7 Calculation of Entropy and the Third Law of Thermodynamics

The Second Law can be used to infer the spontaneity of a process, as long as the entropy of the universe is considered. To do so, we need to remind ourselves that the universe can be divided into a system and its surroundings (environment). When we calculate the entropy of the universe as an indicator of the spontaneity of a process, we need to always consider changes in entropy in both the system (sys) and its surroundings (surr):

\[\begin{equation} \Delta S^{\mathrm{universe}} = \Delta S^{\mathrm{sys}} + \Delta S^{\mathrm{surr}}, \tag{7.1} \end{equation}\]

or, in differential form:

\[\begin{equation} d S^{\mathrm{universe}} = d S^{\mathrm{sys}} + d S^{\mathrm{surr}}, \tag{7.2} \end{equation}\]

7.1 Calculation of \(\Delta S^{\mathrm{sys}}\)

In general \(\Delta S^{\mathrm{sys}}\) can be calculated using either its Definition 6.1, or its differential formula, eq. (6.5). 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:

\[\begin{equation} \begin{aligned} P_i, T_i & \quad \xrightarrow{ \Delta_{\text{TOT}} S^{\text{sys}} } \quad P_f, T_f \\ \scriptstyle{\Delta_1 S^{\text{sys}}} & \searrow \qquad \qquad \nearrow \; \scriptstyle{\Delta_2 S^{\text{sys}}} \\ & \qquad P_i, T_f \\ \\ \Delta_{\text{TOT}} S^{\text{sys}} & = \Delta_1 S^{\text{sys}} + \Delta_2 S^{\text{sys}}, \end{aligned} \tag{7.3} \end{equation}\]

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.

7.1.1 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 eq. (2.18) (or its analogue for \(P\)), we obtain: \[\begin{equation} \Delta S^{\mathrm{sys}} = \int_i^f \frac{đQ_{\mathrm{REV}}}{T} = \frac{-W_{\mathrm{REV}}}{T} = \frac{nRT \ln \frac{V_f}{V_i}}{T} = nR \ln \frac{V_f}{V_i}, \tag{7.4} \end{equation}\] or, similarly: \[\begin{equation} \Delta S^{\mathrm{sys}} = nR \ln \frac{P_i}{P_f}. \tag{7.5} \end{equation}\]

  • 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:

\[\begin{equation} \Delta_{\mathrm{vap}} S = \frac{\Delta_{\mathrm{vap}}H}{T_B}, \tag{7.6} \end{equation}\]

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:

\[\begin{equation} \Delta_{\mathrm{vap}} S \approx 10.5 R, \tag{7.7} \end{equation}\]

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).

Exercise 7.1 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 4.1.

Solution: Using eq. (7.7)—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:

\[\begin{equation} \Delta_{\mathrm{vap}} S_{\mathrm{H}_2\mathrm{O}}^{-\kern-6pt{\ominus}\kern-6pt-}= \frac{44 \times 10^3 \text{J/mol}}{373 \ \text{K}} = 118 \ \text{J} \, \text{mol}^{-1} \, \text{K}^{-1}. \tag{7.8} \end{equation}\]

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.

7.1.2 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:

\[\begin{equation} \begin{aligned} \text{reversible:} \qquad & \frac{đQ_{\mathrm{REV}}}{T} = 0 \longrightarrow \Delta S^{\mathrm{sys}} = 0 \quad \text{(isentropic),}\\ \text{irreversible:} \qquad & \frac{đQ_{\mathrm{IRR}}}{T} = 0 \longrightarrow \Delta S^{\mathrm{sys}} \neq 0. \\ \end{aligned} \tag{7.9} \end{equation}\]

7.1.3 Entropy in isochoric processes

We can calculate the heat exchanged in a process that happens at constant volume, \(Q_V\), using eq. (2.2). Since the heat exchanged at those conditions equals the energy (eq. (3.7)), 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:

\[\begin{equation} \Delta S^{\mathrm{sys}} = \int_i^f \frac{đQ_{\mathrm{REV}}}{T} = \int_i^f nC_V \frac{dT}{T}, \tag{7.10} \end{equation}\]

which, assuming \(C_V\) independent of temperature and solving the integral on the right-hand side, becomes:

\[\begin{equation} \Delta S^{\mathrm{sys}} \approx n C_V \ln \frac{T_f}{T_i}. \tag{7.11} \end{equation}\]

7.1.4 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 eq. (2.4). 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:

\[\begin{equation} \Delta S^{\mathrm{sys}} = \int_i^f \frac{đQ_{\mathrm{REV}}}{T} = \int_i^f nC_P \frac{dT}{T}, \tag{7.12} \end{equation}\]

which, assuming \(C_P\) independent of temperature and solving the integral on the right-hand side, becomes:

\[\begin{equation} \Delta S^{\mathrm{sys}} \approx n C_P \ln \frac{T_f}{T_i}. \tag{7.13} \end{equation}\]

7.2 Calculation of \(\Delta S^{\mathrm{surr}}\)

While the entropy of the system can be broken down into simple cases and calculated using the formulas introduced above, the entropy of the surroundings does not require such a complicated treatment, and it can always be calculated as:

\[\begin{equation} \Delta S^{\mathrm{surr}} = \frac{Q_{\text{surr}}}{T_{\text{surr}}}=\frac{-Q_{\text{sys}}}{T_{\text{surr}}}, \tag{7.14} \end{equation}\]

or, in differential form:

\[\begin{equation} d S^{\mathrm{surr}} = \frac{đQ_{\text{surr}}}{T_{\text{surr}}}=\frac{-đQ_{\text{sys}}}{T_{\text{surr}}}, \tag{7.15} \end{equation}\]

where the substitution \(Q_{\text{surr}}=-Q_{\text{sys}}\) can be performed regardless of whether the transformation is reversible or not. In other words, the surroundings always absorb heat reversibly. To justify this statement, we need to restrict the analysis of the interaction between the system and the surroundings to just the vicinity of the system itself. Outside of a generally restricted region, the rest of the universe is so vast that it remains untouched by anything happening inside the system.28 To facilitate our comprehension, we might consider a system composed of a beaker on a workbench. We can then consider the room that the beaker is in as the immediate surroundings. To all effects, the beaker+room combination behaves as a system isolated from the rest of the universe. The room is obviously much larger than the beaker itself, and therefore every energy production that happens in the system will have minimal effect on the parameters of the room. For example, an exothermal chemical reaction occurring in the beaker will not affect the overall temperature of the room substantially. When we study our reaction, \(T_{\text{surr}}\) will be constant, and the transfer of heat from the reaction to the surroundings will happen at reversible conditions.

Exercise 7.2 Calculate the changes in entropy of the universe for the process of 1 mol of supercooled water, freezing at –10°C, knowing the following data: \(\Delta_{\mathrm{fus}}H = 6 \; \text{kJ/mol}\), \(C_P^{\mathrm{H}_2 \mathrm{O}_{(l)}}=76 \; \text{J} \, \text{mol}^{-1} \, \text{K}^{-1}\), \(C_P^{\mathrm{H}_2 \mathrm{O}_{(s)}}=38 \; \text{J} \, \text{mol}^{-1} \, \text{K}^{-1}\), and assuming both \(C_P\) to be independent on temperature.

Solution: \(\Delta S^{\mathrm{sys}}\) for the process under consideration can be calculated using the following cycle:

\[\begin{equation} \begin{aligned} \mathrm{H}_2 \mathrm{O}_{(l)} & \quad \xrightarrow{\quad \Delta S_{\text{sys}} \quad} \quad \mathrm{H}_2 \mathrm{O}_{(s)} \qquad \quad T=263\;K\\ \scriptstyle{\Delta S_1} \; \bigg\downarrow \quad & \qquad \qquad \qquad \qquad \scriptstyle{\bigg\uparrow \; \Delta S_3} \\ \mathrm{H}_2 \mathrm{O}_{(l)} & \quad \xrightarrow{\quad \Delta S_2 \qquad} \quad \mathrm{H}_2\mathrm{O}_{(s)} \qquad \; T=273\;K\\ \\ \Delta S^{\text{sys}} & = \Delta S_1 + \Delta S_2 + \Delta S_3 \end{aligned} \tag{7.16} \end{equation}\]

\(\Delta S_1\) and \(\Delta S_3\) are the isobaric heating and cooling processes of liquid and solid water, respectively, and can be calculated filling the given data into eq. (7.12). \(\Delta S_2\) is a phase change (isothermal process) and can be calculated translating eq. (7.6) to the freezing transformation. Overall:

\[\begin{equation} \begin{aligned} \Delta S^{\text{sys}} & = \int_{263}^{273} \frac{C_P^{\mathrm{H}_2 \mathrm{O}_{(l)}}}{T}dT+\frac{-\Delta_{\mathrm{fus}}H}{273}+\int_{273}^{263} \frac{C_P^{\mathrm{H}_2 \mathrm{O}_{(s)}}}{T}dT \\ & = 76 \ln \frac{273}{263} - \frac{6 \times 10^3}{273} + 38 \ln \frac{263}{273}= -20.6 \; \text{J/K}. \end{aligned} \tag{7.17} \end{equation}\]

Don’t be confused by the fact that \(\Delta S^{\text{sys}}\) is negative. This is not the entropy of the universe! Hence it tells nothing about spontaneity! We can now calculate \(\Delta S^{\text{surr}}\) from \(Q_{\text{sys}}\), noting that we can calculate the enthalpy around the same cycle in eq. (7.16). To do that, we already have \(\Delta_{\mathrm{fus}}H\) from the given data, and we can calculate \(\Delta H_1\) and \(\Delta H_3\) using eq. (2.4).

\[\begin{equation} \begin{aligned} Q^{\text{sys}} & = \Delta H = \int_{263}^{273} C_P^{\mathrm{H}_2 \mathrm{O}_{(l)}} dT + (-\Delta_{\mathrm{fus}}H) + \int_{273}^{263} C_P^{\mathrm{H}_2 \mathrm{O}_{(s)}}dT \\ & = 76 \times 10^{-3} (273-263) - 6 + 38 \times 10^{-3} (263-273) \\ &= -5.6 \; \text{kJ}. \\ \\ \Delta S^{\text{surr}} & = \frac{-Q_{\text{sys}}}{T}=\frac{5.6 \times 10^3}{263} = + 21.3 \; \text{J/K}. \\ \end{aligned} \tag{7.18} \end{equation}\]

Bringing (7.16) and (7.18) results together, we obtain:

\[\begin{equation} \Delta S^{\text{universe}}=\Delta S^{\text{sys}} + \Delta S^{\text{surr}} = -20.6+21.3=+0.7 \; \text{J/K}. \tag{7.19} \end{equation}\]

Since the entropy changes in the universe are positive, the process is spontaneous, as expected.

7.3 Clausius Theorem

By replacing eq. (7.15) into (7.2) we can write the differential change in the entropy of the system as:

\[\begin{equation} d S^{\mathrm{sys}} = d S^{\mathrm{universe}} - d S^{\mathrm{surr}} = d S^{\mathrm{universe}} + \frac{đQ_{\text{sys}}}{T}. \tag{7.20} \end{equation}\]

According to the second law, for any spontaneous process \(d S^{\mathrm{universe}}\geq0\), and therefore, replacing it into eq. (7.20):

\[\begin{equation} d S^{\mathrm{sys}} \geq \frac{đQ_{\text{sys}}}{T}, \tag{7.21} \end{equation}\]

which is the mathematical expression of the so-called Clausius theorem. Eq. (7.21) distinguishes between three conditions:

\[\begin{equation} \begin{aligned} d S^{\mathrm{sys}} > \frac{đQ_{\text{sys}}}{T} \qquad &\text{spontaneous, irreversible transformation} \\ d S^{\mathrm{sys}} = \frac{đQ_{\text{sys}}}{T} \qquad &\text{reversible transformation} \\ d S^{\mathrm{sys}} < \frac{đQ_{\text{sys}}}{T} \qquad &\text{non-spontaneous, irreversible transformation}, \end{aligned} \tag{7.22} \end{equation}\]

Clausius theorem provides a useful criterion to infer the spontaneity of a process, especially in cases where it’s hard to calculate \(\Delta S^{\mathrm{universe}}\). Eq. (7.21) requires knowledge of quantities that are dependent on the system exclusively, such as the difference in entropy, the amount of heat that crosses the boundaries, and the temperature at which the process happens.29 If a process produces more entropy than the amount of heat that crosses the boundaries divided by the absolute temperature, it will be spontaneous. Vice versa, if the entropy produced is smaller than the amount of heat crossing the boundaries divided by the absolute temperature, the process will be non-spontaneous. The equality holds for systems in equilibrium with their surroundings, or for reversible processes since they happen through a series of equilibrium states. Measuring or calculating these quantities might not always be the simplest of calculations. We will return to the Clausius theorem in the next chapter when we seek more convenient indicators of spontaneity.

7.4 The Third Law of Thermodynamics

In chapter 4, we have discussed how to calculate reaction enthalpies for any reaction, given the formation enthalpies of reactants and products. In this section, we will try to do the same for reaction entropies. In this case, however, our task is simplified by a fundamental law of thermodynamics, introduced by Walther Hermann Nernst (1864–1941) in 1906.30 The statement that was initially known as Nernst’s Theorem is now officially recognized as the third fundamental law of thermodynamics, and it has the following definition:

Definition 7.1 Third Law of Thermodynamics: The entropy of a perfectly ordered, pure, crystalline substance is zero at \(T=0 \; \text{K}\).

This law sets an unambiguous zero of the entropy scale, similar to what happens with absolute zero in the temperature scale. The absolute value of the entropy of every substance can then be calculated in reference to this unambiguous zero. As such, absolute entropies are always positive. This is in stark contrast to what happened for the enthalpy. An unambiguous zero of the enthalpy scale is lacking, and standard formation enthalpies (which might be negative) must be agreed upon to calculate relative differences.

In simpler terms, given a substance \(i\), we are not able to measure absolute values of its enthalpy \(H_i\) (and we must resort to known enthalpy differences, such as \(\Delta_{\mathrm{f}} H^{-\kern-6pt{\ominus}\kern-6pt-}\)). At the same time, for entropy, we can measure \(S_i\) thanks to the third law, and we usually report them as \(S_i^{-\kern-6pt{\ominus}\kern-6pt-}\). A comprehensive list of standard entropies of inorganic and organic compounds is reported in appendix 15. Reaction entropies can be calculated from the tabulated standard entropies as differences between products and reactants, using:

\[\begin{equation} \Delta_{\text{rxn}} S^{-\kern-6pt{\ominus}\kern-6pt-}= \sum_i \nu_i S_i^{-\kern-6pt{\ominus}\kern-6pt-}, \tag{7.23} \end{equation}\]

with \(\nu_i\) being the usual stoichiometric coefficients with their signs given in Definition 4.2.

The careful wording in the definition of the third law 7.1 allows for the fact that some crystal might form with defects (i.e., not as a perfectly ordered crystal). In this case, a residual entropy will be present even at \(T=0 \; \text{K}\). However, this residual entropy can be removed, at least in theory, by forcing the substance into a perfectly ordered crystal.31

An interesting corollary to the third law states that it is impossible to find a procedure that reduces the temperature of a substance to \(T=0 \; \text{K}\) in a finite number of steps.

7.5 Chapter Review

7.5.1 Practice Problems

Problem 1: Entropy Change in an Isothermal Expansion.

An ideal gas expands isothermally from a volume of \(2.0\,\text{L}\) to \(5.0\,\text{L}\) at \(300\,\text{K}\). Calculate the change in entropy of the system if there are \(0.5\,\text{mol}\) of gas.

Solution: For an isothermal expansion of an ideal gas, we use the equation:

\[\Delta S^{\text{sys}} = nR \ln \left(\frac{V_f}{V_i}\right)\]

Given:

\(n = 0.5\,\text{mol}\),

\(R = 8.314\,\text{J} \, \text{mol}^{-1} \, \text{K}^{-1}\),

\(V_i = 2.0\,\text{L}\),

\(V_f = 5.0\,\text{L}\).

Then:

\[\Delta S^{\text{sys}} = 0.5\,\text{mol} \cdot 8.314\,\text{J} \, \text{mol}^{-1} \, \text{K}^{-1} \cdot \ln \frac{5.0\,\text{L}}{2.0\,\text{L}}\]

\[\Delta S^{\text{sys}} = 4.157\,\text{J/K} \cdot 0.916 = 3.808\,\text{J/K}\] Adjusting for the significant digits, we get the final answer.

Answer: The change in entropy of the system is \(3.8\,\text{J/K}\).

Problem 2: Entropy of Vaporization.

Calculate the standard entropy of vaporization for ethanol at its boiling point of \(78.35^\circ\text{C}\), given that its enthalpy of vaporization is \(38.56\,\text{kJ/mol}\).

Solution: We use the equation for entropy of vaporization:

\[\Delta_{\text{vap}} S^{-\kern-6pt{\ominus}\kern-6pt-}= \frac{\Delta_{\text{vap}} H^{-\kern-6pt{\ominus}\kern-6pt-}}{T_B}\]

Given:

\(\Delta_{\text{vap}} H^{-\kern-6pt{\ominus}\kern-6pt-}= 38.56\,\text{kJ/mol}\),

\(T_B = 38.56^\circ\text{C}+273.15=351.51\,\text{K}\).

Then:

\[\Delta_{\text{vap}} S = \frac{38.56\,\text{kJ/mol}}{351.51\,\text{K}} = 0.1097\,\text{kJ/(mol K)}.\]

Answer: The standard entropy of vaporization for ethanol is \(109.7\,\text{J} \, \text{mol}^{-1} \, \text{K}^{-1}\).

Problem 3: Entropy Change in the Surroundings.

A system releases \(500\,\text{J}\) of heat to its surroundings at a constant temperature of \(25^\circ\text{C}\). Calculate the entropy change of the surroundings.

Solution: We use the equation for entropy change in the surroundings:

\[\Delta S^{\text{surr}} = -\frac{Q_{\text{sys}}}{T_{\text{surr}}}\]

Given:

\(Q_{sys} = -500\,\text{J}\) (negative because the system releases heat),

\(T_{surr} = 25\,\text{°C} + 273.15 = 298.15\,\text{K}\).

Then:

\[\Delta S^{\text{surr}} = -\frac{-500\,\text{J}}{298.15\,\text{K}} = 1.677\,\text{J/K}\] Fixing the significant digits gives us the final answer.

Answer: The entropy in the surroundings changes by \(1.7\,\text{J/K}\).

Problem 4: Entropy Change in an Isobaric Process

Calculate the entropy change when \(2.0\,\text{mol}\) of an ideal gas is heated from \(300\,\text{K}\) to \(400\,\text{K}\) at constant pressure. Assume \(C_P = 29\,\text{J} \, \text{mol}^{-1} \, \text{K}^{-1}\) and is independent of temperature.

Solution: For an isobaric process, we use the equation:

\[\Delta S_{sys} \approx nC_P \ln \left(\frac{T_f}{T_i}\right)\]

Given:

\(n = 2.0\,\text{mol}\),

\(C_P = 29\,\text{J} \, \text{mol}^{-1} \, \text{K}^{-1}\),

\(T_i = 300\,\text{K}\),

\(T_f = 400\,\text{K}\).

Then:

\[\Delta S^{\text{sys}} = 2.0\,\text{mol} \cdot 29\,\text{J} \, \text{mol}^{-1} \, \text{K}^{-1} \cdot \ln \left(\frac{400\,\text{K}}{300\,\text{K}}\right)\]

\[\Delta S^{\text{sys}} = 58.\,\text{J/K} \cdot 0.2877 = 16.69\,\text{J/K}\] Adjusting the significant digits gives us the final answer.

Answer: The entropy change in the system is \(17\,\text{J/K}\).

Problem 5: Reaction Eentropy.

Calculate the standard reaction entropy for the combustion of methane:

\[\text{CH}_{4(\text{g})} + 2\text{O}_{2(\text{g})} \rightarrow \text{CO}_{2(\text{g})} + 2\text{H}_2\text{O}_{(\text{l})}\]

Using the standard entropies reported in the tables at the end of the book.

Solution: We can use eq. (7.23) to calculate the reaction entropy:

\[\Delta_{\text{rxn}} S^{-\kern-6pt{\ominus}\kern-6pt-}= \sum_i \nu_i S^{-\kern-6pt{\ominus}\kern-6pt-}_i\]

\[\Delta_{\text{rxn}} S^{-\kern-6pt{\ominus}\kern-6pt-}= [S^{-\kern-6pt{\ominus}\kern-6pt-}_{\text{CO}_{2\text{(g)}}} + 2S^{-\kern-6pt{\ominus}\kern-6pt-}_{\text{H}_2\text{O}_{\text{(l)}}}] - [S^{-\kern-6pt{\ominus}\kern-6pt-}_{\text{CH}_{4 \text{(g)}}} + 2S^{-\kern-6pt{\ominus}\kern-6pt-}_{\text{O}_{2\text{(g)}}}]\]

\[\Delta_{\text{rxn}} S^{-\kern-6pt{\ominus}\kern-6pt-}= [213.8 + 2(70.0)] - [186.3 + 2(205.2)]\]

\[\Delta_{\text{rxn}} S^{-\kern-6pt{\ominus}\kern-6pt-}= 353.8 - 596.7 = -242.9\,\text{J} \, \text{mol}^{-1} \, \text{K}^{-1}\]

Answer: The standard reaction entropy for the combustion of methane is \(-242.9\,\text{J} \, \text{mol}^{-1} \, \text{K}^{-1}\)

7.5.2 Study Questions

1. What is the formula for calculating the change in entropy of the universe?

  1. \(\Delta S_{\text{universe}} = \Delta S^{\text{sys}} - \Delta S^{\text{surr}}\).
  2. \(\Delta S_{\text{universe}} = \Delta S^{\text{sys}} + \Delta S^{\text{surr}}\).
  3. \(\Delta S_{\text{universe}} = \Delta S^{\text{sys}} \cdot \Delta S^{\text{surr}}\).
  4. \(\Delta S_{\text{universe}} = \Delta S^{\text{sys}} / \Delta S^{\text{surr}}\).
  5. \(\Delta S_{\text{universe}} = (\Delta S^{\text{sys}})^2 + (\Delta S^{\text{surr}})^2\).

2. For an ideal gas undergoing an isothermal process, which formula correctly represents the change in entropy?

  1. \(\Delta S^{\text{sys}} = nR \ln \left(\frac{V_f}{V_i}\right)\).
  2. \(\Delta S^{\text{sys}} = nR \ln \left(\frac{P_f}{P_i}\right)\).
  3. \(\Delta S^{\text{sys}} = nR (V_f - V_i)\).
  4. \(\Delta S^{\text{sys}} = nR (P_f - P_i)\).
  5. \(\Delta S^{\text{sys}} = nR (T_f - T_i)\).

3. What is Trouton’s rule for the entropy of vaporization?

  1. \(\Delta_{text{vap}} S \approx 5.5R\).
  2. \(\Delta_{text{vap}} S \approx 15.5R\).
  3. \(\Delta_{text{vap}} S \approx 10.5R\).
  4. \(\Delta_{text{vap}} S \approx 20.5R\).
  5. \(\Delta_{text{vap}} S \approx 1.5R\).

4. Which statement is true for a reversible adiabatic process?

  1. \(\Delta S^{\text{sys}} > 0\).
  2. \(\Delta S^{\text{sys}} < 0\).
  3. \(\Delta S^{\text{sys}} = 0\).
  4. \(Q_{\text{REV}} > 0\).
  5. \(Q_{\text{IRR}} = 0\).

5. How is the entropy change of the surroundings calculated?

  1. \(\Delta S^{\text{surr}} = \frac{-Q_{\text{surr}}}{T_{surr}}\).
  2. \(\Delta S^{\text{surr}} = \frac{-Q_{\text{sys}}}{T_{sys}}\).
  3. \(\Delta S^{\text{surr}} = \frac{-Q_{\text{sys}}}{T_{surr}}\).
  4. \(\Delta S^{\text{surr}} = \frac{Q_{\text{sys}}}{T_{surr}}\).
  5. \(\Delta S^{\text{surr}} = \frac{Q_{\text{surr}}}{T_{sys}}\).

6. Which of the following formula describes what the Clausius theorem state?

  1. \(dS^{\text{sys}} = \frac{đQ}{T}\).
  2. \(dS^{\text{sys}} < \frac{đQ}{T}\).
  3. \(dS^{\text{sys}} > \frac{đQ}{T}\).
  4. \(dS^{\text{sys}} \leq \frac{đQ}{T}\).
  5. \(dS^{\text{sys}} \geq \frac{đQ}{T}\).

7. According to the Third Law of Thermodynamics, what is the entropy of a perfectly ordered, pure, crystalline substance at \(T = 0\,\text{K}\)?

  1. 1 J/K.
  2. 0 J/K.
  3. -1 J/K.
  4. Infinity.
  5. Undefined.

8. How can reaction entropies be calculated using standard entropies?

  1. \(\Delta_{\text{rxn}} S^{-\kern-6pt{\ominus}\kern-6pt-}= \sum_i \nu_i S^{-\kern-6pt{\ominus}\kern-6pt-}_i\).
  2. \(\Delta_{\text{rxn}} S^{-\kern-6pt{\ominus}\kern-6pt-}= \prod_i \nu_i S^{-\kern-6pt{\ominus}\kern-6pt-}_i\).
  3. \(\Delta_{\text{rxn}} S^{-\kern-6pt{\ominus}\kern-6pt-}= \sum_i S^{-\kern-6pt{\ominus}\kern-6pt-}_i / \nu_i\).
  4. \(\Delta_{\text{rxn}} S^{-\kern-6pt{\ominus}\kern-6pt-}= \sum_i (\nu_i S^{-\kern-6pt{\ominus}\kern-6pt-}_i)^2\).
  5. \(\Delta_{\text{rxn}} S^{-\kern-6pt{\ominus}\kern-6pt-}= \ln(\sum_i \nu_i S^{-\kern-6pt{\ominus}\kern-6pt-}_i)\).

9. How is the change in entropy of the system calculated for an isobaric process? Note: Assume \(C_P\) to be independent of temperature

  1. \(\Delta S^{\text{sys}} \approx nC_P \ln \left(\frac{T_f}{T_i}\right)\).
  2. \(\Delta S^{\text{sys}} \approx nC_V \ln \left(\frac{T_f}{T_i}\right)\).
  3. \(\Delta S^{\text{sys}} \approx nC_P (T_f - T_i)\).
  4. \(\Delta S^{\text{sys}} \approx nR \ln \left(\frac{T_f}{T_i}\right)\).
  5. \(\Delta S^{\text{sys}} \approx \frac{nC_P}{\ln (T_f/T_i)}\).

Which of the following is a corollary to the Third Law of Thermodynamics?

  1. It is possible to reach T = 0 K in a single step.
  2. It is impossible to reach T = 0 K in a finite number of steps.
  3. All substances have non-zero entropy at T = 0 K.
  4. The entropy of a substance is always negative at T > 0 K.
  5. The entropy of a substance is always infinity at T < 0 K.

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  1. Even if we think at the most energetic event that we could imagine happening here on earth—such as the explosion of an atomic bomb or the hit of a meteorite from outer space—such an event will not modify the average temperature of the universe by the slightest degree.↩︎

  2. In cases where the temperature of the system changes throughout the process, \(T\) is just the (constant) temperature of its immediate surroundings, \(T_{\text{surr}}\), as explained in section 7.2.↩︎

  3. Walther Nernst was awarded the 1920 Nobel Prize in Chemistry for his work in thermochemistry.↩︎

  4. A procedure that—in practice—might be extremely difficult to achieve.↩︎