7.1: Calculation of ΔSsys
In general \(\Delta S^{\mathrm{sys}}\) can be calculated using either its Definition: Entropy , or its differential formula, Equation 6.1.5 . In practice, it is always convenient to keep in mind that entropy is a state function, and as such it does not depend on the path. For this reason, we can break every transformation into elementary steps, and calculate the entropy on any path that goes from the initial state to the final state, such as, for example:
\[\begin{equation} \begin{aligned} P_i, T_i & \quad \xrightarrow{ \Delta_{\text{TOT}} S_{\text{sys}} } \quad P_f, T_f \\ \scriptstyle{\Delta_1 S^{\text{sys}}} & \searrow \qquad \qquad \nearrow \; \scriptstyle{\Delta_2 S^{\text{sys}}} \\ & \qquad P_i, T_f \\ \\ \Delta_{\text{TOT}} S^{\text{sys}} & = \Delta_1 S^{\text{sys}} + \Delta_2 S^{\text{sys}}, \end{aligned} \end{equation} \nonumber \]
with \(\Delta_1 S^{\text{sys}}\) calculated at constant \(P\), and \(\Delta_2 S^{\text{sys}}\) at constant \(T\). The most important elementary steps from which we can calculate the entropy resemble the prototypical processes for which we calculated the energy in section 3.1 .
Entropy in isothermal processes
- For an ideal gas at constant temperature \(\Delta U =0\), and \(Q_{\mathrm{REV}} = -W_{\mathrm{REV}}\). Using the formula for \(W_{\mathrm{REV}}\) in either Equation ( ?? ) or Equation ( ?? ) , we obtain: \[ \Delta S^{\mathrm{sys}} = \int_i^f \dfrac{đQ_{\mathrm{REV}}}{T} = \dfrac{-W_{\mathrm{REV}}}{T} = \dfrac{nRT \ln \dfrac{V_f}{V_i}}{T} = nR \ln \dfrac{V_f}{V_i}, \nonumber \] or, similarly: \[ \Delta S^{\mathrm{sys}} = nR \ln \dfrac{P_i}{P_f}. \nonumber \]
- A phase change is a particular case of an isothermal process that does not follow the formulas introduced above since an ideal gas never liquefies. The entropy associated with a phase change at constant pressure can be calculated from its definition, remembering that \(Q_{\mathrm{rev}}= \Delta H\). For example for vaporizations:
\[ \Delta_{\mathrm{vap}} S = \dfrac{\Delta_{\mathrm{vap}}H}{T_B}, \nonumber \]
with \(\Delta_{\mathrm{vap}}H\) being the enthalpy of vaporization of a substance, and \(T_B\) its boiling temperature.
It is experimentally observed that the entropies of vaporization of many liquids have almost the same value of:
\[ \Delta_{\mathrm{vap}} S \approx 10.5 R, \label{7.1.7} \]
which corresponds in SI to the range of about 85–88 J/(mol K). This simple rule is named Trouton’s rule , after the French scientist that discovered it, Frederick Thomas Trouton (1863-1922).
Calculate the standard entropy of vaporization of water knowing \(\Delta_{\mathrm{vap}} H_{\mathrm{H}_2\mathrm{O}}^{-\kern-6pt{\ominus}\kern-6pt-}= 44 \ \text{kJ/mol}\), as calculated in Exercise 4.3.1 .
- Answer
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Using Equation \ref{7.1.7}—and knowing that at standard conditions of \(P^{-\kern-6pt{\ominus}\kern-6pt-}= 1 \ \text{bar}\) the boiling temperature of water is 373 K—we calculate:
\[ \Delta_{\mathrm{vap}} S_{\mathrm{H}_2\mathrm{O}}^{-\kern-6pt{\ominus}\kern-6pt-}= \dfrac{44 \times 10^3 \text{J/mol}}{373 \ \text{K}} = 118 \ \text{J/(mol K)}. \nonumber \]
The entropy of vaporization of water is far from Trouton’s rule range of 85–88 J/(mol K) because of the hydrogen bond interactions between its molecules. Other similar exceptions are ethanol, formic acid, and hydrogen fluoride.
Entropy in adiabatic processes
Since adiabatic processes happen without the exchange of heat, \(đQ=0\), it would be tempting to think that \(\Delta S^{\mathrm{sys}} = 0\) for every one of them. A transformation at constant entropy (isentropic) is always, in fact, a reversible adiabatic process. However, the opposite case is not always true, and an irreversible adiabatic transformation is usually associated with a change in entropy. To explain this fact, we need to recall that the definition of entropy includes the heat exchanged at reversible conditions only. Therefore, for irreversible adiabatic processes \(\Delta S^{\mathrm{sys}} \neq 0\). The calculation of the entropy change for an irreversible adiabatic transformation requires a substantial effort, and we will not cover it at this stage. The situation for adiabatic processes can be summarized as follows:
\[\begin{equation} \begin{aligned} \text{reversible:} \qquad & \dfrac{đQ_{\mathrm{REV}}}{T} = 0 \longrightarrow \Delta S^{\mathrm{sys}} = 0 \quad \text{(isentropic),}\\ \text{irreversible:} \qquad & \dfrac{đQ_{\mathrm{IRR}}}{T} = 0 \longrightarrow \Delta S^{\mathrm{sys}} \neq 0. \\ \end{aligned} \end{equation} \nonumber \]
Entropy in isochoric processes
We can calculate the heat exchanged in a process that happens at constant volume, \(Q_V\), using Equation 2.3.2 . Since the heat exchanged at those conditions equals the energy ( Equation 3.1.7 ), and the energy is a state function, we can use \(Q_V\) regardless of the path (reversible or irreversible). The entropy associated with the process will then be:
\[ \Delta S^{\mathrm{sys}} = \int_i^f \dfrac{đQ_{\mathrm{REV}}}{T} = \int_i^f nC_V \dfrac{dT}{T}, \nonumber \]
which, assuming \(C_V\) independent of temperature and solving the integral on the right-hand side, becomes:
\[ \Delta S^{\mathrm{sys}} \approx n C_V \ln \dfrac{T_f}{T_i}. \nonumber \]
Entropy in isobaric processes
Similarly to the constant volume case, we can calculate the heat exchanged in a process that happens at constant pressure, \(Q_P\), using Equation 2.3.4 . Again, similarly to the previous case, \(Q_P\) equals a state function (the enthalpy), and we can use it regardless of the path to calculate the entropy as:
\[ \Delta S^{\mathrm{sys}} = \int_i^f \dfrac{đQ_{\mathrm{REV}}}{T} = \int_i^f nC_P \dfrac{dT}{T}, \nonumber \]
which, assuming \(C_P\) independent of temperature and solving the integral on the right-hand side, becomes:
\[ \Delta S^{\mathrm{sys}} \approx n C_P \ln \dfrac{T_f}{T_i}. \nonumber \]︎