# Second Law of Thermodynamics

The **second law of thermodynamics** states that the entropy of an isolated system or any cyclic process never decreases; it will either increase or remain the same. Because of this, the second law provides a definitive direction in which time must progress by saying that *time may only pass in the direction of increasing entropy*. This addresses a difficulty with determining the direction of time purely by observation. Namely, through the lens of physics, most processes, especially those that take place on a microscopic scale, are symmetric with time: they look the same happening in reverse as they do when time passes normally. Just observing the process isn't enough to determine which direction time is moving in.

Furthermore, the law is bijective; not only does it say that the passage of time always takes the universe from a spatiotemporal plane of lesser entropy to one of equal or greater entropy, but it also says that there cannot exist a process that decreases the entropy of an isolated system. It is the latter consequence that forms the basis for the thermodynamic study of the second law.

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## Statements of the Law

The entropy of an isolated system or any cyclic process never decreases; it will either increase or remain the same.

There are several other ways to express the same idea, all amounting to an equivalent proposition.

Clausius' Statement$$Unless there is some work done, thermal energy never flows from a body at a colder temperature to a body at a hotter temperature.

Kelvin's Statement$$The conversion of thermal energy to work is never perfectly efficient.

## The Carnot Engine

**The Carnot engine** is a conceptual engine that achieves the most efficient conversion of heat to work permitted by Kelvin's statement. In general, efficiency is defined as the ratio of work out to heat in:

$\eta=\dfrac{W}{Q_h}.$

For a Carnot engine, the efficiency is found in terms of the temperature of the reservoirs the engine operates between:

$\eta_{Carnot}=1-\dfrac{T_l}{T_h}.$

Carnot's TheoremIt is impossible for any heat engine to be more efficient than a Carnot engine when operating between two given temperatures:

$\eta_{max}=\eta_{Carnot}.$

Consider a heat engine drawing heat $Q_1$ from a heat reservoir at temperature $T_1$, delivering work $W$ and dumping heat $Q_2$ into a heat sink at temperature $T_2.$

The heat engine operates in

cycles, that is, it takes in heat $Q_1$, does work $W$, dumps heat $Q_2$, and in the end returns back to its original unchanged state.Consider the net change in entropy $\Delta S$ of the universe:

The heat reservoir releases a heat $Q_1$ at a constant temperature $T_1.$ Thus the change in its entropy is

$\color{green}{\Delta S_1 = -\dfrac{Q_1}{T_1}}.$

The heat sink accepts heat $Q_2$ at a constant temperature $T_2.$ Thus the change in its entropy is

$\color{green}{\Delta S_2 = \dfrac{Q_2}{T_2} }.$

Since the heat engine returns back to its

unchangedstate after the process, the change in its entropy iszero.Thus the net change in entropy of the universe is

$\begin{aligned} \Delta S = \Delta S_1 + \Delta S_2 \\ = \dfrac{Q_2}{T_2} - \dfrac{Q_1}{T_1}. \end{aligned}$

Using the second law of thermodynamics, $\Delta S \geq 0,$ which implies

$\begin{aligned} \dfrac{Q_2}{T_2} - \dfrac{Q_1}{T_1} &\geq 0 \\ \dfrac{Q_2}{T_2} &\geq \dfrac{Q_1}{T_1} \\ \dfrac{Q_2}{Q_1} &\geq \dfrac{T_2}{T_1} \\ 1 - \dfrac{Q_2}{Q_1} &\leq 1- \dfrac{T_2}{T_1}. \end{aligned}$

Since the left hand side represents the efficiency of the given heat engine ($\eta$) and the right hand side is the efficiency of a Carnot engine,

$\eta \leq \eta_{carnot} \implies \eta_{max}= \eta_{Carnot}.\ _\square$

## Equivalence of Clausius' and Kelvin's Statements

Interestingly enough, Clausius' and Kelvin's statements are equivalent.

To prove this, we show that the two statements imply each other.

Clausius' statement implies Kelvin's statement.

Assume Kelvin's statement is false.

Now, there exists some engine that efficiently converts thermal energy to work.

Assume there is a heat pump that carries heat from a cold body to a hot body when work is supplied to it. (This is like a refrigerator.)

But we can couple the imagined engine such that it takes thermal energy from the hot body and uses the produced work to drive the heat pump.

Now, think of the coupled engines in green. It transfers thermal energy from hot to a cold body without any other input!

Thus, the falsity of Kelvin's statement implies the falsity of the Clausius' statement. $_\square$

Kelvin's statement implies Clausius' statement.

Assume Clausius' statement is false.

Now, there exists a device which takes thermal energy from a colder to a hotter body without external work.

Couple this strange engine with a normal heat engine in such a way that it recycles the thermal energy lost to the sink back to the source.

These two engines coupled are actually efficiently converting work to thermal energy without loss!

Using a similar argument as above, we have proven the theorem. $_\square$

## Spontaneity and Gibb's Free Energy

A

Spontaneous processis a process that occurs in a system without being induced by some outside agent. For example, a ball will roll down an inclined plane spontaneously.

Gibb's free energy is a measure of the energy available for conversion to work. Another way of interpreting the second law is that Gibb's free energy in a spontaneous process always decreases.

Gibb's free energyis$G = H - TS,$

where $H$ represents enthalpy, $T$ is the temperature, and $S$ is the entropy.

It is often useful to express the spontaneity of a process in terms of Gibb's energy instead of entropy, since it is difficult to measure the entropy changes of the environment.

Here is the relationship between $K_{p}$ (equilibrium constant) and $G$:

$G = -RT \ln K_{p},$

where $R$ is the gas constant and $T$ is the temperature of the surroundings.

The following conclusions can be drawn about the spontaneity of a chemical reaction:

$\Delta$G | |

- | Spontaneous |

0 | Neither |

+ | Nonspontaneous |

### Fluctuation Theorem

This section requires expantion.

This is an attempt to express the second law with quantum mechanical explanations.

## Possible Violations

Due to the strange empirical and asymmetric nature of the second law, there has been several attempts to refute the second law. Needless to say, the second law still stands correct.

**Loschmidt's Paradox**

How do microscopic physical laws that are time symmetric agree with macroscopic physical laws that are not time symmetric?

*Boltzmann's Response*

Boltzmann claims that it is possible for entropy to decrease, but that it is extremely unlikely.

**Maxwell's Demon**

The second law of thermodynamics states that the entropy of a system cannot be decreased without work being done upon it. However, it is easy to conceive of a demon who could violate this principle.

A container of gas molecules at equilibrium is divided into two parts by an insulated wall, with a door that can be opened and closed the "Maxwell's demon." The demon opens the door to allow only the faster than average molecules to flow through to a favored side of the chamber, and only the slower than average molecules to the other side, causing the favored side to gradually heat up while the other side cools down, thus decreasing entropy.

Why can't such a system effectively violate the second law?

Of course, opening the door does some work but that is far too little compared to what the second law predicts and, in theory, the door could be massless, too. Here is a better explanation based on this physics stackexchange post:

The demon consists of (at least) two parts: a sensor to detect when particles are coming, and an actuator to actually move the door. For the demon to work correctly, the actuator must act on the current instruction from the sensor, instead of the previous one, so it must forget instructions as soon as a new one comes in. This takes some work: there is some physical system encoding a bit and it will take some energy cost to flip it.

This work is given by Landauer's principle which states that for a circuit at (absolute) temperature T, it takes a minimum of $k_BT \ln 2$ energy to remove one bit of information, where $k_B$ is the Boltzmann constant.

Calvin is angry because I mentioned him in this sentence, so he wishes to remove it by overwriting all the bits encoding this sentence with 0s in the Brilliant.org servers.

What is the minimum amount of work in Joules Brilliant.org needs to do this?

If your the energy is $n$, enter the answer as $10^{18} n$.

**Details and Assumptions:**

- The operating temperature of the server is $70^{\circ} \text{C}$.
- The Boltzmann constant is $1.3806503 \times 10^{-23} \text{m}^2 \,\text{kg}\, \text{s}^{-2} \,\text{K}^{-1}.$
- All he wishes is to overwrite that sentence only. However, that includes all the characters from $C$ to $.$ (the one after the s).
- He does not want to break the octet stream so he overwrites
*all the bits in each byte*regardless of their current state.

**Black Hole Information Paradox**

This section requires expansion.

**Cite as:**Second Law of Thermodynamics.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/second-law-thermodynamics/