Thermodynamic processes — isothermal (T const), isochoric (V const), isobaric (P const), adiabatic (Q = 0)

1.7.14 · Physics › Thermodynamics

0. Woh foundation jisse hum kabhi nahi hatenge: First Law & Work

1. Isochoric (V = const) — "iso-CHORic = move karne ki koi CHaR (jagah) nahi"

Kyunki $V$ constant hai, $d V = 0$ : $W = \int P d V = 0$ $Δ U = Q - W = Q \Rightarrow Q = n C_{V} Δ T$

$P$ – $V$ diagram par yeh ek vertical line hai.

2. Isobaric (P = const) — "iso-BARic = constant BARometric pressure"

$P$ constant hai, toh integral se bahar aa jaata hai: $W = \int_{V_{1}}^{V_{2}} P d V = P (V_{2} - V_{1}) = P Δ V = n R Δ T$ Yeh step kyun? Constant $P$ par $P V = n R T$ se $P Δ V = n R Δ T$ milta hai.

$Q = n C_{P} Δ T, Δ U = n C_{V} Δ T$ $P$ – $V$ diagram par yeh ek horizontal line hai; $W$ = rectangle ka area.

3. Isothermal (T = const) — "iso-THERMal = constant THERMometer"

$Δ T = 0 \Rightarrow Δ U = 0 \Rightarrow Q = W$ . Ab $W$ nikalo. $T$ constant hone par $P = n R T / V$ : $W = \int_{V_{1}}^{V_{2}} \frac{n R T}{V} d V = n R T \int_{V_{1}}^{V_{2}} \frac{d V}{V} = n R T ln \frac{V _{2}}{V _{1}}$ $W = Q = n R T ln \frac{V _{2}}{V _{1}} = n R T ln \frac{P _{1}}{P _{2}}$ Yeh step kyun? $n R T$ constant hai toh integral se bahar chala jaata hai; $\int d V / V = ln V$ . $P_{1} V_{1} = P_{2} V_{2} \Rightarrow V_{2} / V_{1} = P_{1} / P_{2}$ use kiya.

Yeh curve ek hyperbola $P V = const$ hai.

4. Adiabatic (Q = 0) — "a-DIA-batic = heat ka koi DIArrhea nahi (heat pass nahi ho sakti)"

$Q = 0 \Rightarrow W = - Δ U = - n C_{V} Δ T$ . Curve pane ke liye First Law differential form se shuru karo: $d Q = 0 = d U + d W = n C_{V} d T + P d V$ $P V = n R T$ use karo $\Rightarrow n R d T = P d V + V d P$ , toh $n C_{V} d T = \frac{C _{V}}{R} (P d V + V d P)$ : $\frac{C _{V}}{R} (P d V + V d P) + P d V = 0$ $R / C_{V}$ se multiply karo aur group karo, $γ \equiv C_{P} / C_{V}$ aur $R / C_{V} = γ - 1$ ke saath: $(P d V + V d P) + (γ - 1) P d V = 0 \Rightarrow γ P d V + V d P = 0$ $γ \frac{d V}{V} + \frac{d P}{P} = 0 integrate γ ln V + ln P = const$ $P V^{γ} = const; T V^{γ - 1} = const; P^{1 - γ} T^{γ} = const$

Kyunki $γ > 1$ hai, $P$ – $V$ plot par adiabat isotherm se zyada steep hoti hai.

Figure — Thermodynamic processes — isothermal (T const), isochoric (V const), isobaric (P const), adiabatic (Q = 0)

5. Ek-nazar summary table

Process	Constant	$W$	$Q$	$Δ U$	Curve
Isochoric	$V$	$0$	$n C_{V} Δ T$	$n C_{V} Δ T$	vertical
Isobaric	$P$	$P Δ V = n R Δ T$	$n C_{P} Δ T$	$n C_{V} Δ T$	horizontal
Isothermal	$T$	$n R T ln \frac{V _{2}}{V _{1}}$	$= W$	$0$	hyperbola
Adiabatic	$Q = 0$	$\frac{P _{1} V _{1} - P _{2} V _{2}}{γ - 1}$	$0$	$- W$	steep ( $P V^{γ}$ )

Common mistakes (Steel-manned)

Kis process mein gas dwara kiya gaya work zero hota hai, aur kyun?

Isochoric —

d V = 0

toh

W = \int P d V = 0

Ideal gas ke liye,

Δ U

SABHI processes mein kya hota hai?

n C_{V} Δ T

(kyunki

U

sirf

T

par depend karta hai).

Isothermal work formula?

W = n R T ln (V_{2} / V_{1})

, aur

Q = W

Δ U = 0

C_{P} > C_{V}

kyun hai?

Constant

P

par, extra heat expansion work ke liye pay karti hai; Mayer:

C_{P} = C_{V} + R

P

aur

V

ke beech adiabatic relation?

P V^{γ} =

const,

γ = C_{P} / C_{V}

T

aur

V

ke beech adiabatic relation?

T V^{γ - 1} =

const.

Adiabatic expansion mein gas garam hoti hai ya thandi? Kyun?

Thandi —

Q = 0

, toh work internal energy se aata hai,

T

gir jaata hai.

PV diagram par isotherm aur adiabat mein kaun zyada steep hai?

Adiabat (slope factor

γ > 1

Isobaric work formula?

W = P Δ V = n R Δ T

Adiabatic process kaunsi constraint define karta hai?

Q = 0

(koi heat exchange nahi), aksar insulation ya rapidity ki wajah se.

Isochoric process mein daali gayi heat?

Q = n C_{V} Δ T

(saari internal energy mein).

Adiabatic work states ke terms mein?

W = (P_{1} V_{1} - P_{2} V_{2}) / (γ - 1) = n C_{V} (T_{1} - T_{2})

Recall Feynman: 12-saal ke bacche ko explain karo

Ek bike pump imagine karo jisme gas hai.

Sealed & garam kiya (isochoric): tum handle push nahi kar sakte. Saari heat bas gas ko zyada gussa (garam) karti hai, koi movement nahi, koi work nahi.
Slowly push karo temperature cool rakhte hue (isothermal): jo bhi energy doge woh seedha gentle work ke roop mein wapas aa jaayegi — temperature kabhi nahi badlega.
Upar fixed weight rakhke push karo (isobaric): pressure same rehta hai, gas badhti hai aur garam hoti hai; energy garam hone aur weight uthane mein split hoti hai.
Insulation ke saath super fast pump karo (adiabatic): heat escape karne ka time nahi, toh gas sirf squeeze hone se garam hoti hai — isliye bike pump garam ho jaata hai!

Connections

First Law of Thermodynamics — parent equation $Δ U = Q - W$ .
Ideal Gas Law — $P V = n R T$ har derivation mein use hota hai.
Heat Capacities Cv and Cp — Mayer's relation $C_{P} = C_{V} + R$ .
Carnot Cycle — 2 isothermals + 2 adiabatics se bana hai.
Entropy — adiabatic-reversible = isentropic ( $Δ S = 0$ ).
Work done by gas — PV diagrams — area interpretation.