Question bank — Adiabatic relations — PV^γ = const, TV^(γ−1) = const (derivation)

False — adiabatic ka sirf matlab hai no heat transfer ($\delta Q=0$); vacuum mein fast expansion bhi adiabatic hoti hai. Lekin relations $P{V}^{\gamma }=$const ke liye zaroori hai ki woh reversible/quasi-static ho taaki $P,V,T$ puri process mein defined rahein.

False — yeh Isothermal Process describe karta hai. Adiabatic mein, kyunki koi heat internal energy ko refill nahi karti, work karna $T$ ko badalne par majboor karta hai: expansion se cooling, compression se heating.

False — $\gamma >1$ hai, isliye adiabat ki slope $-\gamma P/V$ har point par isotherm ki $-P/V$ se steeper hai; woh sirf touch karte hain, kabhi coincide nahi karte (figure s02 dekho).

False — ideal gas ke liye $U=U(T)$ akela hai, isliye yeh kisi bhi process ke liye hold karta hai. "$V$" subscript record karta hai ki $C_V$ kaise measure kiya gaya, yeh nahi ki formula kab apply hota hai.

False — first law hai $0=dU+\delta W$, isliye $dU=-\delta W$. Jab gas work karti hai, uski internal energy girti hai chahe $Q=0$ ho.

False — derivation ne ideal gas law ($PV=nRT$) aur $dU=n{C}_{V}dT$ use kiya tha. Real gases ya liquids ke liye exponent aur form bhi badal jaata hai.

False — $\gamma$ active degrees of freedom par depend karta hai, jo $T$ ke saath change hote hain. Thanda diatomic gas sirf translation+rotation use karta hai ($\gamma =7/5$); itna garam karo ki vibration activate ho aur $C_V$ badhta hai, isliye $\gamma$ $9/7$ ki taraf girta hai (figure s03 dekho).

True — ek reversible adiabatic process mein $dS=\delta Q_{\text{rev}}/T=0$ hota hai (koi reversible heat enter nahi karti), isliye $S$ unchanged rehti hai, yaani yeh isentropic hai; yahi deeper reason hai ki curve ek single well-defined line hai.

False — yeh internal energy surroundings par kiye gaye work mein kho deti hai, heat mein nahi. $Q=0$ puri process mein; energy mechanical work ke roop mein jaati hai, heat ke roop mein nahi.

Wrong exponent — Step 5 $dT/T+(\gamma -1)dV/V=0$ ko integrate karta hai jisse $\ln T+(\gamma -1)\ln V=$const milta hai, giving $T{V}^{\gamma -1}=$const. Sirf $P$ pura $\gamma$ carry karta hai; $T$ ko $\gamma$ minus one milta hai kyunki $PV=nRT$ ke zariye $P\leftrightarrow T$ swap karne par $V$ ki ek power shift ho jaati hai.

Wrong substitution — Step 2 ideal gas law $PV=nRT\Rightarrow P=nRT/V$ use karta hai, $n{C}_{V}T/V$ nahi. Yahan $C_V$ use karne se dimensions break ho jaate aur kabhi $\gamma -1$ produce nahi hota.

Step 4 par double error: $R/C_V=(C_P-C_V)/C_V=\gamma -1$ Mayer's Relation se, $\gamma$ nahi. Yahi wajah hai ki $TV$ exponent $\gamma -1$ hai.

Ulta — har ek differentiate karne par (parent's forecast section) adiabat slope $-\gamma P/V$ aur isotherm slope $-P/V$ milti hai; kyunki $\gamma >1$ hai adiabat steeper hai. Adiabatic expansion mein $P$ zyada tezi se girta hai kyunki $T$ bhi girta hai (figure s02).

Defining condition $\delta Q=0$ se contradiction — yeh koi heat absorb nahi karti. Work puri tarah internal energy se pay hota hai: $W=-\Delta U$.

Sign flipped — $W=-\Delta U=-n{C}_V(T_2-T_1)=n{C}_V(T_1-T_2)$ se. Expansion ke liye $T_2<T_1$ giving $W>0$, jaisa hona chahiye ek gas ke liye jo bahar push karti hai.

Nahi — monatomic ($5/3$) adiabats diatomic ($7/5$) se steeper hote hain kyunki larger $\gamma$ ka matlab larger slope factor hai. Degrees of Freedom and $\gamma$ dekho.

Kyunki $\gamma =C_P/C_V$ hai, aur $C_V$ degrees of freedom count karta hai jisme har molecule energy store kar sakta hai — isliye curve ki steepness literally molecule ki internal structure read out karti hai.

Kyunki equation inexact-looking rates of change ko link karta hai, values ko nahi; yeh ek differential relation hai jo har instant valid hai. Har exact differential $dx/x$ ko integrate karne par $\ln x$ milta hai, aur do logs ka sum ek constant of integration ke equal hone ka matlab hai ki unke arguments ek fixed number se multiply hote hain — aise hi $T{V}^{\gamma -1}=$const born hota hai. Constant khud gas ki starting state $(T_0,V_0)$ se fix hota hai, isliye alag initial conditions alag adiabats deti hain, ek per constant.

Fast compression almost adiabatic hoti hai (heat leak hone ka time nahi, isliye $\delta Q\approx 0$), to jo work tum karte ho woh puri tarah internal energy mein jaati hai $dU=-\delta W$ ke zariye, $T$ raise karta hai — exactly $P{V}^{\gamma }=$const kaam kar raha hai.

Compressions aur rarefactions itni tezi se oscillate karti hain ki unke beech heat diffuse hone ka time nahi hota, isliye har parcel $\delta Q=0$ obey karta hai; yahi wajah hai ki Speed of Sound in Gases adiabatic bulk modulus $\gamma P$ use karta hai, $P$ nahi.

Adiabatic expansion mein gas bhi cool hoti hai, isliye uska pressure do reasons se girta hai (zyada volume aur kam temperature), jabki isotherm $T$ ko incoming heat se prop up karke rakhta hai (figure s02).

Kyunki $U$ ideal gas ke liye sirf $T$ ka ek state function hai, isliye iska exact differential $dU$, $dT$ se fixed hota hai chahe $V,P$ simultaneously kuch bhi karein — path $U$ ke liye matter nahi karta.

Heat remove karna $T$ ko fixed nahi rakhta — heat inflow ke bina temperature expansion ke dauraan zaroor girni chahiye, aur wahi extra temperature change hai jo factor $\gamma$ encode karta hai.

Ek rising air parcel expand hota hai (upar low pressure) apne surroundings par work karta hai almost bina heat exchange ke, isliye woh adiabatically cool hota hai — yeh dry adiabatic lapse rate hai, aur yahi wajah hai ki mountaintops aur high clouds thande hote hain (figure s04).

Slope $-\gamma P/V$ isothermal slope $-P/V$ ki taraf approach karta hai; ek gas jo internally bahut zyada energy store karti hai expansion par barely cool hoti hai, isliye uska adiabat almost isotherm se overlap karta hai.

Nahi — yeh adiabatic hai ($Q=0$) lekin irreversible hai aur koi work nahi karta, isliye $\Delta U=0$ aur $T$ ideal gas ke liye constant rehta hai. Relation sirf reversible case ke liye hold karta hai.

Haan, woh ek point share karte hain ($P,V,T$ sab wahan equal hain), lekin unki slopes $\gamma$ ke factor se differ karti hain, isliye woh dono taraf immediately diverge ho jaate hain (figure s02).

Yeh predict karta hai $P\propto V^{-\gamma }\to \infty$ aur $T\propto V^{-(\gamma -1)}\to \infty$. Lekin ideal-gas assumption bahut pehle fail ho jaati hai: molecules finite size ke hote hain aur ek dusre ko attract karte hain, isliye van der Waals picture take over leti hai — gas liquefy ho jaati hai aur $C_V$ khud change ho jaata hai, isliye koi single $\gamma$ survive nahi karta (figure s03 hint deta hai ki $C_V$ fixed kyun nahi hota).

Nahi — tab $P{V}^{\gamma }=PV=$const aur $T{V}^{\gamma -1}=T=$const, dono isothermal case mein collapse ho jaate. Lekin $\gamma >1$ hamesha real gas ke liye hota hai, isliye yeh sirf ek limiting sanity check hai.

$W=0$ — lekin $Q=0$ aur $\Delta T=0$ ke saath kuch hua hi nahi; work wala genuine adiabatic process temperature zaroor change karta hai.