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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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