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

1.7.16 · D4 · Physics › Thermodynamics › Adiabatic relations — PV^γ = const, TV^(γ−1) = const (deriva

Throughout: $γ = C_{P} / C_{V} > 1$ (ratio of specific heats), $n$ = number of moles, $R = 8.314 J mol^{- 1} K^{- 1}$ , aur teen relations jinpar hum depend karte hain: $P V^{γ} = const, T V^{γ - 1} = const, T^{γ} P^{1 - γ} = const .$

Level 1 — Recognition

Recall Solution L1.1

Tumhe $V$ pata hai aur $T$ chahiye, pressure ka koi zikar nahi. Woh relation jo sirf $T$ aur $V$ ko contain karti hai: $T V^{γ - 1} = const .$ Kyun ye wali: ye exactly un do quantities ko link karti hai jo tumhare paas hain aur jo chahiye, isliye koi substitution nahi chahiye. $P V^{γ}$ use karne par pehle pressures nikalne padenge — extra kaam aur galti ke extra chances. Answer ::: $T V^{γ - 1} = const$ .

Recall Solution L1.2

$P V^{γ}$ mein poora $γ$ hai; $T V^{γ - 1}$ mein $γ$ minus one hai. Fark kyun hai: $P \leftrightarrow T$ swap karne ke liye $P V = n R T$ use karte hain, jisse $V$ ka ek factor trade hota hai, isliye $V$ -exponent exactly $1$ se shift ho jaata hai. Answer ::: $P V^{γ}$ , $T V^{γ - 1}$ .

Recall Solution L1.3

Pehle, notation: ek derivative par subscript batata hai ki use lete waqt kya fixed rakha jaata hai. Toh $(\frac{d P}{d V})_{T}$ matlab hai "temperature $T$ fixed rakhte hue $P$ mein $V$ ke saath badlaav" — isotherm ke saath slope. Isi tarah $(\frac{d P}{d V})_{Q}$ matlab hai "…jab $Q$ (heat) zero par fixed ho" — adiabat ke saath slope. Isliye subscript $Q$ sirf no-heat (adiabatic) constraint ko label karta hai. Is se, dono slopes hain: $(\frac{d P}{d V})_{T} = - \frac{P}{V}, (\frac{d P}{d V})_{Q} = - γ \frac{P}{V} .$ Kyunki $γ > 1$ , adiabat $γ$ factor se zyada steep hai. Figure ye concretely dikhata hai: dono curves ek hi violet point se guzarti hain, lekin magenta adiabat orange isotherm se zyada steeply neeche jaati hai. Do annotated slope arrows dekho — adiabat ka arrow zyada neeche jhukta hai kyunki, jab gas expand hoti hai, woh thandi bhi hoti hai, isliye uska pressure isotherm se zyada tezi se girta hai jahan incoming heat $T$ (aur isliye $P$ ) ko support karti hai.

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

Answer ::: adiabat,

γ

factor se zyada steep.

Level 2 — Application

Recall Solution L2.1

$T V^{γ - 1} = const$ use karo: $T_{2} = T_{1} (\frac{V _{1}}{V _{2}})^{γ - 1} = 300 (3)^{0.4} .$ $3^{0.4} = e^{0.4 l n 3} = e^{0.4394} = 1.5518$ . Toh $T_{2} = 300 \times 1.5518 \approx 465.5 K .$ Sanity: compress karne par koi heat escape nahi hoti aur work gas mein jaata hai, isliye uska temperature badhna chahiye — $465 K > 300 K$ . ✔️ Answer ::: $T_{2} \approx 466 K$ .

Recall Solution L2.2

$P V^{γ} = const$ use karo: $P_{2} = P_{1} (\frac{V _{1}}{V _{2}})^{γ} = 10 (1/2)^{1.4} .$ $(1/2)^{1.4} = e^{- 1.4 l n 2} = e^{- 0.9704} = 0.3789$ . Toh $P_{2} = 10 \times 0.3789 \approx 3.79 atm .$ Sanity: expansion se pressure girta hai, aur kyunki gas thandi bhi hoti hai toh girna isothermal se zyada steep hota hai (jo $5 atm$ deta). ✔️ Answer ::: $P_{2} \approx 3.79 atm$ .

Recall Solution L2.3

Hamare paas $P$ aur $T$ hain, volumes nahi — toh $T^{γ} P^{1 - γ} = const$ use karo, yaani $T_{2} = T_{1} (\frac{P _{2}}{P _{1}})^{(γ - 1) / γ}$ . Yahan $\frac{γ - 1}{γ} = \frac{2/3}{5/3} = 0.4$ , aur $\frac{P _{2}}{P _{1}} = \frac{1}{3}$ . $T_{2} = 400 (1/3)^{0.4} = 400 e^{- 0.4 l n 3} = 400 e^{- 0.4394} = 400 \times 0.6444 = 257.8 K .$ Sanity: pressure gira → gas expand hui → thandi hui: $258 K < 400 K$ . ✔️ Answer ::: $T_{2} \approx 258 K$ .

Level 3 — Analysis

Recall Solution L3.1

Adiabatic matlab $Q = 0$ , toh first law $0 = Δ U + W$ deta hai $W = - Δ U = n C_{V} (T_{1} - T_{2})$ . $C_{V} = \frac{5}{2} R = 2.5 \times 8.314 = 20.785 J mol^{- 1} K^{- 1}$ ke saath: $W = 2 \times 20.785 \times (500 - 350) = 2 \times 20.785 \times 150 = 6235.5 J .$ Kyun positive: gas expand karti hai, isliye apne surroundings par work karti hai; woh energy apne internal store se aati hai, isliye woh thandi hui. Positive $W$ aur girta $T$ ek hi fact ke do views hain. ✔️ Answer ::: $W \approx 6236 J$ .

Recall Solution L3.2

Isotherm: $(\frac{d P}{d V})_{T} = - \frac{P}{V} = - \frac{4}{10} = - 0.4 atm/L$ . Adiabat: $(\frac{d P}{d V})_{Q} = - γ \frac{P}{V} = - 1.4 \times 0.4 = - 0.56 atm/L$ . Ratio $= \frac{- 0.56}{- 0.4} = 1.4 = γ$ . Kyun: dono slopes mein $P / V$ ka factor common hai; adiabat mein extra $γ$ isliye hai kyunki jab ye expand hoti hai toh gas thandi bhi hoti hai, pressure zyada tezi se girta hai. Ratio isliye hamesha exactly $γ$ hota hai. ✔️ Answer ::: isotherm $- 0.4$ , adiabat $- 0.56$ , ratio $1.4$ .

Recall Solution L3.3

Compression se gas garm hoti hai, isliye $T_{2} > T_{1}$ , jisse $(T_{1} - T_{2}) < 0$ aur isliye $W < 0$ . $W$ woh work hai jo gas dwara kiya jaata hai; $W < 0$ matlab work gas par kiya jaata hai — exactly wahi jo use compress karne ke liye chahiye. Jo energy push hoti hai woh internal energy badhati hai, isliye $T$ upar jaata hai. Har arrow same direction mein point karta hai. ✔️ Answer ::: $W < 0$ (gas par work kiya jaata hai), heating ke saath consistent.

Level 4 — Synthesis

Recall Solution L4.1

Stage A (adiabatic, $T V^{γ - 1}$ use karo): $γ - 1 = 2/3$ . $T_{1} = T_{0} (\frac{V _{0}}{V _{1}})^{γ - 1} = 300 (2/4)^{2/3} = 300 (0.5)^{0.6667} .$ $(0.5)^{0.6667} = e^{- 0.6667 l n 2} = e^{- 0.4621} = 0.6300$ , toh $T_{1} = 300 \times 0.6300 = 189.0 K$ . A ke baad pressure, $P V^{γ}$ se: $P_{1} = P_{0} (V_{0} / V_{1})^{γ} = 8 (0.5)^{1.6667} = 8 e^{- 1.6667 l n 2} = 8 e^{- 1.1552} = 8 \times 0.3150 = 2.520 atm$ . Stage B (isothermal, $T$ fixed, $P V = const$ use karo): $P_{2} = P_{1} \frac{V _{1}}{V _{2}} = 2.520 \times \frac{4}{8} = 1.260 atm .$ Kyun beech mein relations switch karte hain: Stage A mein koi heat flow nahi isliye $T$ change hota hai ( $T V^{γ - 1}$ rule karta hai); Stage B mein temperature fixed hai, isliye sirf $P V = const$ apply hota hai. Dono ke liye same relation use karna galat physics hogi. ✔️ Answer ::: $T_{1} \approx 189 K$ , $P_{2} \approx 1.26 atm$ .

Recall Solution L4.2

$v = \frac{γ P}{ρ} = \frac{1.4 \times 1.01 \times 1 0 ^{5}}{1.29} = \frac{1.414 \times 1 0 ^{5}}{1.29} = 1.096 \times 1 0^{5} .$ $1.096 \times 1 0^{5} = 331.1 m/s$ . Kyun $γ$ aur sirf $P$ nahi: sound compressions itni fast hain ki heat crests aur troughs ke beech leak nahi kar sakti, isliye har chhoti compression adiabatic hoti hai — pressure extra factor $γ$ ke saath respond karta hai. Isothermal $B = P$ use karo (Newton ki original galti) toh $\approx 280 m/s$ predict hoga, jo bahut slow hai. ✔️ Answer ::: $v \approx 331 m/s$ .

Recall Solution L4.3

$T V^{γ - 1}$ mein exponent $0.4$ ke barabar hai, toh $γ - 1 = 0.4 \Rightarrow γ = 1.4$ . $γ = 1 + R / C_{V}$ se (rearranged from $R / C_{V} = γ - 1$ ): $C_{V} = \frac{R}{γ - 1} = \frac{R}{0.4} = 2.5 R$ . Phir $\frac{f}{2} R = 2.5 R \Rightarrow f = 5$ . Reading it: $γ = 1.4$ , $C_{V} = \frac{5}{2} R$ , $f = 5$ — ek diatomic gas (3 translational + 2 rotational degrees of freedom). Measured curve exponent literally gas ki freedoms count karta hai. ✔️ Answer ::: $γ = 1.4$ , $C_{V} = 2.5 R$ , $f = 5$ (diatomic).

Level 5 — Mastery

Recall Solution L5.1

Adiabat ke saath $P V^{γ} = P_{1} V_{1}^{γ} \equiv K$ , toh $P = K V^{- γ}$ . Phir $W = \int_{V_{1}}^{V_{2}} K V^{- γ} d V = K [\frac{V ^{1 - γ}}{1 - γ}]_{V_{1}}^{V_{2}} = \frac{K ( V _{2}^{1 - γ} - V _{1}^{1 - γ} )}{1 - γ} .$ Ab $K V_{2}^{1 - γ} = P_{2} V_{2}^{γ} V_{2}^{1 - γ} = P_{2} V_{2}$ , aur similarly $K V_{1}^{1 - γ} = P_{1} V_{1}$ . Toh $W = \frac{P _{2} V _{2} - P _{1} V _{1}}{1 - γ} = \frac{P _{1} V _{1} - P _{2} V _{2}}{γ - 1} .$ Energy method se match: $P V = n R T$ use karke, $P_{1} V_{1} - P_{2} V_{2} = n R (T_{1} - T_{2})$ , aur $\frac{n R}{γ - 1} = n C_{V}$ (kyunki $R / C_{V} = γ - 1$ ). Isliye $W = n C_{V} (T_{1} - T_{2})$ . Do routes, ek answer — integral hi internal-energy drop hai. ✔️ Numerics: $W = \frac{500 - 320}{1.4 - 1} = \frac{180}{0.4} = 450 J$ . Answer ::: $W = \frac{P _{1} V _{1} - P _{2} V _{2}}{γ - 1} = 450 J$ .

Recall Solution L5.2

(a) $P V^{γ} = const$ se: $\frac{V _{2}}{V _{1}} = (\frac{P _{1}}{P _{2}})^{1/ γ} = (2)^{1/ (5/3)} = 2^{0.6} = e^{0.6 l n 2} = e^{0.4159} = 1.5157$ . (b) $T P$ relation se, $T_{2} = T_{1} (\frac{P _{2}}{P _{1}})^{(γ - 1) / γ}$ , jahan $(γ - 1) / γ = (2/3) / (5/3) = 0.4$ : $T_{2} = 600 (1/2)^{0.4} = 600 e^{- 0.4 l n 2} = 600 e^{- 0.2773} = 600 \times 0.7579 = 454.7 K .$ (c) Koi heat flow nahi, isliye saara work internal-energy drop se aata hai: $W = n C_{V} (T_{1} - T_{2})$ . $n = 1$ aur $C_{V} = \frac{3}{2} R = 1.5 \times 8.314 = 12.471 J K^{- 1}$ ke saath, $W = 1 \times 12.471 \times (600 - 454.7) = 12.471 \times 145.3 = 1812 J .$ Consistency chain: pressure aadha hua → volume $\approx 1.52 \times$ badhaa → temperature $455 K$ tak gira → positive work $1812 J$ us cooling se aaya. Har relation ( $P V^{γ}$ , $T P$ , energy) agree karta hai. ✔️ Answer ::: $V_{2} / V_{1} \approx 1.516$ , $T_{2} \approx 455 K$ , $W \approx 1812 J$ .

Recall Solution L5.3

Adiabatic expansion se gas zaroor thandi honi chahiye ( $Q = 0$ , gas apni khud ki energy work karne mein spend karti hai), isliye expansion par koi reported temperature rise impossible hai. Quantitatively: $T_{2} = T_{1} (\frac{V _{1}}{V _{2}})^{γ - 1} = 300 (1/2)^{0.4} = 300 e^{- 0.4 l n 2} = 300 \times 0.7579 = 227.4 K .$ Gas ko $340 K$ par warm hone ki bajay lagbhag $227 K$ tak thanda hona chahiye tha. Student ne shayad compression exponent sign use kiya ya $V_{1} / V_{2}$ swap kar diya. ✔️ Answer ::: impossible; correct $T_{2} \approx 227 K$ .

Connections

First Law of Thermodynamics — har work/energy step mein $0 = Δ U + W$ use kiya.
Work Done in Thermodynamic Processes — L5.1 mein direct integration.
Isothermal Process — L4.1 ka Stage B aur L1.3/L3.2 mein slope contrast.
Mayer's Relation — $R / C_{V} = γ - 1$ deta hai, har jagah use hota hai.
Degrees of Freedom and $\gamma$ — L4.3 mein $γ$ ek adiabatic curve se padh ke $f$ count kiya.
Speed of Sound in Gases — L4.2 mein adiabatic bulk modulus $γ P$ use hota hai.