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

The work function (A) is defined as

A = U – 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 }– TS_{1}

& A_{2 }= – TS_{2}

A_{2} – A_{1} = (U_{2} – U_{1}) – T(S_{2 }– S_{1})

ΔA = ΔU – TΔS

or dA = dU – TdS

dA = dU – TdS [∴ TdS = dq]

dA = dU – dq

From first law dq = dU – ω or dU – dq = ω i.e.

dA = ω

**Free energy function Gibb’s free energy function.**

The free energy funct ion G is given by G = H – TS

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

G_{2} – G_{1} = (H_{2} – H_{1}) – T(S_{2 }– S_{1})

ΔG = ΔH – TΔS

dG = dH – TdS

**Variation of free energy with pressure:**

We know that

dG = VdP – SdT

when process is isothermal then

dT = 0

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

⇒

When pressure is constant the

dP = 0

(dG) _{P} = –S(dT)_{P}

⇒

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} – G_{1 }= V(P_{2} – 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 Gibb’s free energy.****Sol.**

= 2.303 x 2 x 8.314 x 300 log

ΔG = –11488.2 J = –11.4882 KJ

**Gibbs Helmholtz equation. **

(A) Temperature depends of free energy we know that

G = H – TS

& dG = VdP – SdT

dG = VdP – SdT

As pressure is constant, therefore

(dG)P = –S(dT)P

∴ dG_{1} = –S_{1}dT

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

dG_{2 }= –S_{2}dT

Subtracting these equation we get

dG_{2} – dG_{1} = –(S_{2} – S_{1})dT

&

We know that

…(1)

This is Gibb’s Helmholtz equation.

ΔG = ΔH - T ΔS

⇒

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

... (2)

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

We know that A = U – TS

ΔA = ΔU – TΔS

& dA = –PdV – SdT

As vo lume is constant, therefore

dA = –SdT dA_{1} = –S_{1}dT

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

dA_{2} – dA_{1} = –(S_{2} – S_{1})dT

dΔA = –ΔSdT

⇒

This is Gibbs Helmholtz equation.

ΔA = ΔU – TΔS

⇒

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

…(2)

**Problem. Prove that**

**Sol.** (a) ΔG = ΔH – TΔS 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 –68.32 and –56.69 K cals respectively.

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

T = 25° = 273 + 25 = 298 K

–56.69 = 68.32 298

⇒

then at 30°C

= –68.32 + 303 × 0.0390

ΔG = –56.495 kcal.

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