Heat and internal energy — microscopic vs macroscopic: Exercises

1.7.3 · D4 · Physics › Thermodynamics › Heat and internal energy — microscopic vs macroscopic

Yahan sab kuch parent note parent topic par based hai. Agar koi step unfamiliar lage, toh neeche linked prerequisites se rebuild karo.

Teen tools jo tum baar baar use karoge

Kisi bhi problem se pehle, sirf teen equations ko pin down karte hain jo poore page ko handle karti hain — har ek plain words mein stated hai taaki koi symbol use hone se pehle earn ho sake.

Neeche pressure–volume diagram woh map hai jis par hum baar baar point karte rahenge. Curve ke neeche area = gas dwara kiya gaya work.

Figure — Heat and internal energy — microscopic vs macroscopic

Har ek ko classify karo — stored energy (U), heat (Q), ya work (W): (a) Expanding gas dwara ek piston bahar dhakela jata hai. (b) Ek garam coffee mug thandi table par thanda hota hai. (c) Ek sealed flask mein saare molecules ki total kinetic energy.

Recall Solution 1.1

(a) Work $W$ — ek ordered macroscopic force (piston) ek distance ke through act karta hai. (b) Heat $Q$ — energy boundary cross karti hai temperature difference ( $Δ T$ ) ki wajah se. (c) Internal energy $U$ — energy andar stored, ek state function.

Problem 1.2

Ek gas ko constant volume par heat kiya jata hai. $Q$ , $W$ , $Δ U$ mein se kaun sa zero hai, aur kyun?

Recall Solution 1.2

$W = 0$ . Gas dwara work hai $W = P Δ V$ ; constant volume par $Δ V = 0$ , toh koi boundary move nahi karti aur koi work nahi hoti. Phir Tool 1 deta hai $Q = Δ U$ : saari heat stored internal energy ban jaati hai.

Problem 1.3

1 mole monatomic ideal gas $T = 300 K$ par rakha hai. $f$ batao aur $U$ compute karo.

Recall Solution 1.3

Monatomic $\Rightarrow f = 3$ (sirf teen translational directions). $U = \frac{3}{2} n R T = \frac{3}{2} (1) (8.314) (300) = 3741.3 J \approx 3.74 kJ .$

L2 — Application

Problem 2.1

Ek gas $Q = 500 J$ heat absorb karta hai aur surroundings par $W = 200 J$ work karta hai. $Δ U$ nikalo.

Recall Solution 2.1

Tool 1: $Δ U = Q - W = 500 - 200 = 300 J$ . Deposit kiye 500 J mein se 200 J work ke roop mein chale gaye; 300 J ruke rahe internal energy (aur temperature) badhane ke liye.

Problem 2.2

2 moles monatomic ideal gas ko constant volume par $290 K$ se $340 K$ tak heat kiya jata hai. $Δ U$ , $W$ , aur $Q$ nikalo.

Recall Solution 2.2

$Δ T = 340 - 290 = 50 K$ . $Δ U = \frac{3}{2} n R Δ T = \frac{3}{2} (2) (8.314) (50) = 1247.1 J .$ Constant volume $\Rightarrow W = 0$ . Tool 1: $Q = Δ U + W = 1247.1 + 0 = 1247.1 J \approx 1.25 kJ$ .

Problem 2.3

Wohi 2 moles ab constant pressure par same $Δ T = 50 K$ se heat kiye jaate hain. $Δ U$ , $W$ , aur $Q$ nikalo. $Q$ ko Problem 2.2 se compare karo aur difference explain karo.

Recall Solution 2.3

$Δ U$ sirf $T$ par depend karta hai, toh yeh same rahega: $Δ U = 1247.1 J$ . Constant pressure par work: $W = n R Δ T = (2) (8.314) (50) = 831.4 J$ . Tool 1: $Q = Δ U + W = 1247.1 + 831.4 = 2078.5 J$ . $Q$ zyada kyun hai: constant pressure par gas ko piston bhi bahar dhakkelna padta hai, isliye extra heat chahiye jo $U$ badhane ke saath-saath woh work bhi supply kare. Yeh extra chunk exactly $W = n R Δ T$ hai — isi liye $C_{P} = C_{V} + R$ .

L3 — Analysis

Problem 3.1

1 mole ideal gas isothermally $T = 400 K$ par volume $V_{i} = 2.0 L$ se $V_{f} = 6.0 L$ tak expand karta hai. $Δ U$ , $W$ , aur $Q$ nikalo.

Recall Solution 3.1

Isothermal ⇒ $T$ constant ⇒ $Δ U = \frac{f}{2} n R Δ T = 0$ . $W = n R T ln \frac{V _{f}}{V _{i}} = (1) (8.314) (400) ln \frac{6.0}{2.0} = 3325.6 ln 3 = 3653.9 J .$ Tool 1 with $Δ U = 0$ : $Q = W = 3653.9 J \approx 3.65 kJ$ . Interpretation: heat andar aati hai lekin internal energy kabhi nahi badhti. Microscopically, molecules apni kinetic energy exactly utni hi rate par receding piston ko work ke roop mein dete hain jitni heat replenish karti hai, isliye $⟨ v^{2} ⟩$ aur $T$ fixed rahte hain. Yeh sabse clean proof hai ki "heat in" $\neq =$ "internal energy up."

Problem 3.2

1 mole monatomic ideal gas adiabatically ( $Q = 0$ ) expand karta hai, $W = 900 J$ work karta hai. Iski initial temperature $500 K$ hai. $Δ U$ aur final temperature $T_{f}$ nikalo.

Recall Solution 3.2

Adiabatic ⇒ $Q = 0$ . Tool 1: $Δ U = Q - W = 0 - 900 = - 900 J$ . Gas ne work apni stored energy se ki, toh $U$ giraa aur woh thanda hua. $Δ U = \frac{3}{2} n R Δ T \Rightarrow Δ T = \frac{Δ U}{\frac{3}{2} n R} = \frac{- 900}{\frac{3}{2} ( 1 ) ( 8.314 )} = \frac{- 900}{12.471} = - 72.17 K .$ $T_{f} = 500 - 72.17 = 427.83 K .$

Problem 3.3

Ek gas ko state A se state B tak do alag paths se le jaaya jata hai (figure dekho). Path 1 par woh $Q_{1} = 800 J$ absorb karta hai aur $W_{1} = 300 J$ work karta hai. Path 2 par woh $W_{2} = 500 J$ work karta hai. Path 2 par woh kitna heat $Q_{2}$ absorb karta hai?

Recall Solution 3.3

$Δ U$ ek state function hai — yeh sirf endpoints A aur B par depend karta hai, isliye dono paths par same hai. Path 1: $Δ U = Q_{1} - W_{1} = 800 - 300 = 500 J$ . Path 2 mein bhi same $Δ U = 500 J$ aana chahiye: $Q_{2} = Δ U + W_{2} = 500 + 500 = 1000 J .$

L4 — Synthesis

Problem 4.1

Ek sealed rigid tank mein $N = 3.0 \times 1 0^{22}$ helium atoms (monatomic) $T = 350 K$ par hain. (a) Ek atom ki average translational kinetic energy nikalo. (b) Gas ki total internal energy $U$ nikalo. (c) Gas ko constant volume par $420 K$ tak heat kiya jata hai. Kitni heat supply ki gayi?

Recall Solution 4.1

(a) Parent ke key bridge $\frac{1}{2} m ⟨ v^{2} ⟩ = \frac{3}{2} k_{B} T$ se: $⟨ KE ⟩ = \frac{3}{2} k_{B} T = \frac{3}{2} (1.381 \times 1 0^{- 23}) (350) = 7.250 \times 1 0^{- 21} J .$ (b) $U = N ⟨ KE ⟩ = (3.0 \times 1 0^{22}) (7.250 \times 1 0^{- 21}) = 217.5 J .$ (c) Rigid tank ⇒ $W = 0$ , toh $Q = Δ U$ . $Δ U = \frac{3}{2} N k_{B} Δ T$ use karo jahan $Δ T = 70 K$ : $Q = \frac{3}{2} (3.0 \times 1 0^{22}) (1.381 \times 1 0^{- 23}) (70) = 43.50 J .$

Problem 4.2

1 mole diatomic ideal gas ( $f = 5$ at room temperature) ko constant volume par $300 K$ se $350 K$ tak heat kiya jata hai. Ek dost claim karta hai ki is range mein required heat monatomic gas ke barabar hai. Dono compute karo aur argument settle karo.

Recall Solution 4.2

Constant volume ⇒ $Q = Δ U = \frac{f}{2} n R Δ T$ , jahan $Δ T = 50 K$ . Monatomic ( $f = 3$ ): $Q_{mono} = \frac{3}{2} (1) (8.314) (50) = 623.55 J .$ Diatomic ( $f = 5$ ): $Q_{di} = \frac{5}{2} (1) (8.314) (50) = 1039.25 J .$ Dost galat hai. Diatomic gas mein extra "storage bins" hain (do rotational degrees of freedom), aur equipartition se har bin $\frac{1}{2} k_{B} T$ per molecule absorb karta hai. Zyada bins ⇒ same $Δ T$ ke liye zyada heat chahiye. Dekho Kinetic Theory of Gases aur Degrees of Freedom and Molecular Structure.

L5 — Mastery

Problem 5.1 — Ek full cycle

1 mole monatomic ideal gas ek closed cycle A→B→C→A mein drive kiya jata hai:

A→B: constant volume par heat diya jata hai, $T$ jaata hai $300 \to 600 K$ .
B→C: $600 K$ par isothermally expand hota hai, volume triple hota hai ( $V_{C} = 3 V_{B}$ ).
C→A: constant pressure par wapas start tak compress hota hai.

Har leg ke liye $Δ U$ , $W$ , $Q$ nikalo, aur verify karo ki poore cycle mein $Δ U_{cycle} = 0$ aur $Q_{cycle} = W_{cycle}$ .

Recall Solution 5.1

Leg A→B (constant volume, $Δ T = 300 K$ ): $W_{A B} = 0$ (koi volume change nahi). $Δ U_{A B} = \frac{3}{2} n R Δ T = \frac{3}{2} (1) (8.314) (300) = 3741.3 J .$ $Q_{A B} = Δ U_{A B} = 3741.3 J .$

Leg B→C (600 K par isothermal, $V_{C} / V_{B} = 3$ ): $Δ U_{B C} = 0$ (temperature constant). $W_{B C} = n R T ln \frac{V _{C}}{V _{B}} = (1) (8.314) (600) ln 3 = 4988.4 ln 3 = 5480.8 J .$ $Q_{B C} = W_{B C} = 5480.8 J .$

Leg C→A (constant pressure, $T$ jaata hai $600 \to 300 K$ , toh $Δ T = - 300 K$ ): $W_{C A} = n R Δ T = (1) (8.314) (- 300) = - 2494.2 J$ (gas par work kiya jata hai kyunki compress ho raha hai). $Δ U_{C A} = \frac{3}{2} n R Δ T = \frac{3}{2} (1) (8.314) (- 300) = - 3741.3 J .$ $Q_{C A} = Δ U_{C A} + W_{C A} = - 3741.3 + (- 2494.2) = - 6235.5 J$ (heat reject hoti hai).

Whole cycle checks: $Δ U_{cycle} = 3741.3 + 0 + (- 3741.3) = 0 ✓$ ( $U$ ek state function hai, aur cycle start par wapas aati hai). $W_{cycle} = 0 + 5480.8 + (- 2494.2) = 2986.6 J .$ $Q_{cycle} = 3741.3 + 5480.8 + (- 6235.5) = 2986.6 J .$ Toh $Q_{cycle} = W_{cycle} = 2986.6 J ✓$ — kisi bhi full cycle mein, net heat in equals net work out kyunki internal energy wahan wapas aati hai jahan se shuru hui thi.

Problem 5.2 — Cycle ki thermal efficiency

Problem 5.1 ke cycle ke liye, efficiency define karo $η = \frac{W _{cycle}}{Q _{in}}$ , jahan $Q_{in}$ total heat hai jo absorb hui (sirf positive- $Q$ legs). $η$ compute karo.

Recall Solution 5.2

Heat A→B aur B→C par absorb hoti hai (dono $Q > 0$ ); C→A par reject hoti hai ( $Q < 0$ ). $Q_{in} = Q_{A B} + Q_{B C} = 3741.3 + 5480.8 = 9222.1 J .$ $η = \frac{W _{cycle}}{Q _{in}} = \frac{2986.6}{9222.1} = 0.3238 = 32.38%.$ Har 100 J heat mein se jo feed ki gayi, lagbhag 32 J useful net work bante hain; baaki compression leg par thande surroundings mein dump ho jaate hain.

Recall Ek-line takeaways

$Δ U$ sirf endpoints par depend karta hai (state function) ::: do states ke beech har path par same hota hai. Constant volume ka matlab hai ::: $W = 0$ , toh $Q = Δ U$ . Isothermal ideal gas ka matlab hai ::: $Δ U = 0$ , toh $Q = W$ . Adiabatic ka matlab hai ::: $Q = 0$ , toh $Δ U = - W$ (gas expand hone par thandi hoti hai). Cycle efficiency hai ::: net work out divided by heat absorbed, net heat nahi.

Connections

Heat and Internal Energy — Microscopic vs Macroscopic (parent)
First Law of Thermodynamics
Kinetic Theory of Gases
Equipartition Theorem
Specific Heats Cv and Cp
Isothermal and Adiabatic Processes
Temperature and the Zeroth Law
Degrees of Freedom and Molecular Structure

Exercises — Heat and internal energy — microscopic vs macroscopic