Thermodynamics - Entropy and Heat Capacities

Heat Capacities and Poison's Ratio of Gases

The heat capacity of a gas depends on its atomicity and molecular structure. For an ideal gas, the molar heat capacity at constant volume is denoted as $C_v$ and at constant pressure as $C_p$. The relationship between them is defined by Mayer's formula: $C_p - C_v = R$. The Poison's Ratio ($\gamma$) is the ratio of these two heat capacities, defined as $\gamma = \frac{C_p}{C_v}$.

For Monoatomic gases (such as $He$, $Ne$, and $Ar$), the molar heat capacity at constant volume is $C_v = \frac{3}{2}R$ and at constant pressure is $C_p = \frac{5}{2}R$. This results in a Poison's Ratio of $\gamma = \frac{5/2 R}{3/2 R} \approx 1.67$.

For Diatomic gases (such as $O_2$, $N_2$, and $H_2$) and Linear Polyatomic gases (such as $CO_2$ and $HCN$), the molar heat capacity at constant volume is $C_v = \frac{5}{2}R$ and at constant pressure is $C_p = \frac{7}{2}R$. This results in a Poison's Ratio of $\gamma = \frac{7/2 R}{5/2 R} = 1.40$.

For Non-linear Polyatomic gases (such as $CH_4$, $H_2O$, and $NH_3$), the molar heat capacity at constant volume is $C_v = 3R$ and at constant pressure is $C_p = 4R$. This results in a Poison's Ratio of $\gamma = \frac{4R}{3R} \approx 1.33$.

Entropy Changes in Temperature and Phase Transformations

The entropy change of a system ($\Delta S_{sys}$) during a temperature change without a phase transition is calculated using the formula: $\Delta S_{sys} = n C_m \text{ln} \frac{T_2}{T_1}$, where $n$ is the number of moles, $C_m$ is the molar heat capacity (either $C_v$ or $C_p$ depending on the process), and $T_1$ and $T_2$ are the initial and final temperatures in Kelvin. The entropy change of the surroundings ($\Delta S_{surr}$) is defined as $\Delta S_{surr} = \frac{-q_{sys}}{T_{surr}}$. In constant pressure processes, this becomes $\Delta S_{surr} = \frac{-\Delta H_{sys}}{T_{surr}}$, where $\Delta H_{sys} = n C_p (T_2 - T_1)$.

Phase transformations occur at constant temperature (the transition temperature, $T_{trans}$). The entropy change for the system during a phase transition is given by $\Delta S_{sys} = \frac{\Delta H_{trans}}{T_{trans}}$. There are three primary types covered: melting (fusion), boiling (vaporization), and sublimation.

For Melting (Solid to Liquid): $\Delta S_{fus} = \frac{\Delta H_{fus}}{T_{mp}}$, where $T_{mp}$ is the melting point temperature.

For Boiling (Liquid to Vapour): $\Delta S_{vap} = \frac{\Delta H_{vap}}{T_{bp}}$, where $T_{bp}$ is the boiling point temperature.

For Sublimation (Solid to Vapour): $\Delta S_{sub} = \frac{\Delta H_{sub}}{T_{sub}}$, where $T_{sub}$ is the sublimation temperature.

Numerical Examples of Entropy and Phase Change

Example 1: Given that $30.4 \times 10^3 J$ of heat is required to melt one mole of sodium chloride and the entropy change at the melting point is $28.4 J K^{-1} mol^{-1}$, find the melting point ($T$). Using the formula $T = \frac{\Delta H}{\Delta S}$, we calculate $T = \frac{30.4 \times 10^3}{28.4}$. This results in a melting point of approximately $1070 K$.

Example 2: Calculate $\Delta S_{sys}$, $\Delta S_{surr}$, and $\Delta S_{univ}$ for the melting of $2 \text{ moles}$ of ice at $0^{\circ} C$ ($273 K$). Given $\Delta H_{fus} = 6.01 kJ/mol$. Since the process is at equilibrium at the melting point: $\Delta S_{sys} = \frac{2 \times 6.01 \times 10^3}{273} J/K$. For a reversible process, $\Delta S_{surr} = \frac{-(2 \times 6.01 \times 10^3)}{273} J/K$. Thus, $\Delta S_{univ} = \Delta S_{sys} + \Delta S_{surr} = 0$.

Example 3: If the same $2 \text{ moles}$ of ice at $0^{\circ} C$ melt in a bath kept at a constant temperature of $3^{\circ} C$ ($276 K$), the $\Delta S_{sys}$ remains the same: $\Delta S_{sys} = \frac{2 \times 6.01 \times 10^3}{273} J/K$. However, the surroundings are at a different temperature, so $\Delta S_{surr} = \frac{-(2 \times 6.01 \times 10^3)}{276} J/K$. In this irreversible case, $\Delta S_{univ} = \frac{2 \times 6.01 \times 10^3}{273} - \frac{2 \times 6.01 \times 10^3}{276} > 0$, indicating a spontaneous process.

Entropy Change for Multi-Step Paths

When a substance undergoes multiple changes (e.g., heating and phase transition), the total entropy change is the sum of individual steps. Consider heating $1 \text{ mole}$ of ice from $-10^{\circ} C$ ($263 K$) to steam at $120^{\circ} C$ ($393 K$).

Step 1: Heating ice from $263 K$ to $273 K$: $\Delta S_1 = n C_{p,ice} \text{ln} \frac{273}{263}$.

Step 2: Phase transition (fusion) at $273 K$: $\Delta S_2 = \frac{n \Delta H_{fus}}{273}$.

Step 3: Heating water from $273 K$ to $373 K$: $\Delta S_3 = n C_{p,water} \text{ln} \frac{373}{273}$.

Step 4: Phase transition (vaporization) at $373 K$: $\Delta S_4 = \frac{n \Delta H_{vap}}{373}$.

Step 5: Heating steam from $373 K$ to $393 K$: $\Delta S_5 = n C_{p,steam} \text{ln} \frac{393}{373}$.

The total entropy change of the system is $\Delta S_{net} = \Delta S_1 + \Delta S_2 + \Delta S_3 + \Delta S_4 + \Delta S_5$. If the surroundings are at a constant temperature of $393 K$, the total entropy change of the surroundings is calculated by dividing the total heat lost by the surroundings ($-\Delta H_{net}$) by the temperature of the surroundings ($393 K$).

Entropy Change in Ideal Gas Processes

For an ideal gas undergoing a process, the change in entropy of the system is derived from the First Law ($dq = du - dw$). Using $\Delta S = \frac{dq_{rev}}{T}$, we arrive at the general state function equation: $\Delta S_{sys} = n C_v \text{ln} \frac{T_2}{T_1} + n R \text{ln} \frac{V_2}{V_1}$. Alternatively, in terms of pressure: $\Delta S_{sys} = n C_p \text{ln} \frac{T_2}{T_1} + n R \text{ln} \frac{P_1}{P_2}$.

In an Isothermal Process ($T_1 = T_2$), the temperature factor becomes zero ($\text{ln} 1 = 0$). Therefore, $\Delta S_{sys} = n R \text{ln} \frac{V_2}{V_1} = n R \text{ln} \frac{P_1}{P_2}$. If the process is reversible, $\Delta S_{surr} = -\Delta S_{sys}$ and $\Delta S_{univ} = 0$. If the process is irreversible, the state function $\Delta S_{sys}$ remains the same, but $\Delta S_{surr} = \frac{-q_{irrev}}{T}$, and $\Delta S_{univ} > 0$.

In an Adiabatic Process, for a reversible path, $dq_{rev} = 0$, so $\Delta S_{sys} = 0$. This is consistent with the adiabatic relationship $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$, which makes the sum of the temperature and volume terms in the entropy equation zero. For an irreversible adiabatic process, $\Delta S_{surr} = 0$ because no heat is exchanged, but $\Delta S_{sys} > 0$. Although $\Delta S$ is a state function and the formula remains the same, the final temperature $T_2$ achieved in an irreversible adiabatic expansion is different from that in a reversible one, leading to a non-zero entropy change.

Advanced Numerical Questions

Question: Calculate the total entropy change for the transition at $368 K$ of $1 \text{ mole}$ of sulphur from monoclinic to rhombic solid state, if $\Delta H = -40.17 J/mol$ for this transition and the surroundings is an ice-water bath at $0^{\circ} C$ ($273 K$).

$\Delta S_{sys} = \frac{\Delta H}{T_{trans}} = \frac{-40.17 J/mol}{368 K} = -0.109 J K^{-1} mol^{-1}$. $\Delta S_{surr} = \frac{-\Delta H_{sys}}{T_{surr}} = \frac{+40.17}{273} = 0.147 J K^{-1} mol^{-1}$. $\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} = -0.109 + 0.147 = 0.38 J K^{-1} mol^{-1}$. The answer is (C).

Question: $1 \text{ mole}$ of an ideal gas expands freely from $1 L$ to $2 L$ at $300 K$ into a vacuum. Calculate $\Delta S_{universe}$. In free expansion, $w = 0$ and $q = 0$, so $T$ remains constant. $\Delta S_{sys} = R \text{ln} \frac{V_2}{V_1} = 8.314 \times \text{ln} 2 \approx 5.76 J/K$. Since $q_{surr} = 0$, $\Delta S_{surr} = 0$. Thus, $\Delta S_{univ} = 5.76 J/K$.

Question: $2 \text{ moles}$ of an ideal gas ($C_v = \frac{3}{2}R$) are heated from $300 K$ to $600 K$ and expanded from $10 L$ to $20 L$. Calculate $\Delta S$. $\Delta S = n C_v \text{ln} \frac{T_2}{T_1} + n R \text{ln} \frac{V_2}{V_1} = 2 \times \frac{3}{2} R \text{ln} \frac{600}{300} + 2 \times R \text{ln} \frac{20}{10}$. $\Delta S = 3 R \text{ln} 2 + 2 R \text{ln} 2 = 5 R \text{ln} 2 = 5 \times 8.314 \times 0.693 \times 2 \times 0.5 \approx 28.8 J/K$.