Section 26.5 The Second Law and Entropy
We show below that entropy change for an isolated system is always positive for all real processes. In the case of interacting system, then we can combine system with environment to get an isolated system. This will give the entropy change of system more then compensated by opposite change in entropy of the environment.
Subsection 26.5.1 (Calculus) Entropy Change of Universe
Clausius proved that, for any cyclic process, whether it be reversible or irreversible, the following inequality must hold true.
Here, the equality holds only when the process is reversible.
Now, let us consider a cyclic process that is made up of an unknown process from state A to state B, and a reversible process from B to A as shown in the figure.
When Clausius's result is applied to this cyclic process it yields the following result.
where the equality will be the case if both processes were reversible. Now, from the definition of entropy, the integral over the reversible process is equal to the difference in entropy of states A and B.
Moving the entropy change on the other side of the inequality, we obtain the following for change of entropy for the A to B process.
Now, we apply this result to a combined system consisting of two subsystems, X and Y, which can interact thermally as shown in Figure 26.5.1, but they are thermally insulated from the rest of the universe. Consider a state change process in which the combined system changes from state A\(_\text{combo}\) to state B\(_\text{combo}\text{.}\)
Since there is no heat exchange with the outside world, the right side of Eq. (26.5.4) is zero. The final entropy of the combined system \(S(B)\) will then be greater than the entropy of the initial state \(S(A)\text{.}\)
where the equality sign holds when A\(_\text{combo}\) to B\(_\text{combo}\) transformation occurs by a reversible process. This means that if the subsystem X is our finite system and Y the rest of the universe, then the net entropy of the universe does not decrease in any process.
If the process is reversible, then the reduction in entropy of the system will be balanced by increase in entropy of the environment and vice-versa. If the process is irreversible, however, then the increase in entropy of the environment will be more than any reduction in the entropy of the system, or vice-versa. This leads to the following statement of second law of thermodynamics in terms of entropy.
The entropy of the universe cannot decrease.
Checkpoint 26.5.2. (Calculus) Entropy Change in Equilibration of a Two Compartment System.
A nonconducting cylinder of total volume 2 L closed at both ends is divided into two parts by a diathermal frictionless circular wall. Initially the piston is clamped in the middle with the Nitrogen gas, assumed to be an ideal gas of \(C_V=\dfrac{5}{2}R\text{,}\) at different temperatures and pressures on the two side. The initial pressure and temperatures on the left and right sides are (3 atm, 330 K) and (2 atm, 250 K) respectively.
(a) Find the temperature and pressure of the final state when the two sides are in thermal and mechanical equilibrium.
(b) Find the net change in the entropy of the gas.
(a) at mechanical equilibrium pressure on the two sides must become equal. (b) replace the actual process between the initial and final states by constant pressure and constant volume processes since computation of entropy on them will be easier to compute.
(a) \(2.51\ \textrm{atm}\text{,}\) \(V_1 = 1.064\ \textrm{L}\text{,}\) \(V_2 = 0.936\ \textrm{L}. \) (b) see the solution.
(a) From the given data on the left and right chambers in the ideal gas law for each we obtain
Now, we find the common pressure \(p\) and temperature \(T\) values in the two chambers in the final state. To find the temperature we note that total energy of the two systems will not change since although the gas on the sides interact with each other, the two together are isolated from other systems.
The change in the internal energies of the two sides separately will be
Therefore,
Let \(V_1\) and \(V_2\) be the volumes on the left and right sides at equilibrium. Let us denote the total volume by \(V_0\text{.}\) Then, we have
The volumes on the two sides are also given by the ideal gas laws on the two sides.
Put these in Eq. (26.5.7) to obtain
Now, applying the ideal gas law on either of the chamber in the final state gives
(b) To compute the entropy change, we envision reversible steps that lead to the same change in state but where the computation of entropy changes are easier to compute. The following figure gives one set of possible easy-to-compute processes in the
pV
-planes of the gases in the two compartments.For a constant-pressure step,
and for a constant-pressure step,
with
For the case of gas in the left chamber we have
with
and
You can work out similar exprssions for the gas in the right tank. Adding the change in left tank and the right tank will give the ne change.