Adiabatic relations — PV^γ = const, TV^(γ−1) = const (derivation): Worked examples

1.7.16 · D3 · Physics › Thermodynamics › Adiabatic relations — PV^γ = const, TV^(γ−1) = const (deriva

Kuch bhi karne se pehle: do housekeeping notes, phir toolbox.

Definition Gas law aur reversibility — is poore page ke ground rules

Yahan har example assume karta hai ek ideal gas, jiske teen state variables hamesha ideal gas law obey karte hain $P V = n R T,$ jahan $P$ = pressure, $V$ = volume, $T$ = absolute temperature (kelvin mein), $n$ = moles ki sankhya, $R = 8.314 J mol^{- 1} K^{- 1}$ = universal gas constant. Hum is law par kaafi depend karte hain (jaise $P \leftrightarrow T$ swap karne ke liye).

Har example ye bhi assume karta hai ki process reversible (quasi-static) hai: itni dheere hoti hai ki $P, V, T$ ka har instant par ek single well-defined value hota hai, matlab gas hamesha "curve par" hai. Yahi woh condition hai jiske under parent note ke relations $P V^{γ} = const$ , $T V^{γ - 1} = const$ , $T^{γ} P^{1 - γ} = const$ derive kiye gaye the — ek fast, turbulent (irreversible) $Q = 0$ process mein bhi $Δ U = - W_{by}$ hota hai, lekin woh teen curve-equations usse describe nahi karti.

Definition Work ke liye sign convention, aur

$δ$ vs $d$ notation Hum physicist ka "work done BY the gas" convention use karte hain, jise $W_{by}$ likhte hain. First law padhta hai $δ Q = d U + δ W_{by}, δ W_{by} = P d V .$ $δ Q, δ W$ kyun lekin $d U$ ? Internal energy $U$ ek state function hai — gas ke paas $(P, V, T)$ ke har point par iska ek definite amount hota hai, isliye $d U$ ka change exact hai: yeh sirf start aur end states par depend karta hai, aur $\int d U = U_{2} - U_{1}$ chahe path koi bhi ho. Heat $Q$ aur work $W$ stored quantities nahi hain — ye amounts hain jo ek path ke dauran flow karte hain. Do alag paths jinke start aur end states same hain, alag heat aur work exchange kar sakte hain. Ye flag karne ke liye ki "yeh ek path-dependent sliver hai, na ki kisi stored quantity ka change," hum $δ$ (ek inexact differential) likhte hain $d$ ki jagah. Symbol ka yahi poora matlab hai: $d$ = "gas ke paas jo kuch hai uska change," $δ$ = "raaste mein boundary cross kiya hua thoda sa."

Is hisaab se: jab gas expand karta hai ( $d V > 0$ ) to wo positive work karta hai. Ideal gas ke liye internal energy sirf temperature par depend karti hai, jo deta hai $Δ U = n C_{V} (T_{2} - T_{1})$ kisi bhi process ke liye (parent note se " $U = U (T)$ only" fact — $C_{V}$ par $V$ -subscript sirf batata hai ki ise kaise measure kiya gaya, kab apply hota hai nahi). Ek adiabatic process mein $Q = 0$ , to $0 = Δ U + W_{by}$ , matlab $W_{by} = - Δ U = n C_{V} (T_{1} - T_{2})$ .

Scenario matrix

Har adiabatic problem inhi cells mein se kisi ek mein aata hai. Neeche har worked example ko us cell ke saath tag kiya gaya hai jo wo cover karta hai.

#	Cell class	Kya cheez isse alag banati hai	Example
A	Compression, $T$ nikalo	$V$ shrinks → $T$ badhni chahiye	Ex 1
B	Expansion, $P$ nikalo	$V$ grows → $P$ girni chahiye	Ex 2
C	Given $T$ & $P$ , dusra $T$ / $P$ nikalo	$T P$ relation use karta hai	Ex 3
D	Work / energy	koi $V$ -ratio nahi diya, $Δ U$ use karo	Ex 4
E	Degenerate $γ \to 1$	adiabat isotherm par collapse ho jaata hai	Ex 5
F	Limiting behaviour (huge compression $V \to 0$ )	$T$ ya $P$ kitni tezi se blow up karta hai?	Ex 6
G	Real-world word problem	rising air parcel cooling (clouds)	Ex 7
H	Exam twist — mixed process	adiabat phir isotherm, ya slope comparison	Ex 8
I	Inverse: ek $V$ -ratio nikalo	given $T$ (ya $P$ ) ratio, $V_{2} / V_{1}$ solve karo	Ex 9

Worked examples

Worked example Ex 1 — Cell A: compression, temperature nikalo

Air ( $γ = 1.4$ ) $T_{1} = 300$ K par start karta hai aur adiabatically compress hoke ek-tehaai volume tak aata hai, $V_{2} = V_{1} /3$ . $T_{2}$ nikalo.

Forecast: volume ghata → gas garam hona chahiye, to expect karo $T_{2} > 300$ K. Aage padhne se pehle ek number guess karo.

Tool pick karo. Hame ek volume ratio pata hai aur temperature chahiye. Yahi kaam hai $T V^{γ - 1} = const$ ka. Yeh step kyun? Koi pressure mention nahi hai, to koi bhi relation jo $P$ contain kare avoid karo — wo ek unknown leke aayega.
Dono states ko equal likho. $T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1} \Rightarrow T_{2} = T_{1} (\frac{V _{1}}{V _{2}})^{γ - 1} .$ Yeh step kyun? "const" matlab value state 1 aur state 2 par same hai — to dono products ko equal karo aur solve karo.
Plug in karo. $\frac{V _{1}}{V _{2}} = 3$ , $γ - 1 = 0.4$ : $T_{2} = 300 \times 3^{0.4} = 300 \times e^{0.4 l n 3} = 300 \times e^{0.4394} = 300 \times 1.5518 \approx 466 K .$ Yeh step kyun? $3^{0.4}$ ko $e^{0.4 l n 3}$ likhte hain kyunki fractional power sirf exponential/log ke through computable hai — yeh symbol ke peeche ka concrete arithmetic hai.

Verify: $466 > 300$ ✔ — compression ne ise garam kiya, forecast se match karta hai. Units: $K \times (dimensionless ratio) = K$ ✔.

Worked example Ex 2 — Cell B: expansion, pressure nikalo

Ek monatomic gas ( $γ = 5/3$ ) $P_{1} = 8$ atm par chaar guna volume tak adiabatically expand karta hai, $V_{2} = 4 V_{1}$ . $P_{2}$ nikalo.

Forecast: volume chaar guna bada → pressure bahut zyada gir jaana chahiye. Expect karo $P_{2} ≪ 8$ atm.

Tool pick karo. $P$ aur ek volume ratio given hai → $P V^{γ} = const$ use karo. Yeh step kyun? Hame pressure chahiye aur sirf volumes hain; $P V$ form mein exactly wahi dono hain.
States equate karo. $P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ} \Rightarrow P_{2} = P_{1} (\frac{V _{1}}{V _{2}})^{γ} .$ Yeh step kyun? Quantity $P V^{γ}$ states 1 aur 2 par same number hai, to dono equate karo aur $P_{2}$ isolate karne ke liye $V_{2}^{γ}$ se divide karo.
Plug in karo. $\frac{V _{1}}{V _{2}} = \frac{1}{4}$ , $γ = \frac{5}{3} = 1.6667$ : $P_{2} = 8 \times (0.25)^{1.6667} = 8 \times e^{1.6667 l n 0.25} = 8 \times e^{- 2.3105} = 8 \times 0.09921 \approx 0.794 atm .$ Yeh step kyun? Ratio $V_{1} / V_{2} = 1/4 < 1$ ek positive power se raise hone par bhi 1 se neeche rehta hai, jo guarantee karta hai $P_{2} < P_{1}$ — direction ki ek built-in check.

Verify: $0.79 < 8$ ✔ — expansion ne pressure ko sharply gira diya. Note karo ki yeh isotherm ke dene wale $1/4$ se zyada tezi se gira ( $8/4 = 2$ atm), kyunki extra $γ$ -power drop ko steep banata hai — parent note mein Isothermal Process contrast se consistent.

Worked example Ex 3 — Cell C: pressure ratio given, temperature nikalo

Ek diatomic gas ( $γ = 1.4$ ) $T_{1} = 350$ K par adiabatically compress hota hai jab tak pressure double na ho jaye, $P_{2} = 2 P_{1}$ . $T_{2}$ nikalo.

Forecast: compressed (pressure up) → temperature up.

Tool pick karo. Hamare paas ek pressure ratio hai aur temperature chahiye — volume ka koi naam nahi. $T^{γ} P^{1 - γ} = const$ use karo. Yeh step kyun? $T V^{γ - 1}$ choose karne se pehle volume ratio nikalna padta; $T P$ relation woh step skip kar deta hai.
$T$ ratio ke liye rearrange karo. Dono states par relation likhte hain: $T_{1}^{γ} P_{1}^{1 - γ} = T_{2}^{γ} P_{2}^{1 - γ} \Rightarrow (\frac{T _{2}}{T _{1}})^{γ} = (\frac{P _{2}}{P _{1}})^{γ - 1} \Rightarrow \frac{T _{2}}{T _{1}} = (\frac{P _{2}}{P _{1}})^{\frac{γ - 1}{γ}} .$ Yeh step kyun? Jo ratio chahiye use isolate karo $P$ ko dusri side le jaake; exponent $(γ - 1) / γ$ ban jaata hai.
Plug in karo. $\frac{γ - 1}{γ} = \frac{0.4}{1.4} = 0.2857$ : $T_{2} = 350 \times 2^{0.2857} = 350 \times e^{0.2857 l n 2} = 350 \times e^{0.1980} = 350 \times 1.2189 \approx 427 K .$ Yeh step kyun? Exponent $(γ - 1) / γ = 0.2857$ positive aur 1 se kam hai, to pressure double karne se $T$ sirf thoda badhta hai ( $\approx 22%$ ) — ek gentle compression ki physical expectation.

Verify: $427 > 350$ ✔ — pressure badha, to temperature bhi badha. Ideal-gas law se cross-check ke liye volumes chahiye honge, lekin sign already physics confirm karta hai.

Worked example Ex 4 — Cell D: expansion mein work done

Do moles ( $n = 2$ ) ek diatomic gas ( $C_{V} = \frac{5}{2} R$ ) ke adiabatically expand karte hain, $T_{1} = 600$ K se $T_{2} = 400$ K tak cool hote hain. Gas dwara kiya gaya work $W_{by}$ nikalo.

Forecast: gas expand hua (work kiya) → $W_{by} > 0$ . Woh work khoi hui internal energy ke barabar honi chahiye.

$P d V$ nahi, energy conservation use karo. Ideal gas ke liye $Δ U = n C_{V} (T_{2} - T_{1})$ (internal energy sirf $T$ par depend karta hai). Kyunki $Q = 0$ hai, first law $0 = Δ U + W_{by}$ deta hai $W_{by} = - Δ U = - n C_{V} (T_{2} - T_{1}) = n C_{V} (T_{1} - T_{2}) .$ Yeh step kyun? Hamare paas $P$ ya $V$ nahi diye, sirf temperatures hain — $\int P d V$ integrate karna mushkil hota; energy bookkeeping direct hai. Minus sign $(T_{2} - T_{1})$ ko $(T_{1} - T_{2})$ mein flip karta hai, jo yahan positive hai kyunki $T_{1} > T_{2}$ .
Plug in karo. $C_{V} = \frac{5}{2} R = \frac{5}{2} (8.314) = 20.785 J mol^{- 1} K^{- 1}$ : $W_{by} = 2 \times 20.785 \times (600 - 400) = 2 \times 20.785 \times 200 = 8314 J \approx 8.31 kJ .$ Yeh step kyun? $C_{V} = \frac{5}{2} R$ (diatomic value Degrees of Freedom and $\gamma$ se) substitute karte hain to poora answer known SI constants mein express hota hai, joules milte hain.

Verify: positive ✔ — expansion positive work karta hai. Units: $mol \cdot \frac{J}{mol K} \cdot K = J$ ✔. Usi number ke liye $\int P d V$ route ke liye Work Done in Thermodynamic Processes dekho.

Worked example Ex 5 — Cell E: degenerate case

$γ \to 1$ Agar $γ = 1$ ho to adiabatic relation kya ban jaata hai? Physically, yeh boundary case kyun hai?

Forecast: $γ = 1$ matlab $C_{P} = C_{V}$ hoga, yani Mayer's relation $C_{P} - C_{V} = R$ demand karti hai $R = 0$ — real gas ke liye impossible, lekin limit ke taur par instructive.

$γ = 1$ ko $P V^{γ}$ mein dalo. $P V^{1} = const \Rightarrow P V = const .$ Yeh step kyun? Bas value substitute karo; exponent 1 tak collapse ho jaata hai.
Pehchano $P V = const$ kya hai. Ideal gas law se $P V = n R T$ , to $P V = const$ matlab $T = const$ — ek isothermal process (Isothermal Process dekho). Yeh step kyun? Algebra ko physically interpret karte hain ye dekhne ke liye ki degenerate case kaunsi real process mimic karta hai.
$T V$ form bhi check karo. $T V^{γ - 1} = T V^{0} = T = const$ ✔ — consistent, $T$ fixed hai. Yeh step kyun? Ek achha limit teeno tools mein self-consistent hona chahiye; agar $T V$ form $P V$ form se disagree karta to hum galti suspect karte. Yahan dono kehte hain " $T$ constant," isotherm par collapse confirm karta hai.

Verify: Jab $γ \to 1^{+}$ to adiabat ka slope $- γ P / V \to - P / V$ , jo exactly isotherm ka slope hai. To adiabat continuously merge ho jaata hai isotherm mein jab $γ \to 1$ — boundary real hai, sirf koi gas wahan actually nahi baithta.

Worked example Ex 6 — Cell F: limiting behaviour, extreme compression

Ek monatomic gas ( $γ = 5/3$ ) adiabatically compress hota hai to uska volume $1000$ factor se shrink ho jaata hai ( $V_{2} = V_{1} /1000$ ). $T_{1} = 300$ K se start karke, $T_{2}$ nikalo aur comment karo ki $V \to 0$ hone par $T$ kitni tezi se diverge karta hai.

Forecast: volume ek hazaar guna down → temperature soars. Kyunki exponent $γ - 1 = \frac{2}{3} < 1$ hai, yeh soars karta hai lekin volume factor se dheere.

$T V$ relation use karo. $T_{2} = T_{1} (\frac{V _{1}}{V _{2}})^{γ - 1} = 300 \times 100 0^{2/3}$ . Yeh step kyun? Volume ratio → temperature, to $T V^{γ - 1}$ .
Calculate karo. $100 0^{2/3} = (1 0^{3})^{2/3} = 1 0^{2} = 100$ : $T_{2} = 300 \times 100 = 30000 K .$ Yeh step kyun? $1000$ ko $1 0^{3}$ likhte hain to fractional power $()^{2/3}$ exponents ko cleanly multiply karta hai ( $3 \times \frac{2}{3} = 2$ ), ek awkward root ko exact integer power mein badal deta hai — calculator ki zaroorat nahi.
Divergence law padhte hain. Generally $T \propto V^{- (γ - 1)}$ , to $V \to 0$ hone par $T \to \infty$ power law ki tarah $γ - 1$ exponent ke saath. Pressure aur bhi tezi se diverge karta hai: $P \propto V^{- γ}$ . Yeh step kyun? Functional form padhna (sirf number nahi) hume poore compressions family ka qualitative behaviour batata hai: $- (γ - 1)$ exponent hi wajah hai ki $T$ $P$ se dheere blow up karta hai, jiska exponent full $- γ$ hai.

Verify: $30000$ K bahut bada hai (Sun ki surface se bhi garam) — yahi reason hai ki diesel engines sirf adiabatic compression se fuel ignite karte hain. Sign ✔ (heating), magnitude $1000 \times$ squeeze ke liye physically sensible.

Neeche figure is heating law ko plot karta hai aur contrast karta hai us (galat) intuition se ki $T$ compression ratio ke saath linearly grow karna chahiye — amber dot is exact example ko mark karta hai.

Figure s01 — Adiabatic heating ek power law hai. Temperature $T$ versus compression ratio $V_{1} / V_{2}$ ek monatomic gas ke liye ( $γ = 5/3$ ). Cyan curve sahi law hai $T = T_{1} (V_{1} / V_{2})^{γ - 1}$ ; dashed white line naive "linear" guess hai $T \propto V_{1} / V_{2}$ ; amber dot Ex 6 ko mark karta hai $(1000, 30000 K)$ par.

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

Cyan curve dekho: yeh steeply utha hua hai lekin dashed white "linear guess" se bahut neeche rehta hai, kyunki exponent $γ - 1 = \frac{2}{3}$ 1 se kam hai. Amber dot $(1000, 30000)$ par exactly Ex 6 hai.

Worked example Ex 7 — Cell G: real-world word problem (cloud formation)

Dry air ( $γ = 1.4$ ) ka ek parcel ground par $T_{1} = 300$ K aur pressure $P_{1} = 1.00$ atm par hai. Woh tab tak utha rehta hai jab tak surrounding pressure $P_{2} = 0.70$ atm tak na gir jaye. Mante hue ki yeh itni tezi se uthta hai ki adiabatic hai, uska temperature $T_{2}$ nikalo.

Forecast: uthta hua air expand karta hai (lower outside pressure) → yeh thanda hoga. Yahi cooling hai jo water vapour ko clouds mein condense karti hai.

Tool pick karo. Ek pressure ratio given hai, temperature chahiye → $T^{γ} P^{1 - γ} = const$ , Ex 3 ki tarah rearrange karke: $\frac{T _{2}}{T _{1}} = (\frac{P _{2}}{P _{1}})^{\frac{γ - 1}{γ}} .$
Plug in karo. $\frac{P _{2}}{P _{1}} = 0.70$ , exponent $= \frac{0.4}{1.4} = 0.2857$ : $T_{2} = 300 \times 0.7 0^{0.2857} = 300 \times e^{0.2857 l n 0.70} = 300 \times e^{- 0.1019} = 300 \times 0.9031 \approx 271 K .$ Yeh step kyun? Kyunki $P_{2} / P_{1} = 0.70 < 1$ aur exponent positive hai, factor $0.7 0^{0.2857} < 1$ , guarantee karta hai $T_{2} < T_{1}$ — algebra automatically encode karta hai "expansion cools," to sign ek built-in physics check hai.

Verify: $271 < 300$ ✔ — uthta parcel lagbhag $29$ K thanda hua, typical dew point se neeche gira, to vapour condense hota hai → cloud banta hai. Yeh $Q = 0$ ka rozmarra ka chehra hai.

Worked example Ex 8 — Cell H: exam twist, adiabat

phir isotherm Ek gas ( $γ = 1.4$ ) state 1 se start karta hai: $P_{1} = 6$ atm, $V_{1} = 1$ L. Step I: adiabatic expansion to $V_{2} = 2$ L. Step II: isothermal compression wapis $V_{3} = 1$ L tak. Final pressure $P_{3}$ nikalo.

Forecast: Step I pressure ko naive $6/2 = 3$ se neeche gira deta hai (adiabat steeper hai). Step II fixed $T$ par compress karta hai, ise wapas utha ta hai — lekin kya yeh 6 atm wapas aata hai? Guess karo.

Step I — adiabatic, $P V^{γ}$ use karo. $P_{2} = P_{1} (\frac{V _{1}}{V _{2}})^{γ} = 6 \times (0.5)^{1.4} = 6 \times e^{1.4 l n 0.5} = 6 \times e^{- 0.9704} = 6 \times 0.3789 \approx 2.273 atm .$ Yeh step kyun? Step I mein koi heat nahi → $γ$ -exponent apply hota hai (Step I reversible adiabatic hai, to $P V^{γ} = const$ hold karta hai).
Step II — isothermal, $P V = const$ use karo (yani exponent 1). $P_{3} = P_{2} (\frac{V _{2}}{V _{3}})^{1} = 2.273 \times \frac{2}{1} = 4.546 atm .$ Yeh step kyun? Step II $T$ fixed rakhta hai, to isotherm law use karo (Isothermal Process), $γ$ nahi.
Start se compare karo. $P_{3} \approx 4.55 atm < P_{1} = 6 atm$ , chahe hum same volume $V_{3} = V_{1}$ par wapas aaye. Yeh step kyun? Dono paths ne alag exponents use kiye ( $γ$ vs $1$ ), to gas start se thanda khatam hota hai — cycle temperature mein closed nahi hai.

Verify: $P_{3} < P_{1}$ ✔ sense karta hai: Step I ne gas ko thanda kiya, aur isothermal return us cooling ko undo nahi kar sakta, to equal volume par pressure kam hai. Trap hai Step II par $γ$ apply karna — wo galat tarike se $6$ atm wapas de deta.

Worked example Ex 9 — Cell I: inverse problem, volume ratio nikalo

Ek diatomic gas ( $γ = 1.4$ ) adiabatically compress hota hai aur uska temperature $T_{1} = 300$ K se $T_{2} = 450$ K tak badhta hai. Uska volume kitne factor se reduce hua — yani $V_{1} / V_{2}$ nikalo?

Forecast: gas garam hua → ise squeeze kiya gaya, to $V_{1} / V_{2} > 1$ (final volume chhota). Yeh Ex 1 ka ulta hai: temperatures pata hain aur volume ratio hunt kar rahe hain.

Tool pick karo. Hamare paas ek temperature ratio hai aur ek volume ratio chahiye → abhi bhi $T V^{γ - 1} = const$ , bas dusri taraf se solve karo. Yeh step kyun? Wahi relation $T$ aur $V$ ko link karta hai; kaunsi quantity "unknown" hai yeh sirf kya isolate karna hai badalta hai, tool nahi.
Volume ratio ke liye solve karo. $T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1} \Rightarrow (\frac{V _{1}}{V _{2}})^{γ - 1} = \frac{T _{2}}{T _{1}} \Rightarrow \frac{V _{1}}{V _{2}} = (\frac{T _{2}}{T _{1}})^{\frac{1}{γ - 1}} .$ Yeh step kyun? $V_{1} / V_{2}$ ko free karne ke liye dono sides ko $1/ (γ - 1)$ power se raise karo — inverse exponent, jo reverse problems ka poora trick hai.
Plug in karo. $\frac{T _{2}}{T _{1}} = \frac{450}{300} = 1.5$ , $\frac{1}{γ - 1} = \frac{1}{0.4} = 2.5$ : $\frac{V _{1}}{V _{2}} = 1. 5^{2.5} = e^{2.5 l n 1.5} = e^{1.0136} = 2.755.$ Yeh step kyun? Exponent $1/ (γ - 1) = 2.5 > 1$ modest temperature rise ko ek larger volume ratio mein amplify karta hai — Ex 1 ka mirror image, jahan chhota exponent $γ - 1$ ek bade volume ratio ko ek modest temperature rise mein shrink kar deta tha.

Verify: $V_{1} / V_{2} \approx 2.76 > 1$ ✔ — gas apne volume ke lagbhag $1/2.76$ tak compress hua, heating se consistent. Sanity cross-check: $V_{1} / V_{2} = 2.755$ ko Ex 1 ke formula mein wapas daalne par $T_{2} = 300 \times 2.75 5^{0.4} = 450$ K milta hai ✔.

Yeh kaun sa cell hai? (self-test)

Recall Har question opener ko uske cell aur tool se match karo

"Half tak compress hua, $T$ nikalo" — kaun sa relation? ::: Cell A → $T V^{γ - 1} = const$ . "Expand karta hai, pressure double, $T$ nikalo" — kaun sa relation? ::: Cell C → $T^{γ} P^{1 - γ} = const$ . " $T_{1}$ se $T_{2}$ tak thanda hota hai, work nikalo" — kaun sa relation? ::: Cell D → $W_{by} = n C_{V} (T_{1} - T_{2})$ . " $γ \to 1$ , kaun sa curve?" ::: Cell E → adiabat isotherm $P V = const$ ban jaata hai. "Volume $1000 \times$ shrinks, $T$ kaise badhta hai?" ::: Cell F → $T \propto V^{- (γ - 1)}$ , ek power law. "Temperature ratio pata hai, $V$ -ratio nikalo" ::: Cell I → $T V^{γ - 1}$ invert karo: $V_{1} / V_{2} = (T_{2} / T_{1})^{1/ (γ - 1)}$ . " $δ Q$ kyun likhte hain lekin $d U$ ?" ::: $U$ ek stored state function hai (exact $d$ ); $Q, W$ path-dependent flows hain (inexact $δ$ ).

Connections

1.7.16 Adiabatic relations — PV^γ = const, TV^(γ−1) = const (derivation) (Hinglish) — parent derivation jise ye examples exercise karte hain.
First Law of Thermodynamics — work example (Ex 4) aur $δ Q = d U + δ W$ sign convention ko power karta hai.
Isothermal Process — degenerate limit (Ex 5) aur Ex 8 ki second leg.
Mayer's Relation — kyun $γ > 1$ hai, to compression hamesha garam karta hai.
Degrees of Freedom and $\gamma$ — $γ = 5/3$ aur $7/5$ kahan se aate hain.
Speed of Sound in Gases — wahi $γ$ govern karta hai ki sound adiabatically air ko kaise compress karta hai.
Work Done in Thermodynamic Processes — Ex 4 ke liye alternative $\int P d V$ route.