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

1.7.16 · Physics › Thermodynamics

Claim kya hai?

Hum teeno ko scratch se derive karenge, kabhi bhi sirf quote nahi karenge.

First principles se derivation

Step 1 — First law pe $Q = 0$ apply karo. $0 = n C_{V} d T + P d V$ Yeh step kyun? Ideal gas ki internal energy sirf temperature pe depend karti hai, isliye $d U = n C_{V} d T$ hamesha valid hai (chahe $V$ change ho). $δ Q = 0$ set karne se kaam ko internal energy se pay karna padta hai.

Step 2 — Ideal gas law use karke explicit $P$ ko hatao. $P V = n R T$ se $P = \frac{n R T}{V}$ milta hai. Substitute karo: $n C_{V} d T + \frac{n R T}{V} d V = 0$ Yeh step kyun? Hamare paas teen variables $P, V, T$ hain; ideal gas law se ek eliminate kar sakte hain. Hum $T$ aur $V$ rakhte hain taaki pehle $T V$ relation tak pahunch sakein.

Step 3 — Variables separate karo. $n C_{V} T$ se divide karo: $\frac{d T}{T} + \frac{R}{C _{V}} \frac{d V}{V} = 0$

Step 4 — $R / C_{V}$ ko $γ - 1$ se replace karo. Ideal gas ke liye Mayer's relation $C_{P} - C_{V} = R$ hoti hai, isliye $\frac{R}{C _{V}} = \frac{C _{P} - C _{V}}{C _{V}} = \frac{C _{P}}{C _{V}} - 1 = γ - 1.$ Yeh step kyun? Yahi hai jo $γ$ ko laata hai — yeh gas ke degrees of freedom ka signature hai jo $C_{V}$ mein chhupa hai.

$\frac{d T}{T} + (γ - 1) \frac{d V}{V} = 0$

Step 5 — Integrate karo. $ln T + (γ - 1) ln V = const \Rightarrow T V^{γ - 1} = const$ Yeh step kyun? Har term integrate hokar ek logarithm deta hai; logs ka sum = const matlab product const hai.

Step 6 — $P V^{γ}$ mein convert karo. $T = \frac{P V}{n R}$ use karo: $\frac{P V}{n R} V^{γ - 1} = const \Rightarrow P V^{γ} = const$ Yeh step kyun? $T$ ko $V^{γ - 1}$ se multiply karo; $V$ ke powers add ho jaate hain: $V^{1 + (γ - 1)} = V^{γ}$ . $n R$ jaisi constants naye const mein absorb ho jaati hain.

Step 7 — $T P$ relation. $P V^{γ} =$ const aur $V = \frac{n R T}{P}$ se: $P (\frac{n R T}{P})^{γ} = const \Rightarrow P^{1 - γ} T^{γ} = const .$

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

Adiabatic vs Isothermal (Pehle forecast, phir verify)

Dono ko differentiate karo:

Isothermal: $P d V + V d P = 0 \Rightarrow (\frac{d P}{d V})_{T} = - \frac{P}{V}$ .
Adiabatic: $V^{γ} d P + γ P V^{γ - 1} d V = 0 \Rightarrow (\frac{d P}{d V})_{Q} = - γ \frac{P}{V}$ .

Verify: kyunki $γ > 1$ hai, adiabatic slope $γ$ times steeper hai. ✔️ Kyun? Adiabatic expansion mein gas thandi bhi hoti hai, isliye $P$ isothermal se zyada fast drop karta hai (jahan $T$ aane wali heat se maintain rehti hai).

Worked examples

Common mistakes (Steel-manned)

Active-recall flashcards

Adiabatic process kya define karta hai?

δ Q = 0

— surroundings ke saath koi heat exchange nahi.

Ideal-gas adiabatic ke liye starting first-law equation?

0 = n C_{V} d T + P d V

Derivation mein

P

ko kaise eliminate karte ho?

Ideal gas law se

P = n R T / V

substitute karo.

R / C_{V}

kya simplify hota hai, aur kyun?

γ - 1

, Mayer's relation

C_{P} - C_{V} = R

use karke.

T V

adiabatic relation?

T V^{γ - 1} = const

P V

adiabatic relation?

P V^{γ} = const

T P

adiabatic relation?

T^{γ} P^{1 - γ} = const

Adiabatic expansion se gas thandi kyun hoti hai?

Kaam internal energy se aata hai (

Q = 0

), isliye

T

drop karta hai.

PV diagram pe adiabat ka slope isotherm se compare karo?

Adiabat

γ

factor steeper hai:

- γ P / V

- P / V

d U = n C_{V} d T

tab bhi valid kyun hai jab

V

change hota hai?

Ideal gas ke liye

U

sirf

T

pe depend karta hai.

Adiabatic process mein temperatures ke terms mein work?

W = n C_{V} (T_{1} - T_{2})

kyunki

W = - Δ U

Recall Feynman: ek 12-saal ke bacche ko explain karo

Socho ek sealed, perfectly insulated balloon — koi heat andar ya bahar nahi ja sakti. Jab andar ki hawa balloon ko bada karne ke liye push karti hai, use push karne ke liye energy kharch karni padti hai. Lekin koi heat andar aa nahi rahi us energy ko refill karne ke liye, isliye hawa apni khud ki warmth use kar leti hai: woh thandi ho jaati hai. Use dabao aur woh garam ho jaati hai — same idea ulta. Formula $P V^{γ} =$ const bas woh rulebook hai jo exactly bataata hai pressure aur size kaise trade off karte hain jab heat ko cheat karne ki permission nahi hai.

Connections

First Law of Thermodynamics — parent equation $δ Q = d U + δ W$ .
Isothermal Process — contrast: $P V =$ const, $T$ fixed.
Mayer's Relation — $C_{P} - C_{V} = R$ provide karta hai, hence $R / C_{V} = γ - 1$ .
Degrees of Freedom and $\gamma$ — kyun $γ = 5/3$ (mono), $7/5$ (diatomic).
Speed of Sound in Gases — adiabatic bulk modulus $γ P$ use karta hai.
Work Done in Thermodynamic Processes — adiabats ke liye $W = n C_{V} (T_{1} - T_{2})$ .