Clausius Inequality and Entropy

The  Clausius Inequality arises when we consider cyclic processes. This leads to the concept of entropy. For this discussion recall that
\( \frac{Q_2}{Q_1}= \frac{T_2}{T_1} \). In particular \( \delta Q_2 = \frac{T_2}{T_1} \delta Q_1 \).  Now imagine there is an engine doing work in a cyclic process, this will be done with help of carnot engines. There ill be one ultimate reservoir from where we will get our supply of hear. The engine will be taken from one temperature \( T_i\) to \( T_{i+1} \) by an auxillary reservoir at temperature \( T_i \) to keep the reservoir repleished it gets a supply of heat \( \delta Q_i \) for a carnot engine \( C_i\) but this engine gets it heat from another auxillary reservoir in the amount \( \delta Q_i \frac{\tilde{T}}{T_i} \) and this auxillary reservoir got its heat \( \delta Q_i \frac{\tilde{T}}{T_i} \) from the big reservoir at temperature \( \tilde{T} \). We use the index because movement from one temperature \( T_i \) to \( T_{i+1} \) requires a carnot engine \(C_i \). Everything is done in a cycle so that ultimately \( \Delta U = 0\). The total heat transferred to the engine in the cycle in \( \sum \frac{\delta Q_i \tilde{T}}{T_i} = Q \).

From first law we have that \( Q=W \). But this is a violation of second law because we have converted all the incoming heat into work. This can be avoided if Q and W are negative i.e all the work we put in is turned into heat. Therefore we have that \( Q=W \leq 0 \) so \( \tilde{T} \sum_{i} \frac{\delta Q_i}{T_i} \leq 0 \).

For infinitesimal \( \delta Q_i \) we have that
\begin{equation}
 \oint \frac{dQ}{T} \leq 0 \hspace{10mm} \text{This is the Clausius Inequality.}
 \end{equation}
 But we can reverse the direction of the cycle to get
\( \int \frac{dQ}{T} \geq 0 \). Therefore for a reversible process \begin{equation}
\oint \frac{dQ}{T}= 0  \hspace{10mm} \text{Reversible process}
\end{equation}

Now for a reversible process consider going in a cycle from initial state to final state and back to the initial state. We this we have
\( \int_i^f \frac{dQ}{T} + \int_f^i \frac{dQ}{T} = 0 \). So we have that
\( \int_i^f \frac{dQ}{T} = \int_i^f \frac{dQ}{T} \)
This implies the existence of a state function where \( S_f - S_i\).

We have finally arrived at the concept of entropy. Now we can show that the change in this state function always increasing or stays the same. So consider a process we go from some initial state to final state by some irreversible process and go back to the initial state by a reversible process. We therefore have
\begin{equation}
 \int_{i(irreversible)}^f \frac{dQ}{T} + \int_{f(reversible)}^i \frac{dQ}{T} \leq 0
\end{equation}
so
\begin{equation}
 \int_{i(irreversible)}^f \frac{dQ}{T} \leq \int_{i(reversible)}^f \frac{dQ}{T} = \Delta S
\end{equation}
 The first integral was over an irreversible process while the second was over a reversible process. So we have that the first integral over the irreversible process is
\begin{equation}
\int _{i(irreversible)}^f \frac{dQ}{T} \leq \Delta S \hspace{10mm} \text{Integral over an irreversible process}
\end{equation} .

Thermally isolate system we have dQ=0 so \( \Delta S \geq 0 \). This is what people call the second law of thermodynamics.