, T and hence U remain constant. From the 1st law of thermo,
$Q_{h}=W_{AB}=n\ R\ T_{h}ln{\frac {V_{2}}{V_{1}}}$
$Q_{c}=-W_{CD}=n\ R\ T_{c}ln{\frac {V_{3}}{V_{4}}}$
* By dividing the equations,
${\frac {Q_{h}}{Q_{c}}}={\frac {T_{h}ln({\frac {V_{2}}{V_{1}}})}{T_{c}ln({\frac {V_{3}}{V_{4}}})}}$
* For BC and DA,
$T_{h}\ V_{2}^{(\gamma -1)}=T_{c}\ V_{3}^{(\gamma -1)}$
$T_{h}V_{1}^{(\gamma -1)}=T_{c}\ V_{4}^{(\gamma -1)}$
* By dividing the equations and taking the $(\gamma -1)$th root,
${\frac {V_{3}}{V_{4}}}={\frac {V_{2}}{V_{1}}}$
* Thus,
${\frac {Q_{h}}{Q_{c}}}={\frac {T_{h}}{T_{c}}}$
$\eta =1-{\frac {T_{c}}{T_{h}}}$
(And this of course applies also to heat pumps and refrigerators. i.e., in their formulas, you can switch the Qh and Qc by Th and Tc respectively)

## The thermodynamic temperature scale (absolute T)

 Efficiency of a Carnot engine is:
η = 1 – (Qc / Qh) = 1 – (Tc / Th)
 The zero point of the thermodynamic scale is fixed as the T of the cold reservoir Tc at which η = 1 and hence Qc = 0 and W = Qh.
 T must be the absolute scale because otherwise it may have –ve values, and hence the engine will perform work more than the heat given by the source. i.e. we would create E. from nothing, which is in contradiction to 1st law of thermo.
 Thus, T = 0 is the lowest T in all scales i.e. the absolute zero.

## Entropy (S)

* Isolated systems and physical processes tend towards disorder, and entropy is a measure of this disorder.
* E.g. If all molecules of a gas in a room move together, this is a very ordered, unlikely state. If the molecules move randomly in all directions, changing speed after collisions, this is a very disordered, likely state.
* In a reversible process between 2 equilibrium states, change in entropy is given by:
dS = dQ$_r$ / T
Where dS = change in entropy, dQ$_r$ = heat absorbed or expelled by the system in reversible process, T = absolute T.
* Heat absorbed by the system = +ve dQ$_r$ and S↑. Heat lost = –ve dQ$_r$ and S↓.
* The most useful statement of the 2nd law of thermodynamics:
* For reversible adiabatic process or reversible reactions, ∆S = 0.
For irreversible process or irreversible reactions, ∆S > 0.
Where ∆S = change in entropy of the system + surroundings (the universe).
∆S = ∫dS = ∫dQ$_r$ / T
For reversible adiabatic process, no heat is transferred between system and surroundings, so ∆S = 0.
For Carnot engine, ∆S = Q$_h$/T$_h$ – Q$_c$/T$_c$. Since Q$_c$/Q$_h$ = T$_c$/T$_h$, then ∆S = 0.

## Quasi–static reversible process for an ideal gas

For an ideal gas undergoing a quasi–static reversible process from Ti Vi to Tf Vf,
dQr = dU + dW ; dW = PdV. Since it’s n ideal gas,
dU = n cv dT ; P = n R T / V. Thus,
dQr = n cv dT + n R T dV / V. Dividing by T,
dQr / T = n cv (dT / T) + n R (dV / V). Assuming cv is constant,
by integration, ***
∆S = ∫dQr / T = n cv ln (Tf / Ti) + n R ln (Vf / Vi)
Thus, ∆S is independent of the reversible path and depends only on initial and final states
For cyclic process, Ti = Tf and Vi = Vf, so ∆S = 0.

## Heat conduction

 When heat transfers from a hot Th to cold Tc reservoirs, entropy of cold reservoir increases by Q/Tc and entropy of hot reservoir decreases by Q/Th.
 Since Tc < Th, total change in entropy of the system (universe) > 0
∆Su = Q / Tc – Q / Th > 0
 Thus, if ∆Su < 0, process cannot occur (e.g. heat transfer from cold to hot body).

## Free expansion

* An ideal gas in an insulated container occupies a volume Vi. A membrane separating it from a vacuum is suddenly broken and the gas expands irreversibly to Vf.
* Work done against the vacuum = 0. Since walls are insulating, Q = 0. Thus, ∆U = 0 and Ui = Uf. Since the gas is ideal, U depends only on T. Thus, Ti = Tf.
* Since ∆S = ∫dS = ∫dQr / T only applies to reversible processes, we cannot use it directly. Thus, we imagine a reversible process with the same initial and final states: an isothermal reversible expansion. Since T is constant, ∆S = ∫dQr / T = 1/T ∫dQr.
* Since it is isothermal, ∫dQr = W = n R T ln (Vf / Vi). Thus,
∆S = n R ln (Vf / Vi)
Since Vf > Vi, ∆Su > 0. This can also be obtained be eqn *** by putting Tf = Ti

## Immortality paradox and contradiction with the exact quantum and classical mechanics

Second law of thermodynamics when considered strictly contradicts both with quantum and the classical mechanics which predict that both quantum and the classical systems when perfectly insulated are evolutionary reversible. Quantum mechanics predicts that a lot of quantum systems undergo full quantum revival after experimentally reasonable time. In the limit of arbitrary real spectrum and the arbitrary long time it recovers the quantum version of the Poincaré recurrence theorem. The best example is what we could call a Schrödinger's mouse in the analogy to Schrödinger's cat that can be dead and alive at the same time as the quantum state before the outcome of the experiment is measured. Unlike the famous (or infamous example since the cat may be measured dead in the action of the experiment) the Schrödinger's mouse is closed in the perfectly insulating quantum cavity for all fields and matter waves, young, alive and with the water, food supplies and air life support system or in sort of eternal spaceship that can last for the near-infinite time can age, erode inside, finally stop working but cannot ever exchange neither particles, information photons or any energy with the outside. It turns out and obviously against the second law of thermodynamics that it is the only way the mouse can stay immortal being periodically alive. As the whole is a thermodynamical system with the low (ordered) life entropy, she will live till she ages and dies, she will undergo natural decomposition but according to Poincaré due to perfect cavity boundaries making the life environment inside a freely evolving system the death processes must reverse and the total entropy is back to lower (even if it took near infinite but finite time) and she will finally wake up alive at the state she was put into cavity in the future. Even without making the detailed calculations and approximating the mouse as the mouse mass particle in the potential well of some meter size for the energy basic spacing of the apparent continuum of quantum states evolving, one may see however that this time is not only many times longer than the current age of the universe but the multiple of a very high power of 10 of it (about $\hbar ^{-1}$ or about $10^{34}$ in seconds or about $10^{8}$ times the current estimated age of the universe). Therefore while this time is longer than the current age of the universe it is unphysical and the only spaceship or the cavity of this kind is till now the whole universe itself. The explanation is therefore that we only observe the closed themodynamical systems at the very early stage of the evolution and they cannot reach the reversed evolution stage while we see them. Also there are no perfectly insulated systems as they always interact residually with the bigger system i.e. the cavity with Q equal to exact infinity. One should however observe a clear violation of the second law in small nano scale quantum systems like microscopic Carnot engine build from quantum dots with variable volume and working on few electron gas. For the mouse of the mass of the proton the estimated awakening time is only about 3 years.