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# Helmholtz Free Energy Or Helmholtz Function Or Work Function - Thermochemistry Chemistry Notes | EduRev

## IIT JAM : Helmholtz Free Energy Or Helmholtz Function Or Work Function - Thermochemistry Chemistry Notes | EduRev

The document Helmholtz Free Energy Or Helmholtz Function Or Work Function - Thermochemistry Chemistry Notes | EduRev is a part of the IIT JAM Course Physical Chemistry.
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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
A1 = U– TS1
&              A=  – TS2
A2 – A1 = (U2 – U1) – T(S– S1)
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

G2 – G1 = (H2 – H1) – T(S– S1)

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 G1 and G2, P1 and Pwe get   we know that P1V1 = P2V2
then  Isothermal change in free energy for solid and liquid The volume occupied by solid and liquid is nearly independent of pressure:

dG = VdP
G2 – G= V(P2 – P1)

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
∴            dG1 = –S1dT

S1 & S2 are entropy in the init ial and final states.

dG= –S2dT

Subtracting these equation we get
dG2 – dG1 = –(S2 – S1)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 dA1 = –S1dT

S1 and S2 are init ial and final entropy. dA2 = –S2dT Subtracting these equation we get
dA2 – dA1 = –(S2 – S1)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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## Physical Chemistry

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