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

1.7.14 · D3 · Physics › Thermodynamics › Thermodynamic processes — isothermal (T const), isochoric (V

Poore note mein, $R = 8.31 J mol^{- 1} K^{- 1}$ . Monatomic gas: $C_{V} = \frac{3}{2} R$ , $C_{P} = \frac{5}{2} R$ , $γ = \frac{5}{3}$ . Diatomic gas: $C_{V} = \frac{5}{2} R$ , $C_{P} = \frac{7}{2} R$ , $γ = \frac{7}{5}$ .

Scenario matrix

Is chapter ka har problem inhi cells mein se ek hoga. Har column ek "sign ya degenerate axis" hai; neeche har worked example us cell ke saath tagged hai jo wo cover karta hai.

Cell	Description	Sign / limiting feature	Example
A	Isochoric heating	$W = 0$ , $Q = Δ U > 0$	Ex 1
B	Isobaric expansion	$W > 0$ , split $Q = Δ U + W$	Ex 2
C	Isothermal expansion	$Δ U = 0$ , $Q = W > 0$	Ex 3
D	Isothermal compression	$Δ U = 0$ , $Q = W < 0$ (sign flip)	Ex 3b
E	Adiabatic expansion	$Q = 0$ , gas cools, $W > 0$	Ex 4
F	Adiabatic compression	$Q = 0$ , gas heats, $W < 0$ (sign flip)	Ex 5
G	Isotherm vs adiabat same start se compare karo	steepness / $γ > 1$ limiting geometry	Ex 6
H	Real-world word problem	atmosphere / diesel	Ex 5 (diesel)
I	Exam twist: two-leg cycle, net $W$	path dependence of $W$ , $Δ U = 0$ over cycle	Ex 7
J	Degenerate limit: $V_{2} \to V_{1}$	$ln 1 = 0$ , sab quantities $\to 0$	Ex 8

Ab har cell ko walk karte hain.

Example 1 — Cell A: Isochoric heating

Worked example Sealed rigid vessel, 3 mol diatomic gas, heated

$290 \to 350$ K. $W$ , $Δ U$ , $Q$ find karo.

Forecast: vessel rigid hai — compute karne se pehle guess karo: kya koi work hoga? Teeno quantities mein se kaun si zero hai?

Step 1. Volume fixed hai, isliye $d V = 0$ aur $W = \int P d V = 0$ . Yeh step kyun? Work $P$ – $V$ curve ke neeche ka area hai; agar koi volume sweep nahi hua toh koi area nahi.

Step 2. $Δ U = n C_{V} Δ T = 3 \cdot \frac{5}{2} (8.31) (60) = 3739.5 J$ . Yeh step kyun? $Δ U = n C_{V} Δ T$ har process mein lagta hai; yahan diatomic gas ke liye $C_{V} = \frac{5}{2} R$ hai.

Step 3. First Law: $Q = Δ U + W = 3739.5 + 0 = 3739.5 J$ . Yeh step kyun? $W = 0$ hone par, heat ka har joule internal energy ban jaata hai.

Verify: Units: $mol \cdot \frac{J}{mol K} \cdot K = J$ ✓. Sanity: sealed can ko heat karne se sirf $T$ aur $P$ badhta hai, koi motion nahi — $W = 0$ correct hai. Sab energy account ho gayi.

Example 2 — Cell B: Isobaric expansion (the energy split)

Worked example 2 mol monatomic gas constant

$P = 1.0 \times 1 0^{5} Pa$ par, heat diya jaaye taaki volume $V_{1} = 0.020 m^{3} \to V_{2} = 0.035 m^{3}$ ho jaaye. $W$ , $Δ T$ , $Δ U$ , $Q$ find karo, aur check karo ki $Q$ , $Δ U + W$ mein split hota hai.

Forecast: gas dono heat hogi aur piston push karegi. Guess karo: kya $Q$ , $Δ U$ se bada hoga ya chhota?

Step 1. $W = P Δ V = (1.0 \times 1 0^{5}) (0.035 - 0.020) = 1500 J$ . Yeh step kyun? $P$ constant hai isliye $\int P d V$ se bahar aa jaata hai, sirf $P Δ V$ bachta hai — horizontal line ke neeche rectangle area (dekho Work done by gas — PV diagrams).

Step 2. $P Δ V = n R Δ T$ se: $Δ T = \frac{P Δ V}{n R} = \frac{1500}{2 \cdot 8.31} = 90.25 K$ . Yeh step kyun? Constant $P$ par $P V = n R T$ deta hai $P Δ V = n R Δ T$ ; $Δ U$ aur $Q$ ke liye $Δ T$ chahiye.

Step 3. $Δ U = n C_{V} Δ T = 2 \cdot \frac{3}{2} (8.31) (90.25) = 2250 J$ .

Step 4. $Q = n C_{P} Δ T = 2 \cdot \frac{5}{2} (8.31) (90.25) = 3750 J$ . Yeh step kyun? Isobar ke along, heat $C_{P}$ se measure hoti hai (from Heat Capacities Cv and Cp), jo expansion cost pehle se bundle kar leta hai.

Verify: $Δ U + W = 2250 + 1500 = 3750 = Q$ ✓. Mayer's split confirm hua: $C_{P} > C_{V}$ isliye ki extra $R$ per mole ne $1500 J$ work pay kiya. $Q > Δ U$ , forecast se match ✓.

Example 3 — Cell C: Isothermal expansion

Worked example 1 mol gas

$T = 320 K$ par, volume $V_{1}$ se $V_{2} = 3 V_{1}$ tak reservoir ke contact mein expand hota hai. $Δ U$ , $W$ , $Q$ find karo.

Forecast: temperature reservoir se pinned hai. Guess karo: $Δ U$ kya hai? Work energy kahan se aati hai?

Step 1. $T$ constant $\Rightarrow Δ U = n C_{V} Δ T = 0$ . Yeh step kyun? $U$ sirf $T$ par depend karta hai; temperature change nahi toh internal energy change bhi nahi.

Step 2. First Law: $0 = Q - W \Rightarrow Q = W$ . Yeh step kyun? $Δ U = 0$ hone par, heat ka har joule seedha work ban ke nikalti hai.

Step 3. $W = n R T ln \frac{V _{2}}{V _{1}} = (1) (8.31) (320) ln 3 = 2921.6 J$ . Yeh step kyun? $P = n R T / V$ ; $n R T$ constant hai isliye integral se bahar aa jaata hai, aur $\int d V / V = ln V$ . Hum $ln$ use karte hain kyunki integrand $1/ V$ exactly $ln V$ ka derivative hai — koi aur function "kis cheez ka slope $1/ V$ hai?" ka jawab nahi deta.

Verify: $Q = W = 2921.6 J$ , $Δ U = 0$ ✓. Positive $W$ ⇒ expansion ⇒ heat absorb hua, isothermal expansion ke liye correct hai.

Example 3b — Cell D: Isothermal compression (sign flip)

Worked example Same gas, same

$T = 320 K$ , ab $V_{1}$ se $V_{2} = \frac{1}{3} V_{1}$ tak compress kiya. $W$ aur $Q$ find karo.

Forecast: ab gas squeeze ho rahi hai. Guess karo: $W$ positive hai ya negative? Heat absorb hogi ya expel?

Step 1. $W = n R T ln \frac{V _{2}}{V _{1}} = (1) (8.31) (320) ln \frac{1}{3} = - 2921.6 J$ . Yeh step kyun? $ln \frac{1}{3} = - ln 3 < 0$ . Compression matlab volume shrink hua, isliye logarithm negative hai — gas negative work karti hai (work us par kiya jaata hai).

Step 2. $Q = W = - 2921.6 J$ . Yeh step kyun? $Δ U$ ab bhi $0$ hai, isliye $T$ fix rakhne ke liye jab work andar aata hai toh heat gas se bahar jaani chahiye.

Verify: Ex 3 ka exact mirror, sign flip ✓. Physically: fixed $T$ par squeeze karne se $2921.6 J$ reservoir mein dump ho jaata hai. Yeh mistake-magnet cell hai — hamesha $ln$ ko sign carry karne do, kabhi khud minus mat lagao.

Example 4 — Cell E: Adiabatic expansion (gas cools)

Worked example 1 mol monatomic gas (

$γ = \frac{5}{3}$ ) adiabatically expand karta hai. Start: $P_{1} = 2.0 \times 1 0^{5} Pa$ , $V_{1} = 0.010 m^{3}$ . $V_{2} = 0.020 m^{3}$ tak expand hota hai. $P_{2}$ , $T_{1}$ , $T_{2}$ , aur $W$ find karo.

Forecast: heat andar nahi aa sakti. Guess karo: expand hone par gas garm hogi ya thandi? Work kahan se aayega?

Step 1. $P_{2} = P_{1} (\frac{V _{1}}{V _{2}})^{γ} = (2.0 \times 1 0^{5}) (\frac{1}{2})^{5/3} = 63000 Pa$ (2 s.f. tak $6.3 \times 1 0^{4}$ ). Yeh step kyun? Adiabats $P V^{γ} = const$ follow karte hain, isliye $P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$ .

Step 2. $T_{1} = \frac{P _{1} V _{1}}{n R} = \frac{( 2.0 \times 1 0 ^{5} ) ( 0.010 )}{( 1 ) ( 8.31 )} = 240.7 K$ .

Step 3. $T_{2} = \frac{P _{2} V _{2}}{n R} = \frac{( 63000 ) ( 0.020 )}{8.31} = 151.6 K$ . Yeh step kyun? $P_{2}$ aur $V_{2}$ maloom hone par, $P V = n R T$ seedha naya temperature deta hai. $T_{2} < T_{1}$ : gas thandi ho gayi, bilkul forecast ke jaisa.

Step 4. $W = n C_{V} (T_{1} - T_{2}) = (1) \frac{3}{2} (8.31) (240.7 - 151.6) = 1111 J$ . Yeh step kyun? $Q = 0$ ⇒ $W = - Δ U = n C_{V} (T_{1} - T_{2})$ ; work poori tarah internal energy se pay hoti hai.

Verify: $W = \frac{P _{1} V _{1} - P _{2} V _{2}}{γ - 1} = \frac{2000 - 1260}{2/3} = 1110 J$ ✓ se cross-check (rounding). $W > 0$ aur $T$ gira — insulated expansion se consistent hai (aise hi clouds bante hain jab hawa upar jaati hai).

Example 5 — Cell F & H: Adiabatic compression (diesel ignition)

Worked example Ek cylinder mein air (

$γ = 1.4$ , diesel ki tarah treat karo) ko adiabatically $V_{1}$ se $V_{2} = \frac{1}{15} V_{1}$ tak compress kiya jaata hai (compression ratio 15). Initial temperature $T_{1} = 300 K$ . $T_{2}$ aur $W$ ka sign find karo.

Forecast: fast squeeze kiya, koi heat escape nahi hui. Guess karo: kya temperature itni badhegi (~600 K) ki diesel ignite ho jaaye?

Step 1. $T V^{γ - 1} = const \Rightarrow T_{2} = T_{1} (\frac{V _{1}}{V _{2}})^{γ - 1} = 300 \cdot 1 5^{0.4}$ . Yeh step kyun? Adiabat ka $T$ – $V$ form temperatures ko directly volume ratio se link karta hai, pressure skip karke.

Step 2. $1 5^{0.4} = 2.954$ , isliye $T_{2} = 300 \times 2.954 = 886.3 K$ . Yeh step kyun? $886 K$ (≈ 613 °C) diesel ke autoignition (~483 K) se kaafi upar hai — fuel bina spark plug ke ignite ho jaata hai. Yahi diesel engine ka poora principle hai.

Step 3. $W$ ka sign: $T_{2} > T_{1}$ isliye $Δ U > 0$ , aur $W = - Δ U < 0$ . Yeh step kyun? Compression ke dauran gas par work kiya jaata hai, isliye $W$ (gas dwara work) negative hai — parent note se classic sign trap.

Verify: $T_{2} = 886 K > T_{1}$ ✓ compression par heating, $W < 0$ ✓. Real-world (Cell H) satisfy hua: adiabatic compression hi diesel ignite karta hai.

Example 6 — Cell G: Same start se isotherm vs adiabat

Worked example Ek isotherm aur ek adiabat (

$γ = \frac{5}{3}$ ) dono $P_{1} = 1.0 \times 1 0^{5} Pa$ , $V_{1} = 0.010 m^{3}$ se start hokar $V_{2} = 0.020 m^{3}$ tak expand karte hain. Har curve par final pressure find karo aur confirm karo kaun lower jaata hai.

Forecast: same start, same expansion. Guess karo: kaun si curve lower pressure par end hogi?

Step 1. Isotherm ( $P V = const$ ): $P_{2}^{iso} = P_{1} \frac{V _{1}}{V _{2}} = (1.0 \times 1 0^{5}) (0.5) = 50000 Pa$ . Yeh step kyun? Isotherm par pressure volume ke simple inverse proportion mein girta hai.

Step 2. Adiabat ( $P V^{γ} = const$ ): $P_{2}^{ad} = P_{1} (\frac{V _{1}}{V _{2}})^{γ} = (1.0 \times 1 0^{5}) (0.5)^{5/3} = 31498 Pa$ . Yeh step kyun? Exponent $γ > 1$ same volume ratio se zyada bada pressure drop produce karta hai.

Step 3. Compare: $31498 < 50000$ . Adiabat lower end hoti hai — yeh steeper hai. Yeh step kyun? Adiabat par gas ka temperature bhi girta hai (koi heat energy replace nahi karti jo work par spend hui), isliye pressure aur zyada sagg karta hai. Figure dekho: $V_{2}$ par violet adiabat magenta isotherm ke neeche hai.

Verify: ratio $\frac{P _{2}^{ad}}{P _{2}^{iso}} = \frac{31498}{50000} = 0.63 = (0.5)^{γ - 1} = (0.5)^{2/3}$ ✓. Steepness factor exactly $(V_{1} / V_{2})^{γ - 1}$ hai, jo $γ > 1$ ki limiting-geometry signature hai.

Example 7 — Cell I: Two-leg cycle, net work (exam twist)

Worked example 1 mol monatomic gas chalata hai:

Leg 1 isobaric expansion $P = 1.0 \times 1 0^{5} Pa$ par $V_{A} = 0.010$ se $V_{B} = 0.020 m^{3}$ tak; Leg 2 isochoric cooling $V = 0.020 m^{3}$ par pressure itne tak ki ek final isothermal leg wapas le jaaye... — simplify: Leg 1 aur Leg 2 ke baad, gas $P_{C} = 0.5 \times 1 0^{5} Pa$ , $V_{C} = 0.020 m^{3}$ par hai. Donon legs par total work find karo.

Forecast: work path par depend karta hai. Guess karo: kya isochoric leg work mein contribute karta hai?

Step 1. Leg 1 (isobaric): $W_{1} = P Δ V = (1.0 \times 1 0^{5}) (0.020 - 0.010) = 1000 J$ . Yeh step kyun? Horizontal line ke neeche rectangle area.

Step 2. Leg 2 (isochoric): $W_{2} = 0$ . Yeh step kyun? $d V = 0$ , koi swept area nahi — Cell A ka cornerstone mid-cycle par bhi laagu hota hai.

Step 3. Total $W = W_{1} + W_{2} = 1000 + 0 = 1000 J$ . Yeh step kyun? Multi-leg path par work har leg ke works ka sum hota hai — $W$ path-dependent hai, $Δ U$ ke unlike.

Verify: sirf expanding leg work karta hai; $1000 J$ ✓. Agar yeh closed cycle hota, $\oint d U = 0$ isliye $Q_{net} = W_{net}$ — Carnot Cycle aur Entropy bookkeeping ka seed.

Example 8 — Cell J: Degenerate limit $V_{2} \to V_{1}$

Recall Self-test: cell ka naam batao

Ek gas bina heat exchange ke compress hoti hai aur uska temperature badhta hai. Kaun sa cell? ::: Cell F (adiabatic compression, $W < 0$ ). Ek gas fixed temperature par expand karti hai; $Δ U$ kya hai? ::: $0$ (Cell C/D). Adiabat same $V_{2}$ par isotherm se neeche kyun jaati hai? ::: Exponent $γ > 1$ zyada bada pressure drop deta hai; gas thandi bhi hoti hai (Cell G). Sealed rigid can ko heat karo, $W$ kya hai? ::: $0$ (Cell A). Multi-leg cycle mein kaun si quantity path-dependent hai? ::: Work $W$ (aur $Q$ ); $Δ U$ nahi (Cell I).

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