Entropy

If we divide a gas volume $V$ with $N=10^{22}$ gas molecules into the same amount of subvolumes, then the probability of each molecule being one of these subvolumes is given as $z$. The maximum number of possibilities is therefore given as $$\Omega = z^{10^{22}}$$ If we increase or decrease the volume from $V$ to $V_1$, then the fraction of $\Omega_1$ and $\Omega$ becomes $$\frac{\Omega_1}{\Omega} = \frac{z_1^N}{z^N} = \frac{\left(z \frac{V_1}{V}\right)^N}{z^N} = \left(\frac{V_1}{V}\right)^N$$ For sufficiently small volume changes $\Delta V$ we can approximate the right side with $$\frac{\Omega_1}{\Omega} = 1+N\frac{\Delta V}{V}f$$ This finally leads to the following formula for variations in $\Delta \Omega$: $$\frac{\Delta \Omega}{\Omega} = \frac{\Omega_1}{\Omega}-1 = N\frac{\Delta V}{V} \qquad (1)$$ Now we take the mechanical work $$W = -pV = Q$$ into account. Inserting this equation into $(1)$ leads to $$\frac{\Delta \Omega}{\Omega} = -N\frac{Q}{pV}$$ In the next we replace $pV$ with the ideal gas equation $pV = nRT$ which results in $$\frac{\Delta \Omega}{\Omega} = \frac{N}{nR}\frac{Q}{T}$$ We define now the term entropy as the number of possibilities $\Omega$. This quantity is a measure of the disorder of a system and indicates whether a process is reversible. If we apply this definition to the isothermal processes in the Carnot cycle, we get the isothermal expansion $$\Delta S_1 = \frac{\Delta Q_1}{T_1} = R\ln\left(\frac{V_2}{V_1}\right)$$ The same applies to the entropy of isothermal expansion $$\Delta S_2 = \frac{\Delta Q_2}{T_2} = -R\ln\left(\frac{V_2}{V_1}\right)$$ This results in the sum $$\Delta S = S_1 + S_2 = 0$$ In general, the change in entropy for a reversible process is always 0. As a counterexample, we now consider two stones with the same mass and heat capacity but different temperatures. If you place them in such a way that they touch each other, then the heat flows from the hot to the cold stone until finally, a mixed temperature $T_\mathrm{M}$ is reached. Since the heat cannot flow back without tools, this process is irreversible. With $\Delta Q = mc\Delta T$ this results in $$\Delta S_1 = \int_{T_1}^{T_\mathrm{M}}mc\frac{\mathrm{d} T}{T} = mc\ln\left(\frac{T_\mathrm{M}}{T_1}\right)$$ Similarly, for $\Delta S_2$ we get: $$\Delta S_2 = \int_{T_2}^{T_\mathrm{M}}mc\frac{\mathrm{d} T}{T} = mc\ln\left(\frac{T_\mathrm{M}}{T_2}\right)$$ This gives the sum for the application of the logarithm laws: $$\Delta S = \Delta S_1 + \Delta S_2 = mc\ln\left(\frac{T_\mathrm{M}^2}{T_1T_2}\right)$$ The right-hand side of this equation must always be greater than 1 due to the definition of mixing temperature. In general, $\Delta S > 0$ always applies to the change in entropy in an irreversible process. In a closed system, the change in entropy can only remain constant or increase but never decrease. If one assumes that the universe is a closed system and most processes are irreversible, then one day a state will be reached in which entropy is maximum and no more work can be done. This hypothetical point in time, which is very far in the future and is currently still under discussion, is referred to as thermal death.