Gibbs-Helmholtz equation

2.4.4 · Physics › Thermodynamics & Statistical Mechanics (Advanced)

YEH HAI KYA?

Helmholtz-energy ka twin (constant volume) yeh hai: $(\frac{\partial ( A / T )}{\partial T})_{V} = - \frac{U}{T ^{2}}$

KAISE: First Principles se Derivation

Hum Gibbs energy ki definition aur fundamental relation se shuru karte hain.

Step 1 — Definition. $G = H - T S$ Yeh step kyun? Yeh $G$ ki definition hai; sab kuch isi se aana chahiye.

Step 2 — Entropy ko $G$ ke derivative ke roop mein nikalo. Differential $d G = - S d T + V d P$ se (jo khud $d G = d H - T d S - S d T$ aur $d H = T d S + V d P$ se aata hai, $T d S$ terms cancel hoke), constant $P$ par: $(\frac{\partial G}{\partial T})_{P} = - S$ Yeh step kyun? Humein $S$ ko $G$ se judi kisi cheez se replace karna hai, taaki sirf $H$ unknown rehe.

Step 3 — $S$ ko definition mein substitute karo. $G = H - T [- (\frac{\partial G}{\partial T})_{P}] = H + T (\frac{\partial G}{\partial T})_{P}$ Rearrange karo: $H = G - T\left(\frac{\partial G}{\partial T}\right)_P \tag{$\star$}$ Yeh step kyun? Yeh pehle se hi $G$ aur uske slope ke terms mein $H$ deta hai — lekin yeh ugly hai. $G / T$ trick ise clean karti hai.

Step 4 — Quotient rule se $G / T$ ko differentiate karo. $\frac{\partial}{\partial T} (\frac{G}{T})_{P} = \frac{T ( \frac{\partial G}{\partial T} ) _{P} - G \cdot 1}{T ^{2}} = \frac{1}{T} (\frac{\partial G}{\partial T})_{P} - \frac{G}{T ^{2}}$ Yeh step kyun? Humne precisely $G / T$ isliye choose kiya kyunki numerator $T \partial_{T} G - G$ hai, jo $(⋆)$ se $- [G - T \partial_{T} G] = - H$ hai.

Step 5 — $(⋆)$ insert karo. Numerator ko $- 1$ se multiply karo: numerator $= - (G - T \partial_{T} G) = - H$ . Isliye $(\frac{\partial ( G / T )}{\partial T})_{P} = - \frac{H}{T ^{2}}$ Yeh step kyun? Poora point yahi hai: $S$ gayab ho gaya, $H$ bach gaya.

Worked Examples

Common Mistakes

Active Recall

Recall Self-test (answers chupaao)

$G$ ki jagah $G / T$ kyun differentiate karte hain? → kyunki $T \partial_{T} G - G = - H$ hai, isliye entropy cancel ho jaata hai.
$(\partial G / \partial T)_{P}$ kya hai? → $- S$ .
GH ko $1/ T$ form mein batao. → $\partial (Δ G / T) / \partial (1/ T) = Δ H$ .
Constant-volume mein kaunsi parallel equation hai? → $\partial (A / T) / \partial (1/ T) = U$ .

Gibbs–Helmholtz equation (T form)

(\frac{\partial ( G / T )}{\partial T})_{P} = - \frac{H}{T ^{2}}

Gibbs–Helmholtz equation (

1/ T

form)

(\frac{\partial ( G / T )}{\partial ( 1/ T )})_{P} = H

(\partial G / \partial T)_{P}

kiske barabar hai?

- S

(entropy),

H

NAHI

Differentiate karne se pehle

T

se kyun divide karte hain?

Taaki

- T S

entropy term cancel ho jaye, sirf

H

bache

Helmholtz (constant-V) analogue

(\frac{\partial ( A / T )}{\partial ( 1/ T )})_{V} = U

Two-point integrated form (const

Δ H

)

\frac{Δ G _{2}}{T _{2}} - \frac{Δ G _{1}}{T _{1}} = Δ H (\frac{1}{T _{2}} - \frac{1}{T _{1}})

GH ko

Δ G^{\circ} = - R T ln K

par apply karne se kya milta hai?

van 't Hoff:

\frac{d ln K}{d ( 1/ T )} = - \frac{Δ H ^{\circ}}{R}

Δ H

extract karne ke liye kya plot karo

Δ G / T

1/ T

; slope

= Δ H

Derivation ke liye starting relation

G = H - T S

aur

d G = - S d T + V d P

Recall Feynman: ek 12-saal ke bachche ko samjhao

Socho tumhara "khushi ka budget" $G$ real money $H$ se bana hai minus ek "fun tax" $T S$ jo temperature $T$ ke saath badhta hai. Agar main sirf dekhunga ki garam hone par budget kaise badalta hai, toh main asliyat mein sirf tax ka badlav dekh raha hounga — yeh real money chhupa deta hai. Lekin agar main "budget per degree" ( $G / T$ ) dekhun aur usse observe karun, toh tax wala part cancel ho jaata hai aur mujhe finally pata chalta hai ki asli money $H$ kitni hai. Gibbs–Helmholtz dekhne ka yahi smart tarika hai.

Connections

Gibbs free energy — woh quantity jo differentiate ki ja rahi hai.
Enthalpy — jo GH extract karta hai.
Entropy — woh term jo $G / T$ trick se cancel ho jaati hai.
Maxwell relations — thermodynamic identities ki same family.
van 't Hoff equation — GH ko equilibrium constants par apply kiya gaya.
Kirchhoff's law (thermochemistry) — wide ranges par $Δ H (T)$ correct karta hai.
Helmholtz free energy — constant-volume twin equation.