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## Maxwell and Calusius-Clapeyron Equation, Chemistry tutorial

:The Maxwell RelationsSo far we have learnt regarding the five functions U, H, S, A and G which are the state functions and extensive variables. The different expressions helpful in estimating the changes in the above functions in a closed system in terms of temperature, pressure, volume and entropy might be represented as:

dU = TdS - pdV

dH = TdS + Vdp

dA = - SdT - pdV

dG = - SdT + Vdp

At constant volume (dV = 0), the equation (dU = TdS - pdV) becomes

(∂U)

_{V }= T(∂S)_{V}Or (∂U/∂S)

_{V}= VAnd at constant entropy (dS = 0), equation (dU = TdS - pdV) becomes

(∂U)

_{S}= P (∂V)_{S}Or (∂U/∂V)

_{V}= - POn differentiating the equation (∂U/∂S)

_{V}= V with respect to volume at constant entropy and equation (∂U/∂V)_{V}= - PWith respect to the entropy at constant volume, we obtain

∂

^{2}U/∂V∂S = - (∂T/∂V)_{S}∂

^{2}U/∂S∂V = - (∂P/∂S)_{V}As 'U' is a State function, it follows that,

∂

^{2}U/∂V∂S = (∂^{2}U/∂V)_{S}Or (∂T/∂V)

_{S}= - (∂P/∂S)_{V}Following the similar mathematical process as illustrated above, the given expressions can be simply derived:

(∂T/∂P)

_{S}= (∂V/∂V)_{S}(∂S/∂V)

_{T}= (∂V/∂T)_{V}(∂S/∂P)

_{T}= - (∂V/∂T)_{P}The above four equation are termed as Maxwell relation. The Maxwell relations are significant as they equate the rate of change of a particular quantity (which can't be found out experimentally) by the rate of change of volume, pressure or temperature (that can be found out experimentally).

:Gibbs-Helmholtz EquationThe dependence of Gibbs free energy no temperature can be represented in several ways. Beginning from the definition:

G = H - TS

Or - S = (G - H)/T

Replacing the expression for -S in equation (∂S/∂P)

_{T }= - (∂V/∂T)_{P}, we get,(∂G/∂T)

_{P}= (G - H)/TAt times it is significant to know the variation of G/T on temperature. By differentiating G/T with temperature at constant pressure, we get

[∂(G/T)/∂T]

_{P}= (1/T)(∂G/∂T)_{P}- (1/T^{2})GReplacing the value of (∂G/∂T)

_{P}from equation (∂G/∂T)_{P}= (G - H)/T into equation [∂(G/T)/∂T]_{P}= (1/T)(∂G/∂T)_{P}- (1/T^{2})G, we get[∂(G/T)/∂T]

_{P }= [(G - H)/T^{2}] - (1/T^{2})GTherefore, [∂(G/T)/∂T]

_{P}= (G - H - G)/T^{2}Or, [∂(G/T)/∂T]

_{P}= H/T^{2}The above equation is termed as Gibbs-Helmholtz equation and it can be put in the other form by remembering that d(1/T) - (1/T

^{2})dT; so on substituting ∂T in [∂(G/T)/∂T]_{P}= H/T^{2}by -T^{2}∂(1/T^{2}), we have,[∂(G/T)/-T

^{2}(1/T)]_{P}= H/T^{2}Or [∂(G/T)/∂(1/T)]

_{P}= HIdentical equation for Helmholtz free energy can as well be derived in the form of:

[∂(A/T)/∂T]

_{V}= - U/T^{2}[∂(A/T)/∂(1/T)]

_{V}= VThe dependence of Gibbs free energy on temperature can be represented in another way as well. Assume that G

_{1}is the Gibbs free energy for a system in the initial state and at temperature T. Suppose the temperature changes to T + dT and the corresponding value of free energy to G_{1}+ dG_{1}.Alike for the final state of the system, suppose the Gibbs free energies be G

_{1}and G_{2}+ dG_{2}at temperature T and T + dT, correspondingly. At constant pressure, equation dG = Vdp - SdT reduces to:(∂G)

_{p}= - S(∂T)_{p}And therefore (∂G

_{1})_{p}= - S_{1}(∂T)_{p}(∂G

_{2})_{p}= - S_{2}(∂T)_{p }Here S

_{1}and S_{2}are the entropies of the system in the initial and the final states, correspondingly. Subtraction of (∂G_{2})_{p}= - S_{2}(∂T)_{p}gives(∂G

_{2}- ∂G_{1})_{p}= - S_{2}(∂T)_{p}- (- S_{1}(∂T)_{p})Or [∂(G

_{2}- G_{1})]_{p }= - (S_{2}- S_{2}) (∂T)_{p})Or [∂(ΔG)]

_{p}= - ΔS (∂T)_{p}That is, [∂(ΔG)/(∂T)]

_{V}= - ΔSAs per equation ΔG = ΔH - TΔS = ΔH + T (- ΔS)

Replacing the equation [∂(ΔG)/(∂T)]

_{V}= - ΔS in ΔG = ΔH + T (- ΔS) we haveΔG = ΔH + T [∂(ΔG)/(∂T)]

_{V}We are as well familiar by ΔA = ΔU + T [∂(ΔA)/∂T]

_{V}The Gibbs-Helmholtz equation allows the computation of ΔU or ΔH given ΔA or ΔG and their corresponding temperature coefficients,

[∂(ΔA)/∂T]

_{V}or [∂(ΔG)/∂T]_{p}are known:Clausius-Clapeyron EquationFrom the equation (∂P/∂T)

_{V}= (∂S/∂V)_{T}The equation above can be applied to any closed system comprising of two phases of the similar substance in equilibrium with each other. Let us take a closed system in which a pure liquid and its vapor are in equilibrium by one other.

A (liquid) ↔ A (vapor)

The vapor pressure of the liquid (A) based on the temperature however is independent of the volume of the liquid and the vapor.

Therefore, (∂p/∂T)

_{V}= dp/dTWhenever one mole of the liquid is vaporized isothermally and reversibly in such a manner that throughout the process, the liquid and the vapor remain in equilibrium, the increase in enthalpy is equivalent to the molar enthalpy of vaporization (ΔH

_{vap}). As the process is reversible, it follows from the equation,ΔS = ΔH

_{vap}/THere, T is the temperature.

Assume that the increase in volume and entropy throughout the vaporization of one mole of liquid be AV and AS, correspondingly. Then we have at constant temperature (by using the above equation)

(∂S/∂V)

_{T}= ΔS/ΔV = ΔH_{vap}/TΔVComparing the equations (∂S/∂V)

_{T}= (∂V/∂T)_{V}, (∂p/∂T)_{V}= dp/dT and (∂S/∂V)_{T}= ΔH_{vap}/TΔV, we have,dp/dT = ΔH

_{vap}/TΔVThe above equation was first introduced by Clapeyron (1834) and is termed as Clapeyron equation. It was later extended by Clausius in the year 1850. This was derived for a system comprising of liquid and vapor in equilibrium. The similar equation can, though, be derived for equilibrium between any two phases like, solid and liquid, solid and vapor, two crystalline forms of the similar Substance and so on. For a system comprising of water in the two phases, liquid and vapor, in equilibrium with one other,

Water (liquid) ↔ Water (vapor)

The equation dp/dT = ΔH

_{vap}/TΔV can be represented as:dp/dT = ΔH

_{vap}/T(V_{g}- V_{l})Here,

ΔH

_{fus}= Molar enthalpy of the vaporization of waterV

_{g}= The molar volume of water vapor at temperature 'T'V

_{l }= The molar volume of liquid water at temperature 'T'For a system comprising of ice at its melting point, the two phases in equilibrium are ice and liquid water.

Water (ice) ↔ Water (liquid)

The equation dp/dT = ΔH

_{vap}/TΔV can be represented as:dp/dT = ΔH

_{fus}/T(V_{l}- V_{s})Here,

ΔH

_{fus}= Molar enthalpy of fusion of iceV

_{s}= Molar volume of water in the solid (ice) phaseV

_{l }= Molar volume of water in the liquid phaseIn the liquid ↔ vapor equilibrium, Clausius supposed that the molar volume of a liquid is much less than the molar volume of its vapor; therefore V

_{g}-V_{l }can be taken roughly as equation above might be represented as,dp/dT = ΔH

_{vap}/TV_{g}Supposing that the vapor behaves ideally,

V

_{g }= RT/pReplacing for the volume V

_{g}in equation dp/dT = ΔH_{vap}/TV_{g}, we havedp/dT = (ΔH

_{vap}/T) x (p/RT) = ΔH_{vap}p/RT^{2}The above equation is termed as Clausius-Clapeyron equation and can be represented as,

dp/dT = (ΔH

_{vap}/R) x (dT/T^{2})The above equation can be integrated between limits p

_{1}→ p_{2}and T_{1}→T_{2}supposing that ΔH_{vap}remains constant over a small range of temperature._{p1}∫^{p2}(dp/p) = (ΔH_{vap}/R)_{T1}∫^{T2}(dT/T^{2})Or, p

_{2}/p_{1}= - ΔH_{vap}/R [(1/T_{1}) - (1/T_{2})]Or, 2.303 log p

_{2}/p_{1}= ΔH_{vap}/R [(1/T_{1}) - (1/T_{2})]Or, log (p

_{2}/p_{1}) = ΔH_{vap}/2.303R [(T_{2 }- T_{1})/T_{1}T_{2}]This is the integrated form of Clausius-Clapeyron equation

The applications of Clausius-Clapeyron equation are illustrated below:

1) If the vapor pressure of a liquid at a different temperatures are known then its molar enthalpy of vaporization can be computed by using log (p

_{2}/p_{1}) = ΔH_{vap}/2.303R [(T_{2}-T_{1})/T_{1}T_{2}]2) If the vapor pressure of a liquid at any one temperature is known then that the other temperature can be computed by using log (p

_{2}/p_{1}) = ΔH_{vap}/2.303R [(T_{2}-T_{1})/T_{1}T_{2}]3) log (p

_{2}/p_{1}) = ΔH_{vap}/2.303R [(T_{2}-T_{1})/T_{1}T_{2}] can as well be employed for computing the effect of pressure on the boiling point of a liquid.Tutorsglobe: A way to secure high grade in your curriculum (Online Tutoring)Expand your confidence, grow study skills and improve your grades.

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