Internal energy of ideal gas U = (f - 2)nRT: Exercises

1.7.10 · D4 · Physics › Thermodynamics › Internal energy of ideal gas U = (f - 2)nRT

Definition Is page par use hone wale symbols

$U$ = internal energy — gas molecules ki total random kinetic energy (joules mein).
$n$ = number of moles of gas. Ek mole matlab $6.022 \times 1 0^{23}$ molecules; $n$ count karta hai ki aapke paas aise kitne packets of molecules hain. Yeh ek pure count hai, "mol" ke alaawa koi units nahi.
$f$ = har molecule ke active degrees of freedom ki sankhya — wo independent tarike jinse ek molecule energy store karta hai (dekho Degrees of freedom).
$T$ = absolute temperature (kelvin); $R$ = gas constant.
$Q$ = heat — wo energy jo temperature difference ki wajah se bahar se gas mein flow karti hai (joules). $Q > 0$ matlab gas absorb karti hai heat.
$W$ = work done by the gas — wo energy jo gas apne surroundings ko piston bahar dhakelte waqt deta hai (joules). $W > 0$ matlab gas bahari duniya par energy kharach karti hai.
$P$ = pressure — gas dwara apne container walls par exert ki gayi force per unit area (units: pascals, $Pa = N m^{- 2}$ ).
$V$ = volume — wo space jo gas occupy karti hai (units: cubic metres, $m^{3}$ ).
$k_{B}$ = Boltzmann constant, $k_{B} = 1.381 \times 1 0^{- 23} J K^{- 1}$ — single molecule ke liye energy-per-kelvin bookkeeping constant (jabki $R$ wahi kaam per mole karta hai; actually $R = N_{A} k_{B}$ ).

Ye sab First law of thermodynamics se jude hain: $Δ U = Q - W$ — "internal energy mein change equals heat in minus work out."

Neeche har jagah jo key result exercise hota hai woh hai $U = \frac{f}{2} n R T$

Yahan sab kuch parent result Internal energy of an ideal gas, U = (f/2)nRT aur uske supporting ideas par tika hai: Degrees of freedom, Equipartition theorem, Kinetic theory of gases, First law of thermodynamics, Ideal gas law, aur Molar specific heats Cv and Cp.

Shuru karne se pehle, yeh ek master picture hai jis par poora page tika hai.

Figure — Internal energy of ideal gas U = (f - 2)nRT

Figure padhna: horizontal axis teen gas types ke naam deta hai; vertical axis per mole internal energy hai, $R T$ ke units mein measured, yaani uski height $\frac{f}{2} R T$ hai. Monatomic gas ( $f = 3$ ) chhota burnt-orange bar hai; diatomic gas ( $f = 5$ ) thoda lamba teal bar hai; nonlinear polyatomic gas ( $f = 6$ ) sabse lamba plum bar hai. Storage modes ka ek pair add karne par ( $f$ 2 se badhta hai) sirf energy ka ek aur slab stack ho jaata hai. Yeh picture dimag mein rakho — zyaadatar problems bas "bar padho" ya "do bars compare karo" wali hain.

Level 1 — Recognition

L1·Q1 — Formula se seedha padhna

Recall Solution

KYA chahiye: formula $U = \frac{f}{2} n R T$ with $n = 1$ mole. KYUN $f = 3$ : ek single atom ek point mass hai, isliye wo sirf teen independent directions ( $x, y, z$ ) mein move kar sakta hai — teen translational Degrees of freedom, koi rotation nahi, koi vibration nahi. $U = \frac{f}{2} n R T = \frac{3}{2} (1) (8.314) (300)$ $U = 1.5 \times 8.314 \times 300 \approx 3741 J \approx 3.74 kJ$

L1·Q2 — $f$ identify karna

Recall Solution

(a) Argon monatomic hai. KYUN $f = 3$ : akela atom essentially ek point hai, isliye energy store karne ka ek hi tarika hai — teen space directions $x, y, z$ mein move karna → $f = 3$ .
(b) $O_{2}$ diatomic hai (ek dumbbell). KYUN $f = 5$ : poore molecule ko move karne ke liye 3 translational modes, plus bond ke perpendicular do axes ke baare mein tumbling ke liye 2 rotational modes. Bond axis ke baare mein spinning count nahi hoti — neeche note dekho kyun.
(c) $H_{2} O$ ek nonlinear polyatomic molecule hai. KYUN $f = 6$ : 3 translational, plus 3 rotational, kyunki ek bent (non-linear) molecule ke teeno axes ke baare mein real, comparable moments of inertia hote hain, isliye teeno tumbling modes energy store karti hain.

Diatomic molecule ki "frozen" teesri rotation ke baare mein: sachchi wajah quantum-mechanical hai, sirf "small moment of inertia" nahi. Rotational energy discrete steps mein aati hai, aur ek axis ke baare mein moment of inertia jitna chhota hoga, pehli excited rotational state tak energy gap utna bada hoga. Bond axis ke baare mein spin karne mein woh gap bahut bada hota hai — moderate temperature par available thermal energy $k_{B} T$ se kaafi zyada (yaad karo $k_{B} = 1.381 \times 1 0^{- 23} J K^{- 1}$ per-molecule energy-per-kelvin constant hai jo symbol list mein define hua hai) — isliye collisions usay excite karne ke liye enough energy supply nahi kar sakti. Woh mode isliye "frozen out" ho jaata hai aur kuch contribute nahi karta, aur $f = 5$ bach jaata hai. (Classical "negligible moment of inertia" phrasing iske liye ek shorthand hai; literally liya jaaye toh misleading hoga.)

Level 2 — Application

L2·Q1 — Internal energy mein change

Recall Solution

KYA: kyunki $U$ sirf $T$ par depend karta hai, change hai $Δ U = \frac{f}{2} n R Δ T$ with $n = 2$ moles. KYUN temperatures subtract karte hain: $U (T)$ origin se ek straight line hai, isliye $U (T_{2}) - U (T_{1})$ sirf $Δ T = 450 - 250 = 200 K$ par depend karta hai. $Δ U = \frac{5}{2} (2) (8.314) (200) = 5 \times 8.314 \times 200 \approx 8314 J \approx 8.31 kJ$

L2·Q2 — Do gases, same energy

Recall Solution

KYA: har gas ke liye $U = \frac{f}{2} n R T$ likho aur equal set karo. KYUN equal set karte hain: question same internal energy maangta hai, isliye dono expressions numerically identical hone chahiye. KYUN $R T$ cancel hota hai: dono gases ka same $T$ hai, aur $R$ ek universal constant hai, isliye common factor $R T$ dono sides se divide ho jaata hai, sirf mole– $f$ balance bachi. $\frac{3}{2} n_{He} R T = \frac{5}{2} (3) R T \Rightarrow 3 n_{He} = 15 \Rightarrow n_{He} = 5 mol$ Bar picture dekho: helium ka bar chhota hai (per mole), isliye same total energy tak pahunchne ke liye zyaada moles chahiye — $5$ vs $3$ , exactly do $f$ values ka ratio $5/3$ .

L2·Q3 — Isothermal expansion

Recall Solution

KYA: $Δ U = \frac{f}{2} n R Δ T$ . KYUN zero hai: isothermal matlab $Δ T = 0$ , isliye $Δ U = 0$ chahe volume $V$ ya pressure $P$ kitna bhi change ho. $Δ U = \frac{f}{2} n R (0) = 0$ Gas phir bhi work $W$ karti hai aur heat $Q$ absorb karti hai — lekin First law of thermodynamics ( $Δ U = Q - W$ ) se, $Δ U = 0$ force karta hai $Q = W$ , aur internal energy kabhi nahi badlti.

Level 3 — Analysis

L3·Q1 — Same heat, different processes

Recall Solution

$Δ T = 60 K$ , $n = 1$ mole.

Internal energy — dono paths ke liye same. KYUN path matter nahi karta: master formula $U = \frac{f}{2} n R T$ dikhata hai ki $U$ sirf $T$ par depend karta hai, isliye uska change $Δ U = \frac{f}{2} n R Δ T$ sirf $Δ T = 60 K$ se fixed hai — chahe hum volume hold karein ya pressure, $Δ U$ ke liye irrelevant hai. $Δ U = \frac{3}{2} (1) (8.314) (60) = 1.5 \times 8.314 \times 60 \approx 748.3 J$

(a) Constant volume: koi work nahi hota ( $W = 0$ ), isliye First law of thermodynamics se $Q = Δ U$ : $Q_{V} = n C_{v} Δ T = Δ U \approx 748.3 J$

(b) Constant pressure: ab gas piston ko bahar bhi dhakelta hai, work $W = n R Δ T$ karta hai, isliye zyaada heat chahiye. $C_{p} = C_{v} + R = \frac{3}{2} R + R = \frac{5}{2} R$ use karte hue: $Q_{P} = n C_{p} Δ T = \frac{5}{2} (1) (8.314) (60) = 2.5 \times 8.314 \times 60 \approx 1247.1 J$

Lesson: $Δ U$ dono baar identical hai ( $\approx 748 J$ ) dono baar, lekin $Q$ alag hai kyunki constant pressure par heat ka kuch hissa work ban jaata hai. Dekho Molar specific heats Cv and Cp.

L3·Q2 — Energy se temperature nikalna

Recall Solution

$U = \frac{f}{2} n R T$ ko $T$ ke liye rearrange karo (yahan $n = 0.5$ mole): $T = \frac{2 U}{f n R} = \frac{2 ( 3120 )}{5 ( 0.5 ) ( 8.314 )} = \frac{6240}{20.785} \approx 300.2 K$

Level 4 — Synthesis

L4·Q1 — Modes freeze out hona

Recall Solution

KYUN vibration do DOF add karta hai: vibrating bond energy store karta hai — stretching ki kinetic energy ke roop mein bhi aur "spring" mein potential energy ke roop mein bhi. Equipartition theorem ke anusaar har quadratic term ko $\frac{1}{2} k_{B} T$ milta hai, isliye do terms → $f$ 2 se badhta hai. $U_{f = 5} = \frac{5}{2} (2) (8.314) (2000) = 5 \times 8.314 \times 2000 = 83140 J$ $U_{f = 7} = \frac{7}{2} (2) (8.314) (2000) = 7 \times 8.314 \times 2000 = 116396 J$ $Δ U = U_{f = 7} - U_{f = 5} = 116396 - 83140 = 33256 J \approx 33.3 kJ$ Equivalently, DOF ke har extra pair se $n R T = 2 \times 8.314 \times 2000 = 33256 J$ add hota hai — bar chart par ek slab.

L4·Q2 — Gases ka mixture

Recall Solution

KYA: har gas ki internal energy uske apne $f$ aur apne moles $n$ se calculate karo, phir add karo. KYUN alag-alag add karte hain: internal energy ek total hai — har molecule independently contribute karta hai, isliye helium molecules ( $3$ mol, $f = 3$ ) aur nitrogen molecules ( $2$ mol, $f = 5$ ) apna-apna share carry karte hain; koi single $f$ nahi hai jo dono species ek saath describe kare. KYUN $R T$ factors out hota hai: dono gases same $T$ par hain, isliye $R T$ common hai aur sum se bahar aa sakta hai. $U = [\frac{3}{2} (3) + \frac{5}{2} (2)] R T = [4.5 + 5] (8.314) (400)$ $U = 9.5 \times 8.314 \times 400 \approx 31593 J \approx 31.6 kJ$ Note: yahan effective $f$ hai $\frac{9.5 \times 2}{5} = 3.8$ , 3 aur 5 ke beech, moles se weighted — lekin aapko yeh value pehle se kabhi assume nahi karni chahiye; yeh sirf species add karne ke baad milta hai.

Level 5 — Mastery

L5·Q1 — Full first-law cycle leg

Recall Solution

(a) First law: $Δ U = Q - W = 2000 - 800 = 1200 J$ . (b) Kyunki $Δ U = \frac{f}{2} n R Δ T$ (with $n = 1$ mole), $Δ T$ ke liye solve karo: $Δ T = \frac{2 Δ U}{f n R} = \frac{2 ( 1200 )}{3 ( 1 ) ( 8.314 )} = \frac{2400}{24.942} \approx 96.2 K$ Sirf internal $1200 J$ temperature badhata hai; jo $800 J$ work ke roop mein gaye woh nahi badhate.

L5·Q2 — $P$ – $V$ triangle cycle ke baare mein reasoning

Figure padhna: horizontal axis volume $V$ hai, vertical axis pressure $P$ hai. Gas corner $A$ se start karti hai, constant pressure par $A \to B$ se garm hoti hai (daayein move karti hai, volume badhta hai), phir $B \to C$ pe pressure badhta hai jabki volume ghatta hai, aur finally $C \to A$ loop close karta hai. Triangle ke enclosed shaded burnt-orange area ke barabar hai net work $W_{cycle}$ jo gas ek loop mein karta hai.

Recall Solution

KYA: internal energy ek state function hai — yeh sirf current $(P, V, T)$ par depend karta hai, naki us path par jo liya gaya. KYUN cycle change zero hai: gas apni starting state $A$ par wapas aati hai, isliye uska temperature start par wapas aata hai, hence $Δ U_{cycle} = \frac{f}{2} n R Δ T_{cycle} = \frac{f}{2} n R (0) = 0$ Consequence (cycle mein first law): $0 = Q_{cycle} - W_{cycle}$ , isliye $Q_{cycle} = W_{cycle}$ Loop mein absorb ki gayi saari net heat, net work ke roop mein bahar aati hai — exactly aisi hi heat engine chalti hai. Triangle ka enclosed area wahi net work $W_{cycle}$ hai: geometrically, loop ke around summed " $P$ times change in $V$ " precisely triangle ka area deta hai, isliye woh shaded area read karne se aapko $W_{cycle}$ milta hai aur (kyunki $Δ U_{cycle} = 0$ ) net heat $Q_{cycle}$ bhi milta hai.

L5·Q3 — Path independence, numerically

Recall Solution

$Δ U$ sirf endpoints ke temperatures par depend karta hai, isliye sirf $T_{A}$ aur $T_{C}$ matter karte hain — beech ka $T_{B}$ aur har leg ki specific nature (isochoric, isobaric, …) irrelevant hain. $Δ U_{A \to C} = \frac{3}{2} (1) (8.314) (500 - 300) = 1.5 \times 8.314 \times 200 \approx 2494.2 J$ Via the detour, do legs add karo: $Δ U_{A \to B} + Δ U_{B \to C} = \frac{3}{2} (8.314) (250 - 300) + \frac{3}{2} (8.314) (500 - 250)$ $= 1.5 (8.314) (- 50) + 1.5 (8.314) (250) = - 623.55 + 3117.75 = 2494.2 J$ Identical — kyunki $U$ ek state function hai; cooling leg par $T_{B}$ dip exactly heating leg par repaid ho jaata hai, isliye cancel ho jaata hai.

Active recall

Recall Har level ki one-line summary

L1: $U = \frac{f}{2} n R T$ mein plug karo; $f$ = 3 / 5 / 6 jaano.
L2: $Δ U = \frac{f}{2} n R Δ T$ ; isothermal $\Rightarrow Δ U = 0$ .
L3: $Δ U$ sabhi paths ke liye same, lekin $Q$ alag hota hai; $Q_{V} = Δ U$ , $Q_{P} = n C_{p} Δ T$ .
L4: species / modes alag-alag add karo; vibration $f$ mein 2 add karta hai.
L5: cycle $\Rightarrow Δ U = 0 \Rightarrow Q = W$ ; $U$ ek state function hai.