Exercises — Maxwell relations — derivation from each potential

KYA dhundhna hai: ek aisi relation jisme $S$ ko $P$ ke saath differentiate kiya gaya ho constant $T$ par, ek taraf. Toolkit scan karo — yeh chauthi relation hai: ${\left(\frac{\partial S}{\partial P}\right)}_T=-{\left(\frac{\partial V}{\partial T}\right)}_P$ YEH KYUN: left side bilkul match karti hai. Yeh Gibbs free energy $G$ se aati hai, jiske natural variables $(T,P)$ hain — exactly wahi do variables jo yahan kaam mein hain ($T$ held, $P$ varied). YEH KYUN USEFUL HAI: right side $-V\alpha$ hai, jise tum thermometer aur ruler se measure kar sakte ho. Toh ${\left(\frac{\partial S}{\partial P}\right)}_T=-V\alpha$.

KYA: dhundho jahan $T$ ko $V$ se differentiate kiya gaya ho. Yeh pehli relation hai (from $U$): ${\left(\frac{\partial T}{\partial V}\right)}_S=-{\left(\frac{\partial P}{\partial S}\right)}_V$ SUBSCRIPT $S$ KYUN HAI: $U$ ke natural variables $(S,V)$ hain. Jab hum $V$ ke saath differentiate karte hain, toh doosra natural variable, $S$, woh hota hai jo held fixed rehta hai. Us subscript ko drop karna ya change karna relation ko galat bana deta hai — neeche mistake box dekho.

Step 1 — relation chunna (KYUN): humein $\partial S/\partial V$ chahiye constant $T$ par. Helmholtz relation ise measurable terms mein deta hai: ${\left(\frac{\partial S}{\partial V}\right)}_T={\left(\frac{\partial P}{\partial T}\right)}_V$ Step 2 — RHS evaluate karo (KYA): $P=RT/V$ se, $T$ ke saath differentiate karo $V$ fixed rakh kar: ${\left(\frac{\partial P}{\partial T}\right)}_V=\frac{R}{V}$ Toh ${\left(\frac{\partial S}{\partial V}\right)}_T=\frac{R}{V}$. Step 3 — fixed $T$ par integrate karo: $S=\int \frac{R}{V}dV=R\ln V+\text{const}(T)$ Sanity check: yeh exactly Sackur–Tetrode entropy ki volume-dependence hai — famous $R\ln V$ term — sirf ek Maxwell relation se recover kiya gaya.

Step 1 (KYUN): Gibbs relation ${\left(\partial S/\partial P\right)}_T=-{\left(\partial V/\partial T\right)}_P$. $\alpha$ ki definition se, ${\left(\partial V/\partial T\right)}_P=V\alpha$, toh seedha ${\left(\frac{\partial S}{\partial P}\right)}_T=-V\alpha .$ Step 2 — ideal gas $\alpha$: $V=RT/P$ ke saath, ${\left(\partial V/\partial T\right)}_P=R/P$, toh $\alpha =\frac{1}{V}\cdot \frac{R}{P}=\frac{1}{T}$ (kyunki $PV=RT$). Ideal gas ke liye $\alpha =1/T$. Step 3 — number: ${\left(\frac{\partial S}{\partial P}\right)}_T=-V\alpha =-\frac{RT}{P}\cdot \frac{1}{T}=-\frac{R}{P}$. $=-\frac{8.314}{1.00\times 10^5}=-8.314\times 10^{-5}{\text{J K}}^{-1}{\text{Pa}}^{-1}.$ Matlab: pressure badhane se entropy ghatti hai (squeeze karne se gas ordered hoti hai) — sign negative hai, exactly jaisa intuition demand karta hai.

Step 1 — entropy se shuru karo (KYUN): heat capacities hain $C_V=T{\left(\partial S/\partial T\right)}_V$ aur $C_P=T{\left(\partial S/\partial T\right)}_P$. Humein unka difference chahiye, toh $S$ ko ek baar $S(T,V)$ ke roop mein express karo aur compare karo. Step 2 — $S(T,V)$ expand karo: $dS={\left(\frac{\partial S}{\partial T}\right)}_{V}dT+{\left(\frac{\partial S}{\partial V}\right)}_{T}dV.$ Constant $P$ par $dT$ se divide karo: ${\left(\frac{\partial S}{\partial T}\right)}_P={\left(\frac{\partial S}{\partial T}\right)}_V+{\left(\frac{\partial S}{\partial V}\right)}_T{\left(\frac{\partial V}{\partial T}\right)}_P.$ Step 3 — $T$ se multiply karo taaki har $\partial S/\partial T$ ek heat capacity ban jaye: $C_P=C_V+T{\left(\frac{\partial S}{\partial V}\right)}_T{\left(\frac{\partial V}{\partial T}\right)}_P.$ Step 4 — Helmholtz Maxwell relation apply karo ${\left(\partial S/\partial V\right)}_T={\left(\partial P/\partial T\right)}_V$: $C_P-C_V=T{\left(\frac{\partial P}{\partial T}\right)}_V{\left(\frac{\partial V}{\partial T}\right)}_P.$ Step 5 — unmeasurable ${\left(\partial P/\partial T\right)}_V$ ko khatam karo cyclic (triple-product) rule se. Kisi bhi teen variables $P,V,T$ ke liye jo equation of state se jude hain, general identity hai: ${\left(\frac{\partial P}{\partial T}\right)}_V{\left(\frac{\partial T}{\partial V}\right)}_P{\left(\frac{\partial V}{\partial P}\right)}_T=-1.$ Jo piece chahiye uske liye solve karo, baaki do factors ko right mein move karke (note karo ki ${\left(\partial T/\partial V\right)}_P=1/{\left(\partial V/\partial T\right)}_P$, reciprocal): ${\left(\frac{\partial P}{\partial T}\right)}_V=\frac{-1}{{\left(\partial T/\partial V\right)}_P{\left(\partial V/\partial P\right)}_T}=-\frac{{\left(\partial V/\partial T\right)}_P}{{\left(\partial V/\partial P\right)}_T}.$ Ab do measurable coefficients insert karo, ${\left(\partial V/\partial T\right)}_P=V\alpha$ aur ${\left(\partial V/\partial P\right)}_T=-V{\kappa }_T$: ${\left(\frac{\partial P}{\partial T}\right)}_V=-\frac{V\alpha }{-V{\kappa }_T}=\frac{\alpha }{{\kappa }_T}.$ Step 6 — assemble karo (ek baar phir ${\left(\partial V/\partial T\right)}_P=V\alpha$ use karke): $\boxed{C_P-C_V=T\frac{\alpha }{{\kappa }_T}\cdot V\alpha =\frac{TV{\alpha }^2}{{\kappa }_T}}$ Step 7 — ideal-gas check: $\alpha =1/T$, ${\kappa }_T=1/P$ (from $V=RT/P$: ${\left(\partial V/\partial P\right)}_T=-RT/P^2$, toh ${\kappa }_T=-\frac{1}{V}\cdot (-RT/P^2)=RT/(P^{2}V)=1/P$). Phir $C_P-C_V=\frac{T\cdot V\cdot (1/T{)}^2}{1/P}=\frac{VP}{T}=\frac{RT}{T}=R.✓$ > Yeh Heat capacities $C_P - C_V$ ka celebrated result hai — note karo ki ise ek general substance ke liye prove karne ke liye Maxwell relation ki zaroorat thi, sirf ideal gases ke liye nahi. Kyunki $T$, $V$, ${\alpha }^2$ aur ${\kappa }_T$ sab $\ge 0$ hain, yeh yeh bhi prove karta hai ki $C_P\ge C_V$ hamesha hota hai.

Step 1 — energy equation (parent note mein prove kiya gaya): ${\left(\partial U/\partial V\right)}_T=T{\left(\partial P/\partial T\right)}_V-P$. Step 2 — $P$ ke liye solve karo: $P=\frac{RT}{V-b}-\frac{a}{V^2}$. Step 3 — constant $V$ par differentiate karo: sirf pehla term $T$ par depend karta hai: ${\left(\frac{\partial P}{\partial T}\right)}_V=\frac{R}{V-b}.$ Step 4 — assemble karo: ${\left(\frac{\partial U}{\partial V}\right)}_T=T\cdot \frac{R}{V-b}-\left(\frac{RT}{V-b}-\frac{a}{V^2}\right)=\frac{a}{V^2}.$ $\boxed{{\left(\frac{\partial U}{\partial V}\right)}_T=\frac{a}{V^2}>0}$ Interpretation: kyunki $a>0$ (attractions), internal energy badhti hai jab gas expand hoti hai — molecules ko unki mutual attraction ke against alag kheenchne ke liye energy feed karni padti hai. Notice karo ki excluded-volume constant $b$ bilkul cancel ho jaata hai: energy ki volume-dependence purely attractions se set hoti hai. Yeh ek real gas ki Joule (free) expansion mein cooling ka microscopic origin hai.

Step 1 — process isenthalpic hai (KYUN): porous plug ke through throttling enthalpy $H$ conserve karta hai, toh hum $T$ ko $P$ ke saath differentiate karte hain constant $H$ par. Step 2 — $(T,P,H)$ par cyclic rule: teen variables $T$, $P$, $H$ linked hain (koi bhi do state fix karte hain), toh Problem 3.1 mein use kiya gaya wahi general cyclic identity unpar apply hoti hai: ${\left(\frac{\partial T}{\partial P}\right)}_H{\left(\frac{\partial P}{\partial H}\right)}_T{\left(\frac{\partial H}{\partial T}\right)}_P=-1.$ Jo piece chahiye uske liye solve karo (aur ${\left(\partial P/\partial H\right)}_T=1/{\left(\partial H/\partial P\right)}_T$ use karke): ${\left(\frac{\partial T}{\partial P}\right)}_H=-\frac{(\partial H/\partial P{)}_T}{(\partial H/\partial T{)}_P}.$ Denominator hai ${\left(\partial H/\partial T\right)}_P=C_P$ by definition. Step 3 — ${\left(\partial H/\partial P\right)}_T$ evaluate karo $dH=TdS+VdP$ se. Constant $T$ par $dP$ se divide karo: ${\left(\frac{\partial H}{\partial P}\right)}_T=T{\left(\frac{\partial S}{\partial P}\right)}_T+V.$ Step 4 — entropy term ko Gibbs Maxwell relation se khatam karo ${\left(\partial S/\partial P\right)}_T=-{\left(\partial V/\partial T\right)}_P=-V\alpha$: ${\left(\frac{\partial H}{\partial P}\right)}_T=-TV\alpha +V=V(1-T\alpha ).$ Step 5 — assemble karo: $\boxed{{\mu }_{JT}=\frac{V}{C_P}\left(T\alpha -1\right)}$ Step 6 — ideal gas: $\alpha =1/T$, toh $T\alpha -1=0$ aur ${\mu }_{JT}=0$. Ideal gas throttling par cool nahi hoti — ${\left(\partial U/\partial V\right)}_T=0$ aur ${\left(\partial H/\partial P\right)}_T=0$ ke consistent. Real gases tab cool hoti hain jab $T\alpha >1$ (inversion temperature se neeche), yahi tarika hai jisse liquefiers kaam karte hain.

Step 1 — $C_V$ ko entropy ke through likho: $C_V=T{\left(\partial S/\partial T\right)}_V$. Constant $T$ par $V$ ke saath differentiate karo: ${\left(\frac{\partial C_V}{\partial V}\right)}_T=T\frac{\partial }{\partial V}{\left(\frac{\partial S}{\partial T}\right)}_V.$ Step 2 — differentiation ka order swap karo (KYUN — Schwarz): kyunki $S$ ek state function hai, mixed partials commute karte hain: $\frac{\partial }{\partial V}{\left(\frac{\partial S}{\partial T}\right)}_V=\frac{\partial }{\partial T}{\left(\frac{\partial S}{\partial V}\right)}_T.$ Step 3 — Helmholtz Maxwell relation apply karo ${\left(\partial S/\partial V\right)}_T={\left(\partial P/\partial T\right)}_V$: $\frac{\partial }{\partial T}{\left(\frac{\partial P}{\partial T}\right)}_V={\left(\frac{{\partial }^{2}P}{\partial T^2}\right)}_V.$ Step 4 — assemble karo: $\boxed{{\left(\frac{\partial C_V}{\partial V}\right)}_T=T{\left(\frac{{\partial }^{2}P}{\partial T^2}\right)}_V}$ Step 5 — ideal gas: $P=RT/V$ fixed $V$ par $T$ mein linear hai, toh ${\left({\partial }^{2}P/\partial T^2\right)}_V=0$. Isliye ${\left(\partial C_V/\partial V\right)}_T=0$: ideal gas ke liye $C_V$ volume par depend nahi kar sakta — phir se derived, assumed nahi.

Step 1 — Maxwell relation lao (KYUN): $\alpha =\frac{1}{V}{\left(\partial V/\partial T\right)}_P$, aur Gibbs relation ise entropy derivative mein convert karta hai: ${\left(\frac{\partial V}{\partial T}\right)}_P=-{\left(\frac{\partial S}{\partial P}\right)}_T.$ Toh $\alpha =-\frac{1}{V}{\left(\frac{\partial S}{\partial P}\right)}_T$. Step 2 — third law invoke karo (Nernst): jab $T\to 0$, har system ki entropy ek constant $S_0$ ke paas jaati hai jo $P$ se independent hoti hai (aur $V$ se bhi). Isliye ${\lim }_{T\to 0}{\left(\frac{\partial S}{\partial P}\right)}_T=0.$ Step 3 — conclude karo: $\boxed{\underset{T\to 0}{\lim }\alpha =-\frac{1}{V}\cdot 0=0}$ Matlab: yeh ek universal prediction hai — koi bhi material absolute zero ke approach karte waqt (pehle order mein $T$ mein) expand ya contract nahi karta. Yeh solids, liquid helium, sab ke liye confirm ho chuka hai. Ek pure thermodynamic theorem jo ek Maxwell relation aur ek law se nikal aaya.

Step 1 — "adiabatic reversible" ka matlab (KYUN): koi heat flow nahi aur reversible $\Rightarrow$ constant entropy, $S=\text{const}$. Hum compression ($dP>0$) ke under ${\left(\partial T/\partial P\right)}_S$ ka sign jaanna chahte hain. Step 2 — $(T,P,S)$ par cyclic rule: ${\left(\frac{\partial T}{\partial P}\right)}_S=-\frac{(\partial S/\partial P{)}_T}{(\partial S/\partial T{)}_P}.$ Step 3 — pieces identify karo. Denominator: ${\left(\partial S/\partial T\right)}_P=C_P/T>0$ (heat capacity aur temperature positive hain). Numerator: Gibbs Maxwell relation use karo ${\left(\frac{\partial S}{\partial P}\right)}_T=-{\left(\frac{\partial V}{\partial T}\right)}_P=-V\alpha .$ Step 4 — assemble karo: ${\left(\frac{\partial T}{\partial P}\right)}_S=-\frac{-V\alpha }{C_P/T}=\frac{TV\alpha }{C_P}.$ Step 5 — sign padho. $T$, $V$, $C_P$ sab positive hain, toh ${\left(\partial T/\partial P\right)}_S$ ka sign $\alpha$ ka sign hai.

• Normal material ($\alpha >0$): compression ise warm karta hai ($dT>0$) — familiar bicycle-pump heating.
• Negative-expansion material ($\alpha <0$): compression ise cool karta hai ($dT<0$). $\boxed{\text{For }\alpha <0:{\left(\frac{\partial T}{\partial P}\right)}_S<0\Rightarrow \text{adiabatic compression lowers }T.}$ Matlab: water ka near-freezing counter-intuitive expansion behaviour uski adiabatic heating ka sign flip kar deta hai — ek striking, testable consequence jo ek Maxwell relation aur cyclic rule se nikali.

L1: recognise karo ki kaun sa potential kaun sa derivative own karta hai (differentiated variable aur subscript dono match karo). L2: apply karo ek relation ko, calculation ko measurable equation-of-state side par push karke. L3: combine karo relations ko cyclic rule ke saath taaki $C_P-C_V$ aur $(\partial U/\partial V{)}_T$ mile. L4: synthesise karo bilkul naye results — Joule–Thomson, $\partial C_V/\partial V$ — Maxwell + Schwarz ko saath use karke. L5: master karo universal theorems ($\alpha \to 0$ at $T=0$; negative-expansion materials ke liye sign-flip) jahan thermodynamics akele real experiments forecast karta hai. Throughout: derive karo, signs kabhi memorize mat karo.