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

1.7.14 · D2 · Physics › Thermodynamics › Thermodynamic processes — isothermal (T const), isochoric (V

Hum First Law of Thermodynamics, Ideal Gas Law, Heat Capacities Cv and Cp, aur Work done by gas — PV diagrams pe lean karte hain. Sab kuch parent note se trace hota hai: the topic note.

Step 1 — "Adiabatic" ka matlab kya hai (ek aisi wall jo heat ko cross nahi karne deti)

WHAT. Hum gas ko ek aisi cylinder mein daalte hain jo perfect insulation se liptii hai. Ek piston usse band karta hai.

WHY. $Q=0$ set karna woh ek constraint hai jo is process ko special banata hai. Har doosri process mein heat andar ya bahar sneak karti hai; yahan gas apne aap par dependent hai — jo bhi work karti hai woh apni khud ki energy se aana chahiye.

WHAT IT LOOKS LIKE. Figure mein, orange hatched wall wavy red heat arrows ko block kar rahi hai: woh wapas bounce ho jaate hain. Sirf piston move kar sakta hai.

Step 2 — First Law, ek energy bucket ke roop mein draw kiya

WHAT. First Law of Thermodynamics kehta hai energy sirf bookkeeping hai: $\Delta U=Q-W$. Tiny-step form mein, $dU=dQ-dW$.

WHY. Humein ek aisi relation chahiye jo gas ke temperature change ko work jo woh karta hai se connect kare, kyunki wahi link curve ko bend karega. First Law hi woh akeela law hai jo $Q$, $W$, aur $U$ ko ek saath tie karta hai.

WHAT IT LOOKS LIKE. Ek bucket imagine karo jis par $U$ likha hai. Heat upar se andar girti hai ($dQ$), work side se leak hoti hai ($dW$). $dQ=0$ ke saath, upar cap laga di gayi hai — bucket sirf drop kar sakti hai jab work leak hoti hai.

Step 3 — Do chhote pieces ko naam dena: $dU$ aur $dW$

WHAT. Hum vague $dU$ aur $dW$ ko concrete formulas se replace karte hain.

• Ideal gas ki internal energy: $dU=n{C}_{V}dT$. Yeh har process mein hold karta hai (yeh parent ka doosra "mistake" hai — $C_V$ bas temperature se internal energy tak ka dial hai).
• Work jab piston tiny volume $dV$ sweep karta hai: $dW=PdV$ (Work done by gas — PV diagrams se — force $\times$ distance ko pressure $\times$ swept volume ke roop mein repackage kiya).

WHY. Unhe combine karne ke liye hum dono ko ek hi language mein bolna hoga — dono ko state variables $P$, $V$, $T$ ke expressions banana hoga.

WHAT IT LOOKS LIKE. Figure mein piston swept volume $dV$ (blue) ki chhoti si sliver se bahar nudge hoti dikhti hai. Gas thodi si cool hui: $dT<0$, jo thermometer ke girne ke roop mein draw kiya gaya hai.

Dono ko $dU=-dW$ mein substitute karo: nC_V\,dT = -\,P\,dV \tag{$\ast$}

Step 4 — Problem: teen variables hain, hum sirf do chahte hain

WHAT. Equation $(\ast )$ mein $T$, $P$, aur $V$ mix hain. Lekin $P$–$V$ diagram par ek curve sirf $P$ aur $V$ jaanta hai. Hum $T$ ko nikalna chahte hain.

WHY. Ideal Gas Law $PV=nRT$ bridge hai. Yeh $T$ ko $P$ aur $V$ se tie karta hai, isliye hum kisi bhi $dT$ ko $dP$ aur $dV$ ke terms mein rewrite kar sakte hain.

WHAT IT LOOKS LIKE. $T$ ko $(P,V)$ point ki shadow ki tarah socho: picture ek point ko $P$–$V$ plane par move karte dikhaati hai, aur $T$ ek colour hai jo uske saath follow karti hai. Hum plane mein equation chahte hain, isliye hum colour ko coordinates se trade karte hain.

Toh $nRdT=PdV+VdP$, jisse $ndT=\frac{PdV+VdP}{R}$ milta hai.

Step 5 — Do equations ko fuse karo aur $\gamma$ appear hone do

WHAT. Step 4 ka $T$-expression $(\ast )$ mein dalo.

WHY. Yahi woh moment hai jab do physical laws (energy aur gas law) sirf $P$ aur $V$ wali ek equation ban jaate hain — exactly wahi jo ek curve ko chahiye.

$(\ast )$ se shuru karo: $C_V(ndT)=-PdV$. $ndT$ replace karo: $C_V\cdot \frac{PdV+VdP}{R}=-PdV$

$R/C_V$ se multiply karo: $PdV+VdP=-\frac{R}{C_V}PdV$

Ab Heat Capacities Cv and Cp Mayer's relation $C_P=C_V+R$ deta hai, toh $\frac{R}{C_V}=\frac{C_P-C_V}{C_V}=\gamma -1$, jahan $\gamma \equiv \frac{C_P}{C_V}$.

WHAT IT LOOKS LIKE. Figure mein teeno variables wali messy relation collapse hoti dikhaati hai, aur $\gamma$ ek single knob ke roop mein drop in hota hai jo steepness control karta hai.

$\gamma -1$ substitute karo: $PdV+VdP=-(\gamma -1)PdV$

Step 6 — Clean separated form mein tidy karo

WHAT. Sab kuch ek side move karo aur $dV$ terms group karo.

WHY. Hum pattern $\text{(stuff)}\frac{dV}{V}+\frac{dP}{P}=0$ chahte hain, kyunki yeh pattern instantly integrable hai — har piece ek logarithm ka tiny change hai.

$PdV+VdP+(\gamma -1)PdV=0$ $\gamma PdV+VdP=0$

$PV$ se divide karo (dono positive — gas ka pressure aur volume hamesha real hota hai): $\gamma \frac{dV}{V}+\frac{dP}{P}=0$

• $\gamma \frac{dV}{V}$ :: volume mein fractional change, $\gamma$ se weighted.
• $\frac{dP}{P}$ :: pressure mein fractional change.
• $=0$ :: dono cancel hone chahiye — jab volume badhta hai, pressure ko balance rakhne ke liye girna padta hai.

WHAT IT LOOKS LIKE. Figure ek seesaw dikhaata hai: $V$ ko ek fraction se upar push karne par $P$ ko $\gamma$ times us fraction se neeche jaana padta hai. Isiliye adiabat isotherm se zyada steep girta hai (jahan weight sirf $1$ hoga, $\gamma$ nahi).

Step 7 — Integrate karo: logs boxed law ban jaate hain

WHAT. Dono terms ko integrate karo.

$\int \gamma \frac{dV}{V}+\int \frac{dP}{P}=\int 0$ $\gamma \ln V+\ln P=\text{const}$

Log rules use karo ($a\ln x=\ln x^a$ aur $\ln x+\ln y=\ln xy$): $\ln \left(P{V}^{\gamma }\right)=\text{const}⟹\boxed{P{V}^{\gamma }=\text{const}}$

WHY. Exponentiate karne se log hat jaata hai, aur pure product $P{V}^{\gamma }$ poore path par frozen reh jaata hai.

WHAT IT LOOKS LIKE. Figure same start point se finished adiabat (orange) ko isotherm (blue) ke against plot karta hai — adiabat uske neeche ghusta hai, exactly woh steepness jo $\gamma$ ne Step 6 mein di thi.

Step 8 — Edge cases: agar $\gamma \to 1$ ho, ya gas bas waise hi baith jaaye?

Har derivation ko apni khud ki extremes survive karni chahiye.

• $\gamma \to 1$ (isothermal limit). Agar $\gamma =1$, toh boxed law $P{V}^1=\text{const}$ ban jaata hai — isothermal process ka plain hyperbola. Physically $\gamma =1$ ka matlab hoga $R=C_V\to \infty$, yaani infinite heat capacity: gas temperature change kiye bina energy soak karti hai. Toh adiabat continuously isotherm mein deform hoti hai jaise $\gamma \to 1$. Do curves cousins hain, strangers nahi.
• $dV=0$ (piston jammed). Tab $dW=PdV=0$, aur $(\ast )$ deta hai $dT=0$: koi work nahi, koi temperature change nahi — adiabatic aur isochoric conditions ek frozen point par coincide karti hain. Koi contradiction nahi, bas ek still frame.
• Compression ($dV<0$). Balance $\gamma \frac{dV}{V}+\frac{dP}{P}=0$ ab $dP>0$ force karta hai aur $(\ast )$ deta hai $dT>0$: andar push karna gas ko garam karta hai aur heat escape karne ki jagah nahi hai. Yahi parent ke sign mistake mein mention kiya gaya diesel-ignition case hai.

WHAT IT LOOKS LIKE. Figure ek point se $\gamma =1$ (=isotherm), $\gamma =1.4$, $\gamma =1.67$ ke liye teen adiabats overlay karta hai — $\gamma$ ke badhne ke saath steeper — plus ek arrow jo compression direction mark karta hai jahan $T$ badhta hai.

Ek-picture summary

Poori story ek canvas par: insulated cylinder ($Q=0$) → energy bucket with capped lid ($dU=-dW$) → pieces ko naam do ($n{C}_{V}dT=-PdV$) → $T$ ko $PV=nRT$ se nikalo → $\gamma$ appear hota hai → separate karo → logs integrate karo → $P{V}^{\gamma }=\text{const}$, isotherm se steeper.

Quick self-check

Reveals fill karo peeking se pehle.