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**CARNOT CYCLE****Carnot cycle is based on 4 reversible process. **

**(1) Reversible isothermal expansion from A to B. **

ΔE_{AB} = 0 ,

**(2) Reversible adiabatic expansion from B to C **

ΔE_{BC} = nC_{V}(T_{2} -T_{1})

W_{BC} = ΔE_{BC}

**(3) Isothermal compression from C to D **

ΔE_{CD} = 0,

**(4) Adiabatic compression from D to A. **

ΔE_{DA} _{ }= nC_{V}(T_{1 }- T_{2})

W_{DA} = ΔE_{DA}

ΔE_{Cycle} _{ }= 0

For BC,

For ΔA,

W_{cycle} = -nR(T_{1} - T_{2})ln V_{2}/V_{1}

**Efficiency of any engine may be given as **

ΔE_{cycle }= q_{cyc} w_{cycle }

This means ΔS is a state function

**Gibb's Free Energy (G)**

G_{system} = H_{system} - TS_{system}

W = W_{expansion} +W_{non-expansion}

W_{non - expansion} = w_{useful} (useful work)

ΔG = W_{non expansion} = W_{useful}

All those energy which is available with the system which is utilized in doing useful work is called Gibb's free energy :

**R _{x} : **

ΔG° = Standard Gibb's free energy change (P = 1 atm, 298 K)

ΔG = Gibb's free energy change at any condition.

Δ

At equilibrium, ΔG = 0 and Q = K_{eq}.

0 = ΔG° 2.303 RT log k_{eq}

ΔG° = -2.303 RT log k_{eq}

å G°(product) - å G° (Reactant) = - 2.303 RT log k_{eq}

ΔH° - T ΔS° = -2.303 RT log k_{eq }

**(2) **W_{cell} = q × E

ΔG = - W_{cell}

ΔG = - q × E_{cell}

Now, one mole e^{-} have charge 96500 coulomb = 1 Faraday (F)

n mole of e^{-} will have charge = n × F or q = n × F

ΔG = - nFE_{cell}

ΔG° = -nFE°_{cell}**THIRD LAW OF THERMODYNAMICS**

limit_{T→0} S = 0

Third law of thermodynamics states that as the temperature approaches absolute zero, the entropy of perfectly crystalline substance also approaches zero.**APPLICATION OF THIRD LAW OF THERMODYNAMICS**

Taking T_{2} = T and T_{1} = 0°k.

For perfectly crystalline substance. The entropy of perfectly crystalline substance can be determined using third law of thermodynamics.**or **

With the help of third law of thermodynamics we can calculate the exact value of entropy.**ΔG AND NON PV WORK (NON EXPANSION/COMPRESSION)**

(ΔG)_{T,P } is a measure of useful work a non PV work (non expansion work) that can be produced by a chemical transformation.e.g.electrical work .

For reversible reaction at constant T & P

dU = dq + dW_{total}

dU = dq + dW_{P, V + }dW_{non P,V }

dU = T.dS - P. dV _{+} dW_{non P,V }

dU + P.dV = T. dS_{ + }dW_{non P,V}

dH = T.dS + dW_{non P,V}

(dG_{sys})_{T,P} = dW_{non P,V }

Useful work done on the system = increase in Gibb's energy of system at constant T & P.

- (ΔG_{sys})_{T,P}= - W_{non P,V}

- (ΔG_{sys})_{T,P}= - W_{by, non P,V}

Useful work done by the system = decrease in Gibb's energy of system at constant T & P.

If (ΔG_{sys})_{T,P} = 0, then system is unable to deliver useful work.**THERMODYNAMIC RELATION**

For reversible process in which non-expansion work is not possible

dU = dq + dW

H = U + PV

dH = dU + P.dV + VdP

dH = T.dS - PdV + PdV + V.dP

dH = T.dS + V.dP

G = H -TS

dG = dH - T.dS -S.dT

dG = T.dS + V.dP - T.dS - S.dT

For a particular system (s/ℓ/g)

**(1) **At constant temperature dG = V.dP

or**(A)** For a system is s/ ℓ phase

**(B) For an ideal gas, expansion/compression:-**

**(2) **At constant pressure : dG =-S.dT

or

For phase transformation/chemical reaction

d(ΔG) = ΔV.dP - ΔS dT

H_{2}O(s) → H_{2}O(*l*)

ΔV = V_{M}(H_{2}O,*l*) - V_{m}(H_{2}O,s)

ΔS = S_{M}(H_{2}O,*l*) - S_{m}(H_{2}O,s)

A(s) → B(g) + 2C(g)

Δ_{r}S = S_{M}(B,g ) 2S_{m}(C,g) - S_{m}(A,s)

C (s, graphite) → C(s, diamond)

ΔV = V_{m}(C, diamond) - V_{m}(C-graphite)

At constant temperature:-

d(ΔG) = ΔV. dP

At constant pressure

Δ_{r}G_{T2 }- Δ_{r}G_{T1} = - Δ_{r}S(T_{2} - T_{1})

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