Third Law Of Thermodynamics And Determine To Absolute Entropies Of Solids, Liquids And Gases

As per the below equation,

\[\displaystyle T\underset{{}}{\overset{{Lt}}{\longrightarrow}}0Cp=0\]

∆Cp is zero at 0 k. It means that at absolute zero, the heat capacities of products and reactants in a solid state are the same. It leads to the idea that all substances have the same heat capacity at absolute zero.

A quantum theory applied to the heat capacities of solids has shown that the heat capacities of solids tend to become zero at 0 k. Therefore, the Nernst heat theorem is written as,

\[\displaystyle T\underset{{}}{\overset{{Lt}}{\longrightarrow}}0Cp=0---(1)\]

As per the equation,

\[\displaystyle T\underset{{}}{\overset{{Lt}}{\longrightarrow}}0\Delta S=0\]

∆S is zero at absolute zero. It means the entropy change of a process involving solids becomes zero at 0 k. Therefore, entropies of all pure solids approach zero at 0 k. i.e.,

\[\displaystyle T\underset{{}}{\overset{{Lt}}{\longrightarrow}}0\Delta S=0---(2)\]

The statement of the third law of thermodynamics is,

“At absolute zero of temperature, the entropy of every substance may become zero, and it does become zero in the case of perfectly crystalline solid.”

Table of Contents

DETERMINES OF ABSOLUTE ENTROPIES OF SOLIDS, LIQUIDS AND GASES

We know that for a small change of the state of a substance or a system, the entropy change is,

\[\displaystyle ds=\frac{{dq}}{T}---(3)\]

If a change takes place at constant pressure, then,

\[\displaystyle {{\left( {\partial S} \right)}_{p}}=\frac{{{{{\left( {\partial q} \right)}}_{p}}}}{T}---(4)\]

\[\displaystyle {{\left( {\frac{{\partial S}}{{\partial T}}} \right)}_{p}}={{\left( {\frac{{\partial q}}{{\partial T}}} \right)}_{p}}\times \frac{1}{T}---(5)\]

As definition,

\[\displaystyle {{\left( {\frac{{\partial q}}{{\partial T}}} \right)}_{p}}=Cp----(6)\]

\[\displaystyle {{\left( {\frac{{\partial S}}{{\partial T}}} \right)}_{p}}=Cp\times \frac{1}{T}---(7)\]

Or at constant pressure,

\[\displaystyle ds=\left( {\frac{{Cp}}{T}} \right)dT----(8)\]

For perfectly crystalline substance, the absolute entropy S =0 at T=0, therefore may write as,

\[\displaystyle \int\limits_{{s=0}}^{{s=s}}{{ds=\int\limits_{{T=0}}^{{T=T}}{{\left( {\frac{{Cp}}{T}} \right)dT---(9)}}}}\]

\[\displaystyle {{S}_{T}}=\int\limits_{0}^{T}{{\frac{{{{C}_{p}}dT}}{T}}}=\int\limits_{0}^{T}{{{{C}_{p}}}}d\left( {\ln T} \right)---(10)\]

Here, S_T = absolute entropy of the crystalline solid under examination at the temperature (T).

The integral equation (10) can be evaluated by measuring Cp at various temperatures between T = 0 and desired temperature T, plotting Cp → lnT and determining the area under the curve between T =0 and required temperature T. It gives a direct value of S_T.

It is impossible to obtain the value of Cp at the absolute zero; heat capacities are measured up to as low a temperature as possible, up to 15 k, and the value at the absolute zero is by extrapolation.

This method determines the substance’s heat capacities under examination at a temperature varying from approximately 15 k to the required temperature (T). A graph of Cp vs ln T is plotted and extrapolated to the absolute zero of temperature, as shown in the figure.

The area under the graph gives the required value of S_T. The absolute entropy of the substance at temperature T and equation (10) are written as follows;

\[\displaystyle {{S}_{T}}=\int\limits_{0}^{{{{T}^{*}}}}{{{{C}_{p}}}}\frac{{dT}}{T}+\int\limits_{{{{T}^{*}}}}^{T}{{{{C}_{p}}}}\frac{{dT}}{T}---(11)\]

Here , 0 < T*<15 k

The first integral is evaluated the heat capacities of crystalline substances to which at very low temperatures (0 < T < 15 K)

\[\displaystyle {{C}_{p}}\approx {{C}_{v}}\approx a{{T}^{3}}---(12)\]

As per equation (11) is written as,

\[\displaystyle {{S}_{T}}=\int\limits_{0}^{{{{T}^{*}}}}{{a{{T}^{3}}}}\frac{{dT}}{T}+\int\limits_{{{{T}^{*}}}}^{T}{{{{C}_{p}}}}\frac{{dT}}{T}=\frac{1}{3}a{{\left( {{{T}^{*}}} \right)}^{3}}+\int\limits_{{{{T}^{*}}}}^{T}{{{{C}_{p}}}}\frac{{dT}}{T}---(13)\]

Combining the heat capacity data with the enthalpy data on phase transformations, the absolute entropy of the substance is solid, liquid or gas at temperature T, can be determined as below,

The absolute entropy of a gas at 25° C under atmospheric pressure would equal the sum of the entropy change involved in the following processes, each of which brought about reversibly.

It assumes that the substance in the solid state exists in two allotropic forms, α and β.

Heating crystalline solid from absolute zero to temperature T*, where 0 < T*<15 k and evaluating the entropy change with Debye’s theory,

Let entropy change be ∆S₁, then,

\[\displaystyle \Delta {{S}_{1}}=\int\limits_{0}^{{{{T}^{*}}}}{{a{{T}^{3}}}}\frac{{dT}}{T}=\frac{1}{3}a{{\left( {{{T}^{*}}} \right)}^{3}}---(14)\]

Heating the crystalline solid from T* to T_tr

Where T_tr= transition temperature at which crystalline solid changes from allotropic from α to allotropic β. Entropy change in the process is,

\[\displaystyle \Delta {{S}_{2}}=\int\limits_{{{{T}^{*}}}}^{{{{T}_{r}}}}{{+{{C}_{{p,s}}}\left( \alpha \right)}}dT---(15)\]

Where Cp,s (α) = heat capacity of the solid in allotropic from α

The transition of the solid from allotropic form α to allotropic β at transition temperature T The entropy change,

\[\displaystyle \Delta {{S}_{3}}=\frac{{\Delta {{H}_{{tr}}}}}{{{{T}_{{tr}}}}}---(16)\]

∆ H_tr = Molar enthalpy of transition

Heating the solid in allotropic form β to fusion point T_fus. The entropy change is,

\[\displaystyle \Delta {{S}_{4}}=\int\limits_{{{{T}_{{tr}}}}}^{{{{T}_{{fus}}}}}{{{{C}_{{p,s}}}}}\left( \beta \right)dT---(17)\]

Here, Cp,s,₍_β)= heat capacity of the solid in allotropic form β.

Changing the solid in allotropic form β into a liquid state at the fusion temperature T_fusis given by,

\[\displaystyle \Delta {{S}_{5}}=\frac{{\Delta {{H}_{{fus}}}}}{{{{T}_{{fus}}}}}---(18)\]

Heating the liquid from its freezing point (T_fus) to its boiling point (T_b), the entropy change is,

\[\displaystyle \Delta {{S}_{6}}=\int\limits_{{{{T}_{{fus}}}}}^{{{{T}_{b}}}}{{{{C}_{{p,l}}}d\ln T----(19)}}\]

Cp,_l= heat capacity of substance in the liquid state

The liquid is changed into a gaseous state at temperature T_b. The entropy change,

\[\displaystyle \Delta {{S}_{7}}=\frac{{\Delta {{H}_{{vap}}}}}{{{{T}_{b}}}}---(20)\]

∆ Hvap = enthalpy of vaporization per mole of the substance

Heating the gas from T_b to the required temperature, i.e. 25° c (298.15 k), the entropy change involved the process given by,

\[\displaystyle \Delta {{S}_{8}}=\int\limits_{{{{T}_{b}}}}^{{298.15}}{{{{C}_{{p,g}}}d\ln T---(21)}}\]

Cp,_g = heat capacity of the substance in the gaseous state at constant pressure

The absolute entropy of the gas at 298.15 k (25° C), S_T = sum of all the entropy changes listed above, so,

\[\displaystyle \Delta {{S}_{T}}=\Delta {{S}_{1}}+\Delta {{S}_{2}}+\Delta {{S}_{3}}+\Delta {{S}_{4}}+\Delta {{S}_{5}}+\Delta {{S}_{6}}+\Delta {{S}_{7}}+\Delta {{S}_{8}}\]

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