Kinetic Theory: Class 11 Physics Chapter 12 Hinglish mein

Physics Part 2 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 22 Oct 2026 · Updated 22 Oct 2026

85 min
1 simulation
18 practice questions
24 MCQs

Is chapter mein aap kya seekhenge

Matter ki molecular picture samjhao aur densities se molecules ke size aur unke beech ki doori ka estimate lagao
Ideal gas equation ko PV = μRT = Nk_BT forms mein use karo aur Boyle's, Charles' aur Dalton's laws lagao
Wall par molecules ki takkar se P = ⅓nm⟨v²⟩ derive karo aur kinetic model ki assumptions batao
Temperature ko ek molecule ki average translational kinetic energy ke roop mein samjho aur rms speeds calculate karo
Law of equipartition of energy batao aur alag-alag molecules ki degrees of freedom gino
Gases aur solids ki molar heat capacities predict karo aur samjhao ki predictions kahan fail hoti hain
Mean free path l = 1/(√2 nπd²) derive karo aur use karo

NCERT Physics Class 11, Chapter 12, pp. 244–258 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

12.1 Introduction

Garmi ki dopahar mein cycle ke tyre mein hawa bharo to woh sakht ho jaata hai. Andar lagbhag $10^{23}$ air molecules saikdon metre per second ki speed se idhar-udhar bhaag rahe hain aur rubber par lagataar thokar maar rahe hain. Kinetic theory (अणुगति सिद्धांत) gas ka behaviour theek isi picture se shuru karke samjhati hai: chhote-chhote molecules ka jhund jo tez, random motion mein hai.

Itni simple picture kaam kyun kare? Atoms ke beech ke forces sirf kuch atomic sizes tak hi pahunchte hain. Gas molecules aam taur par ek doosre se door hote hain, isliye unhe yeh forces sirf chhoti si collisions mein mehsoos hote hain aur beech mein woh bas aage badhte rehte hain.

Yeh model bahut kaam ka hai. Isse hum gas laws derive karenge, yeh jaanenge ki temperature molecular energy ko naapta hai, predict karenge ki ek gas har kelvin par kitni heat absorb karti hai, aur estimate karenge ki ek molecule collisions ke beech kitni door udta hai. Iska zyada hissa unneesvi sadi mein Clausius, Maxwell aur Boltzmann ne nikala tha.

12.2 Matter ka molecular nature

Apne mashhoor lectures (1963) mein Richard Feynman ne ek sochne wala sawaal poocha: agar saara scientific knowledge kho jaaye aur sirf ek sentence aage diya ja sake, to kaun sa sentence sabse zyada jaankari le jaayega? Unka jawaab tha atomic hypothesis. Hamare shabdon mein: sab kuch atoms se bana hai, chhote particles jo kabhi rukte nahi, jo thoda door hone par ek doosre ko kheenchte hain aur bahut paas aane par ek doosre ko door dhakelte hain. Is chapter ka zyada hissa isi sentence se nikalta hai.

History seNCERT Box: Prachin Bharat aur Greece mein Atomic Hypothesis · p. 245

Experiments se pehle atoms

Vaisheshika school ke Indian philosophers, jinki parampara rishi Kanada se jodi jaati hai, do hazaar saal se bhi pehle sirf tark se atoms tak pahunch gaye the. Unki sabse chhoti unit, paramanu (परमाणु), na toda ja sakta tha, na banaya ja sakta tha, na khatam kiya ja sakta tha. Combinations se variety samjhayi gayi: do paramanus milkar dvyanuka banate the, dyads aage milkar tryanuka banate the, aur koi substance kaisa hai yeh is par depend karta tha ki usmein kaun se paramanus hain aur kitne. Earth, water, fire aur air, har ek ka apna paramanu tha; jabki space (akasha) continuous tha, usmein koi atom tha hi nahi.

Greece mein Democritus (lagbhag 460–370 BCE) ne apne teacher Leucippus ke raaste par chalte hue sabse chhote tukdon ko atomos kaha, jiska matlab hai jise kaata na ja sake. Unhone kai shapes aur sizes ke atoms imagine kiye aur yahan tak andaaza lagaya ki nukile atoms taste ko khatta banate hain aur chikne, gol atoms meetha. Alag-alag andaazon mein faisla karne ke liye experiments nahi the, isliye yeh idea do hazaar saal tak philosophy hi raha. Aakhir mein ise science banaya ek seedhe-saade instrument ne: chemist ka balance.

Do chemical regularities ko samjhana zaroori tha. Pehli, ek compound mein uske elements hamesha mass ke hisaab se ek fixed proportion mein hote hain: pure water, chahe Himalaya ke jharne ka ho ya laboratory reaction ka, usmein har $1 g$ hydrogen ke liye $8 g$ oxygen hoti hai. Yeh law of definite proportions hai. Doosri, agar wahi do elements kai compounds bana sakte hain, to ek element ke jo masses doosre ke fixed mass se judte hain, unka ratio chhote whole numbers ka hota hai. $12 g$ carbon ke saath, carbon monoxide mein $16 g$ oxygen hoti hai aur carbon dioxide mein $32 g$ , yaani $1 : 2$ ka ratio. Yeh law of multiple proportions hai.

Dalton ka explanation (1808): ek element ek hi tarah ke atoms ka collection hai, har hydrogen atom ek jaisa aur kisi bhi oxygen atom se alag, aur compound tab banta hai jab atoms chhote whole-number groups mein judte hain. Isse fixed, whole-number mass ratios turant nikal aate hain.

Gases ne volumes ke baare mein ek clue diya. Gay-Lussac ne paaya ki react karne wali gases volume ke hisaab se simple ratios mein judti hain: do litre hydrogen aur ek litre oxygen milkar do litre steam dete hain, sab same temperature aur pressure par naape hue. 1811 mein Amedeo Avogadro ne ise ek bold idea se samjhaya, Avogadro's law: same temperature aur pressure par sabhi gases ke barabar volumes mein molecules ki sankhya barabar hoti hai. Tab volume ratio $2 : 1 : 2$ asal mein molecules ka ratio hai, jaise $2 H_{2} + O_{2} \overset{}{\to} 2 H_{2} O$ mein. Yeh tabhi chalta hai jab hydrogen aur oxygen do-atom wale molecules ke roop mein rehte hon, isliye chemists matter ki molecular theory ki baat karne lage.

Timeline

Andaaze se naape gaye fact tak

c. 400 BCE
Democritus atom ko naam dete hain
Matter ko khaali space mein ghoomte, na kat sakne wale particles ke roop mein socha jaata hai.
1662
Boyle's law
Robert Boyle dikhate hain ki band hawa ka pressure uske volume ke inversely vary karta hai.
1738
Takkar se pressure
Daniel Bernoulli gas pressure ko walls se takrate particles ka result batate hain.
1808–1811
Dalton, Gay-Lussac, Avogadro
Atomic theory, combining volumes aur equal-volumes hypothesis.
1857–1860
Kinetic theory aakaar leti hai
Clausius mean free path laate hain; Maxwell molecular speeds ka phailaav nikalte hain.
1908
Atoms ki pushti
Brownian motion par Jean Perrin ke measurements Einstein ki 1905 wali theory se match karte hain aur bahas khatam kar dete hain.

1900 ke aas-paas bhi kuch bade scientists ko shak tha ki atoms sach mein hote hain. Perrin ke experiments ke baad yeh shak mit gaye. Aaj electron microscopes aur scanning tunnelling microscope (1981 mein invent hua) aisi images banate hain jinmein single atoms alag se pehchaane ja sakte hain.

Modern measurements is picture ko numbers dete hain (Table 12.1). Natural unit ångström hai: 1 Å matlab $10^{- 10} m$ . Dhyaan do ki bada jump kahan aata hai. Gas mein molecules liquid ke muqable sirf lagbhag das guna door hote hain, phir bhi har molecule kisi doosre se milne se pehle is spacing ka lagbhag bees guna udta hai. Yahi udaan ki doori, mean free path, Section 12.7 ka topic hai.

Table 12.1Matter mein typical lengthsNCERT p. 245

Length	Typical value	Isse kya samajh aata hai
Atom ka size	lagbhag 1 Å	basic scale
Solid mein atoms ki spacing	2–3 Å, centre se centre	atoms apni jagah fixed: ek fixed shape
Liquid mein spacing	lagbhag solid jitni	atoms ek doosre ke paas se phisalte hain: liquids behte hain
1 atm par gas mein spacing	das-das Å (tens of Å)	molecules ko forces sirf collisions mein mehsoos hote hain
1 atm par gas mein mean free path	kai sau Å (room ki hawa mein lagbhag 700 Å)	container na roke to gas phail jaati hai

Do atoms ke beech ka force doori ke saath apna nature badalta hai (Fig. 12.1). Apni natural spacing se zyada paas dhakelne par woh zor se repel karte hain. Thoda door hone par woh attract karte hain. Aur door jaane par force khatam ho jaata hai. Yahi chhoti reach samjhati hai ki solids aur liquids kyun jude rehte hain jabki gas aise behave karti hai jaise uske molecules mushkil se interact karte hon.

2/3 $r_{0}$ se thoda aage woh attract karte hain: yahi solids aur liquids ko jode rakhta hai.

Fig. 12.1Do atoms ke beech ka force: dabane par strong repulsion, thoda door hone par weak attraction, bahut door hone par lagbhag kuch nahi.NCERT p. 245

Force vs separation ka graph: positive values ka matlab repulsion, negative values ka matlab attraction. Curve chhoti separation par bade repulsion se tezi se neeche aata hai, natural spacing r0 par zero cross karta hai, ek halke attractive minimum tak girta hai aur phir flat hokar zero ho jaata hai. Pehle step mein do atoms dabe hue hain aur arrows unhe door dhakel rahe hain, aur ek dot curve ke steep repulsive part par hai. Doosre step mein atoms kuch ångström door hain aur arrows unhe paas kheench rahe hain, aur dot attractive dip ke bottom par hai. Aakhri step mein atoms bahut door hain, koi arrow nahi, aur dot flat tail par hai jahan force lagbhag zero hai.

SimulationWahi molecules, teen states

Paani ke molecules ka ek band container, saath mein ek thermometer aur minus 20 se 140 degree Celsius tak ka temperature slider. 0 degree se neeche molecules ek regular pattern mein baithe rehte hain aur apni jagah kaanpte hain; 0 aur 100 degree ke beech woh liquid ki tarah ek doosre ke paas se phisalte hain; 100 degree se upar woh door ud jaate hain aur gas ki tarah poore container mein bhar jaate hain. Ek label state ka naam batata hai.

Fig. 12.2Temperature slider ko khiskao aur dekho ki paani ke molecules ki spacing aur motion kaise badalti hai (sizes aur spacings scale ke hisaab se nahi hain).

Try karo

Paani ko 0 °C se neeche thanda karo. Kya molecules hilna band kar dete hain, ya apni jagah par kaanpte rehte hain?
Ise 100 °C se upar garam karo. Gas mein spacing liquid ki spacing ke muqable kaisi hai?
Kis state mein ek molecule doosre se milne se pehle sabse zyada door jaata hai?

Footbridge se ek bheed bhare railway platform ko dekho: bheed ki overall density mushkil se badalti hai, phir bhi usmein har insaan chal raha hai. Band bottle mein gas bhi aisi hi hai, bas kahin zyada busy. Uske molecules ek second mein arabon baar takrate hain aur har collision par speed badalte hain; sirf pressure aur mean energy jaise averages steady rehte hain. Ise dynamic equilibrium kehte hain.

Aur zoom karo to na kat sakne wala atom bhi parts wala nikalta hai: protons aur neutrons ke nucleus ke chaaron taraf electrons, aur protons-neutrons khud quarks se bane hain, aur abhi tak koi nahi jaanta ki quarks aakhri layer hain ya nahi. Yahan inmein se kuch bhi matter nahi karta. Har molecule ko lagatar motion mein ek chhoti ball maan lena gases ko samjhane ke liye kaafi hai, aur ek extra idea ke saath solids ko bhi.

12.3 Gases ka behaviour

Teeno states mein gases sabse simple rules follow karti hain. Hawa se bhari band syringe ka plunger dabao to pressure badhta hai; syringe ko garam karo to hawa aur zor se peeche dhakelti hai. Jab tak gas patli hai aur apne liquefying temperature se kaafi upar hai, aise saare changes ek hi equation mein fit hote hain (Chapter 10):

P V = K T

(12.1)

jahan

$T$: absolute (kelvin) temperature
$K$: di gayi sample ke liye ek constant

Constant $K$ is par depend karta hai ki gas kitni hai: same $P$ aur $T$ par amount double karo to volume double ho jaata hai, isliye $K$ double hota hai. Molecular terms mein $K$ molecules ki sankhya $N$ ke saath-saath badhta hai, aur hum likhte hain $K = N k_{B}$ . Experiments dikhate hain ki $k_{B}$ universal hai: hydrogen, helium aur carbon dioxide sab ek hi value dete hain. Yeh Boltzmann constant hai.

\frac{P_{1} V_{1}}{N_{1} T_{1}} = \frac{P_{2} V_{2}}{N_{2} T_{2}} = k_{B}

(12.2)

Eq. (12.2) ko ulta padho: alag gases ke do samples jinke $P$ , $V$ aur $T$ same hain, unmein $N$ bhi same hona chahiye. Avogadro ki hypothesis ab gas law ka result ban gayi hai. Ek mole mein molecules ki sankhya Avogadro number $N_{A} \approx 6.02 \times 10^{23}$ hai. $273 K$ aur $1 a t m$ par kisi bhi gas ka ek mole lagbhag $22.4 L$ jagah leta hai, aur grams mein uska mass u mein uske molecular mass ke barabar hota hai.Chemists ab aam taur par standard pressure $1 b a r = 10^{5} P a$ lete hain, jo 1 atm se thoda kam hai; 273 K aur 1 bar par ek mole 22.7 L jagah leta hai. Moles gin lene ke baad gas law apni jaani-pehchaani form leta hai:

ideal gas equation

P V = μ R T

(12.3)

jahan

$μ$: moles ki sankhya
$R = N_{A} k_{B}$: universal gas constant, $8.314 J m o l^{- 1} K^{- 1}$

μ = \frac{M}{M_{0}} = \frac{N}{N_{A}}

(12.4)

jahan

$M$: sample ka mass
$M_{0}$: molar mass
$N$: molecules ki sankhya

Eq. (12.3) mein $μ = N / N_{A}$ rakhne par $P V = N k_{B} T$ milta hai. $V$ se divide karne par number density $n = N / V$ aati hai, yaani unit volume mein molecules ki sankhya:

P = n k_{B} T

jahan

$n$: unit volume mein molecules
$k_{B}$: $1.38 \times 10^{- 23} J / K$

Iski jagah $μ = M / M_{0}$ likho, aur mass density ko $ρ = M / V$ , to ek aisi form milti hai jismein molecules ginne ki zaroorat nahi:

P = \frac{ρ R T}{M_{0}}

(12.5)

jahan

$ρ$: gas ki mass density

Definition

Ideal gas

Ek model gas jiske molecules ka koi size nahi hota aur jo collisions ke alawa koi force nahi lagate, taaki woh har condition mein $P V = μ R T$ follow kare. Real gases iske paas tab aati hain jab woh patli aur garam hon.

Ek real gas kitna paas aati hai? Ek aasaan test hai ratio $Z = P V / μ R T$ , jo ideal gas ke liye bilkul $1$ hota hai. Fig. 12.3 ek real gas ke liye teen temperatures par pressure ke against $Z$ ka sketch dikhata hai. Jab gas thandi aur dabi hui hoti hai, to uske molecules bheed mein hote hain: unke attractions aur unka apna size matter karta hai, aur $Z$ $1$ se hat jaata hai. Pressure girne par har curve wapas $Z = 1$ ki taraf jaata hai, kyunki door-door rehne wale molecules mushkil se interact karte hain.

3/3Garam gases $Z = 1$ ke zyada paas rehti hain, aur $P \to 0$ hone par har curve isse mil jaata hai.

Fig. 12.3Real gases low pressure aur high temperature par ideal behaviour ke paas aati hain (yeh ek sketch hai, kisi khaas gas ka data nahi).NCERT Fig. 12.1 · p. 247

Axes par upar ki taraf ratio Z = PV/μRT aur aade mein pressure hai. Z = 1 par ek dashed horizontal line ideal gas label ki gayi hai. Pehle step mein sirf yahi line bani hai. Doosre step mein low temperature ka ek curve Z = 1 se shuru hota hai, line se kaafi neeche girta hai aur high pressure par phir upar uthta hai. Teesre step mein medium aur high temperature ke curves jode gaye hain: medium wala thoda sa hi girta hai, high wala dheere se 1 se upar uthta hai, aur left mein ek ring dikhata hai ki pressure zero ki taraf jaane par teeno curves ideal line se kahan milte hain.

Gas ka amount aur uska temperature dono steady rakho, to Eq. (12.3) ke hisaab se product $P V$ nahi badalta:

Boyle's law

P V = constant

(12.6)

Room temperature par ek band syringe ko dabakar uska volume aadha karo to andar ka pressure double ho jaata hai: yahi Boyle's law hai, fixed $T$ par $P V$ constant. Ek balloon ko dhoop wali khidki par chhod do to woh phool jaata hai, uska volume kelvin temperature ke saath-saath badhta hai jabki pressure atmospheric ke aas-paas rehta hai: yahi Charles' law hai, fixed $P$ par $V \propto T$ . Real gases patli aur garam hon to dono laws ko kaafi achhe se follow karti hain, aur liquefaction ke paas inse hat jaati hain (Fig. 12.4).

Hawa ek mixture hai, aur uski har gas walls par aise dhakka deti hai jaise baaki gases wahan hon hi nahi. Ek band vessel (volume $V$ , temperature $T$ ) mein ek gas ke $μ_{1}$ moles, doosri ke $μ_{2}$ moles, aur isi tarah aur bhi daalo, jinmein se koi react nahi karti. Gas law ke liye molecule to molecule hai, isliye sirf total amount matter karta hai:

P V = (\sum_{i} μ_{i}) R T

(12.7)

jahan

$\sum_{i} μ_{i}$: $μ_{1} + μ_{2} + \dots$ , moles ki total sankhya

\begin{aligned} P & = μ_{1} \frac{R T}{V} \\ + μ_{2} \frac{R T}{V} + \dots \end{aligned}

(12.8)

Dalton's law of partial pressures

P = P_{1} + P_{2} + \dots

(12.9)

Pehle term ko akela padho: $μ_{1} R T / V$ woh reading hai jo gauge dikhata agar vessel mein sirf gas 1 hoti. Ise hum us gas ka partial pressure $P_{1}$ kehte hain. Mixture ki reading bas un readings ka sum hai jo har gas akele deti: yahi Dalton's law of partial pressures hai.

1/3Fixed $T$ par $P \propto 1 / V$ : har curve ek temperature ka hai, aur garam gas ka curve zyada bahar hota hai.

Fig. 12.4Ideal gas equation se nikalne wale gas laws.NCERT Figs. 12.2, 12.3 · p. 247

Pehla step fixed amount ki gas ke liye pressure vs volume dikhata hai: do curves jo volume badhne par girte hain, bahar wala hotter aur andar wala cooler label kiya gaya hai. Doosra step fixed pressure par volume vs kelvin temperature dikhata hai: do seedhi lines jo dashed lines ke roop mein aage badhane par 0 K se guzarti hain, zyada steep wali lower pressure label ki gayi hai. Teesra step pressure ke teen bars dikhata hai: akeli gas 1, akeli gas 2, aur mixture, jiska bar pehle do bars ko ek ke upar ek rakhne se banta hai.

Agle teen examples densities se estimate karte hain ki molecules asal mein kitni jagah lete hain.

Example 12.1NCERT Example 12.1 · p. 247

Vapour kitna khaali hai?

Liquid ethanol ki density $789 k g / m^{3}$ hai. Normal boiling point ( $78^{\circ} C$ aur $1 a t m$ ) par iski vapour ki density $1.60 k g / m^{3}$ hai. Saare molecules ka volume (molecular volume) utna maano jitna woh liquid banakar ghereinge. Molecules khud vapour ke volume ka kitna fraction bharte hain?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 12.2NCERT Example 12.2 · p. 248

Ek molecule ka size

Example 12.1 ke data se ethanol molecule ( $C_{2} H_{5} O H$ , molar mass $46 g / m o l$ ) ka radius estimate karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 12.3NCERT Example 12.3 · p. 248

Vapour molecules ek doosre se kitni door hain?

$78^{\circ} C$ aur $1 a t m$ par ethanol vapour mein padosi molecules ke beech ki average doori estimate karo, aur use ek molecule ke size se compare karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 12.4NCERT Example 12.4 · p. 248

Welding ka gas mixture

Ek welding cylinder mein argon (atomic mass $39.9 u$ ) aur carbon dioxide (molecular mass $44.0 u$ ) ka non-reacting mixture hai. Unke partial pressures $4 : 1$ ke ratio mein hain. Argon aur carbon dioxide ka ratio nikalo (a) molecules ki sankhya ka aur (b) mass densities ka.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

27^{\circ} C

aur

1.0 a t m

par ek

2.0 L

ki bottle mein hawa ke kitne moles hain?

1 a t m = 1.01 \times 10^{5} P a

lo.

mol

Answer dekho

Answer 0.081 mol

μ = (1.01 \times 10^{5}) (2.0 \times 10^{- 3}) / (8.314 \times 300) \approx 0.081 m o l

. Litres ko cubic metres mein convert karna bhool gaye to answer 1000 guna bada aayega.

12.4 Ideal gas ki kinetic theory

Ab hum gas laws ke peeche ka model banate hain. Iske rules thode se hain:

Bahut badi sankhya. Ek cup bhar hawa mein bhi lagbhag $10^{22}$ molecules hote hain, isliye unke averages bahut reliable hote hain.
Khaali space mein points. Har molecule apne aas-paas ke gaps ke muqable bahut chhota hai, isliye zyadatar time use kuch mehsoos nahi hota aur woh steady speed se seedha udta hai.
Chhoti, elastic mulaqaatein. Ek molecule ki velocity tabhi badalti hai jab woh kisi doosre molecule ya wall se milta hai. Yeh collisions momentum ke saath-saath kinetic energy bhi conserve karte hain, aur collisions ke beech ki udaan ke time ke muqable har collision pal bhar mein khatam ho jaata hai.
Koi favourite direction nahi. Average mein molecules har direction mein barabar chalte hain.

12.4.1 Ideal gas ka pressure

Pressure woh hai jo wall ko mehsoos hota hai jab molecules usse takrakar wapas aate hain. Ise nikalne ke liye hum ginenge ki har second wall ko kitna momentum milta hai (Fig. 12.5).

2/3Sirf woh molecules jo wall se $v_{x} Δ t$ ke andar hain, aur uski taraf ja rahe hain, time $Δ t$ mein pahunchte hain.

Fig. 12.5Pressure us momentum se aata hai jo molecules wall ko dete hain.NCERT Fig. 12.4 · p. 249

Right mein area A ki ek flat wall khadi hai. Pehle step mein ek molecule tirchha wall ki taraf aata hai aur mirror-image tirchhe angle par wapas jaata hai: uski velocity ka sirf wall ki taraf wala part ulatta hai, aur wall ke paas ek arrow use mila impulse 2mvx dikhata hai. Doosre step mein wall se lagi vx times delta t moti ek shaded slab hai; usmein kuch molecules wall ki taraf ja rahe hain (highlighted, right ki taraf arrows) aur kuch usse door (grey, left ki taraf arrows), aur ek label kehta hai ki sirf wall ki taraf jaane wale hi usse takrate hain. Teesre step mein poore container mein bahut saare molecules random directions mein chal rahe hain, aur result P = one third n m times mean square speed dikhaya gaya hai.

DerivationMolecules ki takkar se pressure

Side $l$ wala ek cubical box lo aur $x$ , $y$ aur $z$ axes uske edges ke along rakho. Ek face par dhyaan do, jo $x$ -axis ke right angle par hai; uska area $A = l^{2}$ hai.

Mass $m$ aur velocity $(v_{x}, v_{y}, v_{z})$ wala ek molecule is wall se elastically bounce karta hai. Sirf wall ke normal wala component ulatta hai; $v_{y}$ aur $v_{z}$ nahi badalte. Uska $x$ -momentum $m v_{x}$ se $- m v_{x}$ ho jaata hai, yaani change hai

Δ p_{x} = - 2 m v_{x}

isliye, conservation of momentum se, aisi har takkar par wall ko $2 m v_{x}$ milta hai.

Ek chhote interval $Δ t$ mein kaun se molecules wall tak pahunch paate hain? $v_{x}$ se chalne wala molecule is time mein $v_{x} Δ t$ doori tay karta hai, isliye sirf wahi pahunch sakte hain jo wall ke paas $v_{x} Δ t$ moti patli slab (volume $A v_{x} Δ t$ ) mein shuru hote hain, aur inmein se bhi average mein sirf aadhe, jo wall ki taraf ja rahe hain. Agar unit volume mein $n$ molecules ki yeh velocity hai, to takkaron ki sankhya $\frac{1}{2} n A v_{x} Δ t$ hai aur diya gaya momentum hai

\begin{aligned} Q & = (2 m v_{x}) \\ \times (\frac{1}{2} n A v_{x} Δ t) \end{aligned}

(12.10)

Force per unit time diya gaya momentum hai, aur pressure per unit area force hai:

P = \frac{Q}{A Δ t} = n m v_{x}^{2}

(12.11)

Real molecules ki ek hi velocity nahi hoti: Eq. (12.11) ek velocity group ka contribution hai. Saare groups jodne par $v_{x}^{2}$ ki jagah saare molecules par uska average aa jaata hai, aur $n$ ab total number density hai:

P = n m ⟨ v_{x}^{2} ⟩

(12.12)

Gas ki koi preferred direction nahi hoti (yeh isotropic hai), isliye teeno components ka mean square same hota hai. Kyunki $v^{2} = v_{x}^{2} + v_{y}^{2} + v_{z}^{2}$ ,

\begin{aligned} ⟨ v_{x}^{2} ⟩ & = ⟨ v_{y}^{2} ⟩ = ⟨ v_{z}^{2} ⟩ \\ = \frac{1}{3} ⟨ v^{2} ⟩ \end{aligned}

(12.13)

Ise Eq. (12.12) mein rakhne par:

ideal gas ka pressure

P = \frac{1}{3} n m ⟨ v^{2} ⟩

(12.14)

jahan

$⟨ v^{2} ⟩$: molecules ki mean square speed

Is derivation ke baare mein do sawaal. Pehla, humne cube use kiya; kya gol cylinder ya bottle alag pressure dete? Nahi. Argument mein sirf wall ka ek chhota patch aaya, aur $A$ aur $Δ t$ answer se hat gaye. Yeh Pascal's law (Chapter 9) se match karta hai: ruke hue fluid mein har jagah same pressure hota hai.

Doosra, humne molecules ki aapas ki takkar ko ignore kiya. Collision molecule ki velocity badalta zaroor hai, lekin steady gas mein collisions molecules ko velocity groups ke beech idhar-udhar karte hain bina yeh badle ki har group mein kitne hain, jaise bhari train mein passengers seats badal lete hain. Average $⟨ v_{x}^{2} ⟩$ , jo formula mein bas yahi use hota hai, same rehta hai, bashart har collision collisions ke beech ki free udaan ke muqable chhota ho.

InteractiveBox mein gas

Helium, nitrogen ya carbon dioxide ke molecules ek litre ke box mein chalte hain jo right mein ek piston se band hai. Sliders temperature ko 100 se 600 kelvin, volume ko 0.4 se 1 litre aur dikhaye gaye molecules ki sankhya set karte hain. Piston se takrane par molecules chamakte hain, aur ek dial un takkaron se nikala gaya pressure dikhata hai, saath mein nkBT ki predicted value par ek tick. Ek histogram molecular speeds ka phailaav dikhata hai, predicted curve aur rms speed par ek line ke saath. Ek table measured aur predicted pressure, takkaron ki sankhya, rms speed aur per molecule mean kinetic energy dikhati hai.

Fig. 12.6Play dabao. Har bana hua molecule

10^{20}

asli molecules ko represent karta hai; dial unka piston ko diya gaya momentum jodta hai.

Try karke dekho

Gas box ke saath experiments

Counting ko aadha minute chalne do. Measured pressure predicted $n k_{B} T$ ke kitna paas aata hai?
Fixed temperature par volume aadha karo. Pressure ka kya hota hai, aur piston par takkaron ki rate ka?
Fixed volume par absolute temperature double karo. $v_{rms}$ kitne guna badhta hai? Pressure kitne guna badhta hai?
Same temperature par nitrogen se helium par switch karo. Kaun si readings badalti hain, aur kaun si same rehti hain?

$V$ aadha karne se $n$ double hota hai, isliye hit rate aur pressure double ho jaate hain (Boyle's law). $T$ double karne se $v_{rms}$ $\sqrt{2}$ guna badhta hai lekin $P$ double hota hai: har hit $\sqrt{2}$ guna zyada momentum deta hai aur hits $\sqrt{2}$ guna zyada baar aate hain. Helium molecules nitrogen molecules se lagbhag 2.6 guna tez chalte hain, lekin har ek 7 guna halka hai, isliye pressure aur mean kinetic energy nahi badalte.

Quick check

3 mein se 1

Derivation mein har bounce wall ko

m v_{x}

ki jagah

2 m v_{x}

ka impulse kyun deta hai?

x

-momentum

+ m v_{x}

- m v_{x}

ho jaata haiMolecule wall se do baar takrata haiWall

v_{x}

speed se peeche hatti haiSirf aadhe molecules wall ki taraf chalte hain

Answer dekho

Answer (A) $x$ -momentum $+ m v_{x}$ se $- m v_{x}$ ho jaata hai

Change hai

(- m v_{x}) - (+ m v_{x}) = - 2 m v_{x}

, isliye wall ko

2 m v_{x}

milta hai. Wall ki taraf jaane wale molecules ka factor

\frac{1}{2}

alag se aata hai, jab hum hits ginte hain.

12.4.2 Temperature ka kinetic interpretation

DerivationTemperature aur molecular energy

Eq. (12.14) ko $V$ se multiply karo aur molecules ki sankhya ke liye $N = n V$ likho:

P V = \frac{1}{3} N m ⟨ v^{2} ⟩

(12.15a)

P V = \frac{2}{3} N (\frac{1}{2} m ⟨ v^{2} ⟩)

(12.15b)

Bracket ke andar per molecule mean translational kinetic energy hai. Ideal gas mein molecules collisions ke beech ek doosre par koi force nahi lagate, isliye jodne ke liye koi potential energy nahi hai, aur total translational kinetic energy hai

E = N \times \frac{1}{2} m ⟨ v^{2} ⟩

(12.16)

taaki Eq. (12.15b) ban jaaye

P V = \frac{2}{3} E

(12.17)

Ise $P V = N k_{B} T$ ke roop mein likhe gas law se compare karo:

E = \frac{3}{2} N k_{B} T

(12.18)

ek molecule ki average kinetic energy

\frac{1}{2} m ⟨ v^{2} ⟩ = \frac{3}{2} k_{B} T

(12.19)

Eq. (12.19) par thoda ruko, yeh chapter ka sabse important result hai. Gas mein rakha thermometer asal mein yeh naap raha hai ki uske molecules kitni tezi se chal rahe hain. $300 K$ par har gas, chahe halki helium ho ya bhaari xenon, ki per molecule average translational energy same hoti hai, lagbhag $6.2 \times 10^{- 21} J$ , aur kelvin temperature double karne par yeh double ho jaati hai; pressure aur volume ismein nahi aate. Constant $k_{B}$ thermometer ke scale aur molecular energy ke beech ka exchange rate hai.

Isse do bonus milte hain. Eq. (12.18) dikhati hai ki ideal gas ki internal energy tabhi badalti hai jab uska temperature badalta hai, yeh fact thermodynamics (Chapter 11) mein baar-baar use hota hai. Aur kyunki model ko $P V = N k_{B} T$ se match karne ke liye banaya gaya tha, us equation se nikalne wale har gas law ka ab ek molecular explanation hai.Yahan $E$ sirf ek jagah se doosri jagah jaane ki energy hai. Jo molecules rotate ya vibrate karte hain woh extra energy store karte hain, jise agla section ginta hai.

Non-reacting ideal gases ke mixture mein har gas Eq. (12.14) mein apna hissa jodti hai:

\begin{aligned} P & = \frac{1}{3} n_{1} m_{1} ⟨ v_{1}^{2} ⟩ \\ + \frac{1}{3} n_{2} m_{2} ⟨ v_{2}^{2} ⟩ \\ + \dots \end{aligned}

(12.20)

Equilibrium mein har gas ke molecules ki average kinetic energy same hoti hai, $\frac{3}{2} k_{B} T$ , isliye har term $n_{i} k_{B} T$ hai aur

P = (\sum_{i} n_{i}) k_{B} T

(12.21)

To partial pressures jud jaate hain, bilkul jaise Dalton ne experiment se paaya tha, lekin ab yeh rule molecular motion se nikalta hai.

Eq. (12.19) yeh bhi batati hai ki molecules kitni tez chalte hain. Mean square speed ka square root root mean square (rms) speed hai:

\begin{aligned} v_{rms} & = \sqrt{⟨ v^{2} ⟩} \\ = \sqrt{\frac{3 k_{B} T}{m}} \\ = \sqrt{\frac{3 R T}{M_{0}}} \end{aligned}

$40^{\circ} C$ wali garmi ki dopahar ( $T = 313 K$ ) mein nitrogen molecules ( $M_{0} = 0.028 k g / m o l$ ) ki rms speed lagbhag $528 m / s$ hoti hai, us hawa mein sound ki speed (lagbhag $355 m / s$ ) ki lagbhag dedh guna. Kisi diye gaye temperature par halke molecules zyada tez hote hain, kyunki $v_{rms} \propto 1 / \sqrt{m}$ .

Example 12.5NCERT Example 12.5 · p. 250

Biogas ke molecules

Gaon ke digester ki biogas mein zyadatar methane ( $C H_{4}$ , $16.0 u$ ) aur carbon dioxide ( $C O_{2}$ , $44.0 u$ ) hote hain. Maano $35^{\circ} C$ par ek sample mein volume ke hisaab se 60 % methane hai. Methane aur carbon dioxide ka ratio nikalo (a) per molecule average kinetic energy ka aur (b) rms speeds ka.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 12.6NCERT Example 12.6 · p. 251

Neon atoms ko alag karna

Neon mein $20 u$ aur $22 u$ mass wale atoms hote hain. 1913 mein F. W. Aston ne neon ko ek porous material se baar-baar diffuse karwa kar inhe alag karne ki koshish ki; unhe jo badlaav mila woh bahut chhota tha. Average mein kaun se atoms zyada tez hain, aur unki rms speeds mein kitne percent ka fark hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Fig. 12.7 dikhata hai ki porous wall molecules ko kyun chhaantti hai. Pores itne patle hain ki molecules ek-ek karke nikalte hain, aur har tarah ke molecule apni average speed ke proportional rate se paar hote hain. Isliye halke molecules zyada tezi se leak hote hain: kisi gas ki rate of diffusion uske molecular mass ke square root ke ulte (one over) ke hisaab se vary karti hai (Graham's law). Ek pass mixture ko thoda sa hi badalta hai, isliye process ko kai stages mein repeat kiya jaata hai.

2/3Tez, halke molecules pores tak zyada baar pahunchte hain, isliye paar hone wali gas mein woh zyada hote hain.

Fig. 12.7Halke molecules porous wall se zyada baar nikalte hain (yahan fark ko badha-chadha kar dikhaya gaya hai).NCERT Fig. 12.5 · p. 251

Ek box ek porous wall se do hisson mein bata hai, wall gaps wale chhote blocks ke column jaisi bani hai. Pehle step mein left side mein das halke molecules (filled circles) aur das bhaari molecules (hollow rings) hain aur right side khaali hai. Doosre step mein paanch halke aur do bhaari molecules right mein paar ho gaye hain, left mein paanch halke aur aath bhaari bache hain, kyunki halke molecules tez chalte hain aur pores tak zyada baar pahunchte hain. Teesre step mein neeche ek arrow kehta hai ki enrichment badhane ke liye stage ko baar-baar repeat kiya jaata hai.

Example 12.7NCERT Example 12.7 · p. 251

Chalti wall se takrakar lautna

(a) $10 m / s$ se chalti ek ball ek bahut bhaari wall se seedhi takrati hai aur elastically wapas aati hai. Uski rebound speed nikalo agar wall (i) ruki hui hai, (ii) ball ki taraf $2 m / s$ se aa rahi hai, (iii) ball se door $2 m / s$ se ja rahi hai.

(b) (a) ka use karke samjhao ki pump karte waqt cycle pump ka barrel garam kyun ho jaata hai. (c) Jo gas piston ko bahar dhakelti hai uska kya hota hai? (d) Same speed se ghumane par bhaari hockey stick ball ko zyada tez kyun bhejti hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

2/3 $V$ se andar aati wall ball ko $u + 2 V$ se lautati hai: compress hoti gas garam hoti hai.

Fig. 12.8Chalti wall rebound speed badal deti hai: isi tarah piston gas ko garam ya thanda karta hai.

Ek ball upar wali row mein left se 10 m/s par aati hai aur right mein ek bhaari wall se takrakar neeche wali row mein left ki taraf jaati hai. Pehle step mein wall ruki hai aur ball 10 m/s par jaati hai. Doosre step mein wall 2 m/s se left mein ball ki taraf chalti hai aur ball 14 m/s par jaati hai, lambe arrow aur is note ke saath ki ball tez ho jaati hai. Teesre step mein wall 2 m/s se right mein, ball se door, chalti hai aur ball 6 m/s par jaati hai, chhote arrow aur is note ke saath ki woh dheemi ho jaati hai.

12.5 Law of equipartition of energy

Velocity $(v_{x}, v_{y}, v_{z})$ se chalte molecule ki translational kinetic energy hai

\begin{aligned} ε_{t} & = \frac{1}{2} m v_{x}^{2} + \frac{1}{2} m v_{y}^{2} \\ + \frac{1}{2} m v_{z}^{2} \end{aligned}

(12.22)

Temperature $T$ par thermal equilibrium mein Eq. (12.19) kehti hai ki is sum ka average hai

⟨ ε_{t} ⟩ = \frac{3}{2} k_{B} T

(12.23)

Koi direction special nahi hai, isliye teeno terms ko barabar hissa milta hai:

\begin{aligned} ⟨ \frac{1}{2} m v_{x}^{2} ⟩ & = \frac{1}{2} k_{B} T \\ ⟨ \frac{1}{2} m v_{y}^{2} ⟩ & = \frac{1}{2} k_{B} T \\ ⟨ \frac{1}{2} m v_{z}^{2} ⟩ & = \frac{1}{2} k_{B} T \end{aligned}

(12.24)

Kisi cheez ki jagah batane ke liye kitne numbers chahiye? Seedhe taar par phisalte bead ko ek coordinate chahiye, carrom board par coin ko do, aur kamre mein udti makkhi ko teen. Yeh ginti motion ki degrees of freedom (स्वातंत्र्य कोटि) hai. Gas molecule makkhi jaisa hai: uske teen numbers Eq. (12.22) ke teen squared terms se match karte hain. Har ek poore molecule ke ek jagah se doosri jagah jaane ko describe karta hai, jise hum translation kehte hain, aur Eq. (12.24) pehle hi bata chuki hai ki har ek ka average $\frac{1}{2} k_{B} T$ hai.

Definition

Degree of freedom

Molecule ke chalne ka ek independent tareeka: translation ke liye, uski position fix karne ke liye zaroori ek coordinate. Energy ginte waqt, molecule ki energy ka har squared term ek mode hai jo energy rakh sakta hai.

ExampleBoard par phisalte carrom coin ki do translational degrees of freedom hoti hain.

Hamare kaam ke liye argon aur helium atoms bina kisi feature wale points hain: translation hi sab kuch hai jo woh kar sakte hain. $N_{2}$ ya $O_{2}$ jaise dumbbell-shaped molecule ke paas zyada options hain. Woh apne centre of mass ke around ulat-palat ghoom sakta hai, aur aisa karne ke do independent tareeke hain, bond ke right angle par do axes mein se kisi ke bhi around (Fig. 12.9). Aur bond ke hi around takli ki tarah ghoomna? Iska moment of inertia itna chhota hai ki quantum mechanics ke hisaab se normal temperatures par yeh spin energy nahi le sakta, isliye ise nahi gina jaata. Isliye rigid dumbbell ki energy hai

\begin{aligned} ε & = ε_{t} + ε_{r} \\ = ε_{t} + \frac{1}{2} I_{1} ω_{1}^{2} \\ + \frac{1}{2} I_{2} ω_{2}^{2} \end{aligned}

(12.25)

jahan

$ω_{1}, ω_{2}$: do axes ke around angular speeds
$I_{1}, I_{2}$: unke around moments of inertia

Har rotation ek aur squared term jodta hai, ab angular speed mein.

2/4Bond ke perpendicular axes ke around do rotations: angular speed mein do squared terms.

Fig. 12.9Diatomic molecule ke chalne aur energy store karne ke tareeke.NCERT Fig. 12.6 · p. 252

Bond se jude do atoms beech mein hain. Pehle step mein centre se teen arrows x, y aur z ki taraf point karte hain, label hai 3 translational. Doosre step mein centre se guzarti do dashed axes, ek vertical aur ek tirchhi page ke andar jaati hui, har ek par ek curved arrow, label hai 2 rotational. Teesre step mein bond ke along ek dashed axis hai jis par chhota curved arrow hai, label hai spin about the bond, too small to count. Aakhri step mein bond ek spring ban jaata hai aur atoms bahar ki taraf arrows ke saath door hote hain, label hai vibration: kinetic aur potential energy, 2 terms.

Lekin real bonds springy hote hain: do atoms bond ke along aise oscillate kar sakte hain jaise spring se jude masses. $C O$ jaise halke, stiff molecules mein yeh vibration room temperature par mushkil se active hota hai, lekin $C l_{2}$ jaise bhaari, dheele molecules mein yeh pehle se hi energy ka achha khaasa hissa le leta hai (Table 12.3). Ek vibration do squared terms jodta hai, ek kinetic energy ka aur ek potential energy ka:

ε_{v} = \frac{1}{2} m {(\frac{d y}{d t})}^{2} + \frac{1}{2} k y^{2}

(12.26)

jahan

$y$: bond ka apni natural length se khinchaav
$k$: bond ka force constant
$m$: vibrate karte pair ka effective mass

Tab $ε = ε_{t} + ε_{r} + ε_{v}$ . Fark par dhyaan do: translation aur rotation har degree of freedom par ek squared term dete hain, lekin har vibrational mode do deta hai.

LawLaw of equipartition of energy

Absolute temperature $T$ par thermal equilibrium mein, ek molecule ki energy uski energy ke saare squared terms mein barabar bant jaati hai, aur har term ki average value $\frac{1}{2} k_{B} T$ hoti hai. Isliye har translational aur har rotational degree of freedom average mein $\frac{1}{2} k_{B} T$ rakhti hai, aur har vibrational mode $2 \times \frac{1}{2} k_{B} T = k_{B} T$ rakhta hai.

Quick check

3 mein se 1

C O

molecule translate, rotate aur vibrate karta hai. Uski energy ke kitne squared terms energy rakhte hain?

5678

Answer dekho

Answer (C) 7

3 translational + 2 rotational + 2 vibrational (kinetic aur potential) = 7. 6 answer dena vibration ko sirf ek baar ginna hai.

12.6 Specific heat capacity

Equipartition squared terms ki ginti ko heat capacity ke prediction mein badal deta hai. Chapter 11 se yaad karo ki constant volume par molar specific heat capacity, $C_{v}$ , ek mole ki internal energy mein per kelvin badhotri hai, aur ideal gas ke liye $C_{p} - C_{v} = R$ .

12.6.1 Monatomic gases

Monatomic molecule ki sirf teen translational degrees of freedom hoti hain, isliye uski average energy $\frac{3}{2} k_{B} T$ hai. Ek mole mein $N_{A}$ molecules hote hain, isliye

\begin{aligned} U & = \frac{3}{2} k_{B} T \times N_{A} \\ = \frac{3}{2} R T \end{aligned}

(12.27)

C_{v} = \frac{d U}{d T} = \frac{3}{2} R

(12.28)

Kisi bhi ideal gas ke liye,

C_{p} - C_{v} = R

(12.29)

isliye

C_{p} = \frac{5}{2} R

(12.30)

γ = \frac{C_{p}}{C_{v}} = \frac{5}{3}

(12.31)

12.6.2 Diatomic gases

Rigid diatomic molecule ki paanch degrees of freedom hoti hain, teen translational aur do rotational, isliye

\begin{aligned} U & = \frac{5}{2} k_{B} T \times N_{A} \\ = \frac{5}{2} R T \end{aligned}

(12.32)

C_{v} = \frac{5}{2} R, C_{p} = \frac{7}{2} R

(12.33)

γ = \frac{7}{5}

(12.34)

Agar molecule vibrate bhi karta hai, to vibrational mode per molecule $k_{B} T$ jodta hai, kul milakar $\frac{5}{2} k_{B} T + k_{B} T = \frac{7}{2} k_{B} T$ :

\begin{aligned} U & = \frac{7}{2} k_{B} T \times N_{A} \\ = \frac{7}{2} R T \\ C_{v} & = \frac{7}{2} R \\ C_{p} & = \frac{9}{2} R \\ γ & = \frac{9}{7} \end{aligned}

(12.35)

Kaun si picture sahi hai? Dono, alag-alag temperatures par. Fig. 12.10 hydrogen ke liye $C_{v}$ ka sketch dikhata hai. Bahut thandi hone par yeh monatomic gas jaisa behave karti hai; room temperature ke aas-paas $C_{v} \approx \frac{5}{2} R$ ; sirf bahut garam hone par vibration aakhri $R$ jodta hai. Classical equipartition in steps ko explain nahi kar sakta. Quantum mechanics kar sakta hai: rotations aur vibrations energy sirf packets mein lete hain, aur low temperatures par $k_{B} T$ unhe dene ke liye bahut chhota hota hai.

2/3Rotation on ho gaya hai: $C_{v} \approx \frac{5}{2} R$ .

Fig. 12.10Hydrogen ki molar heat capacity (schematic): temperature badhne par modes on hote hain.

Logarithmic scale par temperature ke against Cv divided by R ka graph, 3/2, 5/2 aur 7/2 par dashed levels ke saath. Pehle step mein 3/2 par ek flat line translation label ki gayi hai. Doosre step mein curve smoothly 5/2 tak uthta hai, label plus rotation, aur is plateau par ek dot room temperature dikhata hai. Teesre step mein yeh bahut high temperatures par phir 7/2 tak uthta hai, label plus vibration.

12.6.3 Polyatomic gases

Ek non-linear polyatomic molecule, jaise $H_{2} O$ ya $C H_{4}$ , teen independent axes ke around ghoom sakta hai. Iski 3 translational aur 3 rotational degrees of freedom hoti hain, aur saath mein kuch sankhya $f$ vibrational modes ki. Tab har molecule average mein $\frac{3}{2} k_{B} T + \frac{3}{2} k_{B} T + f k_{B} T$ rakhta hai, isliye per mole

\begin{aligned} U & = (3 + f) k_{B} T N_{A} \\ = (3 + f) R T \end{aligned}

\begin{aligned} C_{v} & = (3 + f) R \\ C_{p} & = (4 + f) R \\ γ & = \frac{4 + f}{3 + f} \end{aligned}

(12.36)

Relation $C_{p} - C_{v} = R$ har ideal gas ke liye sahi hai, uske molecules kaise bhi hon. $C O_{2}$ jaisa linear molecule ek special case hai: diatomic molecule ki tarah iski sirf do rotational degrees of freedom hoti hain, isliye ek rigid $C O_{2}$ molecule ka $C_{v} = \frac{5}{2} R$ hota. Table 12.2 vibrations ko ignore karne ya ginne par predictions ikattha karta hai, aur Table 12.3 unhe measurements se compare karta hai.

Table 12.2Equipartition se predict ki gayi molar heat capacitiesNCERT Table 12.1 · p. 254

Molecule ka type	Squared terms	$C_{v}$ ( $J m o l^{- 1} K^{- 1}$ )	$C_{p}$ ( $J m o l^{- 1} K^{- 1}$ )	$γ$
Monatomic	3	$\frac{3}{2} R = 12.5$	$\frac{5}{2} R = 20.8$	1.67
Rigid diatomic	5	$\frac{5}{2} R = 20.8$	$\frac{7}{2} R = 29.1$	1.40
Vibration wala diatomic	7	$\frac{7}{2} R = 29.1$	$\frac{9}{2} R = 37.4$	1.29
Rigid non-linear polyatomic	6	$3 R = 24.9$	$4 R = 33.3$	1.33

Table 12.3

298 K

ke aas-paas measured molar heat capacities (NIST data se

C_{p}

;

C_{v} = C_{p} - R

)NCERT Table 12.2 · p. 254

Gas	Type	$C_{p}$ ( $J m o l^{- 1} K^{- 1}$ )	$C_{v}$ ( $J m o l^{- 1} K^{- 1}$ )	$γ$
Krypton, $K r$	monatomic	20.8	12.5	1.67
Carbon monoxide, $C O$	diatomic	29.1	20.8	1.40
Chlorine, $C l_{2}$	diatomic	34.0	25.6	1.32
Carbon dioxide, $C O_{2}$	linear triatomic	37.1	28.8	1.29
Ethane, $C_{2} H_{6}$	polyatomic	52.5	44.2	1.19

Krypton aur carbon monoxide rigid predictions se lagbhag bilkul match karte hain. Chlorine, carbon dioxide aur ethane unse aage nikal jaate hain: har ek mein low-frequency vibrations hain, bhaari atoms ya bending motions ki wajah se, jo room temperature par pehle se hi thode excited hote hain. Phir bhi har vibration ko poora ginne se answer doosri taraf measurements se aage nikal jaayega. Sach beech mein hai, kyunki vibrations sirf thode active hain, bilkul jaisa Fig. 12.10 ki quantum picture batati hai. Normal temperatures par equipartition achha kaam karta hai, bas pata hona chahiye ki kaun se modes poori tarah active hain.

Example 12.8NCERT Example 12.8 · p. 254

Nitrogen ke flask ko garam karna

Ek rigid $5.0 L$ steel flask mein $27^{\circ} C$ aur $2.0 a t m$ par nitrogen hai. Gas ko $87^{\circ} C$ tak garam karne ke liye kitni heat chahiye? Nitrogen ko ideal rigid diatomic gas maano aur $1 a t m = 1.013 \times 10^{5} P a$ lo. Agar gas ko iski jagah constant pressure par rakha jaata to kitni heat chahiye hoti?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

12.6.4 Solids ki specific heat capacity

Equipartition kai solids ke liye bhi kaam karta hai. Socho ki har atom apne padosiyon se spring jaise bonds dwara apni jagah par tika hai (Fig. 12.11). Yeh apni mean position ke around teen independent directions mein vibrate karta hai. Har direction ek oscillator hai jismein ek kinetic aur ek potential term hai, jinki value $2 \times \frac{1}{2} k_{B} T = k_{B} T$ hai, isliye har atom average mein $3 k_{B} T$ rakhta hai. Ek mole ke liye,

\begin{aligned} U & = 3 k_{B} T \times N_{A} \\ = 3 R T \end{aligned}

Jab solid ko atmospheric pressure par garam kiya jaata hai, to lagbhag saari heat $U$ mein jaati hai: solid mushkil se phailta hai, isliye $Δ Q = Δ U + P Δ V$ mein work $P Δ V$ bahut chhota hai. Isliye

solid ki molar heat capacity

\begin{aligned} C & = \frac{Δ Q}{Δ T} = \frac{Δ U}{Δ T} \\ = 3 R \end{aligned}

(12.37)

3/3Teen directions se per atom $3 k_{B} T$ milta hai, isliye per mole $U = 3 R T$ .

Fig. 12.11Solid ka har atom teen directions mein oscillate karta hai.

Pandrah atoms ek square grid par baithe hain, apne padosiyon se seedhi lines se jude hain jo spring jaise bonds ko darshati hain; beech wala atom highlighted hai. Pehle step mein lattice ruka hua hai. Doosre step mein beech wala atom side mein khisakta hai aur ek double arrow use ek direction mein oscillate karta dikhata hai, jismein average mein kBT jitni kinetic aur potential energy hai. Teesre step mein do aur double arrows baaki do directions mein oscillation dikhate hain, jisse per atom 3kBT milta hai.

Table 12.4

25^{\circ} C

ke aas-paas kuch solids ki heat capacitiesNCERT Table 12.3 · p. 254

Solid	Molar mass ( $g / m o l$ )	$c$ ( $J k g^{- 1} K^{- 1}$ )	$C$ ( $J m o l^{- 1} K^{- 1}$ )	$C / R$
Iron	55.85	449	25.1	3.02
Gold	196.97	129	25.4	3.06
Zinc	65.38	388	25.4	3.06
Nickel	58.69	445	26.1	3.14
Silicon	28.09	712	20.0	2.41
Diamond	12.01	508	6.1	0.73

Iron, gold, zinc aur nickel sab $3 R \approx 25 J m o l^{- 1} K^{- 1}$ ke paas hain, jaisa Dulong aur Petit ne 1819 mein experiment se paaya tha. Per kilogram inki specific heats mein teen guna se zyada fark hai, phir bhi per mole yeh lagbhag barabar hain: matter karti hai atoms ki sankhya. Silicon, aur khaas kar diamond, kaafi peeche reh jaate hain. Inke halke atoms aur bahut stiff bonds itni tezi se vibrate karte hain ki quantum mechanics ke hisaab se room temperature par vibrations sirf thode active hote hain. Kaafi thanda karne par har solid ki heat capacity zero ki taraf girti hai.

Quick check

4 mein se 1

Tum kis gas ka

γ \approx 1.67

hone ki ummeed karoge?

nitrogencarbon dioxideargonsteam

Answer dekho

Answer (C) argon

Argon monatomic hai, isliye

C_{v} = \frac{3}{2} R

aur

γ = \frac{5}{3}

. Nitrogen diatomic hai (

\frac{7}{5}

); carbon dioxide aur steam mein aur bhi zyada modes hain aur

γ

chhota hai.

12.7 Mean free path

Shaant classroom ke aage attar ki ek bottle kholo. Khushboo ke molecules kuch sau metre per second se nikalte hain, phir bhi peeche baitha student kai seconds tak smell mehsoos nahi karta. Wajah hai collisions: khushboo ka molecule kabhi bhi zyada door nahi jaata ki koi air molecule usse takra jaata hai, isliye uska raasta zigzag hota hai jo bottle se bahut dheere-dheere door jaata hai. (Asli kamre mein halki hawa ki dhaarayein aam taur par khushboo ko sirf diffusion se zyada tezi se le jaati hain.)

Collisions ke beech ki doori estimate karne ke liye molecules ko diameter $d$ ke hard spheres maano. Do molecules tab takrate hain jab unke centres ek doosre se $d$ ke andar aa jaate hain (Fig. 12.12).

2/3Time $Δ t$ mein ek molecule radius $d$ aur volume $π d^{2} ⟨ v ⟩ Δ t$ ki ek tube cover karta hai.

Fig. 12.12Molecule jo tube banata hai usse hone wali collisions ginna.NCERT Fig. 12.7 · p. 255

Pehle step mein do barabar spheres ek doosre ko chhoo rahe hain, unke centres mark hain aur ek double arrow dikhata hai ki centres ke beech doori d hai. Doosre step mein ek molecule right ki taraf chalta hai aur radius d aur length average speed times delta t ki ek lambi tube banata hai; teen molecules jinke centres tube ke andar hain, highlighted hain kyunki yeh unse takrayega, jabki bahar wale grey mein bane hain. Teesre step mein ek zigzag line ek molecule ka raasta dikhati hai, har collision par ek dot ke saath, label hai free paths between collisions.

DerivationCollisions ke beech kitni doori?

Average speed $⟨ v ⟩$ se chalte ek molecule ko follow karo, aur abhi ke liye maano ki baaki sab ruke hue hain. Time $Δ t$ mein yeh radius $d$ aur length $⟨ v ⟩ Δ t$ ki tube cover karta hai, aur har us molecule se takrata hai jiska centre us tube ke andar hai, jiska volume $π d^{2} ⟨ v ⟩ Δ t$ hai.

Unit volume mein $n$ molecules hon, to yeh $Δ t$ mein $n π d^{2} ⟨ v ⟩ Δ t$ collisions karta hai. Isliye collision rate $n π d^{2} ⟨ v ⟩$ hai, aur collisions ke beech ka average time hai

τ = \frac{1}{n π d^{2} ⟨ v ⟩}

(12.38)

Collisions ke beech tay ki gayi average doori mean free path hai:

l = ⟨ v ⟩ τ = \frac{1}{n π d^{2}}

(12.39)

Asal mein baaki molecules bhi chalte hain, aur matter karti hai relative speed. Speeds par dhyaan se average lene par collision rate $\sqrt{2}$ guna badh jaati hai, isliye

mean free path

l = \frac{1}{\sqrt{2} n π d^{2}}

(12.40)

Yeh kitna lamba hai? $27^{\circ} C$ aur $1 a t m$ par hawa ke liye $n = P / k_{B} T \approx 2.4 \times 10^{25} m^{- 3}$ . Effective molecular diameter $d \approx 3.7 \times 10^{- 10} m$ lo, yeh woh value hai jo hawa ki measured viscosity se fit hoti hai. Tab Eq. (12.40) se $l \approx 6.7 \times 10^{- 8} m$ milta hai, lagbhag $67 n m$ yaani lagbhag 180 molecular diameters. Lagbhag $470 m / s$ ki mean speed ke saath collisions ke beech ka time $τ = l / ⟨ v ⟩ \approx 1.4 \times 10^{- 10} s$ hai: har molecule ek second mein lagbhag saat arab baar takrata hai.

Kisi vessel se zyadatar hawa nikaal do to free udaanein bahut lambi ho jaati hain: $10^{- 4} P a$ aur room temperature par hawa mein mean free path lagbhag $70 m$ hota hai, kisi bhi laboratory chamber se lamba. Aise vacuum mein molecules ek doosre se mile bina wall se wall tak udte hain. Isi wajah se electron microscope ko high vacuum chahiye: uske electrons ko gas molecules se takraye bina column paar karna hota hai. Aam taur par number density aadhi karne se $l$ double hota hai, aur molecular diameter aadha karne se chaar guna.

Example 12.9NCERT Example 12.9 · p. 255

30 km par patli hawa

Ek research balloon $30 k m$ upar tair raha hai, jahan hawa ka pressure lagbhag $1.2 k P a$ aur temperature lagbhag $227 K$ hai. Upar ke sea-level estimate ( $300 K$ aur $101 k P a$ par $l \approx 67 n m$ ) se shuru karke balloon par mean free path nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Yeh numbers gas ki ek aam galat picture ko theek karte hain. Molecules bahut zyada door nahi hote: room ki hawa mein typical spacing, $(k_{B} T / P)^{1 / 3} \approx 3.4 n m$ , solid ya liquid ki spacing ka sirf lagbhag das guna hai. Jo sach mein lamba hai woh mean free path hai, us spacing ka lagbhag bees guna aur lagbhag do sau molecular diameters. Yahi lambi free udaan gas ko uska typical behaviour deti hai: yeh kisi bhi container ko bharne ke liye phail jaati hai aur bina container ke roki nahi ja sakti.

Kinetic theory mean free path ko bulk mein naapi jaane wali properties se bhi jodti hai. Viscosity, thermal conductivity aur diffusion sab is par depend karte hain ki molecules collisions ke beech momentum, energy ya khud ko kitni door le jaate hain. 1865 mein Josef Loschmidt ne hawa ki measured viscosity ko Maxwell ki theory ke saath milakar molecules ke size ka ek sabse pehla estimate lagaya.

InteractiveMean free path lab

Hard-sphere molecules ek aise box mein chalte hain jo har side par repeat hota hai, saamne se dekha gaya, paas wale molecules zyada gehre bane hain. Ek orange molecule zigzag trail chhodta hai jismein har collision par ek mod hai. Sliders molecules ki sankhya, unka diameter aur temperature set karte hain. Ek table formula 1 over root 2 n pi d squared se mean free path ko saari collisions se naapi gayi value se compare karti hai, aur collisions ki sankhya, per molecule per second collisions aur rms speed dikhati hai.

Fig. 12.13Play dabao. Lengths aisi units mein hain jahan box 100 units chauda hai; orange molecule ek trail chhodta hai.

Try karke dekho

Mean free path lab ke saath experiments

Measured mean free path note karo, phir molecules ki sankhya aadhi karo (250 se 125). Kya yeh double hota hai, jaisa Eq. (12.40) predict karti hai?
Diameter double karo, 3 se 6 units. Mean free path kitne guna badalna chahiye? Check karo.
Temperature badhao. Molecules tez ho jaate hain: kya mean free path badalta hai, ya sirf collision rate?

Kyunki $l \propto 1 / n$ , molecules aadhe karne se yeh double hota hai; kyunki $l \propto 1 / d^{2}$ , $d$ double karne se yeh chaar guna chhota ho jaata hai. Fixed $n$ par garam karne se $l$ nahi badalta: molecules collisions ke beech utni hi doori tay karte hain, bas tezi se, isliye collision rate speed ke proportion mein badhti hai. Sabse bheed wali settings mein measured value formula se thodi chhoti aati hai, kyunki molecules ka apna size matter karne lagta hai.

Quick check

3 mein se 1

Constant temperature par ek gas ka pressure double kiya jaata hai. Uska mean free path

double ho jaata haiek chauthai reh jaata haiaadha ho jaata hainahi badalta

Answer dekho

Answer (C) aadha ho jaata hai

n = P / k_{B} T

double hota hai, aur

l \propto 1 / n

, isliye

l

aadha hota hai. Ek chauthai ke liye

d

ko double hona padta.

Recap

Summary aur key points

Matter lagaataar motion mein rehne wale molecules se bana hai. Atoms kuch ångström par attract karte hain aur dabane par repel karte hain; gas mein yeh forces sirf collisions ke dauraan kaam karte hain (Section 12.2).
Ideal gas equation: $P V = μ R T = N k_{B} T$ , jahan $R = 8.314 J m o l^{- 1} K^{- 1}$ aur $k_{B} = R / N_{A} = 1.38 \times 10^{- 23} J / K$ . Real gases ise low pressure aur high temperature par sabse achhe se follow karti hain (Section 12.3).
Boyle's law, Charles' law, Avogadro's hypothesis aur Dalton's law of partial pressures sab isi se nikalte hain (Fig. 12.4).
Walls ko diya gaya momentum $P = \frac{1}{3} n m ⟨ v^{2} ⟩$ deta hai ((12.14)).
Gas law ke saath, $\frac{1}{2} m ⟨ v^{2} ⟩ = \frac{3}{2} k_{B} T$ : temperature ek molecule ki average translational kinetic energy naapta hai, gas koi bhi ho (Section 12.4.2).
$P V = \frac{2}{3} E$ jahan $E = \frac{3}{2} N k_{B} T$ , isliye ideal gas ki internal energy sirf $T$ par depend karti hai ((12.18)).
$v_{rms} = \sqrt{3 k_{B} T / m}$ : ek temperature par halke molecules zyada tez hote hain, isi tarah porous barriers isotopes ko alag karte hain (Example 12.6).
Chalte piston se compress hui gas garam hoti hai; jo gas piston ko bahar dhakelti hai woh thandi hoti hai (Example 12.7).
Equipartition: molecule ki energy ke har squared term ka average $\frac{1}{2} k_{B} T$ hai, aur ek vibrational mode $k_{B} T$ rakhta hai (Law).
Molar heat capacities: monatomic $C_{v} = \frac{3}{2} R$ , $γ = \frac{5}{3}$ ; rigid diatomic $\frac{5}{2} R$ , $\frac{7}{5}$ ; non-linear polyatomic $(3 + f) R$ ; hamesha $C_{p} - C_{v} = R$ (Section 12.6).
Solids: per mole $C \approx 3 R$ (Dulong–Petit), diamond jaise halke, stiff solids peeche reh jaate hain (Table 12.4).
Mean free path $l = 1 / (\sqrt{2} n π d^{2})$ : room ki hawa mein lagbhag $70 n m$ , molecules ke beech ki spacing se kaafi lamba (Section 12.7).

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise 12.1	Practice Q1
NCERT Exercise 12.2	Practice Q2
NCERT Exercise 12.3	Practice Q3
NCERT Exercise 12.4	Practice Q4
NCERT Exercise 12.5	Practice Q5
NCERT Exercise 12.6	Practice Q6
NCERT Exercise 12.7	Practice Q7
NCERT Exercise 12.8	Practice Q8
NCERT Exercise 12.9	Practice Q9
NCERT Exercise 12.10	Practice Q10

Quick check: MCQs

12.2 Matter ka molecular nature3 questions
12.3 Gases ka behaviour4 questions
12.4.1 Ideal gas ka pressure3 questions
12.4.2 Temperature ka kinetic interpretation4 questions
12.5 Law of equipartition of energy3 questions
12.6.4 Solids ki specific heat capacity4 questions
12.7 Mean free path3 questions

Glossary · 10

Key terms (English + Hindi)

Avogadro numberआवोगाद्रो संख्याavogadro sankhya: Ek mole mein particles ki sankhya, $N_{A} \approx 6.02 \times 10^{23}$ .Chapter mein pehli baar yahan aaya →

Boltzmann constantबोल्ट्ज़मान नियतांकboltzmann niyatank: Per molecule gas constant, $k_{B} = R / N_{A} = 1.38 \times 10^{- 23} J / K$ .Chapter mein pehli baar yahan aaya →

degree of freedomस्वातंत्र्य कोटिswatantrya koti: Molecule ke chalne ka ek independent tareeka; energy ginte waqt, uski energy ka har squared term.Chapter mein pehli baar yahan aaya →

ideal gasआदर्श गैसaadarsh gas: Ek model gas jo $P V = μ R T$ ko bilkul follow karti hai; real gases low pressure aur high temperature par iske paas aati hain.

kinetic theoryअणुगति सिद्धांतanugati siddhant: Woh model jo gas ki properties ko uske molecules ki motion se samjhata hai, Newton's laws aur averages ka use karke.Chapter mein pehli baar yahan aaya →

law of equipartition of energyऊर्जा के समविभाजन का नियमurja ke samvibhajan ka niyam: Thermal equilibrium mein molecule ki energy ke har squared term ka average $\frac{1}{2} k_{B} T$ hota hai.

mean free pathमाध्य मुक्त पथmadhya mukt path: Collisions ke beech molecule dwara tay ki gayi average doori, $l = 1 / (\sqrt{2} n π d^{2})$ .Chapter mein pehli baar yahan aaya →
molar specific heat capacityमोलर विशिष्ट ऊष्मा धारिताmolar vishisht ushma dharita: Ek mole ko ek kelvin garam karne ke liye zaroori heat: constant volume par $C_{v}$ , constant pressure par $C_{p}$ .Chapter mein pehli baar yahan aaya →

partial pressureआंशिक दाबaanshik daab: Mixture ki ek gas jo pressure lagati agar same temperature par akele wahi container bharti.Chapter mein pehli baar yahan aaya →

rms speedवर्ग-माध्य-मूल चालvarg madhya mool chaal: Molecular speeds ke squares ke mean ka square root, $v_{rms} = \sqrt{3 k_{B} T / m}$ .Chapter mein pehli baar yahan aaya →

Kinetic Theory: Class 11 Physics Chapter 12 Hinglish mein

Is chapter mein aap kya seekhenge

12.1 Introduction

12.2 Matter ka molecular nature

12.3 Gases ka behaviour

12.4 Ideal gas ki kinetic theory

12.4.1 Ideal gas ka pressure

12.4.2 Temperature ka kinetic interpretation

12.5 Law of equipartition of energy

12.6 Specific heat capacity

12.6.1 Monatomic gases

12.6.2 Diatomic gases

12.6.3 Polyatomic gases

12.6.4 Solids ki specific heat capacity

12.7 Mean free path

Summary aur key points

Practice questions answers ke saath (NCERT “Exercises”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye