# Biological Physics/The Third Law

So far, we have encountered the first two laws of thermodynamics, and now it is time to see the third. For an ideal gas, recall how entropy, energy, and temperature are related ${\displaystyle dS={\frac {dU}{T}}}$ where temperature is held constant. Now when volume is also held constant, work on the system ${\displaystyle W=P\Delta V=0}$, so ${\displaystyle dS={\frac {dQ}{T}}}$. By the definition of heat in terms of heat capacity ${\displaystyle dS={\frac {CdT}{T}}}$. To add up all the small incremental dSs, ${\displaystyle \Delta S=\int _{T_{i}}^{T_{f}}{\frac {CdT}{T}}}$.

Now, what happens as ${\displaystyle T_{i}=0}$? Then, ${\displaystyle \Delta S=\int _{0}^{T_{f}}{\frac {CdT}{T}}}$, where if C is not dependent on temperature, ${\displaystyle \Delta S=C\int _{0}^{T_{f}}{\frac {dT}{T}}}$. Then, ${\displaystyle \Delta S=C*ln(T_{f})-ln(0)}$. Since ln(0) approaches -∞, then this means that ${\displaystyle \Delta S}$ gets infinitely large as ${\displaystyle T\rightarrow 0}$, which is not good! So, there must be some temperature dependence in the heat capacity C.

The Third Law of Thermodynamics states that heat capacity C must go to zero faster than ln(0) goes to infinity, implying ${\displaystyle \Delta S\rightarrow 0}$. So, the multiplicity Ω where ${\displaystyle S=k_{B}ln(\Omega )}$ must be ${\displaystyle ln(\Omega )=S(0K)=0}$ which implies ${\displaystyle \Omega =e^{0}=1}$. So a more intuitive way to state the third law would be

There's only one configuration for a system at 0 K.

### Playing the GameEdit

A fun way to remember the Three Laws of Thermodynamics is with the following way:

0th Law

- You must play the game

1st Law

- You can't win the game

2nd Law

- You can't even break even

3rd Law

- Not even on a cold day

And there you have it: the Three Laws of Thermodynamics!