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**Helmholtz free energy or Helmholtz function or work function.**

The work function (A) is defined as

A = U â€“ TS

Let an isothermal change at temperature T from the initial state indicated by subscript (1) to final state subscript (Z) so that

A_{1} = U_{1 }â€“ TS_{1}

& A_{2 }= â€“ TS_{2}

A_{2} â€“ A_{1} = (U_{2} â€“ U_{1}) â€“ T(S_{2 }â€“ S_{1})

Î”A = Î”U â€“ TÎ”S

or dA = dU â€“ TdS

dA = dU â€“ TdS [âˆ´ TdS = dq]

dA = dU â€“ dq

From first law dq = dU â€“ Ï‰ or dU â€“ dq = Ï‰ i.e.

dA = Ï‰

**Free energy function Gibbâ€™s free energy function.**

The free energy funct ion G is given by G = H â€“ TS

Let us consider the change of system from (1) to (2) at constant temperature, we have

G_{2} â€“ G_{1} = (H_{2} â€“ H_{1}) â€“ T(S_{2 }â€“ S_{1})

Î”G = Î”H â€“ TÎ”S

dG = dH â€“ TdS

**Variation of free energy with pressure:**

We know that

dG = VdP â€“ SdT

when process is isothermal then

dT = 0

(dG)_{T} = V(dP)_{T}

â‡’

When pressure is constant the

dP = 0

(dG) _{P} = â€“S(dT)_{P}

â‡’

Isothermal change in free energy for ideal gas

dG = VdP

Integrating wit h the limit s G_{1} and G_{2}, P_{1} and P_{2 }we get

we know that P_{1}V_{1} = P_{2}V_{2}

then

Isothermal change in free energy for solid and liquid The volume occupied by solid and liquid is nearly independent of pressure:

dG = VdP

G_{2} â€“ G_{1 }= V(P_{2} â€“ P_{1})**Problem. Two mole of an ideal gas are allowed to expand reversibly and isothermally at 300 K from a pressure of 1 atm to a pressure of 0.1 atm. What is the change is Gibbâ€™s free energy.****Sol.**

= 2.303 x 2 x 8.314 x 300 log

Î”G = â€“11488.2 J = â€“11.4882 KJ

**Gibbs Helmholtz equation. **

(A) Temperature depends of free energy we know that

G = H â€“ TS

& dG = VdP â€“ SdT

dG = VdP â€“ SdT

As pressure is constant, therefore

(dG)P = â€“S(dT)P

âˆ´ dG_{1} = â€“S_{1}dT

S_{1} & S_{2} are entropy in the init ial and final states.

dG_{2 }= â€“S_{2}dT

Subtracting these equation we get

dG_{2} â€“ dG_{1} = â€“(S_{2} â€“ S_{1})dT

&

We know that

â€¦(1)

This is Gibbâ€™s Helmholtz equation.

Î”G = Î”H - T Î”S

â‡’

Differentiating this equation w.r.t. T at constant P we get

... (2)

**(B) Temperature dependence of Helmholtz function:**

We know that A = U â€“ TS

Î”A = Î”U â€“ TÎ”S

& dA = â€“PdV â€“ SdT

As vo lume is constant, therefore

dA = â€“SdT dA_{1} = â€“S_{1}dT

S_{1} and S_{2} are init ial and final entropy. dA_{2} = â€“S_{2}dT Subtracting these equation we get

dA_{2} â€“ dA_{1} = â€“(S_{2} â€“ S_{1})dT

dÎ”A = â€“Î”SdT

â‡’

This is Gibbs Helmholtz equation.

Î”A = Î”U â€“ TÎ”S

â‡’

Different iat ing this equation w.r.t. T at constant V we get

â€¦(2)

**Problem. Prove that**

**Sol.** (a) Î”G = Î”H â€“ TÎ”S dividing by T we get

differentiation of by T at constant P, we get

or

Since therefore, we have

i.e.

**Gibbs Helmholtz equation.**

(b) Similarly

**Gibbs Helmholtz equation. Problem. **For the reaction.

The value of enthalpy change and free energy change are â€“68.32 and â€“56.69 K cals respectively.

**Calculate the value of free energy change at 30Â°C. Sol. We know that **

T = 25Â° = 273 + 25 = 298 K

â€“56.69 = 68.32 298

â‡’

then at 30Â°C

= â€“68.32 + 303 Ã— 0.0390

Î”G = â€“56.495 kcal.

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