Work done in each process — derivation: Exercises

1.7.15 · D4 · Physics › Thermodynamics › Work done in each process — derivation

Poore note mein use hone waale constants: gas constant $R = 8.314 J mol^{- 1} K^{- 1}$ , aur $ln 2 = 0.6931$ , $ln 3 = 1.0986$ .

Level 1 — Recognition

Goal: sahi formula choose karo aur plug in karo. Algebra mein koi trap nahi, sirf choose karne mein hai.

Yeh chaar tools hain jinke beech tumhe choose karna hai:

Q1.1 (L1). Ek gas $2.0 \times 1 0^{5} Pa$ ke constant pressure pe $V_{1} = 1.0 \times 1 0^{- 3} m^{3}$ se $V_{2} = 3.0 \times 1 0^{- 3} m^{3}$ tak expand hoti hai. Gas ke dwara kiya gaya work nikalo.

Recall Solution

Kaunsa tool? Pressure constant hai → isobaric → $W = P (V_{2} - V_{1})$ . $W = 2.0 \times 1 0^{5} \times (3.0 \times 1 0^{- 3} - 1.0 \times 1 0^{- 3}) = 2.0 \times 1 0^{5} \times 2.0 \times 1 0^{- 3}$ $W = 400 J .$ Yeh kaisa dikhta hai: P–V diagram pe yeh ek flat horizontal line hai; uske neeche ka area ek rectangle hai jiska height $P$ aur width $Δ V$ hai. Positive hai kyunki gas expand hui ( $V_{2} > V_{1}$ ).

Q1.2 (L1). Ek gas ko ek rigid sealed steel tank ke andar $300 K$ se $500 K$ tak garam kiya jaata hai. Gas ke dwara kiya gaya work nikalo.

Recall Solution

Kaunsa tool? "Rigid tank" matlab volume change nahi ho sakta → isochoric → $d V = 0$ . $W = \int_{V_{1}}^{V_{1}} P d V = 0 J .$ Piston (walls) kabhi move nahi karta, isliye koi distance push nahi hoti — koi work nahi, chahe temperature ya pressure kitni bhi badhey.

Q1.3 (L1). $1$ mol ideal gas $T = 400 K$ pe isothermally expand hoti hai jab tak uska volume teen guna nahi ho jaata. $W$ nikalo.

Recall Solution

Kaunsa tool? Temperature constant → isothermal → $W = n R T ln (V_{2} / V_{1})$ jahan $V_{2} / V_{1} = 3$ . $W = 1 \times 8.314 \times 400 \times ln 3 = 3325.6 \times 1.0986 = 3653.4 J .$ Positive: expansion.

Level 2 — Application

Goal: units convert karo, $n R T$ nikalo, plug in karne se pehle $P V = n R T$ use karo.

Q2.1 (L2). $2$ mol gas ko constant pressure pe $300 K$ se $450 K$ tak garam kiya jaata hai. Temperature form use karke $W$ nikalo.

Recall Solution

Isobaric hai, isliye $W = n R Δ T$ (yeh wahi rectangle hai jaise $P Δ V$ , bas ideal gas law $P V = n R T$ se rewrite kiya gaya hai). $W = 2 \times 8.314 \times (450 - 300) = 2 \times 8.314 \times 150 = 2494.2 J .$

Q2.2 (L2). $0.5$ mol gas $350 K$ pe isothermally $4.0 L$ se $10.0 L$ tak expand hoti hai. $W$ nikalo.

Recall Solution

Kaunsa tool? Isothermal → $W = n R T ln (V_{2} / V_{1})$ . Litres ratio ke andar cancel ho jaate hain, isliye wahan koi unit conversion zaruri nahi. $\frac{V _{2}}{V _{1}} = \frac{10.0}{4.0} = 2.5, ln 2.5 = 0.9163.$ $W = 0.5 \times 8.314 \times 350 \times 0.9163 = 1454.95 \times 0.9163 = 1333.2 J .$

Q2.3 (L2). $1$ mol monatomic gas ( $γ = 5/3$ ) adiabatically expand hoti hai. Uska temperature $500 K$ se $350 K$ tak girta hai. $W$ nikalo.

Recall Solution

Kaunsa tool? $Q = 0$ → adiabatic → $W = \frac{n R ( T _{1} - T _{2} )}{γ - 1}$ . $γ - 1 = \frac{5}{3} - 1 = \frac{2}{3} = 0.6667.$ $W = \frac{1 \times 8.314 \times ( 500 - 350 )}{0.6667} = \frac{1247.1}{0.6667} = 1870.7 J .$ Positive kyunki expand hua ( $T_{1} > T_{2}$ ). Adiabatic work poora gas ki apni internal energy se aaya, isliye woh thanda ho gaya.

Level 3 — Analysis

Goal: first law ke saath combine karo, signs padho, ulta kaam karo.

Q3.1 (L3). Ek isothermal compression ke dauran, $300 K$ pe $1$ mol gas ko $6.0 L$ se $2.0 L$ tak squeeze kiya jaata hai. (a) Gas ke dwara kiya gaya work nikalo. (b) First Law of Thermodynamics use karke heat $Q$ nikalo.

Recall Solution

(a) Isothermal → $W = n R T ln (V_{2} / V_{1})$ jahan $V_{2} / V_{1} = 2/6 = 1/3$ . $W = 1 \times 8.314 \times 300 \times ln (\frac{1}{3}) = 2494.2 \times (- 1.0986) = - 2740.1 J .$ Negative — gas compress ho rahi hai, isliye gas dwara kiya gaya work negative hai (work gas pe kiya ja raha hai). (b) Isothermal $\Rightarrow Δ T = 0 \Rightarrow Δ U = 0$ . First law: $Δ U = Q - W$ , isliye $Q = W = - 2740.1 J$ . $Q < 0$ : $T$ ko fixed rakhne ke liye gas se heat bahar jaani chahiye jab use compress kiya ja raha ho.

Q3.2 (L3). Ek gas adiabatic expansion mein $500 J$ work karta hai. Uski internal energy $Δ U$ mein kya change aata hai? Kya woh garam hoga ya thanda?

Recall Solution

Adiabatic matlab $Q = 0$ . First law: $Δ U = Q - W = 0 - 500 = - 500 J$ . Internal energy $500 J$ girri, isliye gas thanda hota hai. Saara work internal energy se pay kiya gaya — koi heat andar nahi aayi usse refill karne ke liye.

Q3.3 (L3). Neeche diye figure ko dekho. Ek gas state $A$ $(P = 3 \times 1 0^{5} Pa, V = 2 L)$ se state $B$ $(P = 1 \times 1 0^{5} Pa, V = 6 L)$ tak P–V diagram pe straight line ke along jaati hai. Kiya gaya work nikalo.

Figure — Work done in each process — derivation

Recall Solution

Idea: $W$ = curve ke neeche ka area. Path straight line hai, isliye uske neeche ka area ek trapezium hai (ek rectangle plus ek triangle), jo neeche $V$ -axis se bounded hai. Litres ko $m^{3}$ mein convert karo: $2 L = 2 \times 1 0^{- 3} m^{3}$ , $6 L = 6 \times 1 0^{- 3} m^{3}$ . Trapezium area = average height × width: $W = \frac{P _{A} + P _{B}}{2} (V_{B} - V_{A}) = \frac{( 3 + 1 ) \times 1 0 ^{5}}{2} \times (6 - 2) \times 1 0^{- 3}$ $W = 2 \times 1 0^{5} \times 4 \times 1 0^{- 3} = 800 J .$ Positive: path rightward move karta hai ( $V$ badhta hai). Figure mein shaded region dekho — wahi $800 J$ hai.

Level 4 — Synthesis

Goal: kai processes ko chain karo; sahi signs ke saath works add karo.

Q4.1 (L4). Ek gas teen steps se guzarti hai:

$P = 2 \times 1 0^{5} Pa$ pe isobaric expansion $1 L$ se $3 L$ tak.
$V = 3 L$ pe isochoric cooling (pressure $1 \times 1 0^{5} Pa$ tak girta hai).
$P = 1 \times 1 0^{5} Pa$ pe isobaric compression wapas $3 L$ se $1 L$ tak.

Teeno steps mein gas ke dwara kiya gaya total work nikalo.

Recall Solution

Har step ko apne tool se karo, phir add karo. Step 1 (isobaric expand): $W_{1} = P Δ V = 2 \times 1 0^{5} \times (3 - 1) \times 1 0^{- 3} = + 400 J$ . Step 2 (isochoric): $d V = 0 \Rightarrow W_{2} = 0 J$ . Step 3 (isobaric compress): $W_{3} = P Δ V = 1 \times 1 0^{5} \times (1 - 3) \times 1 0^{- 3} = - 200 J$ (negative — volume shrinks). $W_{total} = 400 + 0 - 200 = + 200 J .$ Steps abhi loop close nahi karte (end pe pressure $1 \times 1 0^{5}$ hai, start mein $2 \times 1 0^{5}$ tha), isliye yeh ek open path hai, cycle nahi.

Q4.2 (L4). P–V diagram pe ek complete rectangular cycle (figure dekho) clockwise chalti hai: $A \to B \to C \to D \to A$ jahan corners hain $A (1 \times 1 0^{5} Pa, 2 L)$ , $B (3 \times 1 0^{5} Pa, 2 L)$ , $C (3 \times 1 0^{5} Pa, 5 L)$ , $D (1 \times 1 0^{5} Pa, 5 L)$ . Ek cycle mein net work nikalo.

Recall Solution

Idea: closed loop ke liye, $W_{net} = \oint P d V$ = enclosed area. Clockwise loop positive net work deta hai. Rectangle ka:

height $Δ P = (3 - 1) \times 1 0^{5} = 2 \times 1 0^{5} Pa$ ,
width $Δ V = (5 - 2) \times 1 0^{- 3} = 3 \times 1 0^{- 3} m^{3}$ . $W_{net} = Δ P \times Δ V = 2 \times 1 0^{5} \times 3 \times 1 0^{- 3} = 600 J .$ Sides add karke check karo (sirf do horizontal legs volume move karte hain):
$B \to C$ (isobaric expand at $3 \times 1 0^{5}$ ): $+ 3 \times 1 0^{5} \times 3 \times 1 0^{- 3} = + 900 J$
$D \to A$ (isobaric compress at $1 \times 1 0^{5}$ ): $- 1 \times 1 0^{5} \times 3 \times 1 0^{- 3} = - 300 J$
vertical legs $A \to B$ , $C \to D$ : $Δ V = 0 \Rightarrow 0 J$ Total $= 900 - 300 = 600 J$ . ✓ Enclosed area se match karta hai.

Level 5 — Mastery

Goal: poori derivations, mixed processes, diye gaye work se ulta kaam karo.

Q5.1 (L5). $1$ mol diatomic gas ( $γ = 7/5$ ) $P_{1} = 4 \times 1 0^{5} Pa$ , $V_{1} = 2 \times 1 0^{- 3} m^{3}$ se shuru hoti hai aur adiabatically $V_{2} = 8 \times 1 0^{- 3} m^{3}$ tak expand hoti hai. (a) Final pressure $P_{2}$ , aur (b) kiya gaya work nikalo.

Recall Solution

(a) Adiabatic law: $P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ} \Rightarrow P_{2} = P_{1} (\frac{V _{1}}{V _{2}})^{γ}$ . $\frac{V _{1}}{V _{2}} = \frac{2}{8} = \frac{1}{4}, γ = 1.4.$ $(\frac{1}{4})^{1.4} = 4^{- 1.4} = e^{- 1.4 l n 4} = e^{- 1.4 \times 1.3863} = e^{- 1.9409} = 0.14359.$ $P_{2} = 4 \times 1 0^{5} \times 0.14359 = 5.744 \times 1 0^{4} Pa .$ (b) Adiabatic work $W = \frac{P _{1} V _{1} - P _{2} V _{2}}{γ - 1}$ . $P_{1} V_{1} = 4 \times 1 0^{5} \times 2 \times 1 0^{- 3} = 800 J .$ $P_{2} V_{2} = 5.744 \times 1 0^{4} \times 8 \times 1 0^{- 3} = 459.5 J .$ $W = \frac{800 - 459.5}{1.4 - 1} = \frac{340.5}{0.4} = 851.3 J .$ Positive: expansion. Gas thanda hua kyunki is work ne uski internal energy drain kar li.

Q5.2 (L5). $T = 320 K$ pe $2$ mol ka isothermal expansion $W = 3000 J$ work karta hai. Ratio $V_{2} / V_{1}$ nikalo.

Recall Solution

$W = n R T ln (V_{2} / V_{1})$ se ulta kaam karo. Pehle log solve karo. $ln \frac{V _{2}}{V _{1}} = \frac{W}{n R T} = \frac{3000}{2 \times 8.314 \times 320} = \frac{3000}{5320.96} = 0.56380.$ $ln$ undo karne ke liye exponentiate karo: $\frac{V _{2}}{V _{1}} = e^{0.56380} = 1.757.$ Toh gas apne starting volume ka approximately $1.76 \times$ tak expand hoti hai.

Q5.3 (L5). Isothermal work formula scratch se derive karo aur phir $1$ mol ke liye $300 K$ pe $V_{1}$ se $V_{2} = 2 V_{1}$ tak expansion ke liye evaluate karo. Confirm karo ki tumhara number parent-note ke example se match karta hai.

Recall Solution

Derivation (WHAT / WHY / WHAT IT LOOKS LIKE):

Master formula $W = \int_{V_{1}}^{V_{2}} P d V$ se shuru karo — kyunki work P–V curve ke neeche ka area hai.
$P$ constant nahi hai, isliye ideal gas law $P V = n R T \Rightarrow P = \frac{n R T}{V}$ substitute karo — yeh $P$ ko $V$ ki ek known function mein badal deta hai taaki hum actually integrate kar sakein.
Kyunki $T$ constant hai, $n R T$ ek fixed number hai; use bahar nikalo: $W = n R T \int_{V_{1}}^{V_{2}} \frac{d V}{V} .$
Integral $\int \frac{d V}{V} = ln V$ — yeh woh ek function hai jiska slope $1/ V$ hai, isliye $1/ V$ curve log-shaped area deta hai. $W = n R T [ln V]_{V_{1}}^{V_{2}} = n R T ln \frac{V _{2}}{V _{1}} .$ Evaluate $V_{2} / V_{1} = 2$ ke saath: $W = 1 \times 8.314 \times 300 \times ln 2 = 2494.2 \times 0.6931 = 1728.8 J .$ Yeh parent note ke isothermal example se exactly match karta hai. ✓

Recall Self-test summary (daayein side cover karo)

Constant pressure ke liye tool ::: $W = P (V_{2} - V_{1}) = n R Δ T$ Constant volume ke liye tool ::: $W = 0$ Constant temperature ke liye tool ::: $W = n R T ln (V_{2} / V_{1})$ $Q = 0$ (adiabatic) ke liye tool ::: $W = \frac{n R ( T _{1} - T _{2} )}{γ - 1} = \frac{P _{1} V _{1} - P _{2} V _{2}}{γ - 1}$ Ek cycle mein net work ::: enclosed area; clockwise positive Isothermal $W$ se $V_{2} / V_{1}$ recover karna ::: $V_{2} / V_{1} = e^{W / (n R T)}$

Connections

Parent: Work done in each process — woh master formula jo yeh exercises drill karti hain
First Law of Thermodynamics — Q3.1, Q3.2 mein $Q$ aur $Δ U$ nikalne ke liye use hua
P-V Diagrams — Q3.3, Q4.2 pure "curve ke neeche / andar ka area" hain
Adiabatic Process and PV^gamma — Q2.3, Q5.1
Isothermal Process and Boyle's Law — Q3.1, Q5.2, Q5.3
Internal Energy and Cv — isliye adiabatic work gas ko thanda karta hai
Carnot Cycle — Q4.2 ka cyclic-area idea, aur aage liya gaya