# Consequences of Third Law

Thermodynamics | |
---|---|

Introduction | What is this thing called Thermodynamics??? | Definitions | Thermal Equilibrium and Zeroth Law | Limitations |

First Law | Work, Heat, Energy, and the First Law | Work, Heat, Energy, and the First Law (simplied) | Derivatives | Derivatives Exercise | Reversibility, Enthalpy, and Heat Capacity |

Second Law | Things to Think About | Observations and Second Law of Thermodynamics | Alternative Approach - the Clausis Inequality | Consequences of the Second Law | Consequences of the Second Law (simplified) | Carnot Principle - motivation and examples | Equivalence of Second Law Statements* |

Third Law | Third Law of Thermodynamics | Consequences of Third Law* |

Development of Thermodynamics | The Thermodynamic Network | Network Exercise | Equations of State (EOS) | EOS Example, Reading Tables, and Numerical Analysis | EOS Exercises | Thermochemistry |

* Optional Section |

This section is *Optional*.

There are two important consequences of the Third Law: the behavior of heat capacities as temperature goes to zero and that we cannot get to absolute zero.

## Heat capacities

We wish to know how heat capacities behave as the temperature goes to zero.

Let us consider a reversible path R, according to the second law

[math]dS=\frac{dQ}{T}[/math]

or

[math]dS=\frac{C_R}{T}dT[/math]

where C_{R} is the heat capacity along path R. Integrating from T = 0 to T = T_{1} gives

[math]S_1=\int^{T_1}_{0}\frac{C_R}{T}dT+S\left ( T=0\right )[/math]

S at T = 0 is by the third law equals zero, therefore

[math]S_1=\int^{T_1}_{0}\frac{C_R}{T}dT[/math]

The third law requires that S_{1} → 0 as T>sub>1</sub> → 0. The integral can only go to zero if C_{R} also goes to zero. Otherwise the integral becomes unbounded.

Therefore,

C

_{R}→ 0 as T → 0 C_{P}→ 0 as T → 0 C_{V}→ 0 as T → 0

All substances measured so far have obeyed this property.

## Unattainability of Absolute Zero

The unattainability of absolute zero says that we can ever reach absolute zero experimentally.

To prove this let us consider a process where we vary parameter *X* from an initial state (X_{1}, T_{1}) to a final state (X_{2}, T_{2}). Then by the second law:

[math]S_1\left (T=0\right ) + \int^{T_1}_0 \frac{C_1}{T}dT \leq S_2\left (T=0\right ) + \int^{T_2}_0 \frac{C_2}{T}dT[/math]

By the third law, S_{1} (T = 0) = S_{2} (T = 0), therefore,

[math]\int^{T_1}_0 \frac{C_1}{T} \leq \int^{T_1}_0 \frac{C_2}{T}[/math]

Let us now cool the system from a positive T_{1} to absolute zero, that is T_{2} = 0. Then the integral on the right is zero. Then,

[math]\int^{T_1}_0 \frac{C_1}{T} \leq 0[/math]

However, the integral on the left is positive since T_{1} %neq; 0.

Therefore, we cannot reach absolute zero.