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

1.7.14 · D4 · Physics › Thermodynamics › Thermodynamic processes — isothermal (T const), isochoric (V

Definition Is poore page par use hone wale symbols aur sign conventions

Koi bhi algebra shuru karne se pehle, har symbol ka meaning fix karo taaki kuch bhi assumed na ho:

$n$ = gas ke moles ki sankhya — literally kitne $6 \times 1 0^{23}$ molecules ke packets tumhare paas hain. Jab problem kehti hai " $2 mol$ ," toh woh number hi $n = 2$ hai; use seedha har formula mein daalo.
$R = 8.31 J mol^{- 1} K^{- 1}$ = universal gas constant (moles-times-kelvin aur joules ke beech fixed conversion).
$T$ = temperature (K), $P$ = pressure (Pa), $V$ = volume (m³).
$Q$ = heat. Sign convention: $Q > 0$ matlab heat gas ke andar aa rahi hai; $Q < 0$ matlab heat bahar ja rahi hai.
$W$ = work. Sign convention: $W > 0$ matlab work gas ke dwara kiya gaya hai (woh expand karta hai, bahar push karta hai); $W < 0$ matlab work gas par kiya gaya hai (usse compress kiya ja raha hai).
First Law (energy bookkeeping): $Δ U = Q - W$ . Internal energy utni badhti hai jitni heat andar daali minus jitna work gas ne bahar diya.
Ideal Gas Law: $P V = n R T$ teen state variables ko $n$ aur $R$ ke through link karta hai.

Gas types ke liye constants: Monatomic $C_{V} = \frac{3}{2} R$ , $C_{P} = \frac{5}{2} R$ , $γ = \frac{5}{3}$ . Diatomic $C_{V} = \frac{5}{2} R$ , $C_{P} = \frac{7}{2} R$ , $γ = \frac{7}{5}$ . Yahan $C_{V}, C_{P}$ heat capacities hain aur $γ = C_{P} / C_{V}$ .

Level 1 — Recognition

Ye test karte hain ki kaun sa quantity frozen hai aur woh immediately kya zero karta hai.

L1.1

Ek $P$ – $V$ diagram par ek gas vertical line ke saath neeche move karti hai (pressure girta hai, volume unchanged). Process ka naam batao aur $W$ kya hoga.

Recall Solution

Picture kya dikhata hai: volume kabhi nahi badla ⇒ yeh isochoric process hai. $W = 0$ kyun: work curve ke neeche ka area hai, $W = \int P d V$ . Agar $V$ kabhi nahi badla toh $d V = 0$ har jagah hai, isliye "area" ek zero width wali line hai. $W = \int P d V = 0.$ Saari energy change heat ka internal energy banana hai: $Q = Δ U = n C_{V} Δ T$ (yaad karo $n$ = moles, upar define kiya gaya).

L1.2

Ek gas ko itni tezi se compress kiya jaata hai (ek insulated cylinder mein) ki koi heat flow nahi ho sakti. Yeh kaun sa process hai, aur $Q$ kya hai?

Recall Solution

Koi heat exchange nahi ⇒ adiabatic, isliye $Q = 0$ . First Law phir padhta hai $Δ U = Q - W = - W$ . Kyunki hum gas ko compress kar rahe hain, work us par ho raha hai, toh hamare sign convention se $W < 0$ hai, deta hai $Δ U > 0$ : gas garam ho jaati hai. (Yahi diesel ignition hai.)

L1.3

Har curve ko uske process se match karo: (a) horizontal line, (b) hyperbola $P V =$ const, (c) vertical line, (d) steeper hyperbola $P V^{γ} =$ const.

Recall Solution

(a) horizontal → pressure fixed → isobaric.
(b) hyperbola $P V =$ const → temperature fixed (kyunki $P V = n R T$ ) → isothermal.
(c) vertical → volume fixed → isochoric.
(d) steeper hyperbola → $Q = 0$ → adiabatic (steeper kyunki $γ > 1$ ).

Level 2 — Application

Ab ek-ek formula mein numbers plug karo.

L2.1

$2 mol$ monatomic gas ko ek rigid can mein band karo aur $300 K$ se $450 K$ tak heat karo. $W$ , $Δ U$ , aur $Q$ nikalo.

Recall Solution

Yahan $n = 2$ (do moles) hai. Rigid can ⇒ isochoric ⇒ $W = 0$ . $Δ U = n C_{V} Δ T = 2 \cdot \frac{3}{2} (8.31) (150) = 3739.5 J .$ First Law: $Q = Δ U + W = 3739.5 J$ . Heat ka har joule internal energy ban gaya.

L2.2

$1 mol$ gas isothermally $350 K$ par $V_{1}$ se $V_{2} = 3 V_{1}$ tak expand karti hai. $W$ , $Δ U$ , $Q$ nikalo.

Recall Solution

Yahan $n = 1$ hai. Isothermal ⇒ $Δ U = 0$ (kyunki $U$ sirf $T$ par depend karta hai). $W = n R T ln \frac{V _{2}}{V _{1}} = (1) (8.31) (350) ln 3 = 3195.4 J .$ $Q = W = 3195.4 J$ (saari heat in seedha work ke roop mein bahar aati hai).

L2.3

$3 mol$ gas isobarically $P = 2.0 \times 1 0^{5} Pa$ par $V_{1} = 0.02 m^{3}$ se $V_{2} = 0.05 m^{3}$ tak expand karti hai. $W$ nikalo. Phir, agar gas diatomic hai, $Q$ nikalo.

Recall Solution

Yahan $n = 3$ hai. Work (pressure constant ⇒ integral se bahar aata hai): $W = P Δ V = (2.0 \times 1 0^{5}) (0.05 - 0.02) = 6000 J .$ Temperature change — $Δ T = P Δ V / (n R)$ kyun aata hai (WHY, authority nahi): shuru mein ideal gas law likho, $P V_{1} = n R T_{1}$ , aur ant mein, $P V_{2} = n R T_{2}$ (same $P$ ). Dono subtract karo: $P (V_{2} - V_{1}) = n R (T_{2} - T_{1})$ , yaani $P Δ V = n R Δ T$ . $Δ T$ ke liye solve karo: $Δ T = \frac{P Δ V}{n R} = \frac{6000}{3 ( 8.31 )} = 240.67 K .$ Heat (diatomic $C_{P} = \frac{7}{2} R$ ): $Q = n C_{P} Δ T = 3 \cdot \frac{7}{2} (8.31) (240.67) = 21000 J .$ (Check: $Q = Δ U + W$ , aur $Δ U = n C_{V} Δ T = 3 \cdot \frac{5}{2} (8.31) (240.67) = 15000 J$ ; $15000 + 6000 = 21000$ . ✓)

Level 3 — Analysis

Do-step reasoning, ya adiabat curve.

L3.1

$1 mol$ monatomic gas ( $γ = \frac{5}{3}$ ) $P_{1} = 1.0 \times 1 0^{5} Pa$ , $V_{1} = 0.025 m^{3}$ par start karta hai aur adiabatically $V_{2} = 0.050 m^{3}$ tak expand karta hai. $P_{2}$ aur kiya gaya work nikalo.

Recall Solution

Kya/Kyun: adiabatic ⇒ $Q = 0$ ⇒ curve hai $P V^{γ} =$ const. Isse $P_{2}$ nikalo. $P_{2} = P_{1} (\frac{V _{1}}{V _{2}})^{γ} = 1.0 \times 1 0^{5} (\frac{1}{2})^{5/3} = 3.150 \times 1 0^{4} Pa .$ Work formula kahan se aata hai (WHY, authority nahi): kyunki $Q = 0$ hai, First Law kehta hai gas sirf internal energy kharch karke work kar sakti hai: $W = - Δ U = - n C_{V} (T_{2} - T_{1}) = n C_{V} (T_{1} - T_{2})$ . Ab $n C_{V}$ ko pressures aur volumes se trade karo. $C_{V} = R / (γ - 1)$ se milta hai $n C_{V} T = \frac{n R T}{γ - 1} = \frac{P V}{γ - 1}$ ( $P V = n R T$ use karke). Isliye $W = n C_{V} T_{1} - n C_{V} T_{2} = \frac{P _{1} V _{1}}{γ - 1} - \frac{P _{2} V _{2}}{γ - 1} = \frac{P _{1} V _{1} - P _{2} V _{2}}{γ - 1} .$ Yeh "state formula" hai — yeh bas $- Δ U$ rewrite hai. Plug in karo: $W = \frac{( 1.0 \times 1 0 ^{5} ) ( 0.025 ) - ( 3.150 \times 1 0 ^{4} ) ( 0.050 )}{\frac{5}{3} - 1} .$ Numerator $= 2500 - 1575 = 925$ ; denominator $= \frac{2}{3}$ ; toh $W = 1387.5 J$ . $W > 0$ (gas expands), aur $Δ U = - W = - 1387.5 J$ — gas thandi ho gayi.

L3.2

L3.1 wali hi start state, lekin ab same $V_{2} = 0.050 m^{3}$ tak isothermally expand karo. Final pressure $P_{2}^{'}$ nikalo aur adiabatic $P_{2}$ se compare karo. Kaun sa curve steeper hai? Figure dekho.

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

Recall Solution

Isotherm: $P V =$ const, isliye $P_{2}^{'} = P_{1} \frac{V _{1}}{V _{2}} = 1.0 \times 1 0^{5} \cdot \frac{1}{2} = 5.00 \times 1 0^{4} Pa$ . Compare karo: adiabatic ne $P_{2} = 3.15 \times 1 0^{4} Pa$ diya, jo kam hai. Kyun: dono curves same point se nikle, lekin adiabat $P V^{γ}$ follow karta hai jahan $γ = \frac{5}{3} > 1$ hai, isliye pressure zyada tezi se girta hai. Same expansion ke liye adiabat isotherm ke neeche land karta hai — yeh steeper hai. Figure padhna (words mein, agar load na ho): horizontal axis volume $V$ litres mein hai, vertical axis pressure $P$ $1 0^{5} Pa$ ki units mein hai. Dono curves white dot $(25 L, 1.0)$ par saath start hote hain aur right ki taraf neeche slope karte hain. Cyan curve gentle isotherm $P V =$ const hai, near $P = 0.5$ par khatam hoti hai. Amber curve steeper adiabat $P V^{γ} =$ const hai, jo tezi se girta hai aur neeche khatam hota hai, near $P = 0.315$ par — same volume par cyan curve ke neeche clearly dikhta hai. $V = 50 L$ par dono ke beech ka vertical gap poora point hai: adiabat zyada pressure kho deta hai kyunki use koi heat top-up nahi milta.

L3.3

Ek gas $Q = 800 J$ absorb karti hai jabki uski internal energy $Δ U = 500 J$ badhti hai. Gas ke dwara kiya gaya work nikalo, aur batao ki woh expand hui ya compress hui.

Recall Solution

First Law: $W = Q - Δ U = 800 - 500 = 300 J$ . $W > 0$ ⇒ work gas ke dwara kiya gaya ⇒ woh expand hui.

Level 4 — Synthesis

Legs ko ek poore cycle mein combine karo.

L4.1

$1 mol$ monatomic gas 3-leg cycle chalata hai (figure dekho):

A→B isochoric heating: $P$ $1.0 \times 1 0^{5}$ se $2.0 \times 1 0^{5} Pa$ tak badhta hai $V = 0.020 m^{3}$ par.
B→C isobaric expansion: $V$ $0.020$ se $0.040 m^{3}$ tak badhta hai $P = 2.0 \times 1 0^{5} Pa$ par.
C→A A tak seedhi line wapas.

Har leg ka $W$ aur cycle ka net work nikalo (= enclosed area).

Recall Solution

A→B isochoric: $d V = 0 \Rightarrow W_{A B} = 0$ . B→C isobaric: $W_{B C} = P Δ V = (2.0 \times 1 0^{5}) (0.040 - 0.020) = 4000 J$ . C→A straight line $(V = 0.040, P = 2.0 \times 1 0^{5})$ se $(V = 0.020, P = 1.0 \times 1 0^{5})$ tak. Work = is line ke neeche trapezoid ka area (volume ghatta hai, isliye $W_{C A} < 0$ ): $W_{C A} = - trapezoid \frac{1}{2} (P_{C} + P_{A}) (V_{C} - V_{A}) = - \frac{1}{2} (2.0 \times 1 0^{5} + 1.0 \times 1 0^{5}) (0.020) = - 3000 J .$ Net work = triangle A-B-C se enclosed area: $W_{net} = 0 + 4000 - 3000 = 1000 J .$ Triangle area se cross-check: base $= Δ V = 0.020$ , height $= Δ P = 1.0 \times 1 0^{5}$ , area $= \frac{1}{2} (0.020) (1.0 \times 1 0^{5}) = 1000 J$ . ✓ Positive ⇒ clockwise cycle ⇒ yeh engine ki tarah kaam karta hai. Figure padhna (words mein): axes hain volume $V$ (litres) horizontally aur pressure $P$ ( $1 0^{5} Pa$ units mein) vertically. Point A low-left par $(20, 1.0)$ par baitha hai, B seedha uske upar $(20, 2.0)$ par, aur C right mein $(40, 2.0)$ par. Cyan vertical segment A→B isochoric leg hai (koi width nahi ⇒ koi work nahi). Amber horizontal segment B→C isobaric expansion hai (uske neeche ka rectangle $+ 4000 J$ hai). White slanted segment C→A loop close karta hai. Loop ke andar lightly shaded triangle net work $1000 J$ hai — literally enclosed area. Loop clockwise chalta hai, ek engine ki pehchaan.

L4.2

Usi cycle ke liye, poore cycle par $Δ U$ nikalo, aur isse net heat $Q_{net}$ nikalo.

Recall Solution

$Δ U_{cycle} = 0$ kyun: internal energy $U$ ek state function hai — yeh sirf is baat par depend karta hai ki gas kahan hai (uska $P, V, T$ ), liye gaye path par nahi. Ek cycle bilkul usi state par khatam hota hai jahan se shuru hua, isliye $T$ apni starting value par wapas aata hai, deta hai $Δ T = 0$ . Phir $Δ U_{cycle} = n C_{V} Δ T = n C_{V} (0) = 0$ . Cycle par First Law: $Q_{net} = Δ U_{cycle} + W_{net} = 0 + 1000 = 1000 J$ . Net heat in equals net work out — yahi engine ka kaam hai.

Level 5 — Mastery

Real algebra, subtle traps.

L5.1

$2 mol$ diatomic gas ( $γ = \frac{7}{5}$ ) ko adiabatically $T_{1} = 300 K$ , $V_{1}$ se $V_{2} = V_{1} /4$ tak compress kiya jaata hai. $T_{2}$ aur gas par kiya gaya work nikalo.

Recall Solution

Yahan $n = 2$ hai. $T V^{γ - 1}$ kyun use karein: humein volumes aur ek temperature pata hai, pressures nahi — isliye adiabatic law ka $T$ – $V$ form choose karo. $T_{2} = T_{1} (\frac{V _{1}}{V _{2}})^{γ - 1} = 300 \cdot 4^{2/5} = 300 \cdot 1.7411 = 522.3 K .$ Gas ke dwara work (same $- Δ U$ reasoning L3.1 se): $W = n C_{V} (T_{1} - T_{2}) = 2 \cdot \frac{5}{2} (8.31) (300 - 522.3) = - 9239 J$ . Negative ⇒ work gas par kiya gaya. Gas par kiya gaya work $= + 9239 J$ . (Compression se garam hoti hai: $T$ $300$ se $522 K$ tak jump kiya.)

L5.2

Ek monatomic gas expand karti hai, $W = 1200 J$ work karti hai, jabki $600 J$ heat add ki jaati hai. Kya yeh expansion isothermal, adiabatic, ya kuch aur tha? Justify karo.

Recall Solution

$Δ U = Q - W = 600 - 1200 = - 600 J$ compute karo.

Isothermal ko $Δ U = 0$ chahiye hoga — aise nahi hai.
Adiabatic ko $Q = 0$ chahiye hoga — lekin $Q = 600 J \neq = 0$ hai. Isliye yeh kuch bhi nahi (neither) hai. Internal energy giri ( $Δ U < 0$ , gas thandi hui) even though heat add ki gayi, kyunki gas ne heat supply se zyada work kiya — extra internal energy se aaya.

L5.3

$(P_{1}, V_{1})$ se start karke, ek monatomic gas volume $2 V_{1}$ tak do tarike se pahunch sakti hai: (a) isothermally, (b) adiabatically. Kaun se case mein gas zyada work karti hai, aur kyun? Numbers $P_{1} = 1.0 \times 1 0^{5} Pa$ , $V_{1} = 0.030 m^{3}$ , $T_{1}$ from $n R T_{1} = P_{1} V_{1}$ ke saath dono work formulas se prove karo.

Recall Solution

Isothermal work: $W_{iso} = P_{1} V_{1} ln \frac{V _{2}}{V _{1}} = (1.0 \times 1 0^{5}) (0.030) ln 2 = 2079.4 J$ . ( $n R T_{1} = P_{1} V_{1} = 3000 J$ use kiya.) Adiabatic work (L3.1 se state formula, jo khud $- Δ U$ tha): $P_{2} = P_{1} (V_{1} / V_{2})^{γ} = 1.0 \times 1 0^{5} \cdot (1/2)^{5/3} = 3.150 \times 1 0^{4} Pa$ . $W_{adi} = \frac{P _{1} V _{1} - P _{2} V _{2}}{γ - 1} = \frac{3000 - ( 3.150 \times 1 0 ^{4} ) ( 0.060 )}{2/3} = \frac{3000 - 1890}{0.6667} = 1665 J .$ Isothermal zyada work karta hai ( $2079 > 1665 J$ ). Kyun: isothermal path mein heat continuously feed hoti hai taaki pressure poore raaste upar rahe, isliye gas har volume par zyada push karta hai. Adiabatic path mein koi heat top-up nahi hai, isliye pressure isotherm ke neeche sag jaata hai — kam force, kam area, kam work. (Yeh exactly wahi vertical gap hai jo tumne L3.2 figure mein dekha.)

Self-test recall

Kaun si quantity ek state function hai, isliye kisi bhi cycle mein uska change zero hota hai?

Internal energy

U

(isliye

Δ U_{cycle} = 0

, deta hai

Q_{net} = W_{net}

Jab

T

aur

V

diye hain toh kaun sa adiabatic form use karte ho?

T V^{γ - 1} = const

Same expansion ke liye, kaun zyada work karta hai — isothermal ya adiabatic?

Isothermal (heat top-up pressure ko path mein zyada upar rakhta hai).

Adiabatic state work formula

W = (P_{1} V_{1} - P_{2} V_{2}) / (γ - 1)

kahan se aata hai?

Yeh bas

W = - Δ U = n C_{V} (T_{1} - T_{2})

hai jo

n C_{V} T = P V / (γ - 1)

use karke rewrite kiya gaya hai.

Related: Carnot Cycle do isotherms + do adiabats se engine banata hai; Entropy in legs mein heat/temperature bookkeeping measure karta hai.